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# Algebra

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Paul Dawkins College Algebra Table of Contents Preface iii Outline iv Preliminaries 1 Introduction 1 Integer Exponents 2 Rational Exponents 9 Real Exponents 15 Radicals 16 Polynomials 25 Factoring Polynomials 31 Rational Expressions 41 Complex Numbers 52 Solving Equations and Inequalities 58 Introduction 58 Solutions and Solution Sets 59 Linear Equations 63 Application of Linear Equations 71 Equations With More Than One Variable 81 Quadratic Equations Part I 85 Quadratic Equations Part II 93 Solving Quadratic Equations A Summary 104 Application of Quadratic Equations 107 Equations Reducible to Quadratic Form 111 Equations with Radicals 116 Linear Inequalities 122 Polynomial Inequalities 129 Rational Inequalities 135 Absolute Value Equations 140 Absolute Value Inequalities 147 Graphing and Functions 152 Introduction 152 Graphing 15 3 Lines 159 Circles 169 The Definition of a Function 175 Graphing Functions 188 Combining Functions 192 Inverse Functions 199 Common Graphs 206 Introduction 206 Lines Circles and Piecewise Functions 207 Parabolas 208 Ellipses 218 Hyperbolas 222 Miscellaneous Functions 226 Transformations 229 Symmetry 235 Rational Functions 240 Polynomial Functions 246 Introduction 246 Dividing Polynomials 247 ZeroesRoots of Polynomials 252 2007 Paul Dawkins i httptutorialmathlamaredutermsaspx College Algebra Graphing Polynomials 257 Finding Zeroes of Polynomials 265 Partial Fractions 273 Exponential and Logarithm Functions 281 Introduction 28 1 Exponential Functions 282 Logarithm Functions 287 Solving Exponential Equations 297 Solving Logarithm Equations 304 Applications 3 10 Systems of Equations 317 Introduction 3 17 Linear Systems with Two Variables 318 Linear Systems with Three Variables 326 Augmented Matrices 330 More on the Augmented Matrix 339 Non Linear Systems 345 2007 Paul Dawkins ii httptutorialmathlamaredutermsaspx College Algebra 6166 Here are my online notes for my Algebra course that I teach here at Lamar University although I have to admit that it s been years since I last taught this course At this point in my career I mostly teach Calculus and Differential Equations Despite the fact that these are my class notes they should be accessible to anyone wanting to learn Algebra or needing a refresher for algebra I ve tried to make the notes as self contained as possible and do not reference any book However they do assume that you ve had some eXposure to the basics of algebra at some point prior to this While there is some review of exponents factoring and graphing it is assumed that not a lot of review will be needed to remind you how these topics work Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed 1 Because I wanted to make this a fairly complete set of notes for anyone wanting to learn algebra I have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here You will need to find one of your fellow class mates to see if there is something in these notes that wasn t covered in class 2 Because I want these notes to provide some more examples for you to read through I don t always work the same problems in class as those given in the notes Likewise even if I do work some of the problems in here I may work fewer problems in class than are presented here 3 Sometimes questions in class will lead down paths that are not covered here I try to anticipate as many of the questions as possible in writing these up but the reality is that I can t anticipate all the questions Sometimes a very good question gets asked in class that leads to insights that I ve not included here You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are 4 This is somewhat related to the previous three items but is important enough to merit its own item THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS Using these notes as a substitute for class is liable to get you in trouble As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class 2007 Paul Dawkins iii httptutorialmathlamaredutermsaspX College Algebra Here is a listing of all the material that is currently available in these notes Preliminaries Integer Exponents In this section we will start looking at exponents and their properties Rational Exponents We will define rational exponents in this section and extend the properties from the previous section to rational exponents Real Exponents This is a short acknowledgment that the exponent properties from the previous two sections will hold for any real exponent Radicals Here we will define radical notation and relate radicals to rational exponents We will also give the properties of radicals Polynomials We will introduce the basics of polynomials in this section including adding subtracting and multiplying polynomials Factoring Polynomials This is the most important section of all the preliminaries Factoring polynomials will appear in pretty much every chapter in this course Without the ability to factor polynomials you will be unable to complete this course Rational Expressions In this section we will define rational expressions and discuss adding subtracting multiplying and dividing them Complex Numbers Here is a very quick primer on complex numbers and how to manipulate them Solving Equations and Inequalities Solutions and Solution Sets We introduce some of the basic notation and ideas involved in solving in this section Linear Eguation In this section we will solve linear equations including equations with rational expressions Applications of Linear Equations We will take a quick look at applications of linear equations in this section Equations With More Than One Variable Here we will look at solving equations with more than one variable in them Quadratic Equations Part I In this section we will start looking at solving quadratic equations We will look at factoring and the square root property in this section Quadratic Equations Part II We will finish up solving quadratic equations in this section We will look at completing the square and quadratic formula in this section Quadratic Equations A Summary We ll give a procedure for determining which method to use in solving quadratic equations in this section We will also take a quick look at the discriminant Applications of Quadratic Equations Here we will revisit some of the applications we saw in the linear application section only this time they will involve solving a quadratic equation Equations Reducible to Quadratic Form In this section we will solve equations that can be reduced to quadratic in form Equations with Radicals Here we will solve equations with square roots in them 2007 Paul Dawkins iv httptutorialmathlamaredutermsaspx College Algebra Linear Inequalities We will start solving inequalities in this section by looking at linear inequalities Polynomial Inequalities In this section we will look at solving inequalities that contain polynomials Rational Inequalities Here we will solve inequalities involving rational expressions Absolute Value Equations We will officially define absolute value in this section and solve equations that contain absolute value Absolute Value Inequalities We will solve inequalities that involve absolute value in this section Graphing and Functions Graphing In this section we will introduce the Cartesian coordinate system and most of the basics of graphing equations Lines Here we will review the main ideas from the study of lines including slope and the special forms of the equation of a line Circles We will look at the equation of a circle and graphing circles in this section The Definition of a Function We will discuss the definition of a function in this section We will also introduce the idea of function evaluation Graphing Functions In this section we will look at the basics of graphing functions We will also graph some piecewise functions in this section Combining functions Here we will look at basic arithmetic involving functions as well as function composition Inverse Functions We will define and find inverse functions in this section Common Graphs Lines Circles and Piecewise Functions This section is here only to acknowledge that we ve already talked about graphing these in a previous chapter Parabolas We ll be graphing parabolas in this section Ellipses In this section we will graph ellipses Hyperbolas Here we will be graphing hyperbolas Miscellaneous Functions In this section we will graph a couple of common functions that don t really take all that much work to so We ll be looking at the constant function square root absolute value and a simple cubic function Transformations We will be looking at shifts and re ections of graphs in this section Collectively these are often called transformations Symmetry We will brie y discuss the topic of symmetry in this section Rational Functions In this section we will graph some rational functions We will also be taking a look at vertical and horizontal asymptotes Polynomial Functions Dividing Polynomials We ll review some of the basics of dividing polynomials in this section ZeroesRoots of Polynomials In this section we ll define just what zeroesroots of polynomials are and give some of the more important facts concerning them Graphing Polynomials Here we will give a process that will allow us to get a rough sketch of some polynomials Finding Zeroes of Polynomials We ll look at a process that will allow us to find some of the zeroes of a polynomial and in special cases all of the zeroes Partial Fractions In this section we will take a look at the process of partial fractions and finding the partial fraction decomposition of a rational expression 2007 Paul Dawkins v httptutorialmathlamaredutermsaspx College Algebra Exponential and Logarithm Functions Exponential Functions In this section we will introduce exponential functions We will be taking a look at some of the properties of exponential functions Logarithm Functions Here we will introduce logarithm functions We be looking at how to evaluate logarithms as well as the properties of logarithms Solving Exponential Equations We will be solving equations that contain exponentials in this section Solving Logarithm Equations Here we will solve equations that contain logarithms Application In this section we will look at a couple of applications of exponential functions and an application of logarithms Svstems of Equations Linear Svstems with Two Variables In this section we will use systems of two equations and two variables to introduce two of the main methods for solving systems of equations Linear Svstems with Three Variables Here we will work a quick example to show how to use the methods to solve systems of three equations with three variables Augmented Mgrices We will look at the third main method for solving systems in this section We will look at systems of two equations and systems of three equations More on the Augmented Matrix In this section we will take a look at some special cases to the solutions to systems and how to identify them using the augmented matrix method Nonlinear Svstems We will take a quick look at solving nonlinear systems of equations in this section 2007 Paul Dawkins vi httptutorialmathlamaredutermsaspx College Algebra Preliminaries Introduction The purpose of this chapter is to review several topics that will arise time and again throughout this material Many of the topics here are so important to an Algebra class that if you don t have a good working grasp of them you will find it very difficult to successfully complete the course Also it is assumed that you ve seen the topics in this chapter somewhere prior to this class and so this chapter should be mostly a review for you However since most of these topics are so important to an Algebra class we will make sure that you do understand them by doing a quick review of them here Exponents and polynomials are integral parts of any Algebra class If you do not remember the basic exponent rules and how to work with polynomials you will find it very difficult if not impossible to pass an Algebra class This is especially true with factoring polynomials There are more than a few sections in an Algebra course where the ability to factor is absolutely essential to being able to do the work in those sections In fact in many of these sections factoring will be the first step taken It is important that you leave this chapter with a good understanding of this material If you don t understand this material you will find it difficult to get through the remaining chapters Here is a brief listing of the material covered in this chapter Integer Exponents In this section we will start looking at exponents and their properties Rational Exponents We will define rational exponents in this section and extend the properties from the previous section to rational exponents Real Exponents This is a short acknowledgment that the exponent properties from the previous two sections will hold for any real exponent Radicals Here we will define radical notation and relate radicals to rational exponents We will also give the properties of radicals Polynomials We will introduce the basics of polynomials in this section including adding subtracting and multiplying polynomials Factoring Polynomials This is the most important section of all the preliminaries Factoring polynomials will appear in pretty much every chapter in this course Without the ability to factor polynomials you will be unable to complete this course Rational Expressions In this section we will define rational expressions and discuss adding subtracting multiplying and dividing them Complex Numbers Here is a very quick primer on complex numbers and how to manipulate them 2007 Paul Dawkins l httptutorialmathlamaredutermsaspx College Algebra Integer Exponents We will start off this chapter by looking at integer exponents In fact we will initially assume that the exponents are positive as well We will look at zero and negative exponents in a bit Let s first recall the definition of exponentiation with positive integer exponents If a is any number and n is a positive integer then an a a a no a n times So for example 35 3 3 3 3 3 243 We should also use this opportunity to remind ourselves about parenthesis and conventions that we have in regards to exponentiation and parenthesis This will be particularly important when dealing with negative numbers Consider the following two cases 2 and 24 These will have different values once we evaluate them When performing exponentiation remember that it is only the quantity that is immediately to the left of the exponent that gets the power In the first case there is a parenthesis immediately to the left so that means that everything in the parenthesis gets the power So in this case we get lt2gt 2gtlt 2gtlt 2gtlt 2 16 In the second case however the 2 is immediately to the left of the exponent and so it is only the 2 that gets the power The minus sign will stay out in front and will NOT get the power In this case we have the following 24 24 222 29 126 16 We put in some extra parenthesis to help illustrate this case In general they aren t included and we would write instead 24 222 16 The point of this discussion is to make sure that you pay attention to parenthesis They are important and ignoring parenthesis or putting in a set of parenthesis where they don t belong can completely change the answer to a problem Be careful Also this warning about parenthesis is not just intended for exponents We will need to be careful with parenthesis throughout this course Now let s take care of zero exponents and negative integer exponents In the case of zero exponents we have a0 1 provided a 0 Notice that it is required that a not be zero This is important since 00 is not defined Here is a quick example of this property 1268 1 2007 Paul Dawkins 2 httptutorialmathlamaredutermsaspx College Algebra We have the following definition for negative exponents If a is any nonzero number and n is a positive integer yes positive then Can you see why we required that a not be zero Remember that division by zero is not defined and if we had allowed a to be zero we would have gotten division by zero Here are a couple of quick examples for this definition 2 L L 3 52 25 4 43 64 64 Here are some of the main properties of integer exponents Accompanying each property will be a quick example to illustrate its use We will be looking at more complicated examples after the properties Properties 1 aquotam aquotm Example 2 a 9a4 a94 a 5 2 aquot m aquot39 Example 2 a7 3 a73 azl n m 514 411 7 an 61 T Cl 61 3 m 1 a 0 Example a a W a4 1 1 a y 61114 7 0 4 abl anbn Example 2 ab 4 a4b4 a n aquot a 8 a8 5 b 0 Example 8 9 19quot b b n b n bn 10 10 10 6 E j Example E 2 L10 9 a aquot b a a b n l 20 l 7 a n Example ab 20 ab ab 1 l 8 n an Example 2 a2 a a a n bm 616 917 9 97 an EXEIIIIPIC I F 7 2007 Paul Dawkins 3 httptutorialmathlamaredutermsaspx College Algebra 10 aquotb k aquotquotb quotquot Example 6141793 a43b 93 a12b27 ll in k Exam le 39 a 6 2 a62 bm bmk p b5 910 Notice that there are two possible forms for the third property Which form you use is usually dependent upon the form you want the answer to be in Note as well that many of these properties were given with only two termsfactors but they can be extended out to as many termsfactors as we need For example property 4 can be extended as follows abcd n a b c d We only used four factors here but hopefully you get the point Property 4 and most of the other properties can be extended out to meet the number of factors that we have in a given problem There are several common mistakes that students make with these properties the first time they see them Let s take a look at a couple of them Consider the following case 1 Correct ab a 2 12 b b 2 l Incorrect ab 2 ab In this case only the 9 gets the exponent since it is immediately off to the left of the exponent and so only this term moves to the denominator Do NOT carry the a down to the denominator with the la Contrast this with the following case ab 2 l ab2 In this case the exponent is on the set of parenthesis and so we can just use property 7 on it and so both the a and the 9 move down to the denominator Again note the importance of parenthesis and how they can change an answer Here is another common mistake l l l 1 Correct 45 5 as 3a 3 a 3 3a5 Incorrect 5 3a In this case the exponent is only on the a and so to use property 8 on this we would have to break up the fraction as shown and then use property 8 only on the second term To bring the 3 up with the a we would have needed the following 2007 Paul Dawkins 4 httptutorialmathlamaredutermsaspx College Algebra 1 3a5 says Once again notice this common mistake comes down to being careful with parenthesis This will be a constant refrain throughout these notes We must always be careful with parenthesis Misusing them can lead to incorrect answers Let s take a look at some more complicated examples now Example 1 Simplify each of the following and write the answers with only positive exponents 3 a 4x y5 E b 1oz2y 42z3y 5 E T Oulon c m S 1 t 1 7mquot4n393 T Solution 5 6 9 g 24a3b Z f W J Solution Note that when we say simplify in the problem statement we mean that we will need to use all the properties that we can to get the answer into the required form Also a simplified answer will have as few terms as possible and each term should have no more than a single exponent on it There are many different paths that we can take to get to the final answer for each of these In the end the answer will be the same regardless of the path that you used to get the answer All that this means for you is that as long as you used the properties you can take the path that you find the easiest The path that others find to be the easiest may not be the path that you find to be the easiest That is okay Also we won t put quite as much detail in using some of these properties as we did in the examples given with each property For instance we won t show the actual multiplications anymore we will just give the result of the multiplication a 4x394y5 3 For this one we will use property 10 first 4x 4y5 3 Z 43 x 12y15 Don t forget to put the exponent on the constant in this problem That is one of the more common mistakes that students make with these simplification problems 2007 Paul Dawkins 5 httptutorialmathlamaredutermsaspx College Algebra At this point we need to evaluate the first term and eliminate the negative exponent on the second term The evaluation of the first term isn t too bad and all we need to do to eliminate the negative exponent on the second term is use the definition we gave for negative exponents 3 l 643215 46 4y5 Z yls 7 We further simplified our answer by combining everything up into a single fraction This should always be done The middle step in this part is usually skipped All the definition of negative exponents tells us to do is move the term to the denominator and drop the minus sign in the exponent So from this point on that is what we will do without writing in the middle step Return to Problems 2 5 b 10z2y 4 z3y In this case we will first use property 10 on both terms and then we will combine the terms using property 1 Finally we will eliminate the negative exponents using the definition of negative exponents 10Z2y4 2 Z3y 5 1O2 Z4y 8Z l5yE5 1OOZ 1ly 13 L There are a couple of things to be careful with in this problem First when using the property 10 on the first term make sure that you square the 10 and not just the 10 i e don t forget the minus sign Second in the final step the 100 stays in the numerator since there is no negative exponent on it The exponent of l l is only on the z and so only the z moves to the denominator Return to Problems n392m C 7m394nquot3 This one isn t too bad We will use the definition of negative exponents to move all terms with negative exponents in them to the denominator Also property 8 simply says that if there is a term with a negative exponent in the denominator then we will just move it to the numerator and drop the minus sign So let s take care of the negative exponents first rfzm m4n3m 7m394n393 7112 Now simplify We will use property 1 to combine the m s in the numerator We will use property 3 to combine the n s and since we are looking for positive exponents we will use the first form of this property since that will put a positive exponent up in the numerator rfzm msn 7m4n3 7 Again the 7 will stay in the denominator since there isn t a negative exponent on it It will NOT 2007 Paul Dawkins 6 httptutorialmathlamaredutermsaspx College Algebra move up to the numerator with the m Do not get excited if all the terms move up to the numerator or if all the terms move down to the denominator That will happen on occasion Return to Problems 5x391y394 2 aw x9 This example is similar to the previous one except there is a little more going on with this one The first step will be to again get rid of the negative exponents as we did in the previous example Any terms in the numerator with negative exponents will get moved to the denominator and we ll drop the minus sign in the exponent Likewise any terms in the denominator with negative exponents will move to the numerator and we ll drop the minus sign in the exponent Notice this time unlike the previous part there is a term with a set of parenthesis in the denominator Because of the parenthesis that whole term including the 3 will move to the numerator 1 Here is the work for this part 2 5x 1y 4 53y5 59 y 45 y6 3y5 2 x9 xy4x9 xy4x9 x10 Return to Problems 6 e Z 6 There are several first steps that we can take with this one The first step that we re pretty much always going to take with these kinds of problems is to first simplify the fraction inside the parenthesis as much as possible After we do that we will use property 5 to deal with the exponent that is on the parenthesis 6 6 Z 5 Z2x1 x 6 X6 Z 2x 1 Z5 Z3 Z13 In this case we used the second form of property 3 to simplify the z s since this put a positive exponent in the denominator Also note that we almost never write an exponent of l When we have exponents of l we will drop them 0 24a3b398 Z 6a395b This one is very similar to the previous part The main difference is negative on the outer exponent We will deal with that once we ve simplified the fraction inside the parenthesis 24a3b398 2 4a3a5 2 L8 2 6a395b 9819 199 Now at this point we can use property 6 to deal with the exponent on the parenthesis Doing this gives us Return to Problems 2007 Paul Dawkins 7 httptutorialmathlamaredutermsaspx College Algebra 24a3bquot 2 K 2 I918 6a395b 4a8 166116 Return to Problems Before leaving this section we need to talk brie y about the requirement of positive only exponents in the above set of examples This was done only so there would be a consistent final answer In many cases negative exponents are okay and in some cases they are required In fact if you are on a track that will take you into calculus there are a fair number of problems in a calculus class in which negative exponents are the preferred if not required form 2007 Paul Dawkins 8 httptutorialmathlamaredutermsaspx College Algebra Rational Exponents Now that we have looked at integer exponents we need to start looking at more complicated exponents In this section we are going to be looking at rational exponents That is exponents in the form b where both m and n are integers We will start simple by looking at the following special case 1 b where n is an integer Once we have this figured out the more general case given above will actually be pretty easy to deal with Let s first define just what we mean by exponents of this form 1 a I is equivalent to a b 1 In other words when evaluating I we are really asking what number in this case a did we 1 raise to the n to get I Often I is called the nth root of b Let s do a couple of evaluations Example 1 Evaluate each of the following 1 a 255 1 1 b 32 Solution 1 C 814 Solution d 8 E e 16 1 1 f 164 Solution Solution When doing these evaluations we will do actually not do them directly When first confronted with these kinds of evaluations doing them directly is often very difficult In order to evaluate these we will remember the equivalence given in the definition and use that instead We will work the first one in detail and then not put as much detail into the rest of the problems 1 a 255 So here is what we are asking in this problem 252 2007 Paul Dawkins 9 httptutorialmathlamaredutermsaspx College Algebra Using the equivalence from the definition we can rewrite this as 2 25 So all that we are really asking here is what number did we square to get 25 In this case that is hopefully easy to get We square 5 to get 25 Therefore 1 252 5 Return to Problems 1 b 323 So what we are asking here is what number did we raise to the 5 power to get 32 1 32 2 because 25 32 Return to Problems 1 c 81 What number did we raise to the 4 power to get 81 1 81 3 because 34 81 Return to Problems 1 d 8 We need to be a little careful with minus signs here but other than that it works the same way as the previous parts What number did we raise to the 3 power i e cube to get 8 1 8 3 2 because 2 3 8 Return to Problems 1 e 164 This part does not have an answer It is here to make a point In this case we are asking what number do we raise to the 4 power to get 16 However we also know that raising any number positive or negative to an even power will be positive In other words there is no real number that we can raise to the 4 power to get 16 Note that this is different from the previous part If we raise a negative number to an odd power we will get a negative number so we could do the evaluation in the previous part As this part has shown we can t always do these evaluations Return to Problems 1 f 164 Again this part is here to make a point more than anything Unlike the previous part this one has an answer Recall from the previous section that if there aren t any parentheses then only the part immediately to the left of the exponent gets the exponent So this part is really asking us to evaluate the following term 2007 Paul Dawkins 10 httptutorialmathlamaredutermsaspx College Algebra 1 1 164 164 So we need to determine what number raised to the 4 power will give us 16 This is 2 and so in this case the answer is Return to Problems As the last two parts of the previous example has once again shown we really need to be careful with parenthesis In this case parenthesis makes the difference between being able to get an answer or not Also don t be worried if you didn t know some of these powers off the top of your head They are usually fairly simple to determine if you don t know them right away For instance in the part b we needed to determine what number raised to the 5 will give 32 If you can t see the power right off the top of your head simply start taking powers until you find the correct one In other words compute 25 35 45 until you reach the correct value Of course in this case we wouldn t need to go past the first computation The next thing that we should acknowledge is that all of the properties for exponents that we gave in the previous section are still valid for all rational exponents This includes the more general rational exponent that we haven t looked at yet Now that we know that the properties are still valid we can see how to deal with the more general rational exponent There are in fact two different ways of dealing with them as we ll see Both methods involve using property 2 from the previous section For reference purposes this property 1s 11 m nm a a So let s see how to deal with a general rational exponent We will first rewrite the exponent as follows 5 Z lltmgt In other words we can think of the exponent as a product of two numbers Now we will use the exponent property shown above However we will be using it in the opposite direction than what we did in the previous section Also there are two ways to do it Here they are 19 b391tm OR 19 b Using either of these forms we can now evaluate some more complicated expressions 2007 Paul Dawkins ll httptutorialmathlamaredutermsaspx College Algebra Example 2 Evaluate each of the following 2 a 83 1 3 b 6253 Solution 4 24415 S 1 j 0 11 10H 32 Solution We can use either form to do the evaluations However it is usually more convenient to use the first form as we will see 3 a 83 Let s use both forms here since neither one is too bad in this case Let s take a look at the first form 2 1 2 1 83 83j 2 4 83 2 because 23 8 Now let s take a look at the second form 1 83 82 64 3 4 643 4 because 43 64 So we get the same answer regardless of the form Notice however that when we used the second form we ended up taking the 3 root of a much larger number which can cause problems on occasion Return to Problems 3 b 6254 Again 1et s use both forms to compute this one 3 1 3 1 6254 6254 53 125 6254 5 because 54 625 2 1 1 6254 62534 24414e625Z 125 because 1254 244140625 As this part has shown the second form can be quite difficult to use in computations The root in this case was not an obvious root and not particularly easy to get if you didn t know it right off the top of your head 4 243 3 C In this case we ll only use the first form However before doing that we ll need to first use property 5 of our exponent properties to get the exponent onto the numerator and denominator Return to Problems 2007 Paul Dawkins 12 httptutorialmathlamaredutermsaspx College Algebra 14 4 1 2433 EJ124135 L 1 34 81 32 32 3214 24 16 Return to Problems We can also do some of the simplification type problems with rational exponents that we saw in the previous section Example 3 Simplify each of the following and write the answers with only positive exponents 1 2 4 3 l6v2 2 xzyj b 1 J x5y393 Solution a For this problem we will first move the exponent into the parenthesis then we will eliminate the negative exponent as we did in the previous section We will then move the term to the denominator and drop the minus sign 1 11j 1 1 1 1611 4 2128 2v8w2 Return to Problems b In this case we will first simplify the expression inside the parenthesis 1 2 7 1 7 21 3 2 7 1 5 7 X2 3 x2x2y3 X 2y 3 L J 7 23 1 2 1 xy 3 E Don t worry if after simplification we don t have a fraction anymore That will happen on occasion Now we will eliminate the negative in the exponent using property 7 and then we ll use property 4 to finish the problem up Return to Problems 2007 Paul Dawkins l3 httptutorialmathlamaredutermsaspx College Algebra We will leave this section with a warning about a common mistake that students make in regards to negative exponents and rational exponents Be careful not to confuse the two as they are totally separate topics In other words w L bit and NOT 1 1939 i bquot This is a Very common mistake when students first learn exponent rules 2007 Paul Dawkins 14 httptutorialmathlamaredutermsaspx College Algebra Real Exponents This is a fairly short section It s only real purpose is to acknowledge that the exponent properties we listed in the first section work for any exponent We Ve already used them on integer and rational exponents but we aren t actually restricted to these kinds of exponents The properties will work for any exponent that we want to use Example 1 Simplify each of the following and write the answers with only positive exponents a X82 y 026Z2 05 3 41 3 b quotxy7 J Solution We will not put as much detail into these as we have in the previous sections By this point it is assumed that you re starting to get a good handle on the exponent rules a 82 026 2 05 41 013 x4391Z x y Z x y Z 013 Y x3y41 393 x3x27 3 x57 3 y41 3 y123 X27 y41 F 1 Note that we won t be doing anything like this in the remainder of this course This section is here only to acknowledge that these rules will work for any kind of exponent that we might need to work with 2007 Paul Dawkins l5 httptutorialmathlamaredutermsaspx College Algebra Radicals We ll open this section with the definition of the radical If n is a positive integer that is greater than 1 and a is a real number then 1 3 a where n is called the index a is called the radicand and the symbol is called the radical The left side of this equation is often called the radical form and the right side is often called the exponent form From this definition we can see that a radical is simply another notation for the first rational exponent that we looked at in the rational exponents section Note as well that the index is required in these to make sure that we correctly evaluate the radical There is one exception to this rule and that is square root For square roots we have ZxZ In other words for square roots we typically drop the index Let s do a couple of examples to familiarize us with this new notation Example 1 Write each of the following radicals in exponent form a VB b 10 8x c x2 yz Solution a 7 16 1 b 1 8x 8x 1 c x2 y2 x2 y2 5 As seen in the last two parts of this example we need to be careful with parenthesis When we convert to exponent form and the radicand consists of more than one term then we need to enclose the whole radicand in parenthesis as we did with these two parts To see why this is consider the following 1 Sx From our discussion of exponents in the previous sections we know that only the term immediately to the left of the exponent actually gets the exponent Therefore the radical form of this is 1 8x5 8 19 1 8x So we once again see that parenthesis are very important in this class Be careful with them 2007 Paul Dawkins 16 httptutorialmathlamaredutermsaspx College Algebra Since we know how to evaluate rational exponents we also know how to evaluate radicals as the following set of examples shows Example 2 Evaluate each of the following a E and 3E Solution b 3 Solution C W Solution d 3 E e 33 E Solution To evaluate these we will first convert them to exponent form and then evaluate that since we already know how to do that a These are together to make a point about the importance of the index in this notation Let s take a look at both of these 1 xE162 4 1 VB 164 2 because 24 16 because 42 16 So the index is important Different indexes will give different evaluations so make sure that you don t drop the index unless it is a 2 and hence we re using square roots 1 b 5 243 2435 3 because 35 243 1 c x41296 12964 6 because 64 1296 d x3125 a 125 5 because 53 125 1 e 416 164 As we saw in the integer exponent section this does not have a real answer and so we can t evaluate the radical of a negative number if the index is even Note however that we can evaluate the radical of a negative number if the index is odd as the previous part shows Let s brie y discuss the answer to the first part in the above example In this part we made the claim that 16 4 because 42 16 However 4 isn t the only number that we can square to get 16 We also have 42 16 So why didn t we use 4 instead There is a general rule about evaluating square roots or more generally radicals with even indexes When evaluating square roots we ALWAYS take the positive answer If we want the negative answer we will do the following 2007 Paul Dawkins 17 httptutorialmathlamaredutermsaspx College Algebra 16 4 This may not seem to be all that important but in later topics this can be very important Following this convention means that we will always get predictable values when evaluating roots Note that we don t have a similar rule for radicals with odd indexes such as the cube root in part d above This is because there will never be more than one possible answer for a radical with an odd index We can also write the general rational exponent in terms of radicals as follows a a j D OR a a quot39 quotam We now need to talk about some properties of radicals Properties If n is a positive integer greater than 1 and both a and I9 are positive real numbers then 1 3a quot a 2 M 2 VB 3 n 3 lt5 b Note that on occasion we can allow a or 9 to be negative and still have these properties work When we run across those situations we will acknowledge them However for the remainder of this section we will assume that a and 9 must be positive Also note that while we can break up products and quotients under a radical we can t do the same thing for sums or differences In other words x ab 5E AND x a b 53E If you aren t sure that you believe this consider the following quick number example 5E916 E34 7 If we break up the root into the sum of the two pieces we clearly get different answers So be careful to not make this very common mistake We are going to be simplifying radicals shortly so we should next define simplified radical form A radical is said to be in simplified radical form or just simplified form if each of the following are true All exponents in the radicand must be less than the index Any exponents in the radicand can have no factors in common with the index No fractions appear under a radical No radicals appear in the denominator of a fraction 39gtP 1 2007 Paul Dawkins 18 httptutorialmathlamaredutermsaspx College Algebra In our first set of simplification examples we will only look at the first two We will need to do a little more work before we can deal with the last two Example 3 Simplify each of the following Assume that x y and z are positive a y7 E b 9 X6 Solution c 18x6y E d 4 32x9y5z Lion e 5 X124Z24 Solution 139 V 9x2 x3 6x2 1 Solution a J7 In this case the exponent 7 is larger than the index 2 and so the first rule for simplification is violated To fix this we will use the first and second properties of radicals above So let s note that we can write the radicand as follows 2 y7 y y 33 y So we ve got the radicand written as a perfect square times a term whose exponent is smaller than the index The radical then becomes Now use the second property of radicals to break up the radical and then use the first property of radicals on the first term x7v32 x fx3 This now satisfies the rules for simplification and so we are done Before moving on let s brie y discuss how we figured out how to break up the exponent as we did To do this we noted that the index was 2 We then determined the largest multiple of 2 that is less than 7 the exponent on the radicand This is 6 Next we noticed that 76l Finally remembering several rules of exponents we can rewrite the radicand as 2 y7 y y ymgy y3 y In the remaining examples we will typically jump straight to the final form of this and leave the details to you to check Return to Problems b V This radical violates the second simplification rule since both the index and the exponent have a common factor of 3 To fix this all we need to do is convert the radical to exponent form do some simplification and then convert back to radical form 2007 Paul Dawkins 19 httptutorialmathlamaredutermsaspx College Algebra 6 2 9x6x6 x5 x3 2 3 2 Return to Problems c 18x6 y Now that we ve got a couple of basic problems out of the way let s work some harder ones Although with that said this one is really nothing more than an extension of the first example There is more than one term here but everything works in exactly the same fashion We will break the radicand up into perfect squares times terms whose exponents are less than 2 ie l l8x6y 9x6y1 Zy 9x32 ys 2 Zy Don t forget to look for perfect squares in the number as well Now go back to the radical and then use the second and first property of radicals as we did in the first example Jlsxw 9ltx3gt2lty5gt2lt2ygt JWmsxwa Note that we used the fact that the second property can be expanded out to as many terms as we have in the product under the radical Also don t get excited that there are no x s under the radical in the final answer This will happen on occasion Return to Problems P T 4 P T y5Zl2 This one is similar to the previous part except the index is now a 4 So instead of get perfect squares we want powers of 4 This time we will combine the work in the previous part into one step 432x9y5Z12 lt116xsy4Z12 Zxy lt1E4x24lt4 z3 rm 2x2yZ3ltE Return to Problems Again this one is similar to the previous two parts 5x12y4Z24 x1oZ2ox2y4Z4 5062 5 5Z4 5 5x2y4Z4 x2Z4 5x2y4Z4 In this case don t get excited about the fact that all the y s stayed under the radical That will happen on occasion Return to Problems 0 W x36x2 This last part seems a little tricky Individually both of the radicals are in simplified form However there is often an unspoken rule for simplification The unspoken rule is that we should have as few radicals in the problem as possible In this case that means that we can use the second property of radicals to combine the two radicals into one radical and then we ll see if there is any simplification that needs to be done 2007 Paul Dawkins 20 httptutorialmathlamaredutermsaspx College Algebra 397679x26x2547 Now that it s in this form we can do some simplification x3 9x2 3 6x2 27 2x 327x3 3Z 3x3Z Return to Problems Before moving into a set of examples illustrating the last two simplification rules we need to talk brie y about addingsubtractingmultiplying radicals Performing these operations with radicals is much the same as performing these operations with polynomials If you don t remember how to addsubtractmultiply polynomials we will give a quick reminder here and then give a more in depth set of examples the next section Recall that to addsubtract terms with x in them all we need to do is add subtract the coefficients of the x For example 4x9xf4 9x 13x 3x llxf3 ll x 8x Addingl subtracting radicals works in exactly the same manner For instance 4 9 4a4 9Z l3 3 193 11 3 3 111 3 8 183 We ve already seen some multiplication of radicals in the last part of the previous example If we are looking at the product of two radicals with the same index then all we need to do is use the second property of radicals to combine them then simplify What we need to look at now are problems like the following set of examples Example 4 Multiply each of the following Assume that x is positive a J 2 5 E b 3 2 5 1 c 5 25Z 2 Lion Solution In all of these problems all we need to do is recall how to FOIL binomials Recall 3x 5x23xxk 3x2 5x 52 396E 6 596 16 31962 x 10 With radicals we multiply in exactly the same manner The main difference is that on occasion we ll need to do some simplification after doing the multiplication lt 2gtlt sgt 2 5 Z 5L2 10 x7 3 lO x 35 10 2007 Paul Dawkins 21 httptutorialmathlamaredutermsaspx College Algebra As noted above we did need to do a little simplification on the first term after doing the multiplication Return to Problems b 3 2 5 Don t get excited about the fact that there are two variables here It works the same way 3 2L5 6P15Z 2Z 5 6x 15E 25 5y 6x 175 5y Again notice that we combined up the terms with two radicals in them Return to Problems c 5 25 2 Not much to do with this one 5 25 225xx 2 19 19 4 25x 4 Notice that in this case the answer has no radicals That will happen on occasion so don t get excited about it when it happens Return to Problems The last part of the previous example really used the fact that a ba 19 a2 192 If you don t recall this formula we will look at it in a little more detail in the next section Okay we are now ready to take a look at some simplification examples illustrating the final two rules Note as well that the fourth rule says that we shouldn t have any radicals in the denominator To get rid of them we will use some of the multiplication ideas that we looked at above and the process of getting rid of the radicals in the denominator is called rationalizing the denominator In fact that is really what this next set of examples is about They are really more examples of rationalizing the denominator rather than simplification examples Example 5 Rationalize the denominator for each of the following Assume that x is positive a i Q J l 2 b 5 Solution OUIOII C 1 S1t39 3 E 5 d T 4x x5 Solution There are really two different types of problems that we ll be seeing here The first two parts illustrate the first type of problem and the final two parts illustrate the second type of problem Solution 2007 Paul Dawkins 22 httptutorialmathlamaredutermsaspx College Algebra Both types are worked differently a i J In this case we are going to make use of the fact that quot aquot a We need to determine what to multiply the denominator by so that this will show up in the denominator Once we figure this out we will multiply the numerator and denominator by this term Here is the work for this part 4 4x 4 4E W x2 x Remember that if we multiply the denominator by a term we must also multiply the numerator by the same term In this way we are really multiplying the term by 1 since E 1 and so aren t a changing its value in any way 2 We ll need to start this one off with first using the third property of radicals to eliminate the fraction from underneath the radical as is required for simplification x3 ax Now in order to get rid of the radical in the denominator we need the exponent on the x to be a 5 Return to Problems This means that we need to multiply by 5 x2 so let s do that 5 3 Z 5 5x2 52x2 52x2 X3 5x3 5x2 5X5 X Return to Problems 1 c 3 In this case we can t do the same thing that we did in the previous two parts To do this one we will need to instead to make use of the fact that aba b a2 192 When the denominator consists of two terms with at least one of the terms involving a radical we will do the following to get rid of the radical 1 1 3E 3E 3E 3 Z 3 3 3 3 9 x 2007 Paul Dawkins 23 httptutorialmathlamaredutermsaspx College Algebra So we took the original denominator and changed the sign on the second term and multiplied the numerator and denominator by this new term By doing this we were able to eliminate the radical in the denominator when we then multiplied out Return to Problems 5 d T 4 3 This one works exactly the same as the previous example The only difference is that both terms in the denominator now have radicals The process is the same however 5 5 4 E 54 E 54 E mm4 4 4 4 16x 3 Return to Problems Rationalizing the denominator may seem to have no real uses and to be honest we won t see many uses in an Algebra class However if you are on a track that will take you into a Calculus class you will find that rationalizing is useful on occasion at that level We will close out this section with a more general version of the first property of radicals Recall that when we first wrote down the properties of radicals we required that a be a positive number This was done to make the work in this section a little easier However with the first property that doesn t necessarily need to be the case Here is the property for a general a i e positive or negative quot7 a ifnis even a a ifn is odd where a is the absolute value of a If you don t recall absolute value we will cover that in detail in a section in the next chapter All that you need to do is know at this point is that absolute value always makes a a positive number So as a quick example this means that x 8x AND 11xT1x For square roots this is x2 x This will not be something we need to worry all that much about here but again there are topics in courses after an Algebra course for which this is an important idea so we needed to at least acknowledge it 2007 Paul Dawkins 24 httptutorialmathlamaredutermsaspx College Algebra Polynomials In this section we will start looking at polynomials Polynomials will show up in pretty much every section of every chapter in the remainder of this material and so it is important that you understand them We will start off with polynomials in one variable Polynomials in one variable are algebraic expressions that consist of terms in the form ax where n is a nonnegative i e positive or zero integer and a is a real number and is called the coefficient of the term The degree of a polynomial in one variable is the largest exponent in the polynomial Note that we will often drop the in one variable part and just say polynomial Here are examples of polynomials and their degrees 5x12 2x6 x5 l98x l degree 12 x4 x3x2 xl degree4 56x23 degree 23 5x 7 degree l 8 degree 0 So a polynomial doesn t have to contain all powers of x as we see in the first example Also polynomials can consist of a single term as we see in the third and fifth example We should probably discuss the final example a little more This really is a polynomial even it may not look like one Remember that a polynomial is any algebraic expression that consists of terms in the form axquot Another way to write the last example is 8x Written in this way makes it clear that the exponent on the x is a zero this also explains the degree and so we can see that it really is a polynomial in one variable Here are some examples of things that aren t polynomials 4x6 l5x398 l 5x xx2 2 x3 2 x The first one isn t a polynomial because it has a negative exponent and all exponents in a polynomial must be positive To see why the second one isn t a polynomial let s rewrite it a little 1 5x xx2 5x2 x xnz By converting the root to exponent form we see that there is a rational root in the algebraic expression All the exponents in the algebraic expression must be integers in order for the algebraic expression to be a polynomial As a general rule of thumb if an algebraic expression has a radical in it then it isn t a polynomial 2007 Paul Dawkins 25 httptutorialmathlamaredutermsaspx College Algebra Let s also rewrite the third one to see why it isn t a polynomial 2 x3 22x1 33 2 x So this algebraic expression really has a negative exponent in it and we know that isn t allowed Another rule of thumb is if there are any variables in the denominator of a fraction then the algebraic expression isn t a polynomial Note that this doesn t mean that radicals and fractions aren t allowed in polynomials They just can t involve the variables For instance the following is a polynomial 3 x4 x2 x 5 KW There are lots of radicals and fractions in this algebraic expression but the denominators of the fractions are only numbers and the radicands of each radical are only a numbers Each x in the algebraic expression appears in the numerator and the exponent is a positive or zero integer Therefore this is a polynomial Next let s take a quick look at polynomials in two variables Polynomials in two variables are algebraic expressions consisting of terms in the form axquot y quot The degree of each term in a polynomial in two variables is the sum of the exponents in each term and the degree of the polynomial is the largest such sum Here are some examples of polynomials in two variables and their degrees x2y 6x3y12 l0x2 7yl degree 15 6x4 8y4 xy2 degree 4 x4y2 x3y3 xy x4 degree 6 6x14 l0y3 3x lly degree 14 In these kinds of polynomials not every term needs to have both x s and y s in them in fact as we see in the last example they don t need to have any terms that contain both x s and y s Also the degree of the polynomial may come from terms involving only one variable Note as well that multiple terms may have the same degree We can also talk about polynomials in three variables or four variables or as many variables as we need The vast majority of the polynomials that we ll see in this course are polynomials in one variable and so most of the examples in the remainder of this section will be polynomials in one variable Next we need to get some terminology out of the way A monomial is a polynomial that consists of exactly one term A binomial is a polynomial that consists of exactly two terms Finally a trinomial is a polynomial that consists of exactly three terms We will use these terms off and on so you should probably be at least somewhat familiar with them Now we need to talk about adding subtracting and multiplying polynomials You ll note that we left out division of polynomials That will be discussed in a later section where we will use division of polynomials quite often 2007 Paul Dawkins 26 httptutorialmathlamaredutermsaspx College Algebra Before actually starting this discussion we need to recall the distributive law This will be used repeatedly in the remainder of this section Here is the distributive law abc ab ac We will start with adding and subtracting polynomials This is probably best done with a couple of examples Example 1 Perform the indicated operation for each of the following a Add 6x5 10x2 x 45 to 13x2 9x 4 Solution b Subtract 5x3 9x2 x 3 from x2 x 1 Solution Solution a Add 6x5 l0x2 x 45 to 13x2 9x 4 The first thing that we should do is actually write down the operation that we are being asked to do 6x5 l0x2 x 45 l3x2 9x 4 In this case the parenthesis are not required since we are adding the two polynomials They are there simply to make clear the operation that we are performing To add two polynomials all that we do is combine like terms This means that for each term with the same exponent we will add or subtract the coefficient of that term In this case this is 6x5 lOx2x 45l3x2 9x46x5 10 1339X2 l 9x 45 4 6x5 J3x2 8x 41 Return to Problems b Subtract 5x3 9x2 x 3 from x2 x 1 Again let s write down the operation we are doing here We will also need to be very careful with the order that we write things down in Here is the operation x2xl 5x3 9x2x 3 This time the parentheses around the second term are absolutely required We are subtracting the whole polynomial and the parenthesis must be there to make sure we are in fact subtracting the whole polynomial In doing the subtraction the first thing that we ll do is distribute the minus sign through the parenthesis This means that we will change the sign on every term in the second polynomial Note that all we are really doing here is multiplying a l through the second polynomial using the distributive law After distributing the minus through the parenthesis we again combine like terms Here is the work for this problem x2 xl 5x3 9x2 x 3 x2 Jae 4 5x3 9x2 ac 48 598 1098 4 2007 Paul Dawkins 27 httptutorialmathlamaredutermsaspx College Algebra Note that sometimes a term will completely drop out after combing like terms as the x did here This will happen on occasion so don t get excited about it when it does happen Return to Problems Now let s move onto multiplying polynomials Again it s best to do these in an example Example 2 Multiply each of the following a 4x2 x2 6x 2 i b 3x 5x 10 E c 4x2 x6 3x E d 3x 7 yx 2y E e 2x 3x2 x 1 Solution a 4x2 x2 6x 2 This one is nothing more than a quick application of the distributive law 4x2 x2 6x 2 4x4 24x3 8x2 Return to Problems b 3x 5 x 10 This one will use the FOIL method for multiplying these two binomials 3x 5x 10 3x2 30x 5x 50 3x2 25x 50 mm mm Ar First Terms Outer Terms Irmer Terms Last Terms Recall that the FOIL method will only work when multiplying two binomials If either of the polynomials isn t a binomial then the FOIL method won t work Also note that all we are really doing here is multiplying every term in the second polynomial by every term in the first polynomial The FOIL acronym is simply a convenient way to remember this Return to Problems c 4x2 x 6 3x Again we will just FOIL this one out 4x2 x6 3x24ae2 12953 6x 3992 Jl2x3 27x2 6x Return to Problems d 3x7yx 2y We can still FOIL binomials that involve more than one variable so don t get excited about these kinds of problems when they arise 3x7yx 2y3x2 xy nqxy l4y2 3x2 Jay l4y2 Return to Problems 2007 Paul Dawkins 28 httptutorialmathlamaredutermsaspx College Algebra e 2x3x2 xl In this case the FOIL method won t work since the second polynomial isn t a binomial Recall however that the FOIL acronym was just a way to remember that we multiply every term in the second polynomial by every term in the first polynomial That is all that we need to do here 2x3x2 xl 2x3 Zacz 25c 31962 3 3 EH3 x2 x 3 Return to Problems Let s work another set of examples that will illustrate some nice formulas for some special products We will give the formulas after the example Example 3 Multiply each of the following a 3x 53x 5 E b 2x62 E c 1 7x2 E d 4x 32 1 Solution a 3x 53x 5 We can use FOIL on this one so let s do that 3x 53x 5 9x2 l5x l5x 25 9x2 25 In this case the middle terms drop out Return to Problems b 2x 62 Now recall that 42 4 4 l6 Squaring with polynomials works the same way So in this case we have 2x62 2x 62x 6 4x2 l2x l2x 36 4x2 24x 36 Return to Problems c l 7x2 This one is nearly identical to the previous part l 7x2 7x l 7x 1 7x 7x 4953 1 l4x 49x2 Return to Problems 2007 Paul Dawkins 29 httptutorialmathlamaredutermsaspx College Algebra d 4x32 This part is here to remind us that we need to be careful with coefficients When we Ve got a coefficient we MUST do the exponentiation first and then multiply the coefficient 4x324x ax 3 4x2 695 9 442 24 36 You can only multiply a coefficient through a set of parenthesis if there is an exponent of l on the parenthesis If there is any other exponent then you CAN T multiply the coefficient through the parenthesis Just to illustrate the point 4x32 4xl22 4x l24x 12 l6x2 496x 4444 This is clearly not the same as the correct answer so be careful Return to Problems The parts of this example all use one of the following special products a ba 19 a2 192 2 2 2 ab a 2ab b a b2 a2 2ab 192 Be careful to not make the following mistakes ab2 a2 192 a b2 612 192 These are Very common mistakes that students often make when they first start learning how to multiply polynomials 2007 Paul Dawkins 30 httptutorialmathlamaredutermsaspx College Algebra Factoring Polynomials Of all the topics covered in this chapter factoring polynomials is probably the most important topic There are many sections in later chapters where the first step will be to factor a polynomial So if you can t factor the polynomial then you won t be able to even start the problem let alone finish it Let s start out by talking a little bit about just what factoring is Factoring is the process by which we go about determining what we multiplied to get the given quantity We do this all the time with numbers For instance here are a variety of ways to factor 12 1226 1234 122 23 12Gj24 12 2 6 12 22 3 There are many more possible ways to factor 12 but these are representative of many of them A common method of factoring numbers is to completely factor the number into positive prime factors A prime number is a number whose only positive factors are 1 and itself For example 2 3 5 and 7 are all examples of prime numbers Examples of numbers that aren t prime are 4 6 and 12 to pick a few If we completely factor a number into positive prime factors there will only be one way of doing it That is the reason for factoring things in this way For our example above with 12 the complete factorization is 12lt2gtlt2gtltsgt Factoring polynomials is done in pretty much the same manner We determine all the terms that were multiplied together to get the given polynomial We then try to factor each of the terms we found in the first step This continues until we simply can t factor anymore When we can t do any more factoring we will say that the polynomial is completely factored Here are a couple of examples x2 16 x4x 4 This is completely factored since neither of the two factors on the right can be further factored Likewise x4 16 x2 44x2 4 is not completely factored because the second factor can be further factored Note that the first factor is completely factored however Here is the complete factorization of this polynomial x4 16x2 44x 42x 2 The purpose of this section is to familiarize ourselves with many of the techniques for factoring polynomials 2007 Paul Dawkins 31 httptutorialmathlamaredutermsaspx College Algebra Greatest Common Factor The first method for factoring polynomials will be factoring out the greatest common factor When factoring in general this will also be the first thing that we should try as it will often simplify the problem To use this method all that we do is look at all the terms and determine if there is a factor that is in common to all the terms If there is we will factor it out of the polynomial Also note that in this case we are really only using the distributive law in reverse Remember that the distributive law states that abc ab ac In factoring out the greatest common factor we do this in reverse We notice that each term has an a in it and so we factor it out using the distributive law in reverse as follows abacab 6 Let s take a look at some examples Example 1 Factor out the greatest common factor from each of the following polynomials a 8x4 4x3 10x2 J b x3y2 3x4y 5x5y3 i C 3x6 9362 3x Solution d 9x2 2x 7 l2x2x7 Solution a 8x4 4x3 10x2 First we will notice that we can factor a 2 out of every term Also note that we can factor an x2 out of every term Here then is the factoring for this problem 8x4 4x3lOx22x24x2 2x 5 Note that we can always check our factoring by multiplying the terms back out to make sure we get the original polynomial Return to Problems b x3y2 3x4y 5x5 323 In this case we have both x s and y s in the terms but that doesn t change how the process works Each term contains and x3 and a y so we can factor both of those out Doing this gives x3y2 3x4y 5x5y3 x3yy 3x 5x y2 Return to Problems c 3x6 9x2 3x In this case we can factor a 3x out of every term Here is the work for this one 3x6 9x2 3x 3xx5 3x 1 Notice the l where the 3x originally was in the final term since the final term was the term we factored out we needed to remind ourselves that there was a term there originally To do this we need the l and notice that it is l instead of l because the term was originally a positive term If it had been a negative term originally we would have had to use l One of the more common mistakes with these types of factoring problems is to forget this l 2007 Paul Dawkins 32 httptutorialmathlamaredutermsaspx College Algebra Remember that we can always check by multiplying the two back out to make sure we get the original To check that the l is required let s drop it and then multiply out to see what we get 3xx5 3x 3x6 9x2 3x6 9x2 3x So without the l we don t get the original polynomial Be careful with this It is easy to get in a hurry and forget to add a l or l as required when factoring out a complete term Return to Problems d 9x2 2x7 l2x2x7 This one looks a little odd in comparison to the others However it works the same way There is a 3x in each term and there is also a 2x 7 in each term and so that can also be factored out Doing the factoring for this problem gives 9x2 2x7 l2x2x7 3x2x 73x 4 Return to Problems Factoring By Grouping This is a method that isn t used all that often but when it can be used it can be somewhat useful This method is best illustrated with an example or two Example 2 Factor by grouping each of the following a 3x2 2x 12x 8 Solution b x5 x 2x 2 E c x5 3x3 2x2 6 J Solution a 3x2 2xl2x 8 In this case we group the first two terms and the final two terms as shown here 3x2 2x l2x 8 Now notice that we can factor an x out of the first grouping and a 4 out of the second grouping Doing this gives 3x2 2x 12x 8 x3x 2 4ax 2 We can now see that we can factor out a common factor of 3x 2 so let s do that to the final factored form 3x2 2xl2x 83x 2 x 4 And we re done That s all that there is to factoring by grouping Note again that this will not always work and sometimes the only way to know if it will work or not is to try it and see what you get Return to Problems b x5 x 2x4 2 In this case we will do the same initial step but this time notice that both of the final two terms are negative so we ll factor out a as well when we group them Doing this gives x5 x 2x4 2 Again we can always distribute the back through the parenthesis to make sure we get the 2007 Paul Dawkins 33 httptutorialmathlamaredutermsaspx College Algebra original polynomial At this point we can see that we can factor an x out of the first term and a 2 out of the second term This gives x5 x 2x4 2 xx4 l 2ac4 1 We now have a common factor that we can factor out to complete the problem X5 x 2x4 2x4 l9x 2 Return to Problems c X5 3x3 2x2 6 This one also has a in front of the third term as we saw in the last part However this time the fourth term has a in front of it unlike the last part We will still factor a out when we group however to make sure that we don t lose track of it When we factor the out notice that we needed to change the on the fourth term to a Again you can always check that this was done correctly by multiplying the back through the parenthsis X5 3x3 2x2 6 Now that we ve done a couple of these we won t put the remaining details in and we ll go straight to the final factoring x5 3x3 2x26x3x2 Er L x2 Er xz 3 x3 2 Return to Problems Factoring by grouping can be nice but it doesn t work all that often Notice that as we saw in the last two parts of this example if there is a in front of the third term we will often also factor that out of the third and fourth terms when we group them Factoring Quadratic Polynomials First let s note that quadratic is another term for second degree polynomial So we know that the largest exponent in a quadratic polynomial will be a 2 In these problems we will be attempting to factor quadratic polynomials into two first degree hence forth linear polynomials Until you become good at these we usually end up doing these by trial and error although there are a couple of processes that can make them somewhat easier Let s take a look at some examples Example 3 Factor each of the following polynomials a X2 2X 15 Solution b X2 10X 24 Solution c x2 6x 9 Solution d X2 5X 1 Solution e 3X2 2X 8 Solution f 5X2 17X 6 Solution g 4x2 10x 6 i 2007 Paul Dawkins 34 httptutorialmathlamaredutermsaspx College Algebra Solution a x2 2x 15 Okay since the first term is x2 we know that the factoring must take the form x2 2x l5 x x We know that it will take this form because when we multiply the two linear terms the first term must be x2 and the only way to get that to show up is to multiply x by x Therefore the first term in each factor must be an x To finish this we just need to determine the two numbers that need to go in the blank spots We can narrow down the possibilities considerably Upon multiplying the two factors out these two numbers will need to multiply out to get 15 In other words these two numbers must be factors of 15 Here are all the possible ways to factor 15 using only integers lt1gtlt1sgt lt1gtlt1sgt ltsgtltsgt ltsgtltsgt Now we can just plug these in one after another and multiply out until we get the correct pair However there is another trick that we can use here to help us out The correct pair of numbers must add to get the coefficient of the x term So in this case the third pair of factors will add to 2 and so that is the pair we are after Here is the factored form of the polynomial x2 2x l5 x 3x 5 Again we can always check that we got the correct answer my doing a quick multiplication Note that the method we used here will only work if the coefficient of the x2 term is one If it is anything else this won t work and we really will be back to trial and error to get the correct factoring form Return to Problems b x2 10x 24 Let s write down the initial form again x2 l0x24 x x Now we need two numbers that multiply to get 24 and add to get 10 It looks like 6 and 4 will do the trick and so the factored form of this polynomial is x2 l0x 24 x 4x 6 Return to Problems c x2 6x 9 Again let s start with the initial form x2 6x9 x x This time we need two numbers that multiply to get 9 and add to get 6 In this case 3 and 3 will be the correct pair of numbers Don t forget that the two numbers can be the same number on occasion as they are here Here is the factored form for this polynomial 2007 Paul Dawkins 35 httptutorialmathlamaredutermsaspx College Algebra x26x9x 3x 3 x 3r2 Note as well that we further simplified the factoring to acknowledge that it is a perfect square You should always do this when it happens Return to Problems d x2 5x 1 Once again here is the initial form x2 5xlx x Okay this time we need two numbers that multiply to get 1 and add to get 5 There aren t two integers that will do this and so this quadratic doesn t factor This will happen on occasion so don t get excited about it when it does Return to Problems e 3x2 2x 8 Okay we no longer have a coefficient of l on the x2 term However we can still make a guess as to the initial form of the factoring Since the coefficient of the x2 term is a 3 and there are only two positive factors of 3 there is really only one possibility for the initial form of the factoring 3x2 2x 8 3x 6f f Since the only way to get a 3x2 is to multiply a 3x and an x these must be the first two terms However finding the numbers for the two blanks will not be as easy as the previous examples We will need to start off with all the factors of 8 lt1gtltsgt lt1gtltsgt lt2gtlt4gt lt2gtlt4gt At this point the only option is to pick a pair plug them in and see what happens when we multiply the terms out Let s start with the fourth pair Let s plug the numbers in and see what we get 3x 2x 4 3x2 l0x 8 Well the first and last terms are correct but then they should be since we ve picked numbers to make sure those work out correctly However since the middle term isn t correct this isn t the correct factoring of the polynomial That doesn t mean that we guessed wrong however With the previous parts of this example it didn t matter which blank got which number This time it does Let s ip the order and see what we get 3x 4x2 3x2 Qx 8 So we got it We did guess correctly the first time we just put them into the wrong spot So in these problems don t forget to check both places for each pair to see if either will work Return to Problems f 5x2 17x 6 Again the coefficient of the x2 term has only two positive factors so we ve only got one possible 2007 Paul Dawkins 36 httptutorialmathlamaredutermsaspx College Algebra initial form 5x2 l7x6 5x x Next we need all the factors of 6 Here they are lt1gtlt6gt lt 1gtlt 6gt lt2gtltsgt lt 2gtlt sgt Don t forget the negative factors They are often the ones that we want In fact upon noticing that the coefficient of the x is negative we can be assured that we will need one of the two pairs of negative factors since that will be the only way we will get negative coefficient there With some trial and error we can get that the factoring of this polynomial is 5x2 l7x6 5x 2 x 3 Return to Problems g 4x2 10x 6 In this final step we ve got a harder problem here The coefficient of the x2 term now has more than one pair of positive factors This means that the initial form must be one of the following possibilities 4x2 10x 6 4x x 4x2 l0x 6 2x i2x To fill in the blanks we will need all the factors of 6 Here they are lt1gtlt6gt lt1gtlt6gt lt2gtltsgt lt2gtltsgt With some trial and error we can find that the correct factoring of this polynomial is 4x2 10x 6 2x l 2x 6 Note as well that in the trial and error phase we need to make sure and plug each pair into both possible forms and in both possible orderings to correctly determine if it is the correct pair of factors or not We can actually go one more step here and factor a 2 out of the second term if we d like to This gives 4x2 10x 6 22x 1ex 3 This is important because we could also have factored this as 4x2 l0x 6 4x 2 x 3 which on the surface appears to be different from the first form given above However in this case we can factor a 2 out of the first term to get 4x2 10x 6 22x 1ex 3 This is exactly what we got the first time and so we really do have the same factored form of this polynomial Return to Problems 2007 Paul Dawkins 37 httptutorialmathlamaredutermsaspx College Algebra Special Forms There are some nice special forms of some polynomials that can make factoring easier for us on occasion Here are the special forms Let s work some examples with these Example 4 Factor each of the following a x2 20x 100 Solution b 25x2 9 Solution C 8x3 1 Solution Solution a x2 20x 100 In this case we ve got three terms and it s a quadratic polynomial Notice as well that the constant is a perfect square and its square root is 10 Notice as well that 2102O and this is the coefficient of the x term So it looks like we ve got the second special form above The correct factoring of this polynomial is x2 20x100e 102 To be honest it might have been easier to just use the general process for factoring quadratic polynomials in this case rather than checking that it was one of the special forms but we did need to see one of them worked Return to Problems b 25x2 9 In this case all that we need to notice is that we ve got a difference of perfect squares 25x2 9 5x2 s So this must be the third special form above Here is the correct factoring for this polynomial 25x2 95x 3I5x 3 Return to Problems c 8x3 1 This problem is the sum of two perfect cubes 3 8x3 1 2x3 1 and so we know that it is the fourth special form from above Here is the factoring for this polynomial 8x3 1 2x 14x2 zae 1 Return to Problems 2007 Paul Dawkins 38 httptutorialmathlamaredutermsaspx College Algebra Do not make the following factoring mistake 612 192 ab2 This just simply isn t true for the vast majority of sums of squares so be careful not to make this very common mistake There are rare cases where this can be done but none of those special cases will be seen here Factoring Polynomials with Degree Greater than 2 There is no one method for doing these in general However there are some that we can do so let s take a look at a couple of examples Example 5 Factor each of the following a 3x4 3x3 36x2 E b x4 25 E c x4 x2 20 1 Solution a 3x4 3x3 36x2 In this case let s notice that we can factor out a common factor of 3x2 from all the terms so let s do that first 3x4 3x3 36x2 3x2 x2 x 12 What is left is a quadratic that we can use the techniques from above to factor Doing this gives us 3x4 3x3 36x2 3x2 x 4x 3 Don t forget that the FIRST step to factoring should always be to factor out the greatest common factor This can only help the process Return to Problems b x4 25 There is no greatest common factor here However notice that this is the difference of two perfect squares x4 25 x2 2 52 So we can use the third special form from above x4 25x2 5x2 5 Neither of these can be further factored and so we are done Note however that often we will need to do some further factoring at this stage Return to Problems c x4 x2 20 Let s start this off by working a factoring a different polynomial uz u 2O u 4u 5 We used a different variable here since we d already used x s for the original polynomial 2007 Paul Dawkins 39 httptutorialmathlamaredutermsaspx College Algebra So why did we work this Well notice that if we let u x2 then u2 x2 2 x4 We can then rewrite the original polynomial in terms of u s as follows x4 x2 20u2 Ju 20 and we know how to factor this So factor the polynomial in u s then back substitute using the fact that we know u x2 x4x2 20u2 1 20 lta 4gteu 5 lt a 4w 5 Finally notice that the first term will also factor since it is the difference of two perfect squares The correct factoring of this polynomial is then x4 x2 2Ox 2x2x2 5 Note that this converting to u first can be useful on occasion however once you get used to these this is usually done in our heads Return to Problems We did not do a lot of problems here and we didn t cover all the possibilities However we did cover some of the most common techniques that we are liable to run into in the other chapters of this work 2007 Paul Dawkins 40 httptutorialmathlamaredutermsaspx College Algebra Rational Expressions We now need to look at rational expressions A rational expression is nothing more than a fraction in which the numerator andor the denominator are polynomials Here are some examples of rational expressions 6 z2 l m4l8ml 4x26x lO x l z25 m2 m 6 l The last one may look a little strange since it is more commonly written 4x2 6x 10 However it s important to note that polynomials can be thought of as rational expressions if we need to although they rarely are There is an unspoken rule when dealing with rational expressions that we now need to address When dealing with numbers we know that division by zero is not allowed Well the same is true for rational expressions So when dealing with rational expressions we will always assume that whatever x is it won t give division by zero We rarely write these restrictions down but we will always need to keep them in mind For the first one listed we need to avoid x l The second rational expression is never zero in the denominator and so we don t need to worry about any restrictions Note as well that the numerator of the second rational expression will be zero That is okay we just need to avoid division by zero For the third rational expression we will need to avoid m 3 and m 2 The final rational expression listed above will never be zero in the denominator so again we don t need to have any restrictions The first topic that we need to discuss here is reducing a rational expression to lowest terms A rational expression has been reduced to lowest terms if all common factors from the numerator and denominator have been canceled We already know how to do this with number fractions so let s take a quick look at an example 12 W 3 3 not reduced to lowest terms gt E Kevduced to lowest terms 8f2 With rational expression it works exactly the same way l355x1 x 1 not reduced to lowest terms gt reduced to lowest terms xjgzl3 x We do have to be careful with canceling however There are some common mistakes that students often make with these problems Recall that in order to cancel a factor it must multiply the whole numerator and the whole denominator So the x3 above could cancel since it multiplied the whole numerator and the whole denominator However the x s in the reduced form can t cancel since the x in the numerator is not times the whole numerator To see why the x s don t cancel in the reduced form above put a number in and see what happens Let s plug in x4 2007 Paul Dawkins 41 httptutorialmathlamaredutermsaspx College Algebra 41 3 41 4 4 4 Clearly the two aren t the same number So be careful with canceling As a general rule of thumb remember that you can t cancel something if it s got a or a on one side of it There is one exception to this rule of thumb wit that we ll deal with in an example later on down the road Let s take a look at a couple of examples Example 1 Reduce the following rational expression to lowest terms 2 a Solution x 9x2O 2 x 25 b A S 1 39 5xx2 J x7 2x6 x5 c A 1 x3 xl Solution When reducing a rational expression to lowest terms the first thing that we will do is factor both the numerator and denominator as much as possible That should always be the first step in these problems Also the factoring in this section and all successive section for that matter will be done without explanation It will be assumed that you are capable of doing andor checking the factoring on your own In other words make sure that you can factor x2 2x 8 a T x2 9x 20 We ll first factor things out as completely as possible Remember that we can t cancel anything at this point in time since every term has a or a on one side of it We ve got to factor first x22x8 x 4x2 x2 9x2O x 5x 4 At this point we can see that we ve got a common factor in both the numerator and the denominator and so we can cancel the x4 from both Doing this gives x2 2x 8 x2 x2 9x20 x 5 This is also all the farther that we can go Nothing else will cancel and so we have reduced this expression to lowest terms Return to Problems 2007 Paul Dawkins 42 httptutorialmathlamaredutermsaspx College Algebra 225 b x 5x x2 Again the first thing that we ll do here is factor the numerator and denominator x2 25 x 5x5 5x x2 x5 x At first glance it looks there is nothing that will cancel Notice however that there is a term in the denominator that is almost the same as a term in the numerator except all the signs are the opposite We can use the following fact on the second term in the denominator a b 9 61 OR ab at b This is commonly referred to as factoring a minus sign out because that is exactly what we ve done There are two forms here that cover both possibilities that we are liable to run into In our case however we need the first form Because of some notation issues let s just work with the denominator for a while x5 xxL x 5j xlt 1ltx 5 5lt 1lt5 5 lt1ltxltx 5 we 5 Notice the steps used here In the first step we factored out the minus sign but we are still multiplying the terms and so we put in an added set of brackets to make sure that we didn t forget that In the second step we acknowledged that a minus sign in front is the same as multiplication by l Once we did that we didn t really need the extra set of brackets anymore so we dropped them in the third step Next we recalled that we change the order of a multiplication if we need to so we ipped the x and the l Finally we dropped the l and just went back to a negative sign in the front Typically when we factor out minus signs we skip all the intermediate steps and go straight to the final step Let s now get back to the problem The rational expression becomes x2 25 x 5x5 5x x2 x x 5 At this point we can see that we do have a common factor and so we can cancel the x5 2 x 25 x 5 x 5 5x x2 x x Return to Problems 2007 Paul Dawkins 43 httptutorialmathlamaredutermsaspx College Algebra x7 2x6 x5 x3 x l8 In this case the denominator is already factored for us to make our life easier All we need to do is factor the numerator c x7 2x6 x5 955 952 2X1 x5 xl2 x3 xl8 x3 xl8 x3 xl8 Now we reach the point of this part of the example There are 5 x s in the numerator and 3 in the denominator so when we cancel there will be 2 left in the numerator Likewise there are 2 x 1 s in the numerator and 8 in the denominator so when we cancel there will be 6 left in the denominator Here is the rational expression reduced to lowest terms x7 2x6 x5 x2 x3xl8 xl6 Return to Problems Before moving on let s brie y discuss the answer in the second part of this example Notice that we moved the minus sign from the denominator to the front of the rational expression in the final form This can always be done when we need to Recall that the following are all equivalent 6l 6l 6l 9 b b In other words a minus sign in front of a rational expression can be moved onto the whole numerator or whole denominator if it is convenient to do that We do have to be careful with this however Consider the following rational expression x3 x 1 In this case the on the x can t be moved to the front of the rational expression since it is only on the x In order to move a minus sign to the front of a rational expression it needs to be times the whole numerator or denominator So if we factor a minus out of the numerator we could then move it into the front of the rational expression as follows x3 X3 x 3 xl xl xl The moral here is that we need to be careful with moving minus signs around in rational expressions We now need to move into adding subtracting multiplying and dividing rational expressions Let s start with multiplying and dividing rational expressions The general formulas are as follows 22 b d bd 2007 Paul Dawkins 44 httptutorialmathlamaredutermsaspx College Algebra 11 dbc ampc3 amp wm Note the two different forms for denoting division We will use either as needed so make sure you are familiar with both Note as well that to do division of rational expressions all that we need to do is multiply the numerator by the reciprocal of the denominator i e the fraction with the numerator and denominator switched Before doing a couple of examples there are a couple of special cases of division that we should look at In the general case above both the numerator and the denominator of the rational expression where fractions however what if one of them isn t a fraction So let s look at the following cases Q C amp Students often make mistakes with these initially To correctly deal with these we will turn the numerator first case or denominator second case into a fraction and then do the general division on them Cl LL i 3 3 1 c c d d 3 3 LL li c 3 b c be 1 Be careful with these cases It is easy to make a mistake with these and incorrectly do the division Now let s take a look at a couple of examples Example 2 Perform the indicated operation and reduce the answer to lowest terms x2 5x l4 x2 4 3 x2 3x2 x2 14x49 E m2 9 3 m b m25m6m2 J 2 C 3 Solution y y5 2007 Paul Dawkins 45 httptutorialmathlamaredutermsaspx College Algebra Solution Notice that with this problem we have started to move away from x as the main variable in the examples Do not get so used to seeing x s that you always expect them The problems will work the same way regardless of the letter we use for the variable so don t get excited about the different letters here x2 5x l4 x2 4 a x2 3x2 x2 l4x49 Okay this is a multiplication The first thing that we should always do in the multiplication is to factor everything in sight as much as possible x2 5x l4 x2 4 x 7x2 x 2x2 x2 3x2 x2 l4x49 x 2x l x72 Now recall that we can cancel things across a multiplication as follows c a c a b 6 d 39 b d Note that this ONLY works for multiplication and NOT for division In this case we do have multiplication so cancel as much as we can and then do the multiplication to get the answer x2 5x l4 x2 4 x2x2 x22 x2 3x2 x2 l4x49x l39x 7 x lx 7 Return to Problems b m2 9 L 3 m m2 5m 6 m 2 With division problems it is very easy to mistakenly cancel something that shouldn t be canceled and so the first thing we do here before factoring is do the division Once we ve done the division we have a multiplication problem and we factor as much as possible cancel everything that can be canceled and finally do the multiplication So let s get started on this problem m2 9 3 m m2 9 m2 m25m6 I m2m25m6 3 m m 3m3 m2 m3m2 3 m Now notice that there will be a lot of canceling here Also notice that if we factor a minus sign out of the denominator of the second rational expression Let s do some of the canceling and then do the multiplication m29 3 mm3 1 m3 m25m639m2 l m 3 m 3 Remember that when we cancel all the terms out of a numerator or denominator there is actually 2007 Paul Dawkins 46 httptutorialmathlamaredutermsaspx College Algebra a 1 left over Now we didn t finish the canceling to make a point Recall that at the start of this discussion we said that as a rule of thumb we can only cancel terms if there isn t a or a on either side of it with one exception for the We are now at that exception If there is a if front of the whole numerator or denominator as we ve got here then we can still cancel the term In this case the acts as a 1 that is multiplied by the whole denominator and so is a factor instead of an addition or subtraction Here is the final answer for this part m2 9 3 mi 1 m25m639m2 1 In this case all the terms canceled out and we were left with a number This doesn t happen all that often but as this example has shown it clearly can happen every once in a while so don t get excited about it when it does happen Return to Problems C yz 25 y 4 y l y 5 This is one of the special cases for division So as with the previous part we will first do the division and then we will factor and cancel as much as we can Here is the work for this part yz 5y4 V2 5y4V5 y1y4y5 y4y5 y2 l y2 1 yly l y 1 y5 Return to Problems Okay it s time to move on to addition and subtraction of rational expressions Here are the general formulas 3 2ab a I 6119 C C C C C C As these have shown we ve got to remember that in order to add or subtract rational expression or fractions we MUST have common denominators If we don t have common denominators then we need to first get common denominators Let s remember how do to do this with a quick number example 5 3 6 4 In this case we need a common denominator and recall that it s usually best to use the least common denominator often denoted lcd In this case the least common denominator is 12 So we need to get the denominators of these two fractions to a 12 This is easy to do In the first case we need to multiply the denominator by 2 to get 12 so we will multiply the numerator and denominator of the first fraction by 2 Remember that we ve got to multiply both the numerator and denominator by the same number since we aren t allowed to actually change the problem and 2007 Paul Dawkins 47 httptutorialmathlamaredutermsaspx College Algebra this is equivalent to multiplying the fraction by 1 since E 1 For the second term we ll need to a multiply the numerator and denominator by a 3 ON P ON Q P U3 r 0 r 0 r 0 r lJ Now the process for rational expressions is identical The main difficulty is in finding the least common denominator However there is a really simple process for finding the least common denominator for rational expressions Here is it 1 Factor all the denominators 2 Write down each factor that appears at least once in any of the denominators Do NOT write down the power that is on each factor only write down the factor 3 Now for each factor written down in the previous step write down the largest power that occurs in all the denominators containing that factor 4 The product all the factors from the previous step is the least common denominator Let s work some examples Example 3 Perform the indicated operation a4 15 S1t39 0111011 6x2 3x5 2x3 b Lil Solution zl z2 c 2 y 2 1 y 2yl y l y2 2x 1 2 dx29x3x3 E 0 4 l1 y2 y Solution 4 l 5 a 6x2 3x5 2x3 For this problem there are coefficients on each term in the denominator so we ll first need the least common denominator for the coefficients This is 6 Now x by itself with a power of 1 is the only factor that occurs in any of the denominators So the least common denominator for this part is x with the largest power that occurs on all the x s in the problem which is 5 So the least common denominator for this set of rational expression is lcd 6x5 So we simply need to multiply each term by an appropriate quantity to get this in the denominator and then do the addition and subtraction Let s do that 2007 Paul Dawkins 48 httptutorialmathlamaredutermsaspx College Algebra L s4ltx3gt 12 sltsx2gt 6x2 3x5 2x3 6x2 x3 3x5 2 2x33x2 4x3 2 l5x2 6x5 6x5 6x5 4x3 2l5x2 Return to Problems In this case there are only two factors and they both occur to the first power and so the least common denominator is lcdz1z2 Now in determining what to multiply each part by simply compare the current denominator to the least common denominator and multiply top and bottom by whatever is missing In the first term we re missing a z 2 and so that s what we multiply the numerator and denominator by In the second term we re missing a z 1 and so that s what we ll multiply in that term Here is the work for this problem 2 z1 2z2 Z1z1 2z2QZ1z1 zl z2 zlz2 z2zl zlz2 The final step is to do any multiplication in the numerator and simplify that up as much as possible 2 z l 2z4zz 1 2z4 z2 1 z2 2z5 zl z2 zlz2 zlz2 zlz2 Be careful with minus signs and parenthesis when doing the subtraction Return to Problems y 2 3 yz 2yl y l y2 Let s first factor the denominators and determine the least common denominator y 2 3 y l2 v 1 v2 So there are two factors in the denominators a yI and a y2 So we will write both of those down and then take the highest power for each That means a 2 for the yI and a 1 for the y2 Here is the least common denominator for this rational expression lcd y2y l2 c Now determine what s missing in the denominator for each term multiply the numerator and denominator by that and then finally do the subtraction and addition 2007 Paul Dawkins 49 httptutorialmathlamaredutermsaspx College Algebra y 2 3 y0H 2O4Xy2 i y02 P 2y1 y 1 y2 yny2 o 4xy axy2 OwJfy2 Zyy2 2y ly23y l2 y07y2 Okay now let s multiply the numerator out and simplify In the last term recall that we need to do the multiplication prior to distributing the 3 through the parenthesis Here is the simplification work for this part y 2 3 Zy22y 2y2y 23y2 2yl yz2y1 y1 y2 y l2y2 y22y 2y2 2y43y2 6y3 y07y2 2y2 6y7 y07y2 Return to Problems 2x 1 2 m 2 x 9 x 3 x 3 Again factor the denominators and get the least common denominator 2x 1 2 x 3x3x3 x 3 The least common denominator is lcd x 3x3 Notice that the first rational expression already contains this in its denominator but that is okay In fact because of that the work will be slightly easier in this case Here is the subtraction for this problem 2x 1 2 2x lx 3 2x3 x2 9x3u 3x 3Xx3 x3Xx 3 x 3Xx3 2x x 3 2x3 x 3x4 3 2x x3 2x 6 Lx 3Lx3 x 3 x 3x3 Notice that we can actually go one step further here If we factor a minus out of the numerator we can do some canceling 2007 Paul Dawkins 50 httptutorialmathlamaredutermsaspx College Algebra 2x 1 2 x3 1 2629 x3 x 3x 3x3 x 3 Sometimes this kind of canceling will happen after the addition subtraction so be on the lookout for it Return to Problems 4 1 1 y 2 y The point of this problem is that l sitting out behind everything That isn t really the problem that it appears to be Let s first rewrite things a little here 4 l l y2yl e In this way we see that we really have three fractions here One of them simply has a denominator of one The least common denominator for this part is lcdyy2 Here is the addition and subtraction for this problem 4 1 1 4 y y 2 y y 2 E Iy2y yy2 yy2 4y y2yy2 yy2 2 gt Notice the set of parenthesis we added onto the second numerator as we did the subtraction We are subtracting off the whole numerator and so we need the parenthesis there to make sure we don t make any mistakes with the subtraction Here is the simplification for this rational expression 4 11 4y y 2y22y y25y 2 y2 y l yy2 yy2 Return to Problems 2007 Paul Dawkins 51 httptutorialmathlamaredutermsaspx College Algebra Complex Numbers The last topic in this section is not really related to most of what we ve done in this chapter although it is somewhat related to the radicals section as we will see We also won t need the material here all that often in the remainder of this course but there are a couple of sections in which we will need this and so it s best to get it out of the way at this point In the radicals section we noted that we won t get a real number out of a square root of a negative number For instance 9 isn t a real number since there is no real number that we can square and get a NEGATIVE 9 Now we also saw that if a and 9 were both positive then ab d E For a second let s forget that restriction and do the following Fz9 r 6 3 1 Now l is not a real number but if you think about it we can do this for any square root of a negative number For instance x100 100x LaxT1 3 6 x 290 290 1 etc So even if the number isn t a perfect square we can still always reduce the square root of a negative number down to the square root of a positive number which we or a calculator can deal with times x l So if we just had a way to deal with l we could actually deal with square roots of negative numbers Well the reality is that at this level there just isn t any way to deal with V 1 so instead of dealing with it we will make it go away so to speak by using the following definition i 1 Note that if we square both sides of this we get It will be important to remember this later on This shows that in some way i is the only number that we can square and get a negative value Using this definition all the square roots above become 3 3 100 10z x3 x3 i 290 x290 These are all examples of complex numbers The natural question at this point is probably just why do we care about this The answer is that as we will see in the next chapter sometimes we will run across the square roots of negative 2007 Paul Dawkins 52 httptutorialmathlamaredutermsaspx College Algebra numbers and we re going to need a way to deal with them So to deal with them we will need to discuss complex numbers So let s start out with some of the basic definitions and terminology for complex numbers The standard form of a complex number is a bi where a and b are real numbers and they can be anything positive negative zero integers fractions decimals it doesn t matter When in the standard form a is called the real part of the complex number and b is called the imaginary part of the complex number Here are some examples of complex numbers 35z xE101 z 16i 113 The last two probably need a little more explanation It is completely possible that a or b could be zero and so in 16139 the real part is zero When the real part is zero we often will call the complex number a purely imaginary number In the last example 113 the imaginary part is zero and we actually have a real number So thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers The conjugate of the complex number a bi is the complex number a bi In other words it is the original complex number with the sign on the imaginary part changed Here are some examples of complex numbers and their conjugates complex number conjugate 3 ii 3 ii 2 2 12 Si 12 5i 1 i 1i 45i 45i 101 101 Notice that the conjugate of a real number is just itself with no changes Now we need to discuss the basic operations for complex numbers We ll start with addition and subtraction The easiest way to think of adding andor subtracting complex numbers is to think of each complex number as a polynomial and do the addition and subtraction in the same way that we add or subtract polynomials Example 1 Perform the indicated operation and write the answers in standard form a 4 7i 5 101 b 4 121 3 151 C Si 9 i Solution There really isn t much to do here other than add or subtract Note that the parentheses on the first terms are only there to indicate that we re thinking of that term as a complex number and in 2007 Paul Dawkins 53 httptutorialmathlamaredutermsaspx College Algebra general aren t used a 47i5 10i1 3i b 412i 3 15i 4 121 3 15i 1 27 c 5i 9i5i 49 439 A 4 Next let s take a look at multiplication Again with one small difference it s probably easiest to just think of the complex numbers as polynomials so multiply them out as you would polynomials The one difference will come in the final step as we ll see Example 2 Multiply each of the following and write the answers in standard form a 7i 5 2i 1 b 1 5i 92i E c 4i23i E d 1 8i18i E Solution a So all that we need to do is distribute the 7139 through the parenthesis 7i 5 2 3514 14i2 Now this is where the small difference mentioned earlier comes into play This number is NOT in standard form The standard form for complex numbers does not have an 1392 in it This however is not a problem provided we recall that i2 1 Using this we get 7i 5 21 35 14e1 14 35 We also rearranged the order so that the real part is listed first Return to Problems b In this case we will FOIL the two numbers and we ll need to also remember to get rid of the 1392 1 5i 9 2 9 24 45 1Oi2 9 41 19 1 1 47 Return to Problems c Same thing with this one 4i23i8 12139 2139 3i8 14i 314 5 14 Return to Problems d Here s one final multiplication that will lead us into the next topic 1 8i1 8i 1 8i 8i 64i2 1 64 65 Don t get excited about it when the product of two complex numbers is a real number That can and will happen on occasion Return to Problems 2007 Paul Dawkins 54 httptutorialmathlamaredutermsaspx College Algebra In the final part of the previous example we multiplied a number by its conjugate There is a nice general formula for this that will be convenient when it comes to discussing division of complex numbers abia bi a2 abi abi b2i2 a2 I92 So when we multiply a complex number by its conjugate we get a real number given by a bia bi a2 2 Now we gave this formula with the comment that it will be convenient when it came to dividing complex numbers so let s look at a couple of examples Example 3 Write each of the following in standard form 3 1 a Solut1on 2 3 b J 9 1 c 8 1 1 1 d 6 9l Solution 21 Solution So in each case we are really looking at the division of two complex numbers The main idea here however is that we want to write them in standard form Standard form does not allow for any i39s to be in the denominator So we need to get the i39s out of the denominator This is actually fairly simple if we recall that a complex number times its conjugate is a real number So if we multiply the numerator and denominator by the conjugate of the denominator we will be able to eliminate the i from the denominator Now that we ve figured out how to do these let s go ahead and work the problems 3 3 i 2 7i6 23i7f 123i 1 23 a27i27i2 7i 2272 53 5 51 Notice that to officially put the answer in standard form we broke up the fraction into the real and imaginary parts Return to Problems 3 3 9i273i 27il 91 9 i9i 9212 82 82 Return to Problems 12 2 C 81 81 l8z 161 168z ii 12i12i1 2i 1222 5 U1 Return to Problems 2007 Paul Dawkins 55 httptutorialmathlamaredutermsaspx College Algebra d This one is a little different from the previous ones since the denominator is a pure imaginary number It can be done in the same manner as the previous ones but there is a slightly easier way to do the problem First break up the fraction as follows 6 9139 6 E 2 2 21 21 2139 i 2 Now we want the 139 out of the denominator and since there is only an 139 in the denominator of the first term we will simply multiply the numerator and denominator of the first term by an 139 691 31 9 3i 9 31 9 9 21 11 2 1222721 3 E 3i Return to Problems The next topic that we want to discuss here is powers of 139 Let s just take a look at what happens when we start looking at various powers of 139 1 l 1392 l 1311 1 132 1 1412 1 l2 1 1421 152114 1 151 1612 1 1a1 1 162 1 1711 1 172 1 Can you see the pattern All powers if 139 can be reduced down to one of four possible answers and they repeat every four powers This can be a convenient fact to remember We next need to address an issue on dealing with square roots of negative numbers From the section on radicals we know that we can do the following 645 49 15 23 6 In other words we can break up products under a square root into a product of square roots provided both numbers are positive It turns out that we can actually do the same thing if one of the numbers is negative For instance 61392LT6 49 E5 2013 61 However if BOTH numbers are negative this won t work anymore as the following shows 64 V 49 9 45 21e31612 6 We can summarize this up as a set of rules If a and 9 are both positive numbers then 2007 Paul Dawkins 56 httptutorialmathlamaredutermsaspx College Algebra xExZE x3xZ Tb xEx312 J35 ab Why is this important enough to worry about Consider the following example Example 4 Multiply the following and write the answer in standard form 2 x 1oo1 36 Solution If we where to multiply this out in its present form we would get 2 1oo13 2 2R 1o0 36 lOO Now if we were not being careful we would probably combine the two roots in the final term into one which can t be done So there is a general rule of thumb in dealing with square roots of negative numbers When faced with them the first thing that you should always do is convert them to complex number If we follow this rule we will always get the correct answer So let s work this problem the way it should be worked 2 E13 2 10i16i 2 21 6Oi2 62 21 The rule of thumb given in the previous example is important enough to make again When faced with square roots of negative numbers the first thing that you should do is convert them to complex numbers There is one final topic that we need to touch on before leaving this section As we noted back in the section on radicals even though 3 there are in fact two numbers that we can square to get 9 We can square both 3 and 3 The same will hold for square roots of negative numbers As we saw earlier 9 3i As with square roots of positive numbers in this case we are really asking what did we square to get 9 Well it s easy enough to check that 3139 is correct 3z2 9 9 However that is not the only possibility Consider the following 3z2 3 i2 9 9 and so if we square 31 we will also get 9 So when taking the square root of a negative number there are really two numbers that we can square to get the number under the radical However we will ALWAYS take the positive number for the value of the square root just as we do with the square root of positive numbers 2007 Paul Dawkins 57 httptutorialmathlamaredutermsaspx College Algebra Solving Equations and Inequalities Introduction In this chapter we will look at one of the standard topics in any Algebra class The ability to solve equations andor inequalities is very important and is used time and again both in this class and in later classes We will cover a wide variety of solving topics in this chapter that should cover most of the basic equationsinequalitiestechniques that are involved in solving Here is a brief listing of the material covered in this chapter Solutions and Solution Sets We introduce some of the basic notation and ideas involved in solving in this section Linear Equations In this section we will solve linear equations including equations with rational expressions Applications of Linear Equations We will take a quick look at applications of linear equations in this section Equations With More Than One Variable Here we will look at solving equations with more than one variable in them Quadratic Equations Part I In this section we will start looking at solving quadratic equations We will look at factoring and the square root property in this section Quadratic Eguations Part II We will finish up solving quadratic equations in this section We will look at completing the square and quadratic formula in this section Quadratic Equations A Summary We ll give a procedure for determining which method to use in solving quadratic equations in this section We will also take a quick look at the discriminant Applications of Quadratic Equations Here we will revisit some of the applications we saw in the linear application section only this time they will involve solving a quadratic equation Eguations Reducible to Quadratic Form In this section we will solve equations that can be reduced to quadratic in form Eguations with Radicals Here we will solve equations with square roots in them Linear Inequalities We will start solving inequalities in this section by looking at linear inequalities Polynomial Inequalities In this section we will look at solving inequalities that contain polynomials Rational Inequalities Here we will solve inequalities involving rational expressions Absolute Value Equations We will officially define absolute value in this section and solve equations that contain absolute value Absolute Value Inequalities We will solve inequalities that involve absolute value in this section 2007 Paul Dawkins 5 8 httptutorialmathlamaredutermsaspx College Algebra Solutions and Solution Sets We will start off this chapter with a fairly short section with some basic terminology that we use on a fairly regular basis in solving equations and inequalities First a solution to an equation or inequality is any number that when plugged into the equationinequality will satisfy the equationinequality So just what do we mean by satisfy Let s work an example or two to illustrate this Example 1 Show that each of the following numbers are solutions to the given equation or inequality a x3 in x2 90 J b y 8 in 3yl 4y 5 J C z 1 in 2z 5 S 4z ml d Z 5 in 2Z 5 S 4z Solution Solution a We first plug the proposed solution into the equation 32 9 0 9 9 0 O 0 OK So What we are asking here is does the right side equal the left side after we plug in the proposed solution That is the meaning of the above the equal sign in the first line Since the right side and the left side are the same we say that x 3 satisfies the equation Return to Problems b So we want to see if y 8 satisfies the equation First plug the value into the equation 381 48 5 2727 OK So y 8 satisfies the equation and so is a solution Return to Problems c In this case we ve got an inequality and in this case satisfy means something slightly different In this case we will say that a number will satisfy the inequality if after plugging it in We get a true inequality as a result Let s check z 1 21 54q 834 OK So 8 is less than or equal to 4 in fact it s less than and so we have a true inequality Therefore z 1 will satisfy the inequality and hence is a solution Return to Problems 2007 Paul Dawkins 59 httptutorialmathlamaredutermsaspx College Algebra d This is the same inequality with a different value so let s check that 2 5 54 5 20 S 20 OK In this case 20 is less than or equal to 20 in this case it s equal and so again we get a true inequality and so z 5 satisfies the inequality and so will be a solution Return to Problems We should also do a quick example of numbers that aren t solution so we can see how these will work as well Example 2 Show that the following numbers aren t solutions to the given equation or inequality a y 2 in 3y 1 4y 5 Solution h z 42 in 2z 5S4z mien Solution a In this case we do essentially the same thing that we did in the previous example Plug the number in and show that this time it doesn t satisfy the equation For equations that will mean that the right side of the equation will not equal the left side of the equation 3 21 4 2 5 3 13 NOT OK So 3 is not the same as 13 and so the equation isn t satisfied Therefore y 2 isn t a solution to the equation Return to Problems b This time we ve got an inequality A number will not satisfy an inequality if we get an inequality that isn t true after plugging the number in 2 12 54 12 34g 48 NOT ox In this case 34 is NOT less than or equal to 48 and so the inequality isn t satisfied Therefore z 12 is not a solution to the inequality Return to Problems Now there is no reason to think that a given equation or inequality will only have a single solution In fact as the first example showed the inequality 2z 5 S 4z has at least two solutions Also you might have noticed that x 3 is not the only solution to x2 9 0 In this case x 3 is also a solution We call the complete set of all solutions the solution set for the equation or inequality There is also some formal notation for solution sets although we won t be using it all that often in this course Regardless of that fact we should still acknowledge it 2007 Paul Dawkins 60 httptutorialmathlamaredutermsaspx College Algebra For equations we denote the solution set by enclosing all the solutions is a set of braces For the two equations we looked at above here are the solution sets 3y 1 4y 5 Solution Set 8 x2 9 0 Solution Set 33 For inequalities we have a similar notation Depending on the complexity of the inequality the solution set may be a single number or it may be a range of numbers If it is a single number then we use the same notation as we used for equations If the solution set is a range of numbers as the one we looked at above is we will use something called set builder notation Here is the solution set for the inequality we looked at above zz2 5 This is read as The set of all z such that z is greater than or equal to 5 Most of the inequalities that we will be looking at will have simple enough solution sets that we often just shorthand this as z2 5 There is one final topic that we need to address as far as solution sets go before leaving this section Consider the following equation and inequality x2l0 x2lt0 If we restrict ourselves to only real solutions which we won t always do then there is no solution to the equation Squaring x makes x greater than equal to zero then adding 1 onto that means that the left side is guaranteed to be at least 1 In other words there is no real solution to this equation For the same basic reason there is no solution to the inequality Squaring any real x makes it positive or zero and so will never be negative We need a way to denote the fact that there are no solutions here In solution set notation we say that the solution set is empty and denote it with the symbol Q This symbol is often called the empty set We now need to make a couple of final comments before leaving this section In the above discussion of empty sets we assumed that were only looking for real solutions While that is what we will be doing for inequalities we won t be restricting ourselves to real solutions with equations Once we get around to solving quadratic equations which x2 l 0 is we will allow solutions to be complex numbers and in the case looked at above there are complex solutions to x2 l 0 If you don t know how to find these at this point that is fine we will be covering that material in a couple of sections At this point just accept that x2 l 0 does have complex solutions 2007 Paul Dawkins 61 httptutorialmathlamaredutermsaspx College Algebra Finally as noted above we won t be using the solution set notation much in this course It is a nice notation and does have some use on occasion especially for complicated solutions However for the vast majority of the equations and inequalities that we will be looking at will have simple enough solution sets that it s just easier to write down the solutions and let it go at that Therefore that is what we will not be using the notation for our solution sets However you should be aware of the notation and know what it means 2007 Paul Dawkins 62 httptutorialmathlamaredutermsaspx College Algebra Linear Equations We ll start off the solving portion of this chapter by solving linear equations A linear equation is any equation that can be written in the form ax b 0 where a and 9 are real numbers and x is a variable This form is sometimes called the standard form of a linear equation Note that most linear equations will not start off in this form Also the variable may or may not be an x so don t get too locked into always seeing an x there To solve linear equations we will make heavy use of the following facts 1 If a b then a c b c for any c All this is saying is that we can add a number c to both sides of the equation and not change the equation 2 If a b then a c b c for any c As with the last property we can subtract a number c from both sides of an equation 3 If a b then ac bc for any c Like addition and subtraction we can multiply both sides of an equation by a number c without changing the equation I 4 If a b then E for any nonzero c We can divide both sides of an equation by a c c nonzero number c without changing the equation These facts form the basis of almost all the solving techniques that we ll be looking at in this chapter so it s very important that you know them and don t forget about them One way to think of these rules is the following What we do to one side of an equation we have to do to the other side of the equation If you remember that then you will always get these facts correct In this section we will be solving linear equations and there is a nice simple process for solving linear equations Let s first summarize the process and then we will work some examples Process for Solving Linear Equations 1 If the equation contains any fractions use the least common denominator to clear the fractions We will do this by multiplying both sides of the equation by the LCD Also if there are variables in the denominators of the fractions identify values of the variable which will give division by zero as we will need to avoid these values in our solution 2 Simplify both sides of the equation This means clearing out any parenthesis and combining like terms 3 Use the first two facts above to get all terms with the variable in them on one side of the equations combining into a single term of course and all constants on the other side 4 If the coefficient of the variable is not a one use the third or fourth fact above this will 2007 Paul Dawkins 63 httptutorialmathlamaredutermsaspx College Algebra depend on just what the number is to make the coefficient a one Note that we usually just divide both sides of the equation by the coefficient if it is an integer or multiply both sides of the equation by the reciprocal of the coefficient if it is a fraction 5 VERIFY YOUR ANSWER This is the final step and the most often skipped step yet it is probably the most important step in the process With this step you can know whether or not you got the correct answer long before your instructor ever looks at it We verify the answer by plugging the results from the previous steps into the original equation It is very important to plug into the original equation since you may have made a mistake in the very first step that led you to an incorrect answer Also if there were fractions in the problem and there were values of the variable that give division by zero recall the first step it is important to make sure that one of these values didn t end up in the solution set It is possible as we ll see in an example to have these values show up in the solution set Let s take a look at some examples Example 1 Solve each of the following equations a 3x52 6 96 2x J m 2 3 5 10 y 2y 6 yz 6y9 1 2m 1 7 Solution c Solution 2z 3 d j 2 S1 39 Z3 Z1O out1on Solution In the following problems we will describe in detail the first problem and the leave most of the explanation out of the following problems a 3x5 2 6 96 2x For this problem there are no fractions so we don t need to worry about the first step in the process The next step tells to simplify both sides So we will clear out any parenthesis by multiplying the numbers through and then combine like terms 3x52 6 96 2x 3xl5 12 266 2x 3xl5 12 4x The next step is to get all the x s on one side and all the numbers on the other side Which side the x s go on is up to you and will probably vary with the problem As a rule of thumb we will usually put the variables on the side that will give a positive coefficient This is done simply because it is often easy to lose track of the minus sign on the coefficient and so if we make sure it is positive we won t need to worry about it 2007 Paul Dawkins 64 httptutorialmathlamaredutermsaspx College Algebra So for our case this will mean adding 4x to both sides and subtracting 15 from both sides Note as well that while we will actually put those operations in this time we normally do these operations in our head 3x15 12 4x 3x15 l54x 12 4x4x l5 7x 27 The next step says to get a coefficient of 1 in front of the x In this case we can do this by dividing both sides by a 7 7x 27 7 7 27 x 7 27 Now if we Ve done all of our work correct x 7 is the solution to the equation The last and final step is to then check the solution As pointed out in the process outline we need to check the solution in the original equation This is important because we may have made a mistake in the Very first step and if we did and then checked the answer in the results from that step it may seem to indicate that the solution is correct when the reality will be that we don t have the correct answer because of the mistake that we originally made The problem of course is that with this solution that checking might be a little messy Let s do it anyway 2 9 2 27 3 7 5 i 2 6 7 2 7 7 7 3 2 E 54 7 7 7 24 24 OK 7 7 So we did our work correctly and the solution to the equation is 27 x 7 Note that we didn t use the solution set notation here For single solutions we will rarely do that in this class However if we had wanted to the solution set notation for this problem would be 31 Before proceeding to the next problem let s first make a quick comment about the messiness of this answer Do NOT expect all answers to be nice simple integers While we do try to keep 2007 Paul Dawkins 65 httptutorialmathlamaredutermsaspx College Algebra most answer simple often they won t be so do NOT get so locked into the idea that an answer must be a simple integer that you immediately assume that you ve made a mistake because of the messiness of the answer Return to Problems m 2 2m b 1 7 Okay with this one we won t be putting quite as much explanation into the problem In this case we have fractions so to make our life easier we will multiply both sides by the LCD which is 21 in this case After doing that the problem will be very similar to the previous problem Note as well that the denominators are only numbers and so we won t need to worry about division by zero issues Let s first multiply both sides by the LCD 7m 2 21 2m3 Be careful to correctly distribute the 21 through the parenthesis on the left side Everything inside the parenthesis needs to be multiplied by the 21 before we simplify At this point we ve got a problem that is similar the previous problem and we won t bother with all the explanation this time 7m 2212m3 7m l42l6m 7m7 6m m 7 So it looks like m 7 is the solution Let s verify it to make sure 21 3 7 9 314 7 7 31 2 2 OK So it is the solution Return to Problems 5 10 y 2 y 6 y2 6 y 9 This one is similar to the previous one except now we ve got variables in the denominator So to get the LCD we ll first need to completely factor the denominators of each rational expression c 2007 Paul Dawkins 66 httptutorialmathlamaredutermsaspx College Algebra 5 lO y 203 y32 So it looks like the LCD is 2y 32 Also note that we will need to avoid y 3 since if we plugged that into the equation we would get division by zero Now outside of the y s in the denominator this problem works identical to the previous one so let s do the work G 3 5y 3 2lO y 5y l520 2y 7y35 y5 Now the solution is not y 3 so we won t get division by zero with the solution which is a good thing Finally let s do a quick verification 5 105 25 6 2 2 OK 4 4 So we did the work correctly Return to Problems d Zz 3 z3 ZlO In this case it looks like the LCD is z 3z 10 and it also looks like we will need to avoid z 3 and z 10 to make sure that we don t get division by zero Let s get started on the work for this problem 2z 3 z3zlOjZ10 24z 3z 19 2zz l03z 3 z z l0 2z2 2Oz3z 9 Qz2 7z 30 At this point let s pause and acknowledge that we ve got a z2 in the work here Do not get excited about that Sometimes these will show up temporarily in these problems You should only worry about it if it is still there after we finish the simplification work So let s finish the problem 2007 Paul Dawkins 67 httptutorialmathlamaredutermsaspx College Algebra 2oz3z 9 14z60 202 Hz 51 519z 51 jjz 9 17 Z 3 17 Notice that the z2 did in fact cancel out Now if we did our work correctly z should be the solution since it is not either of the two values that will give division by zero Let s verify this 21 3 1 2 1 73 1110 3 3 E 3 1 3 T 6 2 3 3 3 3 3 2 3 26 13 1711 0K 13 13 The checking can be a little messy at times but it does mean that we KNOW the solution is correct Return to Problems Okay in the last couple of parts of the previous example we kept going on about watching out for division by zero problems and yet we never did get a solution where that was an issue So we should now do a couple of those problems to see how they work Example 2 Solve each of the following equations a 2 x s 1 t39 0 U 1011 x 2 x2 5x 6 b L 4 E Solution xl xl Solution 2 x a x 2 x2 5x 6 The first step is to factor the denominators to get the LCD 2007 Paul Dawkins 68 httptutorialmathlamaredutermsaspx College Algebra 2 Z x x2 x2x3 So the LCD is x 2x 3 and we will need to avoid x 2 and x 3 so We don t get division by zero Here is the work for this problem x2x3xJ2r2j jx2x3 2x3 x 2x6 x 3x 6 x 2 So we get a solution that is in the list of numbers that we need to avoid so we don t get division by zero and so we can t use it as a solution However this is also the only possible solution That is okay This just means that this equation has no solution Return to Problems xl xl The LCD for this equation is x 1 and we will need to avoid x 1 so we don t get division by zero Here is the work for this equation jxl 4 x2Jx 1 24x lB 2 x 24x 44 2x 22x 4 22x lx So we once again arrive at the single value of x that we needed to avoid so we didn t get division by zero Therefore this equation has no solution Return to Problems So as we ve seen we do need to be careful with division by zero issues when we start off with equations that contain rational expressions At this point we should probably also acknowledge that provided we don t have any division by zero issues such as those in the last set of examples linear equations will have exactly one solution We will never get more than one solution and the only time that we won t get any solutions is if we run across a division by zero problems with the solution 2007 Paul Dawkins 69 httptutorialmathlamaredutermsaspx College Algebra Before leaving this section we should note that many of the techniques for solving linear equations will show up time and again as we cover different kinds of equations so it Very important that you understand this process 2007 Paul Dawkins 70 httptutorialmathlamaredutermsaspx College Algebra Application of Linear Equations We now need to discuss the section that most students hate We need to talk about applications to linear equations Or put in other words we will now start looking at story problems or word problems Throughout history students have hated these It is my belief however that the main reason for this is that students really don t know how to work them Once you understand how to work them you ll probably find that they aren t as bad as they may seem on occasion So we ll start this section off with a process for working applications Process for Working StoryWord Problems 1 READ THE PROBLEM 2 READ THE PROBLEM AGAIN Okay this may be a little bit of overkill here However the point of these first two steps is that you must read the problem This step is the MOST important step but it is also the step that most people don t do properly You need to read the problem very carefully and as many times as it takes You are only done with this step when you have completely understood what the problem is asking you to do This includes identifying all the given information and identifying what you being asked to find Again it can t be stressed enough that you Ve got to carefully read the problem Sometimes a single word can completely change how the problem is worked If you just skim the problem you may well miss that very important word 3 Represent one of the unknown quantities with a variable and try to relate all the other unknown quantities if there are any of course to this variable 4 If applicable sketch a figure illustrating the situation This may seem like a silly step but it can be incredibly helpful with the next step on occasion 5 Form an equation that will relate known quantities to the unknown quantities To do this make use of known formulas and often the figure sketched in the previous step can be used to determine the equation 6 Solve the equation formed in the previous step and write down the answer to all the questions It is important to answer all the questions that you were asked Often you will be asked for several quantities in the answer and the equation will only give one of them 7 Check your answer Do this by plugging into the equation but also use intuition to make sure that the answer makes sense Mistakes can often be identified by acknowledging that the answer just doesn t make sense Let s start things off with a couple of fairly basic examples to illustrate the process Note as well that at this point it is assumed that you are capable of solving fairly simple linear equations and so not a lot of detail will be given for the actual solution stage The point of this section is more on the set up of the equation than the solving of the equation 2007 Paul Dawkins 71 httptutorialmathlamaredutermsaspx College Algebra Example 1 In a certain Algebra class there is a total of 350 possible points These points come from 5 homework sets that are worth 10 points each and 3 hour exams that are worth 100 points each A student has received homework scores of 4 8 7 7 and 9 and the first two exam scores are 78 and 83 Assuming that grades are assigned according to the standard scale and there are no weights assigned to any of the grades is it possible for the student to receive an A in the class and if so what is the minimum score on the third exam that will give an A What about a B Solution Okay let s start off by defining p to be the minimum required score on the third exam Now let s recall how grades are set Since there are no weights or anything on the grades the grade will be set by first computing the following percentage actual points total possible points 2 grade percentage Since we are using the standard scale if the grade percentage is 09 or higher the student will get an A Likewise if the grade percentage is between 08 and 09 the student will get a B We know that the total possible points is 350 and the student has a total points including the third exam of 487797883p196 p The smallest possible percentage for an A is 09 and so if p is the minimum required score on the third exam for an A we will have the following equation l96p 09 350 This is a linear equation that we will need to solve for p 196p09350 315 gtp 315 196119 So the minimum required score on the third exam is 119 This is a problem since the exam is worth only 100 points In other words the student will not be getting an A in the Algebra class Now let s check if the student will get a B In this case the minimum percentage is 08 So to find the minimum required score on the third exam for a B we will need to solve 196 p Z O 8 350 Solving this for p gives l96p O8350 80 gt p 289 196 84 So it is possible for the student to get a B in the class All that the student will need to do is get at least an 84 on the third exam 2007 Paul Dawkins 72 httptutorialmathlamaredutermsaspx College Algebra Example 2 We want to build a set of shelves The width of the set of shelves needs to be 4 times the height of the set of shelves and the set of shelves must have three shelves in it If there are 72 feet of wood to use to build the set of shelves what should the dimensions of the set of shelves be Solution We will first define x to be the height of the set of shelves This means that 4x is width of the set of shelves In this case we definitely need to sketch a figure so we can correctly set up the equation Here it is 4 Now we know that there are 72 feet of wood to be used and we will assume that all of it will be used So we can set up the following word equation length of length of 72 vertical pieces horizontal pieces It is often a good idea to first put the equation in words before actually writing down the equation as we did here At this point we can see from the figure there are two vertical pieces each one has a length of x Also there are 4 horizontal pieces each with a length of 4x So the equation is then 44x2x 72 l6x2x72 l8x72 x4 So it looks like the height of the set of shelves should be 4 feet Note however that we haven t actually answered the question however The problem asked us to find the dimensions This means that we also need the width of the set of shelves The width is 44l6 feet So the dimensions will need to be 4x16 feet Pricing Problems The next couple of problems deal with some basic principles of pricing Example 3 A calculator has been marked up 15 and is being sold for 7850 How much did the store pay the manufacturer of the calculator Solution First let s define p to be the cost that the store paid for the calculator The stores markup on the calculator is 15 This means that 0l5p has been added on to the original price p to get the amount the calculator is being sold for In other words we have the following equation p 0l5p 7850 2007 Paul Dawkins 73 httptutorialmathlamaredutermsaspx College Algebra that we need to solve for p Doing this gives ll5p7850gt p 6826087 The store paid 6826 for the calculator Note that since we are dealing with money we rounded the answer down to two decimal places Example 4 A shirt is on sale for 1500 and has been marked down 35 How much was the shirt being sold for before the sale Solution This problem is pretty much the opposite of the previous example Let s start with defining p to be the price of the shirt before the sale It has been marked down by 35 This means that 035p has been subtracted off from the original price Therefore the equation and solution is p 035 p 1500 065 p 1500 1500 j 230169 065 So with rounding it looks like the shirt was originally sold for 2308 DistanceRate Problems These are some of the standard problems that most people think about when they think about Algebra word problems The standard formula that we will be using here is Distance Rate Time All of the problems that we ll be doing in this set of examples will use this to one degree or another and often more than once as we will see Example 5 Two cars are 500 miles apart and moving directly towards each other One car is moving at a speed of 100 mph and the other is moving at 70 mph Assuming that the cars start moving at the same time how long does it take for the two cars to meet Solution Let s let t represent the amount of time that the cars are traveling before they meet Now we need to sketch a figure for this one This figure will help us to write down the equation that we ll need to solve Car 51 Meetmg Cs E IIIIIII mph pmm TI mph Distance ef Cs 5 Distance ef Cs E EIIIIII miles From this figure we can see that the Distance Car A travels plus the Distance Car B travels must equal the total distance separating the two cars 500 miles 2007 Paul Dawkins 74 httptutorialmathlamaredutermsaspx College Algebra Here is the word equation for this problem in two separate forms Distance Distance 500 of Car A of Car B Rate of Time of Rate of Time of 500 Car A Car A Car B Car B We used the standard formula here twice once for each car We know that the distance a car travels is the rate of the car times the time traveled by the car In this case we know that Car A travels at 100 mph for t hours and that Car B travels at 70 mph for t hours as well Plugging these into the word equation and solving gives us 100t 70t 500 170t 500 t 5E 294l176 hrs 170 So they will travel for approximately 294 hours before meeting Example 6 Repeat the previous example except this time assume that the faster car will start 1 hour after slower car starts Solution For this problem we are going to need to be careful with the time traveled by each car Let s let t be the amount of time that the slower travel car travels Now since the faster car starts out 1 hour after the slower car it will only travel for t 1 hours Now since we are repeating the problem from above the figure and word equation will remain identical and so we won t bother repeating them here The only difference is what we substitute for the time traveled for the faster car Instead of t as we used in the previous example we will use t 1 since it travels for one hour less that the slower car Here is the equation and solution for this example 100r 170r 500 100r 10070r 500 170t 600 t 359412 hrs 170 In this case the slower car will travel for 353 hours before meeting while the faster car will travel for 253 hrs 1 hour less than the faster car 2007 Paul Dawkins 75 httptutorialmathlamaredutermsaspx College Algebra Example 7 Two boats start out 100 miles apart and start moving to the right at the same time The boat on the left is moving at twice the speed as the boat on the right Five hours after starting the boat on the left catches up with the boat on the right How fast was each boat moving Solution Let s start off by letting r be the speed of the boat on the right the slower boat This means that the boat to the left the faster boat is moving at a speed of 2r Here is the figure for this situation Ecat Eu Sp eetl r a 1 E3 I Distance Beat E Travels F Distance Beat A Travels Beat 5 Beat 3 Speed Er Catches Ecat E Here From the figure it looks like we ve got the following word equation Distance Distance 100 of Boat B of Boat A Upon plugging in the standard formula for the distance gives 100 Rate of Time of Rate of Time of Boat B Boat B Boat A Boat A For this problem we know that the time each is 5 hours and we know that the rate of Boat A is Zr and the rate of Boat B is r Plugging these into the work equation and solving gives 100 r5 2r5 100 5r lOr 100 5r 20 r So the slower boat is moving at 20 mph and the faster boat is moving at 40 mph twice as fast WorkRate Problems These problems are actually variants of the DistanceRate problems that we just got done working The standard equation that will be needed for these problems is Portion of job Work Time Spent x done in given time Rate Working As you can see this formula is very similar to the formula we used above 2007 Paul Dawkins 76 httptutorialmathlamaredutermsaspx College Algebra Example 8 An office has two envelope stuffing machines Machine A can stuff a batch of envelopes in 5 hours while Machine B can stuff a batch of envelopes in 3 hours How long would it take the two machines working together to stuff a batch of envelopes Solution Let t be the time that it takes both machines working together to stuff a batch of envelopes The word equation for this problem is Portion of job Portion of job 1 J b o done by Machine A done by Machine B Work Rate Time Spent Work Rate Time Spent 1 of Machine A Working of Machine B Working We know that the time spent working is t however we don t know the work rate of each machine To get these we ll need to use the initial information given about how long it takes each machine to do the job individually We can use the following equation to get these rates Work Time Spent 1 Job gtlt Rate Working Let s start with Machine A 1 Job Work Rate of A 5 x Wbark Rate of A Now Machine B 1 Job Work Rate of B 3 x Work Rate of B Plugging these quantities into the main equation above gives the following equation that we need to solve t gt 1 Multiplying both sides by 15 3t 5t 15 St 15 t 18v75 hours So it looks like it will take the two machines working together 1875 hours to stuff a batch of envelopes Example 9 Mary can clean an office complex in 5 hours Working together John and Mary can clean the office complex in 35 hours How long would it take John to clean the office complex by himself Solution Let t be the amount of time it would take John to clean the office complex by himself The basic word equation for this problem is 2007 Paul Dawkins 77 httptutorialmathlamaredutermsaspx College Algebra Portion of job Portion of job 1 Job done by Mary done by John Work Rate Time Spent Work Rate Time Spent of Mary Working of John Working This time we know that the time spent working together is 35 hours We now need to find the work rates for each person We ll start with Mary 1 Job Work Rate of Mary x 5 gt Work Rate of Mary 1 5 Now we ll find the work rate of John Notice however that since we don t know how long it will take him to do the job by himself we aren t going to be able to get a single number for this That is not a problem as we ll see in a second 1 Job Work Rate of Johngtlt t gt Work Rate of John 1 I Notice that we Ve managed to get the work rate of John in terms of the time it would take him to do the job himself This means that once we solve the equation above we ll have the answer that we want So let s plug into the work equation and solve for the time it would take John to do the job by himself 35 35 1 Multiplying both sides by St 35t355 5t 175 15t t gt t k167 hrs So it looks like it would take John 1167 hours to clean the complex by himself Mixing Problems This is the final type of problems that we ll be looking at in this section We are going to be looking at mixing solutions of different percentages to get a new percentage The solution will consist of a secondary liquid mixed in with water The secondary liquid can be alcohol or acid for instance The standard equation that we ll use here will be the following Amount of secondary Percentage of J Volume of J x liquid in the water Solution Solution Note as well that the percentage needs to be a decimal So if we have an 80 solution we will need to use 080 2007 Paul Dawkins 78 httptutorialmathlamaredutermsaspx College Algebra Example 10 How much of a 50 alcohol solution should we mix with 10 gallons of a 35 solution to get a 40 solution Solution Okay let x be the amount of 50 solution that we need This means that there will be x 10 gallons of the 40 solution once we re done mixing the two Here is the basic work equation for this problem Amount of alcohol Amount of alcohol in 50 Solution in 35 Solution Amount of alcohol in 40 Solution Volume of Volume of Volume of 05 035 04 50 Solution 35 Solution 40 Solution Now plug in the Volumes and solve for x 05x 03510 04x 10 05x 35 04x 4 01x 05 x Sgallons 01 So we need 5 gallons of the 50 solution to get a 40 solution Example 11 We have a 40 acid solution and we want 75 liters of a 15 acid solution How much water should we put into the 40 solution to do this Solution Let x be the amount of water we need to add to the 40 solution Now we also don t how much of the 40 solution we ll need However since we know the final Volume 75 liters we will know that we will need 75 x liters of the 40 solution Here is the word equation for this problem in 40 Solution in 15 Solution Amount of acid Amount of acid in the water Amount of acid J Notice that in the first term we used the Amount of acid in the water This might look a little weird to you because there shouldn t be any acid in the water However this is exactly what we want The basic equation tells us to look at how much of the secondary liquid is in the water So this is the correct wording When we plug in the percentages and Volumes we will think of the water as a 0 percent solution since that is in fact what it is So the new word equation is Volume Volume of Volume of 0 04 015 of Water 40 Solution 15 Solution 2007 Paul Dawkins 79 httptutorialmathlamaredutermsaspx College Algebra Do not get excited about the zero in the first term This is okay and will not be a problem Let s now plug in the Volumes and solve for x Ox 0475 x 01575 30 04x 1125 1875 04x x 3975 468 liters 04 So we need to add in 46875 liters of water to 28125 liters of a 40 solution to get 75 liters of a 15 solution 2007 Paul Dawkins 80 httptutorialmathlamaredutermsaspx College Algebra Equations With More Than One Variable In this section we are going to take a look at a topic that often doesn t get the coverage that it deserves in an Algebra class This is probably because it isn t used in more than a couple of sections in an Algebra class However this is a topic that can and often is used extensively in other classes What we ll be doing here is solving equations that have more than one variable in them The process that we ll be going through here is very similar to solving linear equations which is one of the reasons why this is being introduced at this point There is however one exception to that Sometimes as we will see the ordering of the process will be different for some problems Here is the process in the standard order 1 Multiply both sides by the LCD to clear out any fractions 2 Simplify both sides as much as possible This will often mean clearing out parenthesis and the like 3 Move all terms containing the variable we re solving for to one side and all terms that don t contain the variable to the opposite side 4 Get a single instance of the variable we re solving for in the equation For the types of problems that we ll be looking at here this will almost always be accomplished by simply factoring the variable out of each of the terms 5 Divide by the coefficient of the variable This step will make sense as we work problems Note as well that in these problems the coefficient will probably contain things other than numbers It is usually easiest to see just what we re going to be working with and just how they work with an example We will also give the basic process for solving these inside the first example Example 1 Solve A Pl rt for r Solution What we re looking for here is an expression in the form r Equation involving numbers A P and t In other words the only place that we want to see an r is on the left side of the equal sign all by itself There should be no other r s anywhere in the equation The process given above should do that for us Okay let s do this problem We don t have any fractions so we don t need to worry about that To simplify we will multiply the P through the parenthesis Doing this gives A P Bart Now we need to get all the terms with an r on them on one side This equation already has that set up for us which is nice Next we need to get all terms that don t have an r in them to the other side This means subtracting a P from both sides A PP As a final step we will divide both sides by the coefficient of r Also as noted in the process 2007 Paul Dawkins 81 httptutorialmathlamaredutermsaspx College Algebra listed above the coefficient is not a number In this case it is Pt At this stage the coefficient of a variable is simply all the stuff that multiplies the variable A P A P r gt r Pt Pt To get a final answer we went ahead and ipped the order to get the answer into a more standard form We will work more examples in a bit However let s note a couple things first These problems tend to seem fairly difficult at first but if you think about it all we really did was use exactly the same process we used to solve linear equations The main difference of course is that there is more mess in this process That brings us to the second point Do not get excited about the mess in these problems The problems will on occasion be a little messy but the steps involved are steps that you can do Finally the answer will not be a simple number but again it will be a little messy often messier than the original equation That is okay and expected Let s work some more examples Example 2 Solve V 5aR for R m Solution This one is fairly similar to the first example However it does work a little differently Recall from the first example that we made the comment that sometimes the ordering of the steps in the process needs to be changed Well that s what we re going to do here The first step in the process tells us to clear fractions However since the fraction is inside a set of parenthesis let s first multiply the m through the parenthesis Notice as well that if we multiply the m through first we will in fact clear one of the fractions out automatically This will make our work a little easier when we do clear the fractions out V E 5611 I Now clear fractions by multiplying both sides by I We ll also go ahead move all terms that don t have an R in them to the other side Vb m 5abR Vb m 5abR Be careful to not lose the minus sign in front of the 5 It s very easy to lose track of that The final step is to then divide both sides by the coefficient of the R in this case 5ab 1R lb m Vb m Vb m Vbm m Vb 5611 56119 56119 56119 Sab Notice as well that we did some manipulation of the minus sign that was in the denominator so that we could simplify the answer somewhat In the previous example we solved for R but there is no reason for not solving for one of the other variables in the problems For instance consider the following example 2007 Paul Dawkins 82 httptutorialmathlamaredutermsaspx College Algebra Example 3 Solve V 5aR for b m Solution The first couple of steps are identical here First we will multiply the m through the parenthesis and then we will multiply both sides by 9 to clear the fractions We ve already done this work so from the previous example we have Vb m 5abR In this case we ve got b s on both sides of the equal sign and we need all terms with b s in them on one side of the equation and all other terms on the other side of the equation In this case we can eliminate the minus signs if we collect the b s on the left side and the other terms on the right side Doing this gives Vb 5abR m Now both terms on the right side have a 9 in them so if we factor that out of both terms we arrive at bV5aR m Finally divide by the coefficient of I Recall as well that the coefficient is all the stuff that multiplies the la Doing this gives m V5aR Example 4 Solve 1 1 1 I for c a b c Solution First multiply by the LCD which is abc for this problem abc 6 abc bc ac ah Next collect all the c s on one side the left will probably be easiest here factor a c out of the terms and divide by the coefficient bc acab cb aab ab b a C 2007 Paul Dawkins 83 httptutorialmathlamaredutermsaspx College Algebra Example 5 Solve y for x x Solution First we ll need to clear the denominator To do this we will multiply both sides by 5x 9 We ll also clear out any parenthesis in the problem after we do the multiplication y Sx 9 4 5xy 9 y 4 Now we want to solve for x so that means that we need to get all terms without a y in them to the other side So add 9y to both sides and the divide by the coefficient of x 5xy9y 4 x9y4 5y 4 Example 6 Solve y for x 18x Solution This one is very similar to the previous example Here is the work for this problem y18x 4 3x y8xy 4 3x 8xy3x4 y x8y34 y x 4y 8y3 As mentioned at the start of this section we won t be seeing this kind of problem all that often in this class However outside of this class a Calculus class for example this kind of problem shows up with surprising regularity 2007 Paul Dawkins 84 httptutorialmathlamaredutermsaspx College Algebra Quadratic Equations Part I Before proceeding with this section we should note that the topic of solving quadratic equations will be covered in two sections This is done for the benefit of those viewing the material on the web This is a long topic and to keep page load times down to a minimum the material was split into two sections So we are now going to solve quadratic equations First the standard form of a quadratic equation is axz bx c 0 a 0 The only requirement here is that we have an x2 in the equation We guarantee that this term will be present in the equation by requiring a 7k 0 Note however that it is okay if 9 andor c are zero There are many ways to solve quadratic equations We will look at four of them over the course of the next two sections The first two methods won t always work yet are probably a little simpler to use when they work This section will cover these two methods The last two methods will always work but often require a little more work or attention to get correct We will cover these methods in the next section So let s get started Solving by Factoring As the heading suggests we will be solving quadratic equations here by factoring them To do this we will need the following fact If ab 0 then either a 0 andor 9 0 This fact is called the zero factor property or zero factor principle All the fact says is that if a product of two terms is zero then at least one of the terms had to be zero to start off with Notice that this fact will ONLY work if the product is equal to zero Consider the following product ab 6 In this case there is no reason to believe that either a or 9 will be 6 We could have a 2 and b 3 for instance So do not misuse this fact To solve a quadratic equation by factoring we first must move all the terms over to one side of the equation Doing this serves two purposes First it puts the quadratics into a form that can be factored Secondly and probably more importantly in order to use the zero factor property we MUST have a zero on one side of the equation If we don t have a zero on one side of the equation we won t be able to use the zero factor property Let s take a look at a couple of examples Note that it is assumed that you can do the factoring at this point and so we won t be giving any details on the factoring If you need a review of factoring you should go back and take a look at the Factoring section of the previous chapter 2007 Paul Dawkins 85 httptutorialmathlamaredutermsaspx College Algebra Example 1 Solve each of the following equations by factoring a x2 x 12 E b X2 40 l4x Solution c yz l2y36 0 Lion d 4m2 1 0 1 e 3x2 2x 8 i f l0Z2 l9Z 6 0 Solution g 5x2 2x 1 Solution Now as noted earlier we won t be putting any detail into the factoring process so make sure that you can do the factoring here a x2 x 12 First get everything on side of the equation and then factor x2 x 12 0 x 4x3 0 Now at this point we ve got a product of two terms that is equal to zero This means that at least one of the following must be true x 4O OR x30 x4 OR x 3 Note that each of these is a linear equation that is easy enough to solve What this tell us is that we have two solutions to the equation x 4 and x 3 As with linear equations we can always check our solutions by plugging the solution back into the equation We will check x 3 and leave the other to you to check lt sgt2 lt sgt12 93 12 1212 OK So this was in fact a solution Return to Problems b x2 40 14x As with the first one we first get everything on side of the equal sign and then factor x2 40 14x 0 x4xlO 0 Now we once again have a product of two terms that equals zero so we know that one or both of them have to be zero So technically we need to set each one equal to zero and solve However this is usually easy enough to do in our heads and so from now on we will be doing this solving in our head 2007 Paul Dawkins 86 httptutorialmathlamaredutermsaspx College Algebra The solutions to this equation are x 4 AND x 10 To save space we won t be checking any more of the solutions here but you should do so to make sure we didn t make any mistakes Return to Problems c yz l2y36 O In this case we already have zero on one side and so we don t need to do any manipulation to the equation all that we need to do is factor Also don t get excited about the fact that we now have y s in the equation We won t always be dealing with x s so don t expect to always see them So let s factor this equation yz 12 y 36 0 y62O y6y6O In this case we Ve got a perfect square We broke up the square to denote that we really do have an application of the zero factor property However we usually don t do that We usually will go straight to the answer from the squared part The solution to the equation in this case is y 6 We only have a single Value here as opposed to the two solutions we Ve been getting to this point We will often call this solution a double root or say that it has multiplicity of 2 because it came from a term that was squared Return to Problems d 4m2 1 0 As always let s first factor the equation 4m2 1 0 2m l2ml 0 Now apply the zero factor property The zero factor property tells us that 2m l0 OR 2ml0 2ml OR 2m 1 l m OR m 1 2 2 Again we will typically solve these in our head but we needed to do at least one in complete detail So we have two solutions to the equation ml AND m 1 2 2 Return to Problems e 3x2 2x 8 Now that we Ve done quite a few of these we won t be putting in as much detail for the next two problems Here is the work for this equation 2007 Paul Dawkins 87 httptutorialmathlamaredutermsaspx College Algebra 3x2 2x 8O 3x4x 2O gt 4 g and x 2 Return to Problems f l0z2l9z60 Again factor and use the zero factor property for this one lOz2 l9z6 0 5z22z3O gt z and z Return to Problems g 5x2 2x This one always seems to cause trouble for students even though it s really not too bad First off DO NOT CANCEL AN x FROM BOTH SIDES Do you get the idea that might be bad It is If you cancel an x from both sides you WILL miss a solution so don t do it Remember we are solving by factoring here so let s first get everything on one side of the equal sign 5x2 2x 0 Now notice that all we can do for factoring is to factor an x out of everything Doing this gives x5x 20 2 From the first factor we get that x 0 and from the second we get that x g These are the two solutions to this equation Note that is we d canceled an x in the first step we would NOT have gotten x 0 as an answer Return to Problems Let s work another type of problem here We saw some of these back in the Solving Linear Equations section and since they can also occur with quadratic equations we should go ahead and work on to make sure that we can do them here as well Example 2 Solve each of the following equations 1 1 S1 39 3 X 2x4 OLUCHI b x3i4x Solution x l x l Solution Okay just like with the linear equations the first thing that we re going to need to do here is to clear the denominators out by multiplying by the LCD Recall that we will also need to note values of x that will give division by zero so that we can make sure that these aren t included in the solution 2007 Paul Dawkins 88 httptutorialmathlamaredutermsaspx College Algebra 5 a 1 x 1 2x 4 The LCD for this problem is x 1 2x 4 and we will need to avoid x 1 and x 2 to make sure we don t get division by zero Here is the work for this equation x12x4L1x 1 2x up 24 x 2x4x 4l2x 4 5x 1 2x42x2 24 4 5ac y O2x2 9x 5 02x lx 5 So it looks like the two solutions to this equation are x1 and x5 2 Notice as well that neither of these are the values of x that we needed to avoid and so both are solutions Return to Problems 3 4 b x 3 X x l x 1 In this case the LCD is x 1 and we will need to avoid x 1 so we don t get division by zero Here is the work for this problem x 1x3ij4xx 1 x l x l xlx33 4 x x2 2x334 x x2 3x 4 0 xlx4 0 So the quadratic that we factored and solved has two solutions x 1 and x 4 However when we found the LCD we also saw that we needed to avoid x 1 so we didn t get division by zero Therefore this equation has a single solution x 4 Return to Problems Before proceeding to the next topic we should address that this idea of factoring can be used to solve equations with degree larger than two as well Consider the following example Example 3 Solve 5x3 5x2 10x 0 Solution The first thing to do is factor this equation as much as possible In this case that means factoring out the greatest common factor first Here is the factored form of this equation 2007 Paul Dawkins 89 httptutorialmathlamaredutermsaspx College Algebra 5xx2 x 2O 5xx 2xlO Now the zero factor property will still hold here In this case we have a product of three terms that is zero The only way this product can be zero is if one of the terms is zero This means that 5xO gt x 6 x 20 gtx 2 xlO gt x 1 So we have three solutions to this equation So provided we can factor a polynomial we can always use this as a solution technique The problem is of course that it is sometimes not easy to do the factoring Square Root Property The second method of solving quadratics we ll be looking at uses the square root property Ifp2d then p 3 There is a potentially new symbol here that we should define first in case you haven t seen it yet The symbol i is read as plus or minus and that is exactly what it tells us This symbol is shorthand that tells us that we really have two numbers here One is p J5 and the other is p E Get used to this notation as it will be used frequently in the next couple of sections as we discuss the remaining solution techniques It will also arise in other sections of this chapter and even in other chapters This is a fairly simple property to use however it can only be used on a small portion of the equations that we re ever likely to encounter Let s see some examples of this property Example 4 Solve each of the following equations a x2 100 0 J b 25y2 3 0 E c 4z2 49 0 E d Zr 92 5 E e 3xlO2 810 E Solution There really isn t all that much to these problems In order to use the square root property all that we need to do is get the squared quantity on the left side by itself with a coefficient of l and the number on the other side Once this is done we can use the square root property a x2 100 0 This is a fairly simple problem so here is the work for this equation x2 100 ix 100 10 So there are two solutions to this equation x il0 Remember this means that there are really 2007 Paul Dawkins 90 httptutorialmathlamaredutermsaspx College Algebra two solutions here x 10 and x 10 Return to Problems b 25 y2 3 0 Okay the main difference between this one and the previous one is the 25 in front of the squared term The square root property wants a coefficient of one there That s easy enough to deal with however we ll just divide both sides by 25 Here is the work for this equation 25y2 3 2 1 E y 25 Z quot3 25 5 In this case the solutions are a little messy but many of these will do so don t worry about that Also note that since we knew what the square root of 25 was we went ahead and split the square root of the fraction up as shown Again remember that there are really two solutions here one positive and one negative Return to Problems c 4z2 49 0 This one is nearly identical to the previous part with one difference that we ll see at the end of the example Here is the work for this equation 4z2 49 z2 Q gt Pi 4 3i i3 1 4 4 4 2 7 So there are two solutions to this equation z iaz Notice as well that they are complex solutions This will happen with the solution to many quadratic equations so make sure that you can deal with them Return to Problems d 2r 92 5 This one looks different from the previous parts however it works the same way The square root property can be used anytime we have something squared equals a number That is what we have here The main difference of course is that the something that is squared isn t a single variable it is something else So here is the application of the square root property for this equation 2t 9 5 Now we just need to solve for t and despite the plus or minus in the equation it works the same way we would solve any linear equation We will add 9 to both sides and then divide by a 2 2r9 5 55quot 695 75 Note that we multiplied the fraction through the parenthesis for the final answer We will usually do this in these problems Also do NOT convert these to decimals unless you are asked to This 2007 Paul Dawkins 91 httptutorialmathlamaredutermsaspx College Algebra is the standard form for these answers With that being said we should convert them to decimals just to make sure that you can Here are the decimal values of the two solutions t 9 2 561803 and t2 2 338197 2 2 2 2 Return to Problems e 3x1O281O In this final part we 1l not put much in the way of details into the work 3x1o 81 3x10L 9i 3x E 9i x ll9 3i So we got two complex solutions again and notice as well that with both of the previous part we put the plus or minus part last This is usually the way these are written Return to Problems As mentioned at the start of this section we are going to break this topic up into two sections for the benefit of those viewing this on the web The next two methods of solving quadratic equations completing the square and quadratic formula are given in the next section 2007 Paul Dawkins 92 httptutorialmathlamaredutermsaspx College Algebra Quadratic Equations Part II The topic of solving quadratic equations has been broken into two sections for the benefit of those viewing this on the web As a single section the load time for the page would have been quite long This is the second section on solving quadratic equations In the previous section we looked at using factoring and the square root property to solve quadratic equations The problem is that both of these solution methods will not always work Not every quadratic is factorable and not every quadratic is in the form required for the square root property It is now time to start looking into methods that will work for all quadratic equations So in this section we will look at completing the square and the quadratic formula for solving the quadratic equation axz bx c 0 a 0 Completing the Square The first method we ll look at in this section is completing the square It is called this because it uses a process called completing the square in the solution process So we should first define just what completing the square is Let s start with x2 bx and notice that the x2 has a coefficient of one That is required in order to do this Now to this b 2 lets add ia Doing this gives the following factorable quadratic equation 2 2 x2 bx2j x 2 2 2 This process is called completing the square and if we do all the arithmetic correctly we can guarantee that the quadratic will factor as a perfect square Let s do a couple of examples for just completing the square before looking at how we use this to solve quadratic equations Example 1 Complete the square on each of the following a X2 l6x Solution b y2 7y 1 Solution a x2 l6x Here s the number that we ll add to the equation 16 2 j 82 64 2 Notice that we kept the minus sign here even though it will always drop out after we square things The reason for this will be apparent in a second Let s now complete the square 2007 Paul Dawkins 93 httptutorialmathlamaredutermsaspx College Algebra x2 l6x64x 82 Now this is a quadratic that hopefully you can factor fairly quickly However notice that it will always factor as x plus the blue number we computed above that is in the parenthesis in our case that is 8 This is the reason for leaving the minus sign It makes sure that we don t make any mistakes in the factoring process Return to Problems b y2 7y Here s the number we ll need this time 12 2 4 It s a fraction and that will happen fairly often with these so don t get excited about it Also leave it as a fraction Don t convert to a decimal Now complete the square 49 7 2 2 7 3 3 4 Y 2 This one is not so easy to factor However if you again recall that this will ALWAYS factor as y plus the blue number above we don t have to worry about the factoring process Return to Problems It s now time to see how we use completing the square to solve a quadratic equation The process is best seen as we work an example so let s do that Example 2 Use completing the square to solve each of the following quadratic equations a x2 6xlO J b 2x2 6x 7 0 Solution c 3x2 2x lO Solution We will do the first problem in detail explicitly giving each step In the remaining problems we will just do the work without as much explanation a x2 6xl 0 So let s get started Step 1 Divide the equation by the coefficient of the x2 term Recall that completing the square required a coefficient of one on this term and this will guarantee that we will get that We don t need to do that for this equation however Step 2 Set the equation up so that the x s are on the left side and the constant is on the right side x2 6x 1 Step 3 Complete the square on the left side However this time we will need to add the number to both sides of the equal sign instead of just the left side This is because we have to remember the rule that what we do to one side of an equation we need to do to the other side of the equation First here is the number we add to both sides 2007 Paul Dawkins 94 httptutorialmathlamaredutermsaspx College Algebra 6 2 2 3 9 2 4 gt Now complete the square x2 6x k9 19 x 32 8 Step 4 Now at this point notice that we can use the square root property on this equation That was the purpose of the first three steps Doing this will give us the solution to the equation x 3i x3 gt Ai3 x3 And that is the process Let s do the remaining parts now Return to Problems b 2x2 6x7 0 We will not explicitly put in the steps this time nor will we put in a lot of explanation for this equation This that being said notice that we will have to do the first step this time We don t have a coefficient of one on the x2 term and so we will need to divide the equation by that first Here is the work for this equation x23xZO 2 x23x x23x3s 7 2 4 2 4 3 2 5 x 2 4 xEF E gt x 2 1 2 4 2 2 Don t forget to convert square roots of negative numbers to complex numbers Return to Problems c 3x2 2x l 0 Again we won t put a lot of explanation for this problem 2 2 l x x 3 3 2 2 l x x 3 3 At this point we should be careful about computing the number for completing the square since 9 is now a fraction for the first time 2007 Paul Dawkins 95 httptutorialmathlamaredutermsaspx College Algebra Now finish the problem 2 2 l l l x x 3 9 3 9 if 4 x 3 9 l 4 l 2 x gt xL 3 9 3 3 In this case notice that we can actually do the arithmetic here to get two integer andor fractional solutions We should always do this when there are only integers andor fractions in our solution Here are the two solutions 1 2 3 l 2 l x l and x 3 3 3 3 3 3 Return to Problems A quick comment about the last equation that we solved in the previous example is in order Since we received integer and factions as solutions we could have just factored this equation from the start rather than used completing the square In cases like this we could use either method and we will get the same result Now the reality is that completing the square is a fairly long process and it s easy to make mistakes So we rarely actually use it to solve equations That doesn t mean that it isn t important to know the process however We will be using it in several sections in later chapters and is often used in other classes Quadratic Formula This is the final method for solving quadratic equations and will always work Not only that but if you can remember the formula it s a fairly simple process as well We can derive the quadratic formula by completing the square on the general quadratic formula in standard form Let s do that and we ll take it kind of slow to make sure all the steps are clear First we MUST have the quadratic equation in standard form as already noted Next we need to divide both sides by a to get a coefficient of one on the x2 term ax2bxc0 b c x2 x 0 a 61 Next move the constant to the right side of the equation 2 b c x 96 a a 2007 Paul Dawkins 96 httptutorialmathlamaredutermsaspx College Algebra Now we need to compute the number we ll need to complete the square Again this is onehalf the coefficient of x squared b 2 b2 Z 4612 Now add this to both sides complete the square and get common denominators on the right side to simplify things up a little 2 I I92 192 c x x 2 2 61 4a 4a a if b2 4ac 2a 4a2 Now we can use the square root property on this 2 x ii I 4616 2a 4a2 Solve for x and we ll also simplify the square root a little I I92 4ac x 1 2a 2a As a last step we will notice that we Ve got common denominators on the two terms and so we ll add them up Doing this gives 19 i I92 4ac 261 x So summarizing up provided that we start off in standard form axz bx c O and that s Very important then the solution to any quadratic equation is x bJrI92 4ac 261 Let s work a couple of examples Example 3 Use the quadratic formula to solve each of the following equations a x2 2x 7 E b 3q2 11 Sq 1 c7t26 L9t E d L 3 1 Solution y 2 y e 16x x2 0 1 Solution The important part here is to make sure that before we start using the quadratic formula that we have the equation in standard form first 2007 Paul Dawkins 97 httptutorialmathlamaredutermsaspx College Algebra a x2 2x 7 So the first thing that we need to do here is to put the equation in standard form x2 2x 7 0 At this point we can identify the values for use in the quadratic formula For this equation we have a 1 b 2 c 7 Notice the with c It is important to make sure that we carry any minus signs along with the constants At this point there really isn t anything more to do other than plug into the formula 2i Jlt2gt2 4lt1gtlt 7gt X 21 2 i 3 2 2 There are the two solutions for this equation There is also some simplification that we can do We need to be careful however One of the larger mistakes at this point is to cancel two 2 s in the numerator and denominator Remember that in order to cancel anything from the numerator or denominator then it must be multiplied by the whole numerator or denominator Since the 2 in the numerator isn t multiplied by the whole denominator it can t be canceled In order to do any simplification here we will first need to reduce the square root At that point we can do some canceling 2 1i25 X 2i 2162 224E 2 i 2 E That s a much nicer answer to deal with and so we will almost always do this kind of simplification when it can be done Return to Problems b 3q2 ll5q Now in this case don t get excited about the fact that the variable isn t an x Everything works the same regardless of the letter used for the variable So let s first get the equation into standard form 3q2ll 5qO Now this isn t quite in the typical standard form However we need to make a point here so that we don t make a very common mistake that many student make when first learning the quadratic formula Many students will just get everything on one side as we ve done here and then get the values of a b and c based upon position In other words often students will just let a be the first number listed 9 be the second number listed and then c be the final number listed This is not correct 2007 Paul Dawkins 98 httptutorialmathlamaredutermsaspx College Algebra however For the quadratic formula a is the coefficient of the squared term 9 is the coefficient of the term with just the variable in it not squared and c is the constant term So to avoid making this mistake we should always put the quadratic equation into the official standard form 3q2 5q 11 0 Now we can identify the value of a I and c a 3 b 5 c 11 Again be careful with minus signs They need to get carried along with the values Finally plug into the quadratic formula to get the solution 5Jr25 132 6 Six107 6 5107 i 6 As with all the other methods we ve looked at for solving quadratic equations don t forget to convert square roots of negative numbers into complex numbers Also when bis negative be very careful with the substitution This is particularly true for the squared portion under the radical Remember that when you square a negative number it will become positive One of the more common mistakes here is to get in a hurry and forget to drop the minus sign after you square I so be careful Return to Problems c 7 t2 6 1 9t We won t put in quite the detail with this one that we ve done for the first two Here is the standard form of this equation 7 t2 19t 6 0 Here are the values for the quadratic formula as well as the quadratic formula itself a 7 b 19 c 6 2007 Paul Dawkins 99 httptutorialmathlamaredutermsaspx College Algebra 19 192 47 6 27 19Jr361168 14 19Jr579 T 1923 T I Now recall that when we get solutions like this we need to go the extra step and actually determine the integer andor fractional solutions In this case they are t 1923 g t 19 23 14 7 14 3 Now as with completing the square the fact that we got integer andor fractional solutions means that we could have factored this quadratic equation as well Return to Problems 3 l d j 1 y 2 y So an equation with fractions in it The first step then is to identify the LCD LCD y y 2 So it looks like we ll need to make sure that neither y O or y 2 is in our answers so that we don t get division by zero Multiply both sides by the LCD and then put the result in standard form 3 l ltygtlty 2gtE 1ltygtlty 2 3y y 2 yy 2 3y y 2 y2 2y 0y2 4y 2 Okay it looks like we Ve got the following Values for the quadratic formula a l b 4 c 2 Plugging into the quadratic formula gives 2007 Paul Dawkins 100 httptutorialmathlamaredutermsaspx College Algebra 4JEI 2 42JE 2 2 5 Note that both of these are going to be solutions since neither of them are the values that we need to avoid Return to Problems e 16x x2 0 We saw an equation similar to this in the previous section when we were looking at factoring equations and it would definitely be easier to solve this by factoring However we are going to use the quadratic formula anyway to make a couple of points First let s rearrange the order a little bit just to make it look more like the standard form 2 x 16x 0 Here are the constants for use in the quadratic formula a1 b16 c 0 There are two things to note about these values First we ve got a negative a for the first time Not a big deal but it is the first time we ve seen one Secondly and more importantly one of the values is zero This is fine It will happen on occasion and in fact having one of the values zero will make the work much simpler Here is the quadratic formula for this equation 16 i lt16gt2 4lt 1gtltogt 21 16jJ 5 2 16 i 16 2 Reducing these to integersfractions gives x1616 9 O X1616 32 16 2 2 2 2 So we get the two solutions x 0 and x 16 These are exactly the solutions we would have gotten by factoring the equation Return to Problems 2007 Paul Dawkins 101 httptutorialmathlamaredutermsaspx College Algebra To this point in both this section and the previous section we have only looked at equations with integer coefficients However this doesn t have to be the case We could have coefficient that are fractions or decimals So let s work a couple of examples so we can say that we ve seen something like that as well Example 4 Solve each of the following equations 1 1 a 5362 x E 0 Solution b 004962 02336 009 0 Solution Solution a There are two ways to work this one We can either leave the fractions in or multiply by the LCD 10 in this case and solve that equation Either way will give the same answer We will only do the fractional case here since that is the point of this problem You should try the other way to verify that you get the same solution In this case here are the values for the quadratic formula as well as the quadratic formula work for this equation b1 c 0 g 1 5 In these cases we usually go the extra step of eliminating the square root from the denominator so let s also do that If you do clear the fractions out and run through the quadratic formula then you should get exactly the same result For the practice you really should try that Return to Problems b In this case do not get excited about the decimals The quadratic formula works in exactly the same manner Here are the values and the quadratic formula work for this problem a 004 b 023 c 009 023 i 0232 4004009 x 2004 023 i 00529 00144 008 023 x00385 008 2007 Paul Dawkins 102 httptutorialmathlamaredutermsaspx College Algebra Now to this will be the one difference between these problems and those with integer or fractional coefficients When we have decimal coefficients we usually go ahead and figure the two individual numbers So let s do that X 023 x00385 023 019621 008 008 0230l962l 023 0l962l x T and x T 008 008 5327625 and 0422375 Notice that we did use some rounding on the square root Return to Problems Over the course of the last two sections we ve done quite a bit of solving It is important that you understand most if not all of what we did in these sections as you will be asked to do this kind of work in some later sections 2007 Paul Dawkins 103 httptutorialmathlamaredutermsaspx College Algebra Solving Quadratic Equations A Summary In the previous two sections we ve talked quite a bit about solving quadratic equations A logical question to ask at this point is which method should we use to solve a given quadratic equation Unfortunately the answer is it depends If your instructor has specified the method to use then that of course is the method you should use However if your instructor had NOT specified the method to use then we will have to make the decision ourselves Here is a general set of guidelines that may be helpful in determining which method to use 1 Is it clearly a square root property problem In other words does the equation consist ONLY of something squared and a constant If this is true then the square root property is probably the easiest method for use 2 Does it factor If so that is probably the way to go Note that you shouldn t spend a lot of time trying to determine if the quadratic equation factors Look at the equation and if you can quickly determine that it factors then go with that If you can t quickly determine that it factors then don t worry about it 3 If you ve reached this point then you ve determined that the equation is not in the correct form for the square root property and that it doesn t factor or that you can t quickly see that it factors So at this point you re only real option is the quadratic formula Once you ve solve enough quadratic equations the above set of guidelines will become almost second nature to you and you will find yourself going through them almost without thinking Notice as well that nowhere in the set of guidelines was completing the square mentioned The reason for this is simply that it s a long method that is prone to mistakes when you get in a hurry The quadratic formula will also always work and is much shorter of a method to use In general you should only use completing the square if your instructor has required you to use it As a solving technique completing the square should always be your last choice This doesn t mean however that it isn t an important method We will see the completing the square process arise in several sections in later chapters Interestingly enough when we do see this process in later sections we won t be solving equations This process is very useful in many situations of which solving is only one Before leaving this section we have one more topic to discuss In the previous couple of sections we saw that solving a quadratic equation in standard form axz bx c O we will get one of the following three possible solution sets 1 Two real distinct i e not equal solutions 2 A double root Recall this arises when we can factor the equation into a perfect square 3 Two complex solutions These are the ONLY possibilities for solving quadratic equations in standard form Note however that if we start with rational expression in the equation we may get different solution 2007 Paul Dawkins 104 httptutorialmathlamaredutermsaspx College Algebra sets because we may need avoid one of the possible solutions so we don t get division by zero errors Now it turns out that all we need to do is look at the quadratic equation in standard form of course to determine which of the three cases that we ll get To see how this works let s start off by recalling the quadratic formula 19 i I92 4ac 261 X The quantity I92 4ac in the quadratic formula is called the discriminant It is the value of the discriminant that will determine which solution set we will get Let s go through the cases one at a time 1 Two real distinct solutions We will get this solution set if 92 4616 gt 0 In this case we will be taking the square root of a positive number and so the square root will be a real number Therefore the numerator in the quadratic formula will be 19 plus or minus a real number This means that the numerator will be two different real numbers Dividing either one by 2a won t change the fact that they are real not will it change the fact that they are different 2 A double root We will get this solution set if I92 4616 0 Here we will be taking the square root of zero which is zero However this means that the plus or minus part of the numerator will be zero and so the numerator in the quadratic formula will be 19 In other words we will get a single real number out of the quadratic formula which is what we get when we get a double root 3 Two complex solutions We will get this solution set if 92 4616 lt 0 If the discriminant is negative we will be taking the square root of negative numbers in the quadratic formula which means that we will get complex solutions Also we will get two since they have a plus or minus in front of the square root So let s summarize up the results here 1 If I92 4616 gt 0 then we will get two real solutions to the quadratic equation 2 If I92 4616 0 then we will get a double root to the quadratic equation 3 If I92 4616 lt 0 then we will get two complex solutions to the quadratic equation Example 1 Using the discriminant determine which solution set we get for each of the following quadratic equations a 13x2 1 5x Solution b 6q2 20q 3 E c 49t2 l26t 81 0 Solution Solution All we need to do here is make sure the equation is in standard form determine the value of a I and c then plug them into the discriminant 2007 Paul Dawkins 105 httptutorialmathlamaredutermsaspx College Algebra a 13x2 1 5x First get the equation in standard form 13x2 5x l 0 We then have a 13 b 5 c l Plugging into the discriminant gives 192 4ac a 592 413e1 27 The discriminant is negative and so we will have two complex solutions For reference purposes the actual solutions are x5Jr33 26 Return to Problems b 6q2 20q 3 Again we first need to get the equation in standard form 6q2 20q 3 0 This gives a 6 b 20 c 3 The discriminant is then b2 4ac2O2 46 3 472 The discriminant is positive we will get two real distinct solutions Here they are X 2Oi472 lOJrll8 12 6 Return to Problems c 49t2 126t 81 0 This equation is already in standard form so let s jump straight in a 49 b 126 c 81 The discriminant is then 192 4ac 1262 4f4981 0 In this case we ll get a double root since the discriminant is zero Here it is 9 x 7 Return to Problems For practice you should verify the solutions in each of these examples 2007 Paul Dawkins 106 httptutorialmathlamaredutermsaspx College Algebra Application of Quadratic Equations In this section we re going to go back and revisit some of the applications that we saw in the Linear Applications section and see some examples that will require us to solve a quadratic equation to get the answer Note that the solutions in these cases will almost always require the quadratic formula so expect to use it and don t get excited about it Also we are going to assume that you can do the quadratic formula work and so we won t be showing that work We will give the results of the quadratic formula we just won t be showing the work Also as we will see we will need to get decimal answer to these and so as a general rule here we will round all answers to 4 decimal places Example 1 We are going to fence in a rectangular field and we know that for some reason we want the field to have an enclosed area of 75 ft2 We also know that we want the width of the field to be 3 feet longer than the length of the field What are the dimensions of the field Solution So we ll let x be the length of the field and so we know that x 3 will be the width of the field Now we also know that area of a rectangle is length times width and so we know that xx3 75 Now this is a quadratic equation so let s first write it in standard form x2 3x 75 x2 3x 75 0 Using the quadratic formula gives 3 i 309 2 x Now at this point we ve got to deal with the fact that there are two solutions here and we only want a single answer So let s convert to decimals and see what the solutions actually are x 3309 3 x309 2 72892 and x f l02892 So we have one positive and one negative From the stand point of needing the dimensions of a field the negative solution doesn t make any sense so we will ignore it Therefore the length of the field is 72892 feet The width is 3 feet longer than this and so is 102892 feet Notice that the width is almost the second solution to the quadratic equation The only difference is the minus sign Do NOT expect this to always happen In this case this is more of a function of the problem For a more complicated set up this will NOT happen 2007 Paul Dawkins 107 httptutorialmathlamaredutermsaspx College Algebra Now from a physical standpoint we can see that we should expect to NOT get complex solutions to these problems Upon solving the quadratic equation we should get either two real distinct solutions or a double root Also as the previous example has shown when we get two real distinct solutions we will be able to eliminate one of them for physical reasons Let s work another example or two Example 2 Two cars start out at the same point One car starts out driving north at 25 mph Two hours later the second car starts driving east at 20 mph How long after the first car starts traveling does it take for the two cars to be 300 miles apart Solution We ll start off by letting t be the amount of time that the first car let s call it car A travels Since the second car let s call that car B starts out two hours later then we know that it will travel for t 2 hours Now we know that the distance traveled by an object or car since that s what we re dealing with here is its speed times time traveled So we have the following distances traveled for each car distance of car A 25t distance of car B 20t 2 At this point a quick sketch of the situation is probably in order so we can see just what is going on In the sketch we will assume that the two cars have traveled long enough so that they are 300 miles apart Car5 0 25 2L J 2j llar So we have a right triangle here That means that we can use the Pythagorean Theorem to say 25r2 20r 22 3002 This is a quadratic equation but it is going to need some fairly heavy simplification before we can solve it so let s do that 625t2 20 402 90000 625t2 400 1600t 1600 90000 1025 1600t 88400 0 2007 Paul Dawkins 108 httptutorialmathlamaredutermsaspx College Algebra Now the coefficients here are quite large but that is just something that will happen fairly often with these problems so don t worry about that Using the quadratic formula and simplifying that answer gives t 1600 i x365000000 1600 i l000365 32 i 200365 2050 2050 41 Again we have two solutions and we re going to need to determine which one is the correct one so let s convert them to decimals t 32 204365 41 85390ll 1009998 and r T32 201 365 4 As with the previous example the negative answer just doesn t make any sense So it looks like the car A traveled for 1009998 hours when they were finally 300 miles apart Also even though the problem didn t ask for it the second car will have traveled for 809998 hours before they are 300 miles apart Notice as well that this is NOT the second solution without the negative this time unlike the first example Example 3 An office has two envelope stuffing machines Working together they can stuff a batch of envelopes in 2 hours Working separately it will take the second machine 1 hour longer than the first machine to stuff a batch of envelopes How long would it take each machine to stuff a batch of envelopes by themselves Solution Let t be the amount of time it takes the first machine Machine A to stuff a batch of envelopes by itself That means that it will take the second machine Machine B t 1 hours to stuff a batch of envelopes by itself The word equation for this problem is then Portion of job Portion of job 1 J b o done by Machine A done by Machine B Work Rate Time Spent Work Rate Time Spent of Machine A Working of Machine B Working We know the time spent working together 2 hours so we need to work rates of each machine Here are those computations 1 Job Work Rate of Machine A x t gt Machine A 1 I 1 Job Work Rate of Machine Bgtltt 1 gt Machine B 1 t Note that it s okay that the work rates contain t In fact they will need to so we can solve for it Plugging into the word equation gives 2007 Paul Dawkins 109 httptutorialmathlamaredutermsaspx College Algebra 2 2 1 t t1 So to solve we ll first need to clear denominators and get the equation in standard form 3iltrgtltr1gtlt1gtltrgtltr 1 t t 2r12rr2r 4t2t2 t 0t2 3t 2 Using the quadratic formula gives t3JrW 2 Converting to decimals gives 3xE 3 E 2 2 356l6 and t 95616 Again the negative doesn t make any sense and so Machine A will work for 35616 hours to stuff a batch of envelopes by itself Machine B will need 45616 hours to stuff a batch of envelopes by itself Again unlike the first example note that the time for Machine B was NOT the second solution from the quadratic without the minus sign 2007 Paul Dawkins 110 httptutorialmathlamaredutermsaspx College Algebra Equations Reducible to Quadratic Form In this section we are going to look at equations that are called quadratic in form or reducible to quadratic in form What this means is that we will be looking at equations that if we look at them in the correct light we can make them look like quadratic equations At that point we can use the techniques we developed for quadratic equations to help us with the solution of the actual equation It is usually best with these to show the process with an example so let s do that Example 1 Solve x4 7x2 12 0 Solution Now let s start off here by noticing that 2 x4 x2 In other words we can notice here that the Variable portion of the first term i e ignore the coefficient is nothing more than the Variable portion of the second term squared Note as well that all we really needed to notice here is that the exponent on the first term was twice the exponent on the first term This along with the fact that third term is a constant means that this equation is reducible to quadratic in form We will solve this by first defining ux2 Now this means that Therefore we can write the equation in terms of u s instead of x s as follows 4 2 2 x 7x l2O gtu 7u 12 0 The new equation the one with the u s is a quadratic equation and we can solve that In fact this equation is factorable so the solution is u2 7ul2u 4u 3 0 gt u 3u 4 So we get the two solutions shown above These aren t the solutions that we re looking for We want Values of x not Values of u That isn t really a problem once we recall that we Ve defined ux2 To get Values of x for the solution all we need to do is plug in u into this equation and solve that for x Let s do that u z 3 32 x ix5 L124 4 x2 gt iir Z 2 So we have four solutions to the original equation x i2 and x ix5 2007 Paul Dawkins lll httptutorialmathlamaredutermsaspx College Algebra So the basic process is to check that the equation is reducible to quadratic in form then make a quick substitution to turn it into a quadratic equation We solve the new equation for u the Variable from the substitution and then use these solutions and the substitution definition to get the solutions to the equation that we really want In most cases to make the check that it s reducible to quadratic in form all that we really need to do is to check that one of the exponents is twice the other There is one exception to this that we ll see here once we get into a set of examples Also once you get good at these you often don t really need to do the substitution either We will do them to make sure that the work is clear However these problems can be done without the substitution in many cases Example 2 Solve each of the following equations 2 1 a X 2x5 15 0 Solution b y6 9y3 8 0 Solution c z 9xZ 14 0 Solution d t4 40 1 Solution 2 1 a x 2x 15 0 Okay in this case we can see that Hi and so one of the exponents is twice the other so it looks like we Ve got an equation that is reducible to quadratic in form The substitution will then be 1 12 2 ux3 u2 x3 x3 Substituting this into the equation gives uz 2u 15 0 u 5u30 gt u 3 u 5 Now that we Ve gotten the solutions for u we can find Values of x 1 u 3 xi 3 gt Ex 33 27 1 u5 x35 gt 99 53 125 So we have two solutions here x 27 and x 125 Return to Problems b y396 9y3 8 0 For this part notice that 2007 Paul Dawkins 112 httptutorialmathlamaredutermsaspx College Algebra 6 2 3 and so we do have an equation that is reducible to quadratic form The substitution is u y 3 uz y 3 2 y 6 The equation becomes uz 9u 8 0 u 8u lO ulu 8 Now going back to y s is going to take a little more work here but shouldn t be too bad 1 l 1 ulgt 33gt3 1 gty3 11 y 13 1 3 1 l l l 3 l u8gt 393 3 gt 3 y3 3 8 Y 8 2 l The two solutions to this equation are 2 1 and y 5 Return to Problems c z 9Z140 This one is a little trickier to see that it s quadratic in form yet it is To see this recall that the exponent on the square root is onehalf then we can notice that the exponent on the first term is twice the exponent on the second term So this equation is in fact reducible to quadratic in form Here is the substitution uxE u2E2 z The equation then becomes u2 9ul40 u 7u 20 u2u se Now go back to z s u2 gt 4 2 gt z e22 4 u7 gt T4 7 gt z 2 49 The two solutions for this equation are z 4 and z 49 Return to Problems d r4 4 0 Now this part is the exception to the rule that we Ve been using to identify equations that are reducible to quadratic in form There is only one term with a t in it However notice that we can write the equation as f 40 So if we use the substitution 2007 Paul Dawkins 113 httptutorialmathlamaredutermsaspx College Algebra the equation becomes uz 4 O and so it is reducible to quadratic in form Now we can solve this using the square root property Doing that gives an 441 2 Now going back to t s gives us u 2 gt 2 r E u 2 gt t2 2 gt 4 3 xii In this case we get four solutions and two of them are complex solutions Getting complex solutions out of these are actually more common that this set of examples might suggest The problem is that to get some of the complex solutions requires knowledge that we haven t and won t cover in this course So they don t show up all that often Return to Problems All of the examples to this point gave quadratic equations that were factorable or in the case of the last part of the previous example was an equation that we could use the square root property on That need not always be the case however It is more than possible that we would need the quadratic formula to do some of these We should do an example of one of these just to make the point Example 3 Solve 2x10 x5 4 0 Solution In this case we can reduce this to quadratic in form by using the substitution ux5 u2x1o Using this substitution the equation becomes 2u2 u 40 This doesn t factor and so we ll need to use the quadratic formula on it From the quadratic formula the solutions are u 1 i 33 4 Now in order to get back to x s we are going to need decimals values for these so u15 u1 5 4 4 l68614 l186l4 Now using the substitution to get back to x s gives the following 2007 Paul Dawkins 114 httptutorialmathlamaredutermsaspx College Algebra 1 u168614 x5168614 x168614E 111014 1 u 118614 a 1k8614 x 118614 103473 Of course we had to use a calculator to get the final answer for these This is one of the reasons that you don t tend to see too many of these done in an Algebra class The work andor answers tend to be a little messy 2007 Paul Dawkins 115 httptutorialmathlamaredutermsaspx College Algebra Equations with Radicals The title of this section is maybe a little misleading The title seems to imply that we re going to look at equations that involve any radicals However we are going to restrict ourselves to equations involving square roots The techniques we are going to apply here can be used to solve equations with other radicals however the work is usually significantly messier than when dealing with square roots Therefore we will work only with square roots in this section Before proceeding it should be mentioned as well that in some Algebra textbooks you will find this section in with the equations reducible to quadratic form material The reason is that we will in fact end up solving a quadratic equation in most cases However the approach is significantly different and so we re going to separate the two topics into different sections in this course It is usually best to see how these work with an example Example 1 Solve x xx 6 Solution In this equation the basic problem is the square root If that weren t there we could do the problem The whole process that we re going to go through here is set up to eliminate the square root However as we will see the steps that we re going to take can actually cause problems for us So let s see how this all works Let s notice that if we just square both sides we can make the square root go away Let s do that and see what happens x2x 6L x2 x 60 x 3x2O gt 4 x 2 Upon squaring both sides we see that we get a factorable quadratic equation that gives us two solutions x 3 and x 2 Now for no apparent reason let s do something that we haven t actually done since the section on solving linear equations Let s check our answers Remember as well that we need to check the answers in the original equation That is very important Let s first check x 3 3 3 6 3J OK So x 3 is a solution Now let s check x 2 2007 Paul Dawkins 116 httptutorialmathlamaredutermsaspx College Algebra 2 2 6 2 Z2 NOTOK We have a problem Recall that square roots are ALWAYS positive and so x 2 does not work in the original equation One possibility here is that we made a mistake somewhere We can go back and look however and we ll quickly see that we haven t made a mistake So what is the deal Remember that our first step in the solution process was to square both sides Notice that if we plug x 2 into the quadratic we solved it would in fact be a solution to that When we squared both sides of the equation we actually changed the equation and in the process introduced a solution that is not a solution to the original equation With these problems it is vitally important that you check your solutions as this will often happen When this does we only take the values that are actual solutions to the original equation So the original equation had a single solution x 3 Now as this example has shown us we have to be very careful in solving these equations When we solve the quadratic we will get two solutions and it is possible both of these one of these or none of these values to be solutions to the original equation The only way to know is to check your solutions Let s work a couple more examples that are a little more difficult Example 2 Solve each of the following equations a y y 4 4 E b 1 EH 22 3 Solution C 5Z 6 2 Z Solution Solution my y44 In this case let s notice that if we just square both sides we re going to have problems yy 42 42 y22yy 4y 4l6 Before discussing the problem we ve got here let s make sure you can do the squaring that we did above since it will show up on occasion All that we did here was use the formula 61 b2 a2 2ab b2 with a y and b J y 4 You will need to be able to do these because while this may not have worked here we will need to this kind of work in the next set of problems Now just what is the problem with this Well recall that the point behind squaring both sides in the first problem was to eliminate the square root We haven t done that There is still a square root in the problem and we ve made the remainder of the problem messier as well 2007 Paul Dawkins 117 httptutorialmathlamaredutermsaspx College Algebra So what we re going to need to do here is make sure that We Ve got a square root all by itself on one side of the equation before squaring Once that is done we can square both sides and the square root really will disappear Here is the correct way to do this problem y 4 4 y now square both sides 2 Y 4 4 W y 4 16 8 y y2 0 y2 9 y 20 O y 5 y 4 gt 2y 4 5 As with the first example we will need to make sure and check both of these solutions Again make sure that you check in the original equation Once we Ve square both sides we Ve changed the problem and so checking there won t do us any good In fact checking there could well lead us into trouble First y 4 4 4 4 4 4 4 OK So that is a solution Now 2 5 5J5 4 4 51 4 6 4 NOTOK So as with the first example we worked there is in fact a single solution to the original equation y 4 Return to Problems b 1 EH 2t 3 Okay so we will again need to get the square root on one side by itself before squaring both sides 1 t 2t 2 2 1 t 2t 3 1 2tt2 2t 3 z2 m40 t 22 0 gt t 2 So we have a double root this time Let s check it to see if it really is a solution to the original equation 2007 Paul Dawkins 118 httptutorialmathlamaredutermsaspx College Algebra a z 22 3 12 ill 1 3 So t 2 isn t a solution to the original equation Since this was the only possible solution this means that there are no solutions to the original equation This doesn t happen too often but it does happen so don t be surprised by it when it does Return to Problems c x5z6 2Z This one will work the same as the previous two x5z6z2 x5z62aL 22 5z6z2 4z 4 0z2 z 2 0Ez 2z1gt z1z 2 Let s check these possible solutions start with z 1 5 16 2 1 xI2 1 1 1 OK So that s was a solution Now let s check z 2 526 2 2 OK This was also a solution So in this case we ve now seen an example where both possible solutions are in fact solutions to the original equation as well Return to Problems So as we ve seen in the previous set of examples once we get our list of possible solutions anywhere from none to all of them can be solutions to the original equation Always remember to check your answers Okay let s work one more set of examples that have an added complexity to them To this point all the equations that we ve looked at have had a single square root in them However there can be more than one square root in these equations The next set of examples is designed to show us how to deal with these kinds of problems 2007 Paul Dawkins 119 httptutorialmathlamaredutermsaspx College Algebra Example 3 Solve each of the following equations a 2961X42 E b t 7 23 Solution Solution In both of these there are two square roots in the problem We will work these in basically the same manner however The first step is to get one of the square roots by itself on one side of the equation then square both sides At this point the process is different so we ll see how to proceed from this point once we reach it in the first example a 2x l x 4 2 So the first thing to do is get one of the square roots by itself It doesn t matter which one we get by itself We ll end up the same solutions in the end 2x 2 x74 x2x l2 2 c 42 2x l4 4x 4 x 4 2x l4x 4 x Now we still have a square root in the problem but we have managed to eliminate one of them Not only that but what we Ve got left here is identical to the examples we worked in the first part of this section Therefore we will continue now work this problem as we did in the previous sets ofexamples x l24r2 x2 2xll6x 4 x2 2xll6x 64 x2 l8x650 x l3x 50 gt x 135x 5 Now let s check both possible solutions in the original equation We ll start with x 13 213 1 13 4 2 E 542 5 3 2 OK So the one is a solution Now let s check x 5 25 1 E z l2 3 l2 OK So they are both solutions to the original equation Return to Problems 2007 Paul Dawkins 120 httptutorialmathlamaredutermsaspx College Algebra b xt72x3 In this case we Ve already got a square root on one side by itself so we can go straight to squaring both sides 2 2 xt7 2 V3 ta t74t7 4 3 t t114t7 3 t Next get the remaining square root back on one side by itself and square both sides again 4Js 2 4x2 8 2t2 16r76432t4 16t11264 32r 42 0 4t2 l6t 48 0402 a La 04t 6t 2 gt t 6 t 2 Now check both possible solutions starting with t 2 27 23 2 x5 2l 3 2 7amp1 NOT OK So that wasn t a solution Now let s check t 6 x 67 2 3 6 l 2 12 3 OK It looks like in this case we Ve got a single solution t 6 Return to Problems So when there is more than one square root in the problem we are again faced with the task of checking our possible solutions It is possible that anywhere from none to all of the possible solutions will in fact be solutions and the only way to know for sure is to check them in the original equation 2007 Paul Dawkins l2l httptutorialmathlamaredutermsaspx College Algebra Linear Inequalities To this point in this chapter we ve concentrated on solving equations It is now time to switch gears a little and start thinking about solving inequalities Before we get into solving inequalities we should go over a couple of the basics first At this stage of your mathematical career it is assumed that you know that a lt b means that a is some number that is strictly less that I9 It is also assumed that you know that a 2 b means that a is some number that is either strictly bigger than 9 or is exactly equal to la Likewise it is assumed that you know how to deal with the remaining two inequalities gt greater than and S less than or equal to What we want to discuss is some notational issues and some subtleties that sometimes get students when the really start working with inequalities First remember that when we say that a is less than 9 we mean that a is to the left of 9 on a number line So 1000 lt 0 is a true inequality Next don t forget how to correctly interpret S and 2 Both of the following are true inequalities 434 634 In the first case 4 is equal to 4 and so it is less than or equal to 4 In the second case 6 is strictly less than 4 and so it is less than or equal to 4 The most common mistake is to decide that the first inequality is not a true inequality Also be careful to not take this interpretation and translate it to lt andor gt For instance 4 lt 4 is not a true inequality since 4 is equal to 4 and not strictly less than 4 Finally we will be seeing many double inequalities throughout this section and later sections so we can t forget about those The following is a double inequality 9lt5 6 In a double inequality we are saying that both inequalities must be simultaneously true In this case 5 is definitely greater than 9 and at the same time is less than or equal to 6 Therefore this double inequality is a true inequality On the other hand 10 S 5 lt 20 is not a true inequality While it is true that 5 is less than 20 so the second inequality is true it is not true that 5 is greater than or equal to 10 so the first inequality is not true If even one of the inequalities in a double inequality is not true then the whole inequality is not true This point is more important than you might realize at this point In a later section we will run across situations where many students try to combine two inequalities into a double inequality that simply can t be combined so be careful 2007 Paul Dawkins 122 httptutorialmathlamaredutermsaspx College Algebra The next topic that we need to discuss is the idea of interval notation Interval notation is some very nice shorthand for inequalities and will be used extensively in the next few sections of this chapter The best way to define interval notation is the following table There are three columns to the table Each row contains an inequality a graph representing the inequality and finally the interval notation for the given inequality Inequality Graph Interval Notation a s x s b E A ab a lt x lt 9 i E ab a S x lt 9 if E t ab a lt x S 9 i A ab x gt a i aoo x 2 a E E aoo x lt 19 L oob x 3 9 L J oob Remember that a bracket or means that we include the endpoint while a parenthesis or means we don t include the endpoint Now with the first four inequalities in the table the interval notation is really nothing more than the graph without the number line on it With the final four inequalities the interval notation is almost the graph except we need to add in an appropriate infinity to make sure we get the correct portion of the number line Also note that infinities NEVER get a bracket They only get a parenthesis We need to give one final note on interval notation before moving on to solving inequalities Always remember that when we are writing down an interval notation for an inequality that the number on the left must be the smaller of the two It s now time to start thinking about solving linear inequalities We will use the following set of facts in our solving of inequalities Note that the facts are given for lt We can however write down an equivalent set of facts for the remaining three inequalities 1 If altb then aCltbC and a cltb c for any number c In otherwords we can add or subtract a number to both sides of the inequality and we don t change the inequality itself 2007 Paul Dawkins 123 httptutorialmathlamaredutermsaspx College Algebra I 2 If a lt b and c gt 0 then ac lt I96 and E lt So provided c is a positive number we c c can multiply or divide both sides of an inequality by the number without changing the inequality 19 3 If a lt b and c lt 0 then ac gt I96 and E gt In this case unlike the previous fact if c c c is negative we need to ip the direction of the inequality when we multiply or divide both sides by the inequality by c These are nearly the same facts that we used to solve linear equations The only real exception is the third fact This is the important fact as it is often the most misused andor forgotten fact in solving inequalities If you aren t sure that you believe that the sign of c matters for the second and third fact consider the following number example 3lt5 I hope that we would all agree that this is a true inequality Now multiply both sides by 2 and by 2 3lt5 3lt5 32 lt52 3 2 gt5 2 6lt10 6gt J0 Sure enough when multiplying by a positive number the direction of the inequality remains the same however when multiplying by a negative number the direction of the inequality does change Okay let s solve some inequalities We will start off with inequalities that only have a single inequality in them In other words we ll hold off on solving double inequalities for the next set of examples The thing that we ve got to remember here is that we re asking to determine all the values of the variable that we can substitute into the inequality and get a true inequality This means that our solutions will in most cases be inequalities themselves Example 1 Solving the following inequalities Give both inequality and interval notation forms of the solution a 2m 3lt5m1 12 E b 2l x5 S 32x l Solution Solution Solving single linear inequalities follow pretty much the same process for solving linear equations We will simplify both sides get all the terms with the variable on one side and the numbers on the other side and then multiplydivide both sides by the coefficient of the variable to get the solution The one thing that you ve got to remember is that if you multiplydivide by a negative number then switch the direction of the inequality 2007 Paul Dawkins 124 httptutorialmathlamaredutermsaspx College Algebra a 2m 3 lt5mi i2 There really isn t much to do here other than follow the process outlined above 2m 3 lt5mi i2 2m6 lt5m5 l2 7m lt 13 13 mgt You did catch the fact that the direction of the inequality changed here didn t you We divided 1 by a 7 and so we had to change the direction The inequality form of the solution is m gt l The interval notation for this solution is 73 00 Return to Problems b21 x5 32x 1 Again not much to do here 2l x5 S32x l 2 2x5 S6x 3 l0S8x Esx 8 Esx 4 Now with this inequality we ended up with the variable on the right side when it more traditionally on the left side So let s switch things around to get the variable onto the left side Note however that we re going to need also switch the direction of the inequality to make sure that we don t change the answer So here is the inequality notation for the inequality 5 x2 5 The interval notation for the solution is oo Return to Problems Now let s solve some double inequalities The process here is similar in some ways to solving single inequalities and yet very different in other ways Since there are two inequalities there isn t any way to get the variables on one side of the inequality and the numbers on the other It is easier to see how these work if we do an example or two so let s do that 2007 Paul Dawkins 125 httptutorialmathlamaredutermsaspx College Algebra Example 2 Solve each of the following inequalities Give both inequality and interval notation forms for the solution a 6S2x 5lt7 E b 3lt2 x S5 Solution c 14 lt 7 3x 2 lt1 Solution Solution a 6 S 2x 5 lt 7 The process here is fairly similar to the process for single inequalities but we will first need to be careful in a couple of places Our first step in this case will be to clear any parenthesis in the middle term 6S2x l0lt7 Now we want the x all by itself in the middle term and only numbers in the two outer terms To do this we will addsubtractmultiplydivide as needed The only thing that we need to remember here is that if we do something to middle term we need to do the same thing to BOTH of the out terms One of the more common mistakes at this point is to add something for example to the middle and only add it to one of the two sides Okay we ll add 10 to all three parts and then divide all three parts by two 4 S 2x lt17 2Sxlt1 7 2 That is the inequality form of the answer The interval notation form of the answer is 2 Return to Problems h 3lt 2 xs5 In this case the first thing that we need to do is clear fractions out by multiplying all three parts by 2 We will then proceed as we did in the first part 6lt32 xS10 6 lt 6 3x S10 12 lt 3x S 4 Now we re not quite done here but we need to be very careful with the next step In this step we need to divide all three parts by 3 However recall that whenever we divide both sides of an inequality by a negative number we need to switch the direction of the inequality For us this means that both of the inequalities will need to switch direction here 4gtx2 3 So there is the inequality form of the solution We will need to be careful with the interval 2007 Paul Dawkins 126 httptutorialmathlamaredutermsaspx College Algebra 4 notation for the solution First the interval notation is NOT 4 Remember that in interval notation the smaller number must always go on the left side Therefore the correct 4 interval notation for the solution is g 4 Note as well that this does match up with the inequality form of the solution as well The 4 4 inequality is telling us that x is any number between 4 and g or possibly itself and this is exactly what the interval notation is telling us Also the inequality could be ipped around to get the smaller number on the left if we d like to Here is that form Sxlt4 3 When doing this make sure to correctly deal with the inequalities as well Return to Problems c 14lt 73x2lt1 Not much to this one We ll proceed as we ve done the previous two 14 lt 21x 14 lt1 0 lt 21x lt15 Don t get excited about the fact that one of the sides is now zero This isn t a problem Again as with the last part we ll be dividing by a negative number and so don t forget to switch the direction of the inequalities Ogtxgt 21 Ogtxgt OR ltxltO 7 7 Either of the inequalities in the second row will work for the solution The interval notation of 5 the solution is 7Oj Return to Problems When solving double inequalities make sure to pay attention to the inequalities that are in the original problem One of the more common mistakes here is to start with a problem in which one of the inequalities is lt or gt and the other is S or 2 as we had in the first two parts of the previous example and then by the nal answer they are both lt or gt or they are both S or 2 In other words it is easy to all of a sudden make both of the inequalities the same Be careful with this There is one final example that we want to work here 2007 Paul Dawkins 127 httptutorialmathlamaredutermsaspx College Algebra Example 3 If 1 lt x lt 4 then determine a and bin a lt 2x 3 lt 19 Solution This is easier than it may appear at first All we are really going to do is start with the given inequality and then manipulate the middle term to look like the second inequality Again we ll need to remember that whatever we do to the middle term we ll also need to do to the two outer terms So first we ll multiply everything by 2 2 lt 2x lt 8 Now add 3 to everything 1lt2x3lt11 We ve now got the middle term identical to the second inequality in the problems statement and so all we need to do is pick off a and I From this inequality we can see that a l and b ll 2007 Paul Dawkins 128 httptutorialmathlamaredutermsaspx College Algebra Polynomial Inequalities It is now time to look at solving some more difficult inequalities In this section we will be solving single inequalities that involve polynomials of degree at least two Or to put it in other words the polynomials won t be linear any more Just as we saw when solving equations the process that we have for solving linear inequalities just won t work here Since it s easier to see the process as we work an example let s do that As with the linear inequalities we are looking for all the values of the variable that will make the inequality true This means that our solution will almost certainly involve inequalities as well The process that we re going to go through will give the answers in that form Example 1 Solve x2 10 lt 3x Solution There is a fairly simple process to solving these If you can remember it you ll always be able to solve these kinds of inequalities Step 1 Get a zero on one side of the inequality It doesn t matter which side has the zero however we re going to be factoring in the next step so keep that in mind as you do this step Make sure that you ve got something that s going to be easy to factor x2 3x l0lt0 Step 2 If possible factor the polynomial Note that it won t always be possible to factor this but that won t change things This step is really here to simplify the process more than anything Almost all of the problems that we re going to look at will be factorable x 5x2lt0 Step 3 Determine where the polynomial is zero Notice that these points won t make the inequality true in this case because 0 lt 0 is NOT a true inequality That isn t a problem These points are going to allow us to find the actual solution In our case the polynomial will be zero at x 2 and x 5 Now before moving on to the next step let s address why we want these points We haven t discussed graphing polynomials yet however the graphs of polynomials are nice smooth functions that have no breaks in them This means that as we are moving across the number line in any direction if the value of the polynomial changes sign say from positive to negative then it MUST go through zero So that means that these two numbers x 5 and x 2 are the ONLY places where the polynomial can change sign The number line is then divided into three regions In each region if the inequality is satisfied by one point from that region then it is satisfied for ALL points in that region If this wasn t true ie it was positive at one point in the region and negative at another then it must also be zero somewhere in that region but that can t happen as we ve already determined all the places where the polynomial can be zero Likewise if the inequality isn t 2007 Paul Dawkins 129 httptutorialmathlamaredutermsaspx College Algebra satisfied for some point in that region that it isn t satisfied for ANY point in that region This leads us into the next step Step 4 Graph the points where the polynomial is zero i e the points from the previous step on a number line and pick a test point from each of the regions Plug each of these test points into the polynomial and determine the sign of the polynomial at that point This is the step in the process that has all the work although it isn t too bad Here is the number line for this problem J3 E xC E x6 13j1 I CI E 52CI E lj3 HI I I I I I I I I I I I 4 3 E 1 I 1 E 3 4 5 I5 Now let s talk about this a little When we pick test points make sure that you pick easy numbers to work with So don t choose large numbers or fractions unless you are forced to by the problem Also note that we plugged the test points into the factored from of the polynomial and all we re really after here is whether or not the polynomial is positive or negative Therefore we didn t actually bother with values of the polynomial just the sign and we can get that from the product shown The product of two negatives is a positive etc We are now ready for the final step in the process Step 5 Write down the answer Recall that we discussed earlier that if any point from a region satisfied the inequality then ALL points in that region satisfied the inequality and likewise if any point from a region did not satisfy the inequality then NONE of the points in that region would satisfy the inequality This means that all we need to do is look up at the number line above If the test point from a region satisfies the inequality then that region is part of the solution If the test point doesn t satisfy the inequality then that region isn t part of the solution Now also notice that any value of x that will satisfy the original inequality will also satisfy the inequality from Step 2 and likewise if an x satisfies the inequality from Step 2 then it will satisfy the original inequality So that means that all we need to do is determine the regions in which the polynomial from Step 2 is negative For this problem that is only the middle region The inequality and interval notation for the solution to this inequality are 2007 Paul Dawkins 130 httptutorialmathlamaredutermsaspx College Algebra 2ltxlt5 25 Notice that we do need to exclude the endpoints since we have a strict inequality lt in this case in the inequality Okay that seems like a long process however it really isn t There was lots of explanation in the previous example The remaining examples won t be as long because we won t need quite as much explanation in them Example 2 Solve x2 5x 2 6 Solution Okay this time we ll just go through the process without all the explanations and steps The first thing to do is get a zero on one side and factor the polynomial if possible x2 5x620 x 3x 220 So the polynomial will be zero at x 2 and x 3 Notice as well that unlike the previous example these will be solutions to the inequality since we ve got a greater than or equal to in the inequality Here is the number line for this example x25 ZC5jI5jIII E39l LLLLLLLLLLLL H39 jj LLLLLLLLLLLL Notice that in this case we were forced to choose a decimal for one of the test points Now we want regions where the polynomial will be positive So the first and last regions will be part of the solution Also in this case we ve got an or equal to in the inequality and so we ll need to include the endpoints in our solution since at this points we get zero for the inequality and 0 2 0 is a true inequality Here is the solution in both inequality and interval notation form 00 lt x S 2 and 3 S x lt oo oo2 and 3o0 2007 Paul Dawkins 131 httptutorialmathlamaredutermsaspx College Algebra Example 3 Solve x4 4x3 l2x2 S 0 Solution Again we ll just jump right into the problem We ve already got zero on one side so we can go straight to factoring x44x3 l2x2 0 x2x24x l2S0 x2x6x 2SO So this polynomial is zero at x 6 x 0 and x 2 Here is the number line for this problem 339I II I F I ar39 F J ar39 F I I ar39 quot39 3939 339I II LIJ F I5339 ar39 F I5339 ar39 F I ar39 Hquot39 E I I 1 I I J quotquotH I1 395339 quot39I aIquot I quotquotH I I quot39I aIquot quotquotH I 395339 quot39I aIquot Hquot39 392 39 jj LLLLLLLLLLLL M L41 LLLLLLLLLLLL Ip39I L First notice that unlike the first two examples these regions do NOT alternate between positive and negative This is a common mistake that students make You really do need to plug in test points from each region Don t ever just plug in for the first region and then assume that the other regions will altemate from that point Now for our solution we want regions where the polynomial will be negative that s the middle two here or zero that s all three points that divide the regions So we can combine up the middle two regions and the three points into a single inequality in this case The solution in both inequality and interval notation form is 6SxS2 L62 Example 4 Solve xlx 32 gt 0 Solution The first couple of steps have already been done for us here So we can just straight into the work This polynomial will be zero at x l and x 3 Here is the number line for this problem 2007 Paul Dawkins 132 httptutorialmathlamaredutermsaspx College Algebra I 391 j LLLLLLLLLLLL LHJLLI LLLLLLLLLLLL Again note that the regions don t alternate in sign For our solution to this inequality we are looking for regions where the polynomial is positive that s the last two in this case however we don t want values where the polynomial is zero this time since we ve got a strict inequality gt in this problem This means that we want the last two regions but not x 3 So unlike the previous example we can t just combine up the two regions into a single inequality since that would include a point that isn t part of the solution Here is the solution for this problem lltxlt3 and 3ltxltoo l3 and 3oo Now all of the examples that we ve worked to this point involved factorable polynomials However that doesn t have to be the case We can work these inequalities even if the polynomial doesn t factor We should work one of these just to show you how they work Example 5 Solve 3x2 2x 11 gt 0 Solution In this case the polynomial doesn t factor so we can t do that step However we do still need to know where the polynomial is zero We will have to use the quadratic formula for that Here is what the quadratic formula gives us x li 34 3 In order to work the problem we ll need to reduce this to decimals x134 x1 3 4 3 3 227698 l6lO32 From this point on the process is identical to the previous examples In the number line below the dashed lines are at the approximate values of the two decimals above and the inequalities show the value of the quadratic evaluated at the test points shown 2007 Paul Dawkins 133 httptutorialmathlamaredutermsaspx College Algebra J2 I I I xU 33 5u I nD 100 I I I I I So it looks like we need the two outer regions for the solution Here is the inequality and interval notation for the solution 1 T 1 ooltxlt and jltxltoo 3 3 L r J Zj rJ Z J T W T 2007 Paul Dawkins 134 httptutorialmathlamaredutermsaspx College Algebra Rational Inequalities In this section we will solve inequalities that involve rational expressions The process for solving rational inequalities is nearly identical to the process for solving polynomial inequalities with a few minor differences Let s just jump straight into some examples x1s0 Example 1 Solve x Solution Before we get into solving these we need to point out that these DON T solve in the same way that we ve solve equations that contained rational expressions With equations the first thing that we always did was clear out the denominators by multiplying by the least common denominator That won t work with these however Since we don t know the value of x we can t multiply both sides by anything that contains an x Recall that if we multiply both sides of an inequality by a negative number we will need to switch the direction of the inequality However since we don t know the value of x we don t know if the denominator is positive or negative and so we won t know if we need to switch the direction of the inequality or not In fact to make matters worse the denominator will be both positive and negative for values of x in the solution and so that will create real problems So we need to leave the rational expression in the inequality Now the basic process here is the same as with polynomial inequalities The first step is to get a zero on one side and write the other side as a single rational inequality This has already been done for us here The next step is to factor the numerator and denominator as much as possible Again this has already been done for us in this case The next step is to determine where both the numerator and the denominator are zero In this case these values are numerator x l denominator x 5 Now we need these numbers for a couple of reasons First just like with polynomial inequalities these are the only numbers where the rational expression may change sign So we ll build a number line using these points to define ranges out of which to pick test points just like we did with polynomial inequalities There is another reason for needing the value of x that make the denominator zero however No matter what else is going on here we do have a rational expression and that means we need to avoid division by zero and so knowing where the denominator is zero will give us the values of x to avoid for this Here is the number line for this inequality 2007 Paul Dawkins 135 httptutorialmathlamaredutermsaspx College Algebra I I I I J 2 I IU I x5 I I 1 1390 I ltiU I 3 rI 7 I 5 I 1 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I 3 E 1 I 1 E 3 I 5 E 7quot So we need regions that make the rational expression negative That means the middle region Also since we ve got an or equal to part in the inequality we also need to include where the inequality is zero so this means we include x 1 Notice that we will also need to avoid x 5 since that gives division by zero The solution for this inequality is 1 S x lt 5 1 5 2 Example 2 Solve w gt O x 1 Solution We ve got zero on one side so let s first factor the numerator and determine where the numerator and denominator are both zero xlx3 gt0 x l numerator x 1 x 3 denominator x 1 Here is the number line for this one x 4 2 2 9 II x 2 5 3 X D L I I53 I33 I T LLLLLLLLLLLL FI H U FI H U quotHquot39 392 39 L41 111111111111 FI Lh39 39II39 I quot H U lil U quotH 392 I LL 111111111111 H U In the problem we are after values of x that make the inequality strictly positive and so that looks like the second and fourth region and we won t include any of the endpoints here The solution is then 3ltxlt 1 and 1ltxltoo 3 1 and Loo 2007 Paul Dawkins 136 httptutorialmathlamaredutermsaspx College Algebra x2 l6 x 1 2 lt0 Example 3 Solve Solution There really isn t too much to this example We ll first need to factor the numerator and then determine where the numerator and denominator are zero x 4x4 Hf numerator x 4 x 4 denominator x 1 lt0 The number line for this problem is 35 E xC E 32 E 35 E31E119 I 3333 U I 31 Di 1 3D 33 I 1 E 1 g 13 I I I I I I I I I I I I I 3 3 4 3 3 1 3 1 3 3 4 3 3 So as with the polynomial inequalities we can just assume that the regions will always alternate in sign Also note that while the middle two regions do give negative values in the rational expression we need to avoid x 1 to make sure we don t get division by zero This means that we will have to write the answer as two inequalities andor intervals 4 lt x lt1 and 1 lt x lt 4 43 and 14 Once again it s important to note that we really do need to test each region and not just assume that the regions will alternate in sign Next we need to take a look at some examples that don t already have a zero on one side of the inequality Example 4 Solve 3x 1 21 x4 Solution The first thing that we need to do here is subtract 1 from both sides and then get everything into a single rational expression 2007 Paul Dawkins 137 httptutorialmathlamaredutermsaspx College Algebra 3xl x4 3xlx4 20 x4 x4 3x1 x4 x4 2x 3gt x4 120 20 O In this case there is no factoring to do so we can go straight to identifying where the numerator and denominator are zero 3 numerator x E denominator x 4 Here is the number line for this problem 39 I I f 5 E xC E x2 I I IquotJ I 3ltj I lj I 4 I 5 I I 39 I 39 I I I I I I I I I I I I I I I I I I I I I I I I5 5 4 3 2 1 El 1 2 3 I Okay we want values of x that give positive andor zero in the rational expression This looks 3 like the outer two regions as Well as x 5 As with the first example we will need to avoid x 4 since that will give a division by zero error The solution for this problem is then ooltxlt 4 and Sxltoo and Ed x8S3 x Example 5 Solve x Solution So again the first thing to do is to get a zero on one side and then get everything into a single rational expression 2007 Paul Dawkins 138 httptutorialmathlamaredutermsaspx College Algebra x 8 x x 8 xx 3 x x x 8x2 3x x 3SO S0 SO x 2 33 S0 x42 X S0 We also factored the numerator above so we can now determine where the numerator and denominator are zero numerator x k x 4 denominator x 0 Here is the number line for this problem I I I I I 39 I 3 I 1 3939 1 E I 5 I II jI U 31 5 mi 3j3j g E II 1II1 H 3 I 1 I 1 I 5 I I I I I I I I I I I I I I I I I I I I I I I I I I I I 4 3 2 1 III 1 2 3 4 5 I5 The solution for this inequality is then ooltxS 2 and 0ltxS4 oo 2 and O4 2007 Paul Dawkins 139 httptutorialmathlamaredutermsaspx College Algebra Absolute Value Equations In the final two sections of this chapter we want to discuss solving equations and inequalities that contain absolute values We will look at equations with absolute value in them in this section and we ll look at inequalities in the next section Before solving however we should first have a brief discussion of just what absolute value is The notation for the absolute value of p is p Note as well that the absolute value bars are NOT parentheses and in many cases don t behave as parentheses so be careful with them There are two ways to define absolute value There is a geometric definition and a mathematical definition We will look at both Geometric Definition In this definition we are going to think of p as the distance of p from the origin on a number line Also we will always use a positive value for distance Consider the following number line 339 Distance cf E Distance cf 3 Distance cf 2 I I I 5 4 3 2 1 I3 1 2 3 4 5 From this we can get the following values of absolute value 9 22 3 3 2 9 2 All that we need to do is identify the point on the number line and determine its distance from the origin Note as well that we also have 0 0 Mathematical Definition We can also give a strict mathematicalformula definition for absolute value It is p p20 p p plt0 This tells us to look at the sign of p and if it s positive we just drop the absolute value bar If p is negative we drop the absolute value bars and then put in a negative in front of it So let s see a couple of quick examples 2007 Paul Dawkins 140 httptutorialmathlamaredutermsaspx College Algebra 4 4 because 4 2 0 8 G 8 8 because 8 lt O 0 0 because 0 2 0 Note that these give exactly the same value as if we d used the geometric interpretation One way to think of absolute value is that it takes a number and makes it positive In fact we can guarantee that p20 regardless of the value of p We do need to be careful however to not misuse either of these definitions For example we can t use the definition on Ixl because we don t know the value of x Also don t make the mistake of assuming that absolute value just makes all minus signs into plus signs In other words don t make the following mistake 4x 3 4x3 This just isn t true If you aren t sure that you believe that plug in a number for x For example if x l we would get 7 A l4 1 3 41E3 1 There are a couple of problems with this First the numbers are clearly not the same and so that s all we really need to prove that the two expressions aren t the same There is also the fact however that the right number is negative and we will never get a negative value out of an absolute value That also will guarantee that these two expressions aren t the same The definitions above are easy to apply if all we ve got are numbers inside the absolute value bars However once we put variables inside them we ve got to start being very careful It s now time to start thinking about how to solve equations that contain absolute values Let s start off fairly simple and look at the following equation P4 Now if we think of this from a geometric point of view this means that whatever p is it must have a distance of 4 from the origin Well there are only two numbers that have a distance of 4 from the origin namely 4 and 4 So there are two solutions to this equation p 4 or p 4 2007 Paul Dawkins 141 httptutorialmathlamaredutermsaspx College Algebra Now if you think about it we can do this for any positive number not just 4 So this leads to the following general formula for equations involving absolute value If pgtbb 0 then p borp b Notice that this does require the 9 be a positive number We will deal with what happens if b is zero or negative in a bit Let s take a look at some examples Example 1 Solve each of the following a 2x 5 9 Solution b 1 3t 20 E C 5 y 8 1 Solution Solution Now remember that absolute value does not just make all minus signs into plus signs To solve these we ve got to use the formula above since in all cases the number on the right side of the equal sign is positive a 2x 5 9 There really isn t much to do here other than using the formula from above as noted above All we need to note is that in the formula above p represents whatever is on the inside of the absolute value bars and so in this case we have 2x 5 9 or 2x 59 At this point we ve got two linear equations that are easy to solve 2x 4 or 2x 14 x 2 or x7 So we ve got two solutions to the equation x 2 and x 7 Return to Problems b 1 3r 20 This one is pretty much the same as the previous part so we won t put as much detail into this one 1 3t 20 or 1 3t2O 3t 21 or 3t19 1 t 7 or t 9 3 The two solutions to this equation are t 3 and t 7 Return to Problems 2007 Paul Dawkins 142 httptutorialmathlamaredutermsaspx College Algebra c 5y 8l Again not much more to this one 5y 8 l or 5y 8l 5y7 or 53229 1 O 2 5 5 7 9 In this case the two solutions are y g and y p Y Return to Problems Now let s take a look at how to deal with equations for which 9 is zero or negative We ll do this with an example Example 2 Solve each of the following a l0x 3 0 b 5x9 E 3 Solution a Let s approach this one from a geometric standpoint This is saying that the quantity in the absolute value bars has a distance of zero from the origin There is only one number that has the property and that is zero itself So we must have lOx 3O gtx 3 10 In this case we get a single solution b Now in this case let s recall that we noted at the start of this section that pl 2 0 In other words we can t get a negative value out of the absolute value That is exactly what this equation is saying however Since this isn t possible that means there is no solution to this equation So summarizing we can see that if bis zero then we can just drop the absolute value bars and solve the equation Likewise if bis negative then there will be no solution to the equation To this point we ve only looked at equations that involve an absolute value being equal to a number but there is no reason to think that there has to only be a number on the other side of the equal sign Likewise there is no reason to think that we can only have one absolute value in the problem So we need to take a look at a couple of these kinds of equations Example 3 Solve each of the following a x 2 3x H E b 4x 3 3 x E c 2x 14x 91 1 Solution At first glance the formula we used above will do us no good here It requires the right side of the 2007 Paul Dawkins 143 httptutorialmathlamaredutermsaspx College Algebra equation to be a positive number It turns out that we can still use it here but we re going to have to be careful with the answers as using this formula will on occasion introduce an incorrect answer So while we can use the formula we ll need to make sure we check our solutions to see if they really work a x 23x 1 So we ll start off using the formula above as we have in the previous problems and solving the two linear equations x 2 395 1 3x 1 or x 233e 1 4x21 or 2x3 1 3 x or x 4 Okay we ve got two potential answers here There is a problem with the second one however If we plug this one into the equation we get 2 pG pG 1 2 7 1 Z 2 2 2 Z 1 NOT OK 2 2 We get the same number on each side but with opposite signs This will happen on occasion when we solve this kind of equation with absolute values Note that we really didn t need to plug the solution into the whole equation here All we needed to do was check the portion without the absolute value and if it was negative then the potential solution will NOT in fact be a solution and if it s positive or zero it will be solution We ll leave it to you to verify that the first potential solution does in fact work and so there is a single solution to this equation x Z and notice that this is less than 2 as our assumption required and so is a solution to the equation with the absolute value in it 1 So all together there is a single solution to this equation x Z Return to Problems b 4x3 3 x This one will work in pretty much the same way so we won t put in quite as much explanation 4x3 3 x 3 x or 4x33 x 3x 6 or 5x0 x 2 or xO Now before we check each of these we should give a quick warning Do not make the assumption that because the first potential solution is negative it won t be a solution We only 2007 Paul Dawkins 144 httptutorialmathlamaredutermsaspx College Algebra exclude a potential solution if it makes the portion without absolute value bars negative In this case both potential solutions will make the portion without absolute value bars positive and so both are in fact solutions So in this case unlike the first example we get two solutions x 2 and x 0 Return to Problems c 2x 1 4x 91 This case looks very different from any of the previous problems we ve worked to this point and in this case the formula we ve been using doesn t really work at all However if we think about this a little we can see that we ll still do something similar here to get a solution Both sides of the equation contain absolute values and so the only way the two sides are equal will be if the two quantities inside the absolute value bars are equal or equal but with opposite signs Or in other words we must have 2x l 46 9 4ac 9 or 2x l4a 9 6x 8 or 2xlO xs 1 or xs 5 6 3 Now we won t need to verify our solutions here as we did in the previous two parts of this problem Both with be solutions provided we solved the two equations correctly However it will probably be a good idea to verify them anyway just to show that the solution technique we used here really did work properly 4 Let s first check x 4139 i 3 3 11 11 OK 3 3 In the case the quantities inside the absolute value were the same number but opposite signs 4 However upon takrng the absolute value we got the same number and so x 3 1s a solution Now let s check x 5 12lt sgte1H4lt sgt 9 IuIiI 111 1111 OK In the case we got the same value inside the absolute value bars 2007 Paul Dawkins 145 httptutorialmathlamaredutermsaspx College Algebra So as suggested above both answers did in fact work and both are solutions to the equation Return to Problems So as we Ve seen in the previous set of examples we need to be a little careful if there are Variables on both sides of the equal sign If one side does not contain an absolute Value then we need to look at the two potential answers and make sure that each is in fact a solution 2007 Paul Dawkins 146 httptutorialmathlamaredutermsaspx College Algebra Absolute Value Inequalities In the previous section we solved equations that contained absolute values In this section we want to look at inequalities that contain absolute values We will need to examine two separate cases Inequalities Involving lt and S As we did with equations let s start off by looking at a fairly simple case pS4 This says that no matter what p is it must have a distance of no more than 4 from the origin This means that p must be somewhere in the range 4 S p S 4 We could have a similar inequality with the lt and get a similar result In general we have the following formulas to use here If pSbbgt0 then bSpSb If pltbbgtO then bltpltb Notice that this does require 9 to be positive just as we did with equations Let s take a look at a couple of examples Example 1 Solve each of the following a 2x 4 lt10 1 b 9m 2 S 1 Solution c 3 2z s 5 1 Solution a 2x 4 lt10 There really isn t much to do other than plug into the formula As with equations p simply represents whatever is inside the absolute value bars So with this first one we have 10 lt 2x 4 lt 10 Now this is nothing more than a fairly simple double inequality to solve so let s do that 6 lt 2x lt14 3 lt x lt 7 The interval notation for this solution is 3 7 Return to Problems 2007 Paul Dawkins 147 httptutorialmathlamaredutermsaspx College Algebra b 9m 2 3 1 Not much to do here 1 2 S l l SmS 3 9 l l The interval notation is 3 9 Return to Problems c 3 2zS5 We ll need to be a little careful with solving the double inequality with this one but other than that it is pretty much identical to the previous two parts 5 S 3 2z S 5 8S 2zS2 42z2 1 In the final step don t forget to switch the direction of the inequalities since we divided everything by a negative number The interval notation for this solution is 1 4 Return to Problems Inequalities Involving gt and 2 Once again let s start off with a simple number example p24 This says that whatever p is it must be at least a distance of 4 from the origin and so p must be in one of the following two ranges pS 4 or p24 Before giving the general solution we need to address a common mistake that students make with these types of problems Many students try to combine these into a single double inequality as follows 4 2 p 2 4 While this may seem to make sense we can t stress enough that THIS IS NOT CORRECT Recall what a double inequality says In a double inequality we require that both of the inequalities be satisfied simultaneously The double inequality above would then mean that p is a number that is simultaneously smaller than 4 and larger than 4 This just doesn t make sense There is no number that satisfies this These solutions must be written as two inequalities Here is the general formula for these 2007 Paul Dawkins 148 httptutorialmathlamaredutermsaspx College Algebra If p2bbgtO then 193 or pzb If pgtbbgtO then plt b or pgtb Again we will require that 9 be a positive number here Let s work a couple of examples Example 2 Solve each of the following a 2x 3 gt 7 E b 6t 10 2 3 Solution c I26yI gt10 mien Solution a 2x 3 gt 7 Again p represents the quantity inside the absolute value bars so all we need to do here is plug into the formula and then solve the two linear inequalities 2x 3lt 7 or 2x 3gt7 2xlt 4 or 2xgt1O xlt 2 or xgt5 The interval notation for these are 00 2 or 5 00 Return to Problems b 6t 10 2 3 Let s just plug into the formulas and go here 6t10 3 or 6t1023 6t 3 13 or 6t 2 7 1 t S 3 or t 2 1 13 7 The interval notation for these are 00 or g oo Return to Problems c 2 6y gt10 Again not much to do here 2 6ylt 1O or 2 6ygt1O 6ylt 12 or 6ygt8 4 y gt 2 or y lt E Notice that we had to switch the direction of the inequalities when we divided by the negative 4 number The interval notation for these solutions is 2 00 or 00 j Return to Problems 2007 Paul Dawkins 149 httptutorialmathlamaredutermsaspx College Algebra Okay we next need to take a quick look at what happens if bis zero or negative We ll do these with a set of examples and let s start with zero Example 3 Solve each of the following a 3x 2 lt O J b x 9 S 0 E c 2x 4 2 0 E d 3x 9 gt 0 E Solution These four examples seem to cover all our bases a Now we know that pl 2 O and so can t ever be less than zero Therefore in this case there is no solution since it is impossible for an absolute value to be strictly less than zero i e negative b This is almost the same as the previous part We still can t have absolute value be less than zero however it can be equal to zero So this will have a solution only if xq0 and we know how to solve this from the previous section x 9 0 gt x 9 c In this case let s again recall that no matter what p is we are guaranteed to have p 2 0 This means that no matter what x is we can be assured that 2x 4 2 0 will be true since absolute values will always be positive or zero The solution in this case is all real numbers or all possible values of x In inequality notation this wouldbe o0ltxlto0 d This one is nearly identical to the previous part except this time note that we don t want the absolute value to ever be zero So we don t care what value the absolute value takes as long as it isn t zero This means that we just need to avoid values of x for which we get 3x 90 gt3x 9 9 x 3 The solution in this case is all real numbers except x 3 Now let s do a quick set of examples with negative numbers Example 4 Solve each of the following a 4x 15 lt 2 and 4x 15 s 2 b 2x 9 2 8 and 2x 9 gt 8 Solution Notice that we re working these in pairs because this time unlike the previous set of examples the solutions will be the same for each 2007 Paul Dawkins 150 httptutorialmathlamaredutermsaspx College Algebra Both all four of these will make use of the fact that no matter what p is we are guaranteed to have p 2 O In other words absolute values are always positive or zero a Okay if absolute values are always positive or zero there is no way they can be less than or equal to a negative number Therefore there is no solution for either of these b In this case if the absolute value is positive or zero then it will always be greater than or equal to a negative number The solution for each of these is then all real numbers 2007 Paul Dawkins 151 httptutorialmathlamaredutermsaspx College Algebra Graphing and Functions Introduction In this chapter we will be introducing two topics that are very important in an algebra class We will start off the chapter with a brief discussion of graphing This is not really the main topic of this chapter but we need the basics down before moving into the second topic of this chapter The next chapter will contain the remainder of the graphing discussion The second topic that we ll be looking at is that of functions This is probably one of the more important ideas that will come out of an Algebra class When first studying the concept of functions many students don t really understand the importance or usefulness of functions and function notation The importance andor usefulness of functions and function notation will only become apparent in later chapters and later classes In fact there are some topics that can only be done easily with function and function notation Here is a brief listing of the topics in this chapter Graphing In this section we will introduce the Cartesian coordinate system and most of the basics of graphing equations Lines Here we will review the main ideas from the study of lines including slope and the special forms of the equation of a line Circles We will look at the equation of a circle and graphing circles in this section The Definition of a Function We will discuss the definition of a function in this section We will also introduce the idea of function evaluation Graphing Functions In this section we will look at the basics of graphing functions We will also graph some piecewise functions in this section Combining functions Here we will look at basic arithmetic involving functions as well as function composition Inverse Functions We will define and find inverse functions in this section 2007 Paul Dawkins 152 httptutorialmathlamaredutermsaspX College Algebra Graphing In this section we need to review some of the basic ideas in graphing It is assumed that you ve seen some graphing to this point and so we aren t going to go into great depth here We will only be reviewing some of the basic ideas We will start off with the Rectangular or Cartesian coordinate system This is just the standard axis system that we use when sketching our graphs Here is the Cartesian coordinate system with a few points plotted v QuatIra111 II IE3 Quadraxit I 323 22 131 quotI 23939 3 2 I 1 2 3 II 1 2 1 3 33 Quat1ra11t III Quat1ra111 III The horizontal and vertical axes typically called the xaxis and the yaxis respectively divide the coordinate system up into quadrants as shown above In each quadrant we have the following signs for x and y Quadrant I x gt O or x positive y gt 0 or y positive Q11ad1 antH x lt 0 or x negative y gt O or y positive Q11ad1 antHI x lt O or x negative y lt O or y negative Quadrant IV x gt O or x positive y lt 0 or y negative Each point in the coordinate system is defined by an ordered pair of the form x y The first number listed is the xcoordinate of the point and the second number listed is the ycoordinate of the point The ordered pair for any given point x y is called the coordinates for the point The point where the two axes cross is called the origin and has the coordinates 0 0 Note as well that the order of the coordinates is important For example the point 21 is the point that is two units to the right of the origin and then 1 unit up while the point 1 2 is the point that is 1 unit to the right of the origin and then 2 units up 2007 Paul Dawkins 153 httptutorialmathlamaredutermsaspx College Algebra We now need to discuss graphing an equation The first question that we should ask is what exactly is a graph of an equation A graph is the set of all the ordered pairs whose coordinates satisfy the equation For instance the point 2 3 is a point on the graph of y x l2 4 while 15 isn t on the graph How do we tell this All we need to do is take the coordinates of the point and plug them into the equation to see if they satisfy the equation Let s do that for both to verify the claims made above 2 3 In this case we have x 2 and y 2 so plugging in gives 3 3 OK So the coordinates of this point satisfies the equation and so it is a point on the graph 15 Here we have x 1 and y 5 Plugging these in gives 5 r 1 4 9 5mf 4 5 4 NOTOK The coordinates of this point do NOT satisfy the equation and so this point isn t on the graph Now how do we sketch the graph of an equation Of course the answer to this depends on just how much you know about the equation to start off with For instance if you know that the equation is a line or a circle we ve got simple ways to determine the graph in these cases There are also many other kinds of equations that we can usually get the graph from the equation without a lot of work We will see many of these in the next chapter However let s suppose that we don t know ahead of time just what the equation is or any of the ways to quickly sketch the graph In these cases we will need to recall that the graph is simply all the points that satisfy the equation So all we can do is plot points We will pick values of x compute y from the equation and then plot the ordered pair given by these two values How do we determine which values of x to choose Unfortunately the answer there is we guess We pick some values and see what we get for a graph If it looks like we ve got a pretty good sketch we stop If not we pick some more Knowing the values of x to choose is really something that we can only get with experience and some knowledge of what the graph of the equation will probably look like Hopefully by the end of this course you will have gained some of this knowledge 2007 Paul Dawkins 154 httptutorialmathlamaredutermsaspx College Algebra Let s take a quick look at a graph Example 1 Sketch the graph of y x l2 4 Solution Now this is a parabola and after the next chapter you will be able to quickly graph this without much effort However we haVen t gotten that far yet and so we will need to choose some Values of x plug them in and compute the y Values As mentioned earlier it helps to have an idea of what this graph is liable to look like when picking Values of x So don t worry at this point why we chose the Values that we did After the next chapter you would also be able to choose these Values of x Here is a table of Values for this equation x y x y 2 5 25 1 0 10 0 3 0 3 1 4 1 4 2 3 2 3 3 0 30 4 5 45 Let s Verify the first one and we ll leave the rest to you to Verify For the first one we simply plug x 2 into the equation and compute y y 2 l2 4 4 32 4 9 4 5 Here is the graph of this equation 2007 Paul Dawkins 155 httptutorialmathlamaredutermsaspX College Algebra Z39I39I J2 Iquot39 IIIIIIIIIIIIIII 1 III II i39I IILI Notice that when we set up the axis system in this example we only set up as much as we needed For example since we didn t go past 2 with our computations we didn t go much past that with our axis system Also notice that we used a different scale on each of the axes With the horizontal axis we incremented by l s while on the vertical axis we incremented by 2 This will often be done in order to make the sketching easier The final topic that we want to discuss in this section is that of intercepts Notice that the graph in the above example crosses the xaxis in two places and the yaxis in one place All three of these points are called intercepts We can and often will be more specific however We often will want to know if an intercept crosses the x or yaxis specifically So if an intercept crosses the xaxis we will call it an xintercept Likewise if an intercept crosses the yaxis we will call it a yintercept Now since the xintercept crosses xaxis then the y coordinates of the xintercepts will be zero Also the x coordinate of the yintercept will be zero since these points cross the yaxis These facts give us a way to determine the intercepts for an equation To find the xintercepts for an equation all that we need to do is set y O and solve for x Likewise to find the yintercepts for an equation we simply need to set x 0 and solve for y Let s take a quick look at an example Example 2 Determine the xintercepts and yintercepts for each of the following equations a yx2 4x 6 E b y x2 2 E C y X 1 Solution Solution As verification for each of these we will also sketch the graph of each function We will leave the details of the sketching to you to verify Also these are all parabolas and as mentioned earlier we 2007 Paul Dawkins 156 httptutorialmathlamaredutermsaspx College Algebra will be looking at these in detail in the next chapter a y x2 Jae 6 Let s first find the yintercepts Again we do this by setting x 0 and solving for y This is usually the easier of the two So let s find the yintercepts 2 y O 0 6r 6 So there is a single yintercept 0 6 The work for the xintercepts is almost identical except in this case we set y O and solve for x Here is that work 0x2 Jae 6 0sFx 3x 2 x 3x 2 For this equation there are two xintercepts 3 0 and 2 0 Oh and you do remember how to solve quadratic equations right For Verification purposes here is sketch of the graph for this equation 2 4 E Return to Problems b y x2 2 First the yintercepts y 02 2 2 gt 02 So we Ve got a single yintercepts Now the xintercepts O x2 2 2 x2 gt x F x2139 Okay we got complex solutions from this equation What this means is that we will not have any xintercepts Note that it is perfectly acceptable for this to happen so don t worry about it when it does happen Here is the graph for this equation 2007 Paul Dawkins 157 httptutorialmathlamaredutermsaspx College Algebra 5 1 139 Ll E 1 I I 2 Sure enough it doesn t cross the xaxis Return to Problems c y X 1 Here is the yintercept work for this equation y 0 12 1 gt 01 Now the xintercept work egtx 19 xgtl 10 In this case we have a single xintercept Here is a sketch of the graph for this equation 3 2 1 Now notice that in this case the graph doesn t actually cross the xaxis at x l This point is still called an xintercept however Return to Problems We should make one final comment before leaving this section In the previous set of examples all the equations were quadratic equations This was done only because the exhibited the range of behaviors that we were looking for and we would be able to do the work as well You should not walk away from this discussion of intercepts with the idea that they will only occur for quadratic equations They can and do occur for many different equations 2007 Paul Dawkins 158 httptutorialmathlamaredutermsaspX College Algebra Lines Let s start this section off with a quick mathematical definition of a line Any equation that can be written in the form Ax By C where we can t have both A and B be zero simultaneously is a line It is okay if one of them is zero we just can t have both be zero Note that this is sometimes called the standard form of the line Before we get too far into this section it would probably be helpful to recall that a line is defined by any two points that are on the line Given two points that are on the line we can graph the line andor write down the equation of the line This fact will be used several times throughout this section One of the more important ideas that we ll be discussing in this section is that of slope The slope of a line is a measure of the steepness of a line and it can also be used to measure whether a line is increasing or decreasing as we move from left to right Here is the precise definition of the slope of a line Given any two points on the line say xl yl and x2 y2 the slope of the line is given by y2 y1 x2 x1 m In other words the slope is the difference in the y values divided by the difference in the x values Also do not get worried about the subscripts on the variables These are used fairly regularly from this point on and are simply used to denote the fact that the variables are both x or y values but are in all likelihood different When using this definition do not worry about which point should be the first point and which point should be the second point You can choose either to be the first andor second and we ll get exactly the same value for the slope There is also a geometric definition of the slope of the line as well You will often hear the slope as being defined as follows rise 11111 The two definitions are identical as the following diagram illustrates The numerators and denominators of both definitions are the same 2007 Paul Dawkins 159 httptutorialmathlamaredutermsaspx College Algebra Note as well that if we have the slope written as a fraction and a point on the line say xl yl then we can easily find a second point that is also on the line Before seeing how this can be done let s take the convention that if the slope is negative we will put the minus sign on the numerator of the slope In other words we will assume that the rise is negative if the slope is negative Note as well that a negative rise is really a fall So we have the slope written as a fraction and a point on the line x1 y1 To get the coordinates of the second point x2 yz all that we need to do is start at xl yl then move to the right by the run or denominator of the slope and then updown by rise or the numerator of the slope depending on the sign of the rise We can also write down some equations for the coordinates of the second point as follows x2 x1 run 372 Y1 lse Note that if the slope is negative then the rise will be a negative number Let s compute a couple of slopes Example 1 Determine the slope of each of the following lines Sketch the graph of each line a The line that contains the two points 2 3 and 31 Solution 5 and O 2 Solution 32 and 52 Solution b The line that contains the two points 1 d The line that contains the two points 4 3 and 4 2 Solution c The line that contains the two points Solution Okay for each of these all that we ll need to do is use the slope formula to find the slope and then plot the two points and connect them with a line to get the graph a The line that contains the two points 2 3 and 31 Do not worry which point gets the subscript of l and which gets the subscript of 2 Either way 2007 Paul Dawkins 160 httptutorialmathlamaredutermsaspX College Algebra will get the same answer Typically we ll just take them in the order listed So here is the slope for this part quotru 1g 3 2 3 2 5 Be careful with minus signs in these computations It is easy to lose track of them Also when the slope is a fraction as it is here leave it as a fraction Do not convert to a decimal unless you absolutely have to Here is a sketch of the line IIIIIIIIIIIIIIIIIII II JIIIIIII Notice that this line increases as we move from left to right Return to Problems b The line that contains the two points 1 5 and 0 2 Here is the slope for this part 2 5 7 0 1 1 Again watch out for minus signs Here is a sketch of the graph 15 E 45 23 E I I 1 2 2I2 K This line decreases as we move from left to right Return to Problems 2007 Paul Dawkins 161 httptutorialmathlamaredutermsaspX College Algebra c The line that contains the two points 3 2 and 5 2 Here is the slope for this line 2 2 O 39 5 3 8 We got a slope of zero here That is okay it will happen sometimes Here is the sketch of the line 39I39 IIIII II 1 M IIIIIIIII 392 Fl IIIIIIIIIIIIIIIIIIIIIII 4121245 K I 4 I ll In this case we Ve got a horizontal line Return to Problems d The line that contains the two points 4 3 and 4 2 The final part Here is the slope computation 2 3 5 m undefined 4 4 O In this case we get division by zero which is undefined Again don t worry too much about this it will happen on occasion Here is a sketch of this line 1 3 5143 2 1 llll lllllllllllllllllll lllll 1 1 1 2 3 1 E 2 H2 3 This final line is a Vertical line Return to Problems We can use this set of examples to see some general facts about lines First we can see from the first two parts that the larger the number ignoring any minus signs the steeper the line So we can use the slope to tell us something about just how steep a line is 2007 Paul Dawkins 162 httptutorialmathlamaredutermsaspX College Algebra Next we can see that if the slope is a positive number then the line will be increasing as we move from left to right Likewise if the slope is a negative number then the line will decrease as we move from left to right We can use the final two parts to see what the slopes of horizontal and vertical lines will be A horizontal line will always have a slope of zero and a vertical line will always have an undefined slope We now need to take a look at some special forms of the equation of the line We will start off with horizontal and vertical lines A horizontal line with a yintercept of b will have the equation yb Likewise a vertical line with an xintercept of a will have the equation So if we go back and look that the last two parts of the previous example we can see that the equation of the line for the horizontal line in the third part is y 2 while the equation for the vertical line in the fourth part is x 4 The next special form of the line that we need to look at is the pointslope form of the line This form is very useful for writing down the equation of a line If we know that a line passes through the point x1 y1 and has a slope of m then the pointslope form of the equation of the line is yy1mx 3 Sometimes this is written as yyt1 mac xl The form it s written in usually depends on the instructor that is teaching the class As stated earlier this form is particularly useful for writing down the equation of a line so let s take a look at an example of this Example 2 Write down the equation of the line that passes through the two points 2 4 and 3 5 Solution At first glance it may not appear that we ll be able to use the pointslope form of the line since this requires a single point we ve got two and the slope which we don t have However that fact that we ve got two points isn t really a problem in fact we can use these two points to determine the missing slope of the line since we do know that we can always find that from any two points on the line 2007 Paul Dawkins 163 httptutorialmathlamaredutermsaspx College Algebra So let s start off my finding the slope of the line 5 4 9 rnzg 5 32 Now which point should we use to write down the equation of the line We can actually use either point To show this we will use both First we ll use 2 4 Now that we Ve gotten the point all that we need to do is plug into the formula We will use the second form y4 x2gt 9 ltx gt Now let s use 3 5 y 5 x 3 Okay we claimed that it wouldn t matter which point we used in the formula but these sure look like different equations It turns out however that these really are the same equation To see this all that we need to do is distribute the slope through the parenthesis and then simplify Here is the first equation 9 y 4 gkx 2 4 2 E 5 5 9 2 x 5 5 Here is the second equation 9 y 5 gx 3 5 5 9 2 5 5 So sure enough they are the same equation The final special form of the equation of the line is probably the one that most people are familiar with It is the slopeintercept form In this case if we know that a line has slope m and has a y intercept of 0 b then the slopeintercept form of the equation of the line is y mx 9 This form is particularly useful for graphing lines Let s take a look at a couple of examples 2007 Paul Dawkins 164 httptutorialmathlamaredutermsaspX College Algebra Example 3 Determine the slope of each of the following equations and sketch the graph of the line a 2y 6x 2 E b 3 496 6 Solution Okay to get the slope we ll first put each of these in slopeintercept form and then the slope will simply be the coefficient of the x including sign To graph the line we know the yintercept of the line that s the number without an x including sign and as discussed above we can use the slope to find a second point on the line At that point there isn t anything to do other than sketch the line a 2 y 6x 2 First solve the equation for y Remember that we solved equations like this back in the previous chapter 2y6x 2 y3x 1 So the slope for this line is 3 and the yintercept is the point 0 1 Don t forget to take the sign when determining the yintercept Now to find the second point we usually like the slope written as a fraction to make it clear what the rise and run are So WQ3 2 2 rise 3 run 1 1 run The second point is then x01 1 y 1I 2 2 gt 12 Here is a sketch of the graph of the line Return to Problems 2007 Paul Dawkins 165 httptutorialmathlamaredutermsaspX College Algebra b 3y4x6 Again solve for y 3y 49C 6 4 2 3 395 4 In this case the slope is g and the yintercept is 0 2 As with the previous part let s first determine the rise and the run 4 rise gt E1186 4 run 3 3 run Again remember that if the slope is negative make sure that the minus sign goes with the numerator The second point is then x2033 y2264 2 gt 3r 2 wIHgt Here is the sketch of the graph for this line i39I IIIIIIIIILI 2 1 l r391 39 IIIIIIIIIIII IIIIIIIIII IIIIIIIII in Return to Problems The final topic that we need to discuss in this section is that of parallel and perpendicular lines Here is a sketch of parallel and perpendicular lines L13911e 1 Line 1 Parallel Lines Ferpezridicillar Lines 2007 Paul Dawkins 166 httptutorialmathlamaredutermsaspX College Algebra Suppose that the slope of Line 1 is ml and the slope of Line 2 is m2 We can relate the slopes of parallel lines and we can relate slopes of perpendicular lines as follows parallel ml m2 perpendicular mlm2 1or ml Note that there are two forms of the equation for perpendicular lines The second is the more common and in this case we usually say that m2 is the negative reciprocal of ml Example 4 Determine if the line that passes through the points 2 10 and 6 1 is parallel perpendicular or neither to the line given by 7 y 9x 15 Solution Okay in order to do answer this we ll need the slopes of the two lines Since we have two points for the first line we can use the formula for the slope 10 2 ml 6 2 8 We don t actually need the equation of this line and so we won t bother with it Now to get the slope of the second line all we need to do is put it into slopeintercept form 7y9x 15 p 3 m 2 7 7 7 Okay since the two slopes aren t the same they re close but still not the same the two lines are not parallel Also one it 8 7 56 so the two lines aren t perpendicular either Therefore the two lines are neither parallel nor perpendicular Example 5 Determine the equation of the line that passes through the point 8 2 and is a parallel to the line given by 10y 396 2 Solution b perpendicular to the line given by 10 y 396 2 Solution Solution In both parts we are going to need the slope of the line given by 10 y 336 2 so let s actually find that before we get into the individual parts 10y 3ac 2 y ix gt 3 10 5 ml 10 Now let s work the example 2007 Paul Dawkins 167 httptutorialmathlamaredutermsaspX College Algebra a parallel to the line given by 10y 3x 2 In this case the new line is to be parallel to the line given by 10 y 336 2 and so it must have the same slope as this line Therefore we know that m2 10 Now we Ve got a point on the new line 8 2 and we know the slope of the new line E so we can use the pointslope form of the line to write down the equation of the new line Here is the equation 3 2 x 8 ya 2 ix E 10 10 3 44 x j 10 10 I3 x 2 10 5 Return to Problems b perpendicular to the line given by 10y 3x 2 For this part we want the line to be perpendicular to 10y 336 2 and so we know we can find the new slope as follows E Then just as we did in the previous part we can use the pointslope form of the line to get the equation of the new line Here it is 10 yeIa2 x 8 2 Ex Q 3 Q E 3 3 Return to Problems 2007 Paul Dawkins 168 httptutorialmathlamaredutermsaspX College Algebra Circles In this section we are going to take a quick look at circles However before we do that we need to give a quick formula that hopefully you ll recall seeing at some point in the past Given two points x1 y1 and x2 y2 the distance between them is given by d x2 12 Y2 3 12 So why did we remind you of this formula Well let s recall just what a circle is A circle is all the points that are the same distance r called the radius from a point h k called the center In other words if x y is any point that is on the circle then it has a distance of r from the center h k If we use the distance formula on these two points we would get rJltx hf w W Or if we square both sides we get x h2 y k2 r2 This is the standard form of the equation of a circle with radius r and center h k Example 1 Write down the equation of a circle with radius 8 and center 4 7 Solution Okay in this case we have r 8 h 4 and k 7 so all we need to do is plug them into the standard form of the equation of the circle x lt 4gt2lty 7gt2s2 x42y 7264 Do not square out the two terms on the left Leaving these terms as they are will allow us to quickly identify the equation as that of a circle and to quickly identify the radius and center of the circle Graphing circles is a fairly simple process once we know the radius and center In order to graph a circle all we really need is the right most left most top most and bottom most points on the circle Once we know these it s easy to sketch in the circle 2007 Paul Dawkins 169 httptutorialmathlamaredutermsaspx College Algebra Nicely enough for us these points are easy to find Since these are points on the circle we know that they must be a distance of r from the center Therefore the points will have the following coordinates right most point h r k left most point h r k top most point h k r bottom most point h k r In other words all we need to do is add r on to the x coordinate or y coordinate of the point to get the right most or top most point respectively and subtract r from the x coordinate or y coordinate to get the left most or bottom most points Let s graph some circles Example 2 Determine the center and radius of each of the following circles and sketch the graph of the circle a x2y21 E b x2y 324 1 cx 12y4216 E Solution In all of these all that we really need to do is compare the equation to the standard form and identify the radius and center Once that is done find the four points talked about above and sketch in the circle a x2 yz 1 In this case it s just x and y squared by themselves The only way that we could have this is to have both h and k be zero So the center and radius is center 00 radius I 1 Don t forget that the radius is the square root of the number on the other side of the equal sign Here is a sketch of this circle 2007 Paul Dawkins 170 httptutorialmathlamaredutermsaspX College Algebra 15 EI3939ll 052 H J 05 1 05 053 153 A circle centered at the origin with radius 1 ie this circle is called the unit circle The unit circle is very useful in a Trigonometry class Return to Problems b x2 y 32 4 In this part it looks like the x coordinate of the center is zero as with the previous part However this time there is something more with the y term and so comparing this term to the standard form of the circle we can see that the y coordinate of the center must be 3 The center and radius of this circle is then center 0 3 radius Z 2 Here is a sketch of the circle The center is marked with a red cross in this graph 5I393395I I 3 I39339lI 21010 Return to Problems c x l2 y42 16 For this part neither of the coordinates of the center are zero By comparing our equation with the standard form it s fairly easy to see hopefully that the x coordinate of the center is l The y coordinate isn t too bad either but we do need to be a little careful In this case the term is y 42 and in the standard form the term is y k2 Note that the signs are different The only way that this can happen is if k is negative So the y coordinate of the center must be 4 2007 Paul Dawkins l7l httptutorialmathlamaredutermsaspX College Algebra The center and radius for this circle are center 1 4 radius E 4 Here is a sketch of this circle with the center marked with a red cross 2 I I11 02 2 4 3 1an Return to Problems So we ve seen how to deal with circles that are already in the standard form However not all circles will start out in the standard form So let s take a look at how to put a circle in the standard form Example 3 Determine the center and radius of each of the following a x2 y2 8x7 0 J b x2 y2 3xl0y l 0 J Solution Neither of these equations are in standard form and so to determine the center and radius we ll need to put it into standard form We actually already know how to do this Back when we were solving quadratic equations we saw a way to turn a quadratic polynomial into a perfect square The process was called completing the square This is exactly what we want to do here although in this case we aren t solving anything and we re going to have to deal with the fact that we ve got both x and y in the equation Let s step through the process with the first part a x2 y2 8x7 0 We ll go through the process in a step by step fashion with this one Step 1 First get the constant on one side by itself and at the same time group the x terms together and the y terms together x2 8x y2 7 In this case there was only one term with a y in it and two with x s in them Step 2 For each variable with two terms complete the square on those terms 2007 Paul Dawkins 172 httptutorialmathlamaredutermsaspX College Algebra So in this case that means that we only need to complete the square on the x terms Recall how this is done We first take half the coefficient of the x and square it We then add this to both sides of the equation x2 8xl6 y2 7 16 9 Now the first three terms will factor as a perfect square 2 x 4 y2 9 Step 3 This is now the standard form of the equation of a circle and so we can pick the center and radius right off this They are center 4 0 radius x5 3 Return to Problems b x2 y2 3xl0y l 0 In this part we ll go through the process a little quicker First get terms properly grouped and placed x2 3x y2 l0y 1 V V complete the square complete the square Now as noted above we ll need to complete the square twice here once for the x terms and once for the y terms Let s first get the numbers that we ll need to add to both sides 3 2 9 10 2 2 El 7 El 5 25 Now add these to both sides of the equation x2 3x2y210y251 2 25 E 4 W 4 4 factor this factor this When adding the numbers to both sides make sure and place them properly This means that we need to put the number from the coefficient of the x with the x terms and the number from the coefficient of the y with the y terms This placement is important since this will be the only way that the quadratics will factor as we need them to factor Now factor the quadratics as show above This will give the standard form of the equation of the circle 113 x 2y52 7 This looks a little messier than the equations that we ve seen to this point However this is something that will happen on occasion so don t get excited about it Here is the center and radius for this circle 2007 Paul Dawkins 173 httptutorialmathlamaredutermsaspX College Algebra 3 113 113 center 5 radius 2 4 2 Do not get excited about the messy radius or fractions in the center coordinates Return to Problems 2007 Paul Dawkins 174 httptutorialmathlamaredutermsaspX College Algebra The Definition of a Function We now need to move into the second topic of this chapter Before we do that however we need a quick definition taken care of Definition of Relation I A relation is a set of ordered pairs This seems like an odd definition but we ll need it for the definition of a function which is the main topic of this section However before we actually give the definition of a function let s see if we can get a handle on just what a relation is Think back to Example 1 in the Graphing section of this chapter In that example we constructed a set of ordered pairs we used to sketch the graph of y x l2 4 Here are the ordered pairs that we used 25 L0 03 14 23 30 45 Any of the following are then relations because they consist of a set of ordered pairs lt 2sgt lt Logt lt2sgt 03 lt2sgt 30 lt45 5 Logt era 14 23 30 lt45 There are of course many more relations that we could form from the list of ordered pairs above but we just wanted to list a few possible relations to give some examples Note as well that we could also get other ordered pairs from the equation and add those into any of the relations above if we wanted to 0 30 4 5 l 2 Now at this point you are probably asking just why we care about relations and that is a good question Some relations are very special and are used at almost all levels of mathematics The following definition tells us just which relations are these special relations Definition of a Function A function is a relation for which each value from the set the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair Okay that is a mouth full Let s see if we can figure out just what it means Let s take a look at the following example that will hopefully help us figure all this out Example 1 The following relation is a function 2007 Paul Dawkins 175 httptutorialmathlamaredutermsaspx College Algebra lt Logt 03 23 30 lt45 Solution From these ordered pairs we have the following sets of first components i e the first number from each ordered pair and second components ie the second number from each ordered pair 1 components 1 0 2 34 2 components O 3 05 For the set of second components notice that the 3 occurred in two ordered pairs but we only listed it once To see why this relation is a function simply pick any Value from the set of first components Now go back up to the relation and find every ordered pair in which this number is the first component and list all the second components from those ordered pairs The list of second components will consist of exactly one Value For example let s choose 2 from the set of first components From the relation we see that there is exactly one ordered pair with 2 as a first component 2 3 Therefore the list of second components i e the list of Values from the set of second components associated with 2 is exactly one number 3 Note that we don t care that 3 is the second component of a second ordered par in the relation That is perfectly acceptable We just don t want there to be any more than one ordered pair with 2 as a first component We looked at a single Value from the set of first components for our quick example here but the result will be the same for all the other choices Regardless of the choice of first components there will be exactly one second component associated with it Therefore this relation is a function In order to really get a feel for what the definition of a function is telling us we should probably also check out an example of a relation that is not a function Example 2 The following relation is not a function 640 73 04 6 4 Solution Don t worry about where this relation came from It is just one that we made up for this example Here is the list of first and second components 1 components 6 70 2nd components l03 4 4 From the set of first components let s choose 6 Now if we go up to the relation we see that there are two ordered pairs with 6 as a first component 6lO and 6 4 The list of second components associated with 6 is then 10 4 2007 Paul Dawkins 176 httptutorialmathlamaredutermsaspx College Algebra The list of second components associated with 6 has two values and so this relation is not a function Note that the fact that if we d chosen 7 or 0 from the set of first components there is only one number in the list of second components associated with each This doesn t matter The fact that we found even a single value in the set of first components with more than one second component associated with it is enough to say that this relation is not a function As a final comment about this example let s note that if we removed the first andor the fourth ordered pair from the relation we would have a function So hopefully you have at least a feeling for what the definition of a function is telling us Now that we ve forced you to go through the actual definition of a function let s give another working definition of a function that will be much more useful to what we are doing here The actual definition works on a relation However as we saw with the four relations we gave prior to the definition of a function and the relation we used in Example 1 we often get the relations from some equation It is important to note that not all relations come from equations The relation from the second example for instance was just a set of ordered pairs we wrote down for the example and didn t come from any equation This can also be true with relations that are functions They do not have to come from equations However having said that the functions that we are going to be using in this course do all come from equations Therefore let s write down a definition of a function that acknowledges this fact Before we give the working definition of a function we need to point out that this is NOT the actual definition of a function that is given above This is simply a good working definition of a function that ties things to the kinds of functions that we will be working with in this course Working Definition of Function A function is an equation for which any x that can be plugged into the equation will yield exactly one y out of the equation There it is That is the definition of functions that we re going to use and will probably be easier to decipher just what it means Before we examine this a little more note that we used the phrase x that can be plugged into in the definition This tends to imply that not all x s can be plugged into an equation and this is in fact correct We will come back and discuss this in more detail towards the end of this section however at this point just remember that we can t divide by zero and if we want real numbers out of the equation we can t take the square root of a negative number So with these two examples it is clear that we will not always be able to plug in every x into any equation Further when dealing with functions we are always going to assume that both x and y will be real numbers In other words we are going to forget that we know anything about complex numbers for a little bit while we deal with this section 2007 Paul Dawkins 177 httptutorialmathlamaredutermsaspx College Algebra Okay with that out of the way let s get back to the definition of a function and let s look at some examples of equations that are functions and equations that aren t functions Example 3 Determine which of the following equations are functions and which are not functions a y 5x 1 E b y x2 1 E c y2 x H E d x2 y2 4 1 Solution The working definition of function is saying is that if we take all possible values of x and plug them into the equation and solve for y we will get exactly one value for each value of x At this stage of the game it can be pretty difficult to actually show that an equation is a function so we ll mostly talk our way through it On the other hand it s often quite easy to show that an equation isn t a function a y 5x 1 So we need to show that no matter what x we plug into the equation and solve for y we will only get a single value of y Note as well that the value of y will probably be different for each value of x although it doesn t have to be Let s start this off by plugging in some values of x and see what happens x 4 y 5 4 4 20 1 19 x0 y5O 4 0 1 1 xl0 y l0 l5amp l 51 So for each of these value of x we got a single value of y out of the equation Now this isn t sufficient to claim that this is a function In order to officially prove that this is a function we need to show that this will work no matter which value of x we plug into the equation Of course we can t plug all possible value of x into the equation That just isn t physically possible However let s go back and look at the ones that we did plug in For each x upon plugging in we first multiplied the x by 5 and then added 1 onto it Now if we multiply a number by 5 we will get a single value from the multiplication Likewise we will only get a single value if we add 1 onto a number Therefore it seems plausible that based on the operations involved with plugging x into the equation that we will only get a single value of y out of the equation So this equation is a function Return to Problems b y x2 1 Again let s plug in a couple of values of x and solve for y to see what happens 2007 Paul Dawkins 178 httptutorialmathlamaredutermsaspX College Algebra x 1 y 1 111 2 x3 y32 15 1 10 Now let s think a little bit about what we were doing with the evaluations First we squared the value of x that we plugged in When we square a number there will only be one possible value We then add 1 onto this but again this will yield a single value So it seems like this equation is also a function Note that it is okay to get the same y value for different x s For example x 3 F 32 191 10 We just can t get more than one y out of the equation after we plug in the x Return to Problems c y2 X B As we ve done with the previous two equations let s plug in a couple of value of x solve for y and see what we get x3 y2a 1 4 gt at 2 x y2 4 1 0 gt 2y 0 xl0 y21o 1 11 gt 31 H Now remember that we re solving for y and so that means that in the first and last case above we will actually get two different y values out of the x and so this equation is NOT a function Note that we can have values of x that will yield a single y as we ve seen above but that doesn t matter If even one value of x yields more than one value of y upon solving the equation will not be a function What this really means is that we didn t need to go any farther than the first evaluation since that gave multiple values of y Return to Problems d x2 y2 4 With this case we ll use the lesson learned in the previous part and see if we can find a value of x that will give more than one value of y upon solving Because we ve got a y2 in the problem this shouldn t be too hard to do since solving will eventually mean using the square root property which will give more than one value of y xO 923311 y2gt4i y 2 So this equation is not a function Recall that from the previous section this is the equation of a circle Circles are never functions Return to Problems Hopefully these examples have given you a better feel for what a function actually is 2007 Paul Dawkins 179 httptutorialmathlamaredutermsaspx College Algebra We now need to move onto something called function notation Function notation will be used heavily throughout most of the remaining chapters in this course and so it is important to understand it Let s start off with the following quadratic equation y x2 5x 3 We can use a process similar to what we used in the previous set of examples to convince ourselves that this is a function Since this is a function we will denote it as follows fxx2 5x3 So we replaced the y with the notation f x This is read as f of x Note that there is nothing special about the f we used here We could just have easily used any of the following gxx2 5x3 hxx2 5x3 Rxx2 5x3 The letter we use does not matter What is important is the x part The letter in the parenthesis must match the variable used on the right side of the equal sign It is very important to note that f x is really nothing more than a really fancy way of writing y If you keep that in mind you may find that dealing with function notation becomes a little easier Also this is NOT a multiplication of f by x This is one of the more common mistakes people make when they first deal with functions This is just a notation used to denote functions Next we need to talk about evaluating functions Evaluating a function is really nothing more than asking what its value is for specific values of x Another way of looking at it is that we are asking what the y value for a given x is Evaluation is really quite simple Let s take the function we were looking at above fx x2 5x3 and ask what its value is for x 4 In terms of function notation we will ask this using the notation f 4 So when there is something other than the variable inside the parenthesis we are really asking what the value of the function is for that particular quantity Now when we say the value of the function we are really asking what the value of the equation is for that particular value of x Here is f 4 f443 54 3 is 320 3 1 Notice that evaluating a function is done in exactly the same way in which we evaluate equations All we do is plug in for x whatever is on the inside of the parenthesis on the left Here s another evaluation for this function f 6 62 5 6 3 36 30 3 69 2007 Paul Dawkins 180 httptutorialmathlamaredutermsaspx College Algebra So again whatever is on the inside of the parenthesis on the left is plugged in for x in the equation on the right Let s take a look at some more examples Example 4 Given f x x2 2x 8 and g x J evaluate each of the following a f 3 and 33 E b f l0 and g l0 i c f 0 E d f t E e ftl and fxl E I f x3 E g gx2 5 E Solution a f3 and g3 Okay we ve got two function evaluations to do here and we ve also got two functions so we re going to need to decide which function to use for the evaluations The key here is to notice the letter that is in front of the parenthesis For f 3 we will use the function f x and for g 3 we will use g x In other words we just need to make sure that the variables match up Here are the evaluations for this part f 3 8 3 32 2389 6811 E353 Return to Problems h f 10 and g 10 This one is pretty much the same as the previous part with one exception that we ll touch on when we reach that point Here are the evaluations 2 f 10 10 2 10 8 100 20 8 128 Make sure that you deal with the negative signs properly here Now the second one g 10 10 6 4 We ve now reached the difference Recall that when we first started talking about the definition of functions we stated that we were only going to deal with real numbers In other words we only plug in real numbers and we only want real numbers back out as answers So since we would get a complex number out of this we can t plug 10 into this function Return to Problems 6 f 0 Not much to this one 2007 Paul Dawkins 181 httptutorialmathlamaredutermsaspx College Algebra fO02 20 8 8 Again don t forget that this isn t multiplication For some reason students like to think of this one as multiplication and get an answer of zero Be careful Return to Problems d f r The rest of these evaluations are now going to be a little different As this one shows we don t need to just have numbers in the parenthesis However evaluation works in exactly the same way We plug into the x s on the right side of the equal sign whatever is in the parenthesis In this case that means that we plug in t for all the x s Here is this evaluation ftt2 2t8 Note that in this case this is pretty much the same thing as our original function except this time we re using t as a variable Return to Problems e ftl and fxl Now let s get a little more complicated or at least they appear to be more complicated Things aren t as bad as they may appear however We ll evaluate f t 1 first This one works exactly the same as the previous part did All the x s on the left will get replaced with t 1 We will have some simplification to do as well after the substitution ft1t12 2t18 t2 at 41 2t 2 8 t2 7 Be careful with parenthesis in these kinds of evaluations It is easy to mess up with them Now let s take a look at f x 1 With the exception of the x this is identical to f t 1 and so it works exactly the same way fx1 x 12 2x 1 8 x2x2 7 x2 x2 Do not get excited about the fact that we reused x s in the evaluation here In many places where we will be doing this in later sections there will be x s here and so you will need to get used to seeing that Return to Problems 0 fx3 2007 Paul Dawkins 182 httptutorialmathlamaredutermsaspx College Algebra Again don t get excited about the x s in the parenthesis here Just evaluate it as if it were a number Return to Problems ggx25 One more evaluation and this time we ll use the other function Return to Problems Function evaluation is something that we ll be doing a lot of in later sections and chapters so make sure that you can do it You will find several later sections very difficult to understand andor do the work in if you do not have a good grasp on how function evaluation works While we are on the subject of function evaluation we should now talk about piecewise functions We ve actually already seen an example of a piecewise function even if we didn t call it a function or a piecewise function at the time Recall the mathematical definition of absolute value x x20 lxl x xlt0 This is a function and if we use function notation we can write it as follows fxxx ifx20 xlt0 This is also an example of a piecewise function A piecewise function is nothing more than a function that is broken into pieces and which piece you use depends upon value of x So in the absolute value example we will use the top piece if x is positive or zero and we will use the bottom piece if x is negative Let s take a look at evaluating a more complicated piecewise function Example 5 Given 24 ts 4 gt 10 if 4 t 15 1 6t iftgt15 evaluate each of the following a g6 E b g4 1 2007 Paul Dawkins 183 httptutorialmathlamaredutermsaspx College Algebra c g1 E d g15 E e g21 1 Solution Before starting the evaluations here let s notice that we re using different letters for the function and variable than the ones that we ve used to this point That won t change how the evaluation works Do not get so locked into seeing f for the function and x for the variable that you can t do any problem that doesn t have those letters Now to do each of these evaluations the first thing that we need to do is determine which inequality the number satisfies and it will only satisfy a single inequality When we determine which inequality the number satisfies we use the equation associated with that inequality So let s do some evaluations a 3 6 In this case 6 satisfies the top inequality and so we ll use the top equation for this evaluation g 6 3 62 4 112 Return to Problems b g 4 Now we ll need to be a little careful with this one since 4 shows up in two of the inequalities However it only satisfies the top inequality and so we will once again use the top function for the evaluation g 4 3 42 4 52 Return to Problems 0 8 1 In this case the number 1 satisfies the middle inequality and so we ll use the middle equation for the evaluation This evaluation often causes problems for students despite the fact that it s actually one of the easiest evaluations we ll ever do We know that we evaluate functionsequations by plugging in the number for the variable In this case there are no variables That isn t a problem Since there aren t any variables it just means that we don t actually plug in anything and we get the following gl 10 Return to Problems d g 15 Again like with the second part we need to be a little careful with this one In this case the number satisfies the middle inequality since that is the one with the equal sign in it Then like the previous part we just get g 15 10 Don t get excited about the fact that the previous two evaluations were the same value This will 2007 Paul Dawkins 184 httptutorialmathlamaredutermsaspX College Algebra happen on occasion Return to Problems e 3 21 For the final evaluation in this example the number satisfies the bottom inequality and so we ll use the bottom equation for the evaluation g2 l 4 e21 125 Return to Problems Piecewise functions do not arise all that often in an Algebra class however the do arise in several places in later classes and so it is important for you to understand them if you are going to be moving on to more math classes As a final topic we need to come back and touch on the fact that we can t always plug every x into every function We talked brie y about this when we gave the definition of the function and we saw an example of this when we were evaluating functions We now need to look at this in a little more detail First we need to get a couple of definitions out of the way Domain and Range The domain of an equation is the set of all x s that we can plug into the equation and get back a real number for y The range of an equation is the set of all y s that we can ever get out of the equation Note that we did mean to use equation in the definitions above instead of functions These are really definitions for equations However since functions are also equations we can use the definitions for functions as well Determining the range of an equationfunction can be pretty difficult to do for many functions and so we aren t going to really get into that We are much more interested here in determining the domains of functions From the definition the domain is the set of all x s that we can plug into a function and get back a real number At this point that means that we need to avoid division by zero and taking square roots of negative numbers Let s do a couple of quick examples of finding domains Example 6 Determine the domain of each of the following functions 3 a gx x23x lO b f x 5 3x E V7x8 C hXw xlOx 5 d Rx 2 16 1 Solution Solution 2007 Paul Dawkins 185 httptutorialmathlamaredutermsaspx College Algebra The domains for these functions are all the values of x for which we don t have division by zero or the square root of a negative number If we remember these two ideas finding the domains will be pretty easy 3 8 x 3 x 3x 10 So in this case there are no square roots so we don t need to worry about the square root of a negative number There is however a possibility that we ll have a division by zero error To determine if we will we ll need to set the denominator equal to zero and solve x23x lOkx 56x 0 x 5x 2 So we will get division by zero if we plug in x 5 or x 2 That means that we ll need to avoid those two numbers However all the other values of x will work since they don t give division by zero The domain is then Domain All real numbers except x 5and x 2 Return to Problems b fx5 3x In this case we won t have division by zero problems since we don t have any fractions We do have a square root in the problem and so we ll need to worry about taking the square root of a negative numbers This one is going to work a little differently from the previous part In that part we determined the values of x to avoid In this case it will be just as easy to directly get the domain To avoid square roots of negative numbers all that we need to do is require that 5 3x20 This is a fairly simple linear inequality that we should be able to solve at this point 5 2 3x gt x 3 The domain of this function is then Domain x 3 Return to Problems V7x8 C h X 2 x 4 In this case we ve got a fraction but notice that the denominator will never be zero for any real number since x2 is guaranteed to be positive or zero and adding 4 onto this will mean that the denominator is always at least 4 In other words the denominator won t ever be zero So all we need to do then is worry about the square root in the numerator To do this we ll require 2007 Paul Dawkins 186 httptutorialmathlamaredutermsaspX College Algebra 7x820 7x2 8 x2 7 Now we can actually plug in any value of x into the denominator however since we ve got the square root in the numerator we ll have to make sure that all x s satisfy the inequality above to avoid problems Therefore the domain of this function is Domain x 2 7 Return to Problems xlOx 5 d Rx x2 16 In this final part we ve got both a square root and division by zero to worry about Let s take care of the square root first since this will probably put the largest restriction on the values of x So to keep the square root happy i e no square root of negative numbers we ll need to require that 10x 5 2 O 10x 2 5 1 x2 2 1 So at the least we ll need to require that x 2 5 in order to avoid problems with the square root Now let s see if we have any division by zero problems Again to do this simply set the denominator equal to zero and solve x2 16ex 4x 4 0 x 4x 4 Now notice that x 4 doesn t satisfy the inequality we need for the square root and so that value of x has already been excluded by the square root On the other hand x 4 does satisfy the inequality This means that it is okay to plug x 4 into the square root however since it would give division by zero we will need to avoid it The domain for this function is then 1 Domain x 2 5 except x 4 Return to Problems 2007 Paul Dawkins 187 httptutorialmathlamaredutermsaspX College Algebra Graphing Functions Now we need to discuss graphing functions If we recall from the previous section we said that f x is nothing more than a fancy way of writing y This means that we already know how to graph functions We graph functions in exactly the same way that we graph equations If we know ahead of time what the function is a graph of we can use that information to help us with the graph and if we don t know what the function is ahead of time then all we need to do is plug in some x s compute the Value of the function which is really a y Value and then plot the points Example 1 Sketch the graph of f x x l3 1 Solution Now as we talked about when we first looked at graphing earlier in this chapter we ll need to pick Values of x to plug in and knowing the Values to pick really only comes with experience Therefore don t worry so much about the Values of x that we re using here By the end of this chapter you ll also be able to correctly pick these Values Here are the function evaluations x y 7 00 L1 22 39 39 22 x fx 1 7 0 0 1 1 2 2 3 9 Here is the sketch of the graph 83 45 E 1 1 I1 I I I I 4 II1 7 Jibt393quot3939Iquotquot2 3 1K So graphing functions is pretty much the same as graphing equations 2007 Paul Dawkins 188 httptutorialmathlamaredutermsaspX College Algebra There is one function that we ve seen to this point that we didn t really see anything like when we were graphing equations in the first part of this chapter That is piecewise functions So we should graph a couple of these to make sure that we can graph them as well Example 2 Sketch the graph of the following piecewise function x2 4 if x lt 1 3 x 2x 1 1f x 2 1 Solution Okay now when we are graphing piecewise functions we are really graphing several functions at once except we are only going to graph them on very specific intervals In this case we will be graphing the following two functions x24 on xlt1 2x l on x21 We ll need to be a little careful with what is going on right at x 1 since technically that will only be valid for the bottom function However we ll deal with that at the very end when we actually do the graph For now we will use x 1 in both functions The first thing to do here is to get a table of values for each function on the specified range and again we will use x 1 in both even though technically it only should be used with the bottom function x x2 4 xy 2 0 20 1 3 1 0 4 04 1 3 13 x 2x 1 2632 1 1 11 2 3 23 3 5 35 Here is a sketch of the graph and notice how we denoted the points at x 1 For the top function we used an open dot for the point at x 1 and for the bottom function we used a closed dot at x 1 In this way we make it clear on the graph that only the bottom function really has a point at x 1 2007 Paul Dawkins 189 httptutorialmathlamaredutermsaspx College Algebra I 39 lllllllll EII 2 1 J 1 2 3 3 rL1 IIII I Notice that since the two graphs didn t meet at x 1 we left a blank space in the graph Do NOT connect these two points with a line There really does need to be a break there to signify that the two portions do not meet at x l Sometimes the two portions will meet at these points and at other times they won t We shouldn t ever expect them to meet or not to meet until we ve actually sketched the graph Let s take a look at another example of a piecewise function Example 3 Sketch the graph of the following piecewise function x 3 if x lt 2 hx x2 if 2 x l x2 ifxzl Solution In this case we will be graphing three functions on the ranges given above So as with the previous example we will get function values for each function in its specified range and we will include the endpoints of each range in each computation When we graph we will acknowledge which function the endpoint actually belongs with by using a closed dot as we did previously Also the top and bottom functions are lines and so we don t really need more than two points for these two We ll get a couple more points for the middle function 75 X3 x y 3 0 30 2 1 21 2007 Paul Dawkins 190 httptutorialmathlamaredutermsaspx College Algebra Here is the sketch of the graph 0 0 00 1 1 11 x x 2 x y 1 1 11 2 0 20 F39 2 1 L 1 2 I 3939 Note that in this case two of the portions met at the breaking point x 1 and at the other breaking point x 2 they didn t meet up As noted in the previous example sometimes they meet up and sometimes they won t 2007 Paul Dawkins 191 httptutorialmathlamaredutermsaspX College Algebra Combining Functions The topic with functions that we need to deal with is combining functions For the most part this means performing basic arithmetic addition subtraction multiplication and division with functions There is one new way of combing functions that we ll need to look at as well Let s start with basic arithmetic of functions Given two functions f x and g x we have the following notation and operations f3xfx 8x f8xfx 8x ltrggtltxgt rltxgtgltxgt 1ltxgt f 8 8 Sometimes we will drop the x part and just write the following f8fx 8x f8fx 8626 L f x rg rxgx g gm Note as well that we put x s in the parenthesis but we will often put in numbers as well Let s take a quick look at an example Example 1 Given f x 2 33 x2 and g x 2x 1 evaluate each of the following a fg4 E b g f E c fgx E d 06 Solution Solution By evaluate we mean one of two things depending on what is in the parenthesis If there is a number in the parenthesis then we want a number If there is an x or no parenthesis since that implies and x then we will perform the operation and simplify as much as possible lt3 f 84 In this case we ve got a number so we need to do some function evaluation f 34 f4 8 4 2t 34 42 24 1 2 7 5 Return to Problems 2007 Paul Dawkins 192 httptutorialmathlamaredutermsaspx College Algebra wgf Here we don t have an x or a number so this implies the same thing as if there were an x in parenthesis Therefore we ll subtract the two functions and simplify Note as well that this is written in the opposite order from the definitions above but it works the same way 3f8x fix 2x 1 62 339X 3amp2 2x 1 2 3x 9amp2 x2 ac 3 Return to Problems c fgx As with the last part this has an x in the parenthesis so we ll multiply and then simplify f8x f x 8 x 23x x 2x 1 4x 6x2 2x3 2 3x 962 zxi 7x 36 2 Return to Problems W law In this final part we ve got a number so we ll once again be doing function evaluation U 23a 2 1 20 Return to Problems Now we need to discuss the new method of combining functions The new method of combining functions is called function composition Here is the definition Given two functions f x and g x we have the following two definitions 1 The composition of f x and g x note the order here is f 8xf8xl 2 The composition of g x and f x again note the order is 80fx8fxl 2007 Paul Dawkins 193 httptutorialmathlamaredutermsaspX College Algebra We need to note a couple of things here about function composition First this is NOT multiplication Regardless of what the notation may suggest to you this is simply not multiplication Second the order we ve listed the two functions is very important since more often than not we will get different answers depending on the order we ve listed them Finally function composition is really nothing more than function evaluation All we re really doing is plugging the second function listed into the first function listed In the definitions we used for the function evaluation instead of the standard to avoid confusion with too many sets of parenthesis but the evaluation will work the same Let s take a look at a couple of examples Example 2 Given f x 2 33 x2 and g x 2x 1 evaluate each of the following a fgx E b f ltgtgx E c g ofx E Solution a These are the same functions that we used in the first set of examples and we ve already done this part there so we won t redo all the work here It is here only here to prove the point that function composition is NOT function multiplication Here is the multiplication of these two functions fgx 2x3r 7x3r 96 2 Return to Problems b Now for function composition all you need to remember is that we are going to plug the second function listed into the first function listed If you can remember that you should always be able to write down the basic formula for the composition Here is this function composition f 8xf8xl f2x 1 Now notice that since we ve got a formula for g x we went ahead and plugged that in first Also we did this kind of function evaluation in the first section we looked at for functions At the time it probably didn t seem all that useful to be looking at that kind of function evaluation yet here it is Let s finish this problem out 2007 Paul Dawkins 194 httptutorialmathlamaredutermsaspx College Algebra fO8xf8xl f2x 1 2 32x 1 2x 12 2 6 3 4x1 494 1 1 6x 4x2 4x 1 4Iac2 10x 2 Notice that this is very different from the multiplication Remember that function composition is NOT function multiplication Return to Problems c We ll not put in the detail in this part as it works essentially the same as the previous part 8 O fx 8 f xl g 2 3x 952 22 ax aez 1 4 69 2x5 1 2x2 6x 3 Notice that this is NOT the same answer as that from the second part In most cases the order in which we do the function composition will give different answers Return to Problems The ideas from the previous example are important enough to make again First function composition is NOT function multiplication Second the order in which we do function composition is important In most case we will get different answers with a different order Note however that there are times when we will get the same answer regardless of the order Let s work a couple more examples Example 3 Given f x x2 3 and hx W evaluate each of the following a f ohx E b hofx E c f ltgtfx E d hoh8 E e f ltgth4 1 Solution 3 f hx Not much to do here other than run through the formula 2007 Paul Dawkins 195 httptutorialmathlamaredutermsaspX College Algebra f hx 1 hxl fJT1 ltmf s x 1 3 x 2 Return to Problems 0 h0fx Again not much to do here x2 31 2 J Return to Problems lt6 f fx Now in this case do not get excited about the fact that the two functions here are the same Composition works the same way f gtfxffxl fx2 3 x2 32 3 x4 6x2 9 3 x4 6x4 6 Return to Problems d h 0 h 8 In this case unlike all the previous examples we Ve got a number in the parenthesis instead of an x but it works in exactly the same manner h h8quotquot8l hE h 5 h3 31 2 Return to Problems 2007 Paul Dawkins 196 httptutorialmathlamaredutermsaspx College Algebra lt6 f 0h4 Again we ve got a number here This time there are actually two ways to do this evaluation The first is to simply use the results from the first part since that is a formula for the general function composition If we do the problem that way we get foh44 22 We could also do the evaluation directly as we did in the previous part The answers should be the same regardless of how we get them So to get another example down of this kind of evaluation let s also do the evaluation directly f oh4 f h4l fJT1 f So sure enough we got the same answer although it did take more work to get it Return to Problems 2 Example 4 Given f x 3x 2 and g x 3 evaluate each of the following X 3 a fltgtgx E b gofx 1 Solution a Hopefully by this point these aren t too bad f 8xf8xl Looks like things simplified down considerable here Return to Problems 2007 Paul Dawkins 197 httptutorialmathlamaredutermsaspX College Algebra b All we need to do here is use the formula so let s do that 8 fX8fxl g3x 2 3x 2 2 2 x 3 3 x 2 3 So in this case we get the same answer regardless of the order we did the composition in Return to Problems So as we Ve seen from this last example it is possible to get the same answer from both compositions on occasion In fact when the answer from both composition is x as it is in this case we know that these two functions are Very special functions In fact they are so special that we re going to devote the whole next section to these kinds of functions So let s move onto the next section 2007 Paul Dawkins 198 httptutorialmathlamaredutermsaspx College Algebra Inverse Functions In the last example from the previous section we looked at the two functions f x 3x 2 and 2 g x 5 and saw that f 8x8 fx x and as noted in that section this means that these are very special functions Let s see just what makes them so special Consider the following evaluations f 13 1 2 5 gt g 5 75 3 3 1 g2 3 f 3 24 22 In the first case we plugged x 1 into f x and got a value of 5 We then turned around and plugged x 5 into g x and got a value of l the number that we started off with In the second case we did something similar Here we plugged x 2 into g x and got a value 4 of E we turned around and plugged this into f x and got a value of 2 which is again the number that we started with Note that we really are doing some function composition here The first case is really 8 f18f 1H 8E5 1 and the second case is really r 0 gm r 8 21 r E 2 Note as well that these both agree with the formula for the compositions that we found in the previous section We get back out of the function evaluation the number that we originally plugged into the composition So just what is going on here In some way we can think of these two functions as undoing what the other did to a number In the first case we plugged x 1 into f x and then plugged the result from this function evaluation back into g x and in some way g x undid what f x had done to x l and gave us back the original x that we started with Function pairs that exhibit this behavior are called inverse functions Before formally defining inverse functions and the notation that we re going to use for them we need to get a definition out of the way 2007 Paul Dawkins 199 httptutorialmathlamaredutermsaspX College Algebra A function is called onetoone if no two values of x produce the same y This is a fairly simple definition of onetoone but it takes an example of a function that isn t onetoone to show just what it means Before doing that however we should note that this definition of onetoone is not really the mathematically correct definition of onetoone It is identical to the mathematically correct definition it just doesn t use all the notation from the formal definition Now let s see an example of a function that isn t onetoone The function f x x2 is not onetoone because both f 2 4 and f 2 4 In other words there are two different values of x that produce the same value of y Note that we can turn f x x2 into a onetoone function if we restrict ourselves to 0 S x lt oo This can sometimes be done with functions Showing that a function is onetoone is often a tedious and often difficult For the most part we are going to assume that the functions that we re going to be dealing with in this section are one toone We did need to talk about onetoone functions however since only onetoone functions can be inverse functions Now let s formally define just what inverse functions are Inverse Functions Given two onetoone functions f x and g x if fOgxx AND gOfxx then we say that f x and g x are inverses of each other More specifically we will say that g x is the inverse of f x and denote it by 8 x f T1 x Likewise we could also say that f x is the inverse of g x and denote it by f X 8x The notation that we use really depends upon the problem In most cases either is acceptable For the two functions that we started off this section with we could write either of the following two sets of notation fx3x 2 f1x gx gquot1x3x 2 Now be careful with the notation for inverses The l is NOT an exponent despite the fact that is sure does look like one When dealing with inverse functions we ve got to remember that 1 l r xv W 2007 Paul Dawkins 200 httptutorialmathlamaredutermsaspx College Algebra This is one of the more common mistakes that students make when first studying inverse functions The process for finding the inverse of a function is a fairly simple one although there is a couple of steps that can on occasion be somewhat messy Here is the process Finding the Inverse of a Function Given the function f x we want to find the inverse function f 1 x First replace f x with y This is done to make the rest of the process easier 2 Replace every x with a y and replace every y with an x 3 Solve the equation from Step 2 for y This is the step where mistakes are most often made so be careful with this step 4 Replace y with f 1 x In other words we ve managed to find the inverse at this point 5 Verify your work by checking that e4 o f1x x and f1 o 4 x are both true This work can sometimes be messy making it easy to make mistakes so again be careful That s the process Most of the steps are not all that bad but as mentioned in the process there are a couple of steps that we really need to be careful with In the verification step we technically really do need to check that both f o f quot1 x x and f quot1 o f x x are true For all the functions that we are going to be looking at in this section if one is true then the other will also be true However there are functions they are far beyond the scope of this course however for which it is possible for only of these to be true This is brought up because in all the problems here we will be just checking one of them We just need to always remember that technically we should check both Let s work some examples Example 1 Given fx 3x 2 find f391 1 Solution Now we already know what the inverse to this function is as we ve already done some work with it However it would be nice to actually start with this since we know what we should get This will work as a nice verification of the process So let s get started We ll first replace f x with y y 3x 2 Next replace all x s with y and all y s with x x3y 2 Now solve for y 2007 Paul Dawkins 201 httptutorialmathlamaredutermsaspX College Algebra x23y x 2 y x 2 y Finally replace y with f 1 x W ltxgt W V Now we need to verify the results We already took care of this in the previous section however we really should follow the process so we ll do that here It doesn t matter which of the two that we check we just need to check one of them This time we ll check that f o f quot1 x x is tI3911C Example2 Given gxx 3 find g391x x20 Solution Now the fact that we re now using g x instead of f x doesn t change how the process y ac x y Now to solve for y we will need to first square both sides and then proceed as normal works Here are the first few steps U3 U3 3 x x2 y 3 x2 3 y This inverse is then gquot1 x x2 3 Finally let s verify and this time we ll use the other one just so we can say that we ve gotten both down somewhere in an example 2007 Paul Dawkins 202 httptutorialmathlamaredutermsaspX College Algebra 91 O 8x 8 8 xl F 3 2 x 3 3 x 3 3 x So we did the work correctly and we do indeed have the inverse Before we move on we should also acknowledge the restrictions of x 2 0 that we gave in the problem statement but never apparently did anything with Note that this restriction is required to make sure that the inverse g1 x given above is in fact onetoone Without this restriction the inverse would not be onetoone as is easily seen by a couple of quick evaluations gl 1 12 3 4 g1 1 12 3 4 Therefore the restriction is required in order to make sure the inverse is onetoone The next example can be a little messy so be careful with the work here 4 Example 3 Given hx x find h 1x 2x 5 Solution The first couple of steps are pretty much the same as the previous examples so here they are x 4 y 2x 5 4 x y 2y 5 Now be careful with the solution step With this kind of problem it is very easy to make a mistake here x2y 5y 4 2xy 5xy 4 2xy y45x 2x ly44 5x y45x 2x l So if We ve done all of our work correctly the inverse should be 2007 Paul Dawkins 203 httptutorialmathlamaredutermsaspx College Algebra 1 45X h x 2x l Finally we ll need to do the verification This is also a fairly messy process and it doesn t really matter which one we work with 1 1 hoh x hh x h45x 2x l 45x 2x l 4e 2 5x 5 2x l Okay this is a mess Let s simplify things up a little bit by multiplying the numerator and denominator by 2x l 4 45x hohltxgt 2quot 1 2x l 245xj5 2x l 2x l51C4j x 2x 12 1 j 5j 45x42x l 245x 52x l 45x8x 4 8l0x l0x5 l3x quotF x 4 Wow That was a lot of work but it all worked out in the end We did all of our work correctly and we do in fact have the inverse There is one final topic that we need to address quickly before we leave this section There is an interesting relationship between the graph of a function and its inverse Here is the graph of the function and inverse from the first two examples We ll not deal with the final example since that is a function that we haven t really talked about graphing yet 2007 Paul Dawkins 204 httptutorialmathlamaredutermsaspX College Algebra 1 B E III 53 3 4 2 El II J 1 E 3 11 IIIIIIIIIIIIIIIIIIIIIII 3 III 1 2 3 4 5 El 15 2 Ezarnplel Etarnplel In both cases we can see that the graph of the inverse is a re ection of the actual function about the line y x This will always be the case with the graphs of a function and its inverse 2007 Paul Dawkins 205 httptutorialmathlamaredutermsaspX College Algebra Common Graphs Introduction We started the process of graphing in the previous chapter However since the main focus of that chapter was functions we didn t graph all that many equations or functions In this chapter we will now look at graphing a wide variety of equations and functions Here is a listing of the topics that we ll be looking at in this chapter Lines Circles and Piecewise Functions This section is here only to acknowledge that we ve already talked about graphing these in a previous chapter Parabolas We ll be graphing parabolas in this section Ellipses In this section we will graph ellipses Hyperbolas Here we will be graphing hyperbolas Miscellaneous Functions In this section we will graph a couple of common functions that don t really take all that much work to do We ll be looking at the constant function square root absolute value and a simple cubic function Transformations We will be looking at shifts and re ections of graphs in this section Collectively these are often called transformations Symmetry We will brie y discuss the topic of symmetry in this section Rational Functions In this section we will graph some rational functions We will also be taking a look at vertical and horizontal asymptotes 2007 Paul Dawkins 206 httptutorialmathlamaredutermsaspX College Algebra Lines Circles and Piecewise Functions We re not really going to do any graphing in this section In fact this section is here only to acknowledge that We Ve already looked at these equations and functions in the previous chapter Here are the appropriate sections to see for these Lines Graphing and Functions Lines Circles Graphing and Functions Circles Piecewise Functions Graphing and Functions Graphing Functions 2007 Paul Dawkins 207 httptutorialmathlamaredutermsaspX College Algebra Parabolas In this section we want to look at the graph of a quadratic function The most general form of a quadratic function is f x axz Jbx E The graphs of quadratic functions are called parabolas Here are some examples of parabolas All parabolas are vaguely U shaped and they will have a highest or lowest point that is called the vertex Parabolas may open up or down and may or may not have xintercepts and they will always have a single yintercept Note as well that a parabola that opens down will always open down and a parabola that opens up will always open up In other words a parabola will not all of a sudden turn around and start opening up if it has already started opening down Similarly if it has already started opening up it will not turn around and start opening down all of a sudden The dashed line with each of these parabolas is called the axis of symmetry Every parabola has an axis of symmetry and as the graph shows the graph to either side of the axis of symmetry is a mirror image of the other side This means that if we know a point on one side of the parabola we will also know a point on the other side based on the axis of symmetry We will see how to find this point once we get into some examples We should probably do a quick review of intercepts before going much farther Intercepts are the points where the graph will cross the x or yaxis We also saw a graph in the section where we introduced intercepts where an intercept just touched the axis without actually crossing it Finding intercepts is a fairly simple process To find the yintercept of a function y f x all we need to do is set x 0 and evaluate to find the y coordinate In other words the yintercept is the point 0 f We find xintercepts in pretty much the same way We set y O and solve the resulting equation for the x coordinates So we will need to solve the equation fx0 Now let s get back to parabolas There is a basic process we can always use to get a pretty good sketch of a parabola Here it is 2007 Paul Dawkins 208 httptutorialmathlamaredutermsaspx College Algebra Sketching Parabolas 1 Find the Vertex We ll discuss how to find this shortly It s fairly simple but there are several methods for finding it and so will be discussed separately 2 Find the y intercept O f p 3 Solve f x O to find the x coordinates of the x intercepts if they exist As we will see in our examples we can have 0 1 or 2 x intercepts 4 Make sure that you ve got at least one point to either side of the vertex This is to make sure we get a somewhat accurate sketch If the parabola has two x intercepts then we ll already have these points If it has 0 or 1 x intercept we can either just plug in another x value or use the y intercept and the axis of symmetry to get the second point 5 Sketch the graph At this point we ve gotten enough points to get a fairly decent idea of what the parabola will look like Now there are two forms of the parabola that we will be looking at This first form will make graphing parabolas very easy Unfortunately most parabolas are not in this form The second form is the more common form and will require slightly and only slightly more work to sketch the graph of the parabola Let s take a look at the first form of the parabola 2 f x a x h k There are two pieces of information about the parabola that we can instantly get from this function First if a is positive then the parabola will open up and if a is negative then the parabola will open down Secondly the vertex of the parabola is the point h k Be very careful with signs when getting the vertex here So when we are lucky enough to have this form of the parabola we are given the vertex for free Let s see a couple of examples here Example 1 Sketch the graph of each of the following parabolas a 2x 32 8 Solution bgx be 2amp2 1 E c hxx2 4 Solution Solution Okay in all of these we will simply go through the process given above to find the needed points and the graph 2007 Paul Dawkins 209 httptutorialmathlamaredutermsaspx College Algebra a fx 2x 32 8 First we need to find the vertex We will need to be careful with the signs however Comparing our equation to the form above we see that we must have h 3 and k 8 since that is the only way to get the correct signs in our function Therefore the vertex of this parabola is 8 Now let s find the yintercept This is nothing more than a quick function evaluation fO2O 32 29 29 8 i0 yintercept 010 Next we need to find the xintercepts This means we ll need to solve an equation However before we do that we can actually tell whether or not we ll have any before we even start to solve the equation In this case we have a 2 which is positive and so we know that the parabola opens up Also the vertex is a point below the xaxis So we know that the parabola will have at least a few points below the xaxis and it will open up Therefore since once a parabola starts to open up it will continue to open up eventually we will have to cross the xaxis In other words there are x intercepts for this parabola To find them we need to solve the following equation 0 2x 32 43 We solve equations like this back when we were solving quadratic equations so hopefully you remember how to do them 2x328 x324 x3 xZ 2 x 3t2 gt x 1x 5 The two xintercepts are then 50 and 10 Now at this point we ve got points on either side of the vertex so we are officially done with finding the points However let s talk a little bit about how to find a second point using the y intercept and the axis of symmetry since we will need to do that eventually First notice that the yintercept has an x coordinate of 0 while the vertex has an x coordinate of 3 This means that the yintercept is a distance of 3 to the right of the axis of symmetry since that will move straight up from the vertex Now the left part of the graph will be a mirror image of the right part of the graph So since there is a point at y 10 that is a distance of 3 to the right of the axis of symmetry there must also be a point at 2 10 that is a distance of 3 to the left of the axis of symmetry So since the x coordinate of the vertex is 3 and this new point is a distance of 3 to the left its x 2007 Paul Dawkins 210 httptutorialmathlamaredutermsaspx College Algebra coordinate must be 6 The coordinates of this new point are then 610 We can verify this by evaluating the function at x 6 If we are correct we should get a value of 10 Let s verify this f 62 6 32 43 a 392 s 29 8 1e So we were correct Note that we usually don t bother with the verification of this point Okay it s time to sketch the graph of the parabola Here it is 39T EII in 39 39 Note that we included the axis of symmetry in this graph and typically we won t It was just included here since we were discussing it earlier Return to Problems ltbgtgx 6 2 1 Okay with this one we won t put in quite a much detail First let s notice that a l which is negative and so we know that this parabola will open downward Next by comparing our function to the general form we see that the vertex of this parabola is 2 1 Again be careful to get the signs correct here Now let s get the yintercept 2 gO 0 2 1 22 1 4 1 5 The yintercept is then 0 5 Now we know that the vertex starts out below the xaxis and the parabola opens down This means that there can t possibly be xintercepts since the x axis is above the vertex and the parabola will always open down This means that there is no reason in general to go through the solving process to find what won t exist However let s do it anyway This will show us what to look for if we don t catch right away that they won t exist from the vertex and direction the parabola opens We ll need to solve 2007 Paul Dawkins 211 httptutorialmathlamaredutermsaspx College Algebra 0 ac 22 1 x 22 1 x 2 1 x2 ii So we got complex solutions Complex solutions will always indicate no xintercepts Now we do want points on either side of the Vertex so we ll use the yintercept and the axis of symmetry to get a second point The yintercept is a distance of two to the left of the axis of symmetry and is at y 5 and so there must be a second point at the same y Value only a distance of 2 to the right of the axis of symmetry The coordinates of this point must then be 4 5 Here is the sketch of this parabola Return to Problems c hx x2 4 This one is actually a fairly simple one to graph We ll first notice that it will open upwards Now the Vertex is probably the point where most students run into trouble here Since we have x2 by itself this means that we must have h 0 and so the Vertex is O 4 Note that this means there will not be any xintercepts with this parabola since the Vertex is above the xaxis and the parabola opens upwards Next the yintercept is hO 02 4 4 y intercept 04 The yintercept is exactly the same as the Vertex This will happen on occasion so we shouldn t get too worried about it when that happens Although this will mean that we aren t going to be able to use the yintercept to find a second point on the other side of the Vertex this time In fact we don t even have a point yet that isn t the Vertex 2007 Paul Dawkins 212 httptutorialmathlamaredutermsaspx College Algebra So we ll need to find a point on either side of the vertex In this case since the function isn t too bad we ll just plug in a couple of points h 2 22 4 8 gt 28 h222 4 8 gt 28 Note that we could have gotten the second point here using the axis of symmetry if we d wanted to Here is a sketch of the graph 2 Fl J 3 2 I 1 2 3 I l IIIIIIIII IIIIIIIIIIIIIIIIIIII Return to Problems Okay we ve seen some examples now of this form of the parabola However as noted earlier most parabolas are not given in that form So we need to take a look at how to graph a parabola that is in the general form fx ax2 Jbx E In this form the sign of a will determine whether or not the parabola will open upwards or downwards just as it did in the previous set of examples Unlike the previous form we will not get the vertex for free this time However it is will easy to find Here is the vertex for a parabola in the general form To get the vertex all we do is compute the x coordinate from a and I9 and then plug this into the function to get the y coordinate Not quite as simple as the previous form but still not all that difficult Note as well that we will get the yintercept for free from this form The yintercept is 2 f0aO 199 c IC gt 06 so we won t need to do any computations for this one Let s graph some parabolas I Example 2 Sketch the graph of each of the following parabolas 2007 Paul Dawkins 213 httptutorialmathlamaredutermsaspx College Algebra a gx3x2 6x 5 1 bfx 263 8x E c fxx24x4 J Solution a For this parabola we ve got a 3 b 6 and c 5 Make sure that you re careful with signs when identifying these values So we know that this parabola will open up since a is positive Here are the evaluations for the vertex 6 6 23 6 ygl3l2 6l5 3 6 5 2 The vertex is then 1 2 Now at this point we also know that there won t be any xintercepts for this parabola since the vertex is above the xaxis and it opens upward The yintercept is 0 5 and using the axis of symmetry we know that 25 must also be on the parabola Here is a sketch of the parabola 1 Return to Problems b In this case a l b 8 and c 0 From these we see that the parabola will open downward since a is negative Here are the vertex evaluations 9 8 1 2 1 2 y f4 4 84 16 So the vertex is 4 16 and we also can see that this time there will be xintercepts In fact let s go ahead and find them now 2007 Paul Dawkins 214 httptutorialmathlamaredutermsaspX College Algebra 0 62 8x Ox as 8 gt xOx So the xintercepts are 00 and 80 Notice that 00 is also the yintercept This will happen on occasion so don t get excited about it when it does At this point we Ve got all the information that we need in order to sketch the graph so here it is 39n 39T I 4 39T I III IIIIIIIIIIIIII I39 JIII 2 4 E Return to Problems c In this final part we have a 1 9 4 and c 4 So this parabola will open up Here are the Vertex evaluations 4 T yf 2 22 4f 2 4 0 So the Vertex is 2 0 Note that since the y coordinate of this point is zero it is also an x 4 x j 2 intercept In fact it will be the only xintercept for this graph This makes sense if we consider the fact that the Vertex in this case is the lowest point on the graph and so the graph simply can t touch the xaxis anywhere else The fact that this parabola has only one xintercept can be Verified by solving as we Ve done in the other examples to this point 0x24x4 0x 2 x 2 Sure enough there is only one xintercept Note that this will mean that we re going to have to use the axis of symmetry to get a second point from the yintercept in this case Speaking of which the yintercept in this case is 0 4 This means that the second point is 4 4 2007 Paul Dawkins 215 httptutorialmathlamaredutermsaspx College Algebra Here is a sketch of the graph 4 E2 I I I I I I I I I I I I I I I 39Iquot I I I I I I I I I I I 5 4 3 2 1 u 1 Return to Problems As a final topic in this section we need to brie y talk about how to take a parabola in the general form and convert it into the form fxax h2 k This will use a modified completing the square process It s probably best to do this with an example Example 3 Convert each of the following into the form f x a x h2 k a fx 2x2 12x 3 E b fx X3 10 1 1 Solution Okay as we pointed out above we are going to complete the square here However it is a slightly different process than the other times that we ve seen it to this point a The thing that we ve got to remember here is that we must have a coefficient of 1 for the x2 term in order to complete the square So to get that we will first factor the coefficient of the x2 term out of the whole right side as follows fx 2x2 6 Note that this will often put fractions into the problem that is just something that we ll need to be able to deal with Also note that if we re lucky enough to have a coefficient of 1 on the x2 term we won t have to do this step Now this is where the process really starts differing from what we ve seen to this point We still take onehalf the coefficient of x and square it However instead of adding this to both sides we do the following with it 2007 Paul Dawkins 216 httptutorialmathlamaredutermsaspX College Algebra er u 9 fx2Ex2 6x9I 9 We add and subtract this quantity inside the parenthesis as shown Note that all we are really doing here is adding in zero since 990 The order listed here is important We MUST add first and then subtract The next step is to factor the first three terms and combine the last two as follows 15 f x 2x 392 3 As a final step we multiply the 2 back through fx 2x 32 15 And there we go Return to Problems b Be careful here We don t have a coefficient of 1 on the x2 term we Ve got a coefficient of 1 So the process is identical outside of that so we won t put in as much detail this time fx 362 19x 1 10j2 5 25 92 10x25 Q5 1 x 52 24 ec 52 24 Return to Problems 2007 Paul Dawkins 217 httptutorialmathlamaredutermsaspX College Algebra Ellipses In a previous section we looked at graphing circles and since circles are really special cases of ellipses we ve already got most of the tools under our belts to graph ellipses All that we really need here to get us started is then standard form of the ellipse and a little information on how to interpret it Here is the standard form of an ellipse ltx hgt2 lty kgt2 2 2 1 a 19 Note that the right side MUST be a l in order to be in standard form The point h k is called the center of the ellipse To graph the ellipse all that we need are the right most left most top most and bottom most points Once we have those we can sketch in the ellipse Here are formulas for finding these points right most point h a k left most point h a k top most point h k 19 bottom most point h k 19 Note that a is the square root of the number under the x term and is the amount that we move right and left from the center Also 9 is the square root of the number under the y term and is the amount that we move up or down from the center Let s sketch some graphs Example 1 Sketch the graph of each of the following ellipses 2 2 x 2 y 4 9 25 L2 1 49 1 Solution a Solution c 4xl2 y32 1 i Solution a So the center of this ellipse is 2 4 and as usual be careful with signs here Also we have a 3 and b 5 So the points are right most point 1 4 left most point 5 4 top most point 29 bottom most point 2 1 2007 Paul Dawkins 218 httptutorialmathlamaredutermsaspX College Algebra Here is a sketch of this ellipse uquot39r391quotquot439quotquotubquot 39I39 4 Return to Problems b The center for this part is 0 3 and we have a 7 and b 2 The points we need are right most point 7 3 left most point 7 3 top most point 05 bottom most point 01 Here is the sketch of this ellipse 542024 35 Return to Problems c Now with this ellipse we re going to have to be a little careful as it isn t quite in standard form yet Here is the standard form for this ellipse xl2 y32 1 1 4 2007 Paul Dawkins 219 httptutorialmathlamaredutermsaspx College Algebra Note that in order to get the coefficient of 4 in the numerator of the first term we will need to 1 have a Z in the denominator Also note that we don t even have a fraction for the y term This implies that there is in fact a 1 in the denominator We could put this in if it would be helpful to see what is going on here 1 2 3 2 y 1 X l 1 4 1 So in this form we can see that the center is 1 3 and that a and b 1 The points for this ellipse are right most point 3 left most point 3 3 top most point 1 2 bottom most point 1 4 Here is this ellipse Ln I 5 I391quotI 39 39quotquot395 EEC I39T39I 4 39I39 quot2 IIIIIIIIIIIIIIIIIIIIIIII l 39 Return to Problems Finally let s address a comment made at the start of this section We said that circles are really nothing more than a special case of an ellipse To see this let s assume that a b In this case we have 2 2 ltx2hgt lty2kgt 2 Cl Cl 2007 Paul Dawkins 220 httptutorialmathlamaredutermsaspX College Algebra Note that We acknowledged that a b and used a in both cases Now if we clear denominators we get x h2y k2 a2 This is the standard form of a circle with center h k and radius a So circles really are special cases of ellipses 2007 Paul Dawkins 221 httptutorialmathlamaredutermsaspX College Algebra Hyperbolas The next graph that we need to look at is the hyperbola There are two basic forms of a hyperbola Here are examples of each Hyperbolas consist of two vaguely parabola shaped pieces that open either up and down or right and left Also just like parabolas each of the pieces has a vertex Note that they aren t really parabolas they just resemble parabolas There are also two lines on each graph These lines are called asymptotes and as the graphs show as we make x large in both the positive and negative sense the graph of the hyperbola gets closer and closer to the asymptotes The asymptotes are not officially part of the graph of the hyperbola However they are usually included so that we can make sure and get the sketch correct The point where the two asymptotes cross is called the center of the hyperbola There are two standard forms of the hyperbola one for each type shown above Here is a table giving each form as well as the information we can get from each one 2007 Paul Dawkins 222 httptutorialmathlamaredutermsaspX College Algebra ltxhgt1ltykgt1 Form a2 b2 1 b2 a2 1 Center h k h k Opens Opens left and right Opens up and down Vertices hak and hak hkb and hk b Slope of Asymptotes i 2 i 2 a a b 9 Equations of Asymptotes y rk x h y rk x h a a Note that the difference between the two forms is which term has the minus sign If the y term has the minus sign then the hyperbola will open left and right If the x term has the minus sign then the hyperbola will open up and down We got the equations of the asymptotes by using the pointslope form of the line and the fact that we know that the asymptotes will go through the center of the hyperbola Let s take a look at a couple of these Example 1 Sketch the graph of each of the following hyperbolas 3 x32yl2 1 25 49 2 b y x 22 1 Solution Solution Solution a Now notice that the y term has the minus sign and so we know that we re in the first column of the table above and that the hyperbola will be opening left and right The first thing that we should get is the center since pretty much everything else is built around that The center in this case is 3 1 and as always watch the signs Once we have the center we can get the Vertices These are 8 1 and 2 1 Next we should get the slopes of the asymptotes These are always the square root of the number under the y term divided by the square root of the number under the x term and there will always 7 be a positive and a negative slope The slopes are then i Now that we ve got the center and the slopes of the asymptotes we can get the equations for the asymptotes They are 2007 Paul Dawkins 223 httptutorialmathlamaredutermsaspX College Algebra y H x 3 and 322 1 x 3 We can now start the sketching We start by sketching the asymptotes and the Vertices Once these are done we know what the basic shape should look like so we sketch it in making sure that as x gets large we move in closer and closer to the asymptotes Here is the sketch for this hyperbola Return to Problems b In this case the hyperbola will open up and down since the x term has the minus sign Now the center of this hyperbola is 2 0 Remember that since there is a y2 term by itself we had to have k 0 At this point we also know that the Vertices are 2 3 and 2 3 In order to see the slopes of the asymptotes let s rewrite the equation a little y2x22 1 9 l 3 So the slopes of the asymptotes are i 3 The equations of the asymptotes are then y0F 3X5l 2 396 6 and y0 3 l39 2 3 6 Here is the sketch of this hyperbola 2007 Paul Dawkins 224 httptutorialmathlamaredutermsaspX College Algebra Return to Problems 2007 Paul Dawkins 225 httptutorialmathlamaredutermsaspX College Algebra Miscellaneous Functions The point of this section is to introduce you to some other functions that don t really require the work to graph that the ones that we ve looked at to this point in this chapter For most of these all that we ll need to do is evaluate the function as some x s and then plot the points Constant Function This is probably the easiest function that we ll ever graph and yet it is one of the functions that tend to cause problems for students The most general form for the constant function is f x c where c is some number Let s take a look at f x 4 so we can see what the graph of constant functions look like Probably the biggest problem students have with these functions is that there are no x s on the right side to plug into for evaluation However all that means is that there is no substitution to do In other words no matter what x we plug into the function we will always get a value of 4 or c in the general case out of the function So every point has a y coordinate of 4 This is exactly what defines a horizontal line In fact if we recall that f x is nothing more than a fancy way of writing y we can rewrite the function as y 4 And this is exactly the equation of a horizontal line Here is the graph of this function Z39I39I 39T I I quot2 d F39 39L39 IIIIIIIIIIIIIIIIIII IIIIIIIIII Fl Ll 4 2 El 2 1 Square Root Next we want to take a look at f x J First note that since we don t want to get complex numbers out of a function evaluation we have to restrict the values of x that we can plug in We 2007 Paul Dawkins 226 httptutorialmathlamaredutermsaspx College Algebra can only plug in value of x in the range x 2 0 This means that our graph will only exist in this range as Well To get the graph we ll just plug in some values of x and then plot the points The graph is then 3 2 1 II I I I I I I I I I I I I I I I I I I I I I I I I I I I I I III 2 1 E E III Absolute Value We ve dealt with this function several times already It s now time to graph it First let s remind ourselves of the definition of the absolute value function f x ifx20 x x ifxlt0 This is a piecewise function and we ve seen how to graph these already All that we need to do is get some points in both ranges and plot them Here are some function evaluations fx Ix lJlJr39 r O tJtJr o Here is the graph of this function 2007 Paul Dawkins 227 httptutorialmathlamaredutermsaspX College Algebra 3 25 2 15 1 5 139 IIIILL 3 2 1 n 1 2 3 So this is a V shaped graph Cubic Function We re not actually going to look at a general cubic polynomial here We ll do some of those in the next chapter Here we are only going to look at f x x3 There really isn t much to do here other than just plugging in some points and plotting x fx 0 0 1 1 1 1 2 8 2 8 Here is the graph of this function 15 III 55 DEIIIIIIIIIIIII ll 1 2 5 102 15 We will need some of these in the next section so make sure that you can identify these when you see them and can sketch their graphs fairly quickly 2007 Paul Dawkins 228 httptutorialmathlamaredutermsaspX College Algebra Transformations In this section we are going to see how knowledge of some fairly simple graphs can help us graph some more complicated graphs Collectively the methods we re going to be looking at in this section are called transformations Vertical Shifts The first transformation we ll look at is a vertical shift Given the graph of f x the graph of g x f x C will be the graph of f x shifted up by c units if c is positive and or down by c units if c is negative So if we can graph f x getting the graph of g x is fairly easy Let s take a look at a couple of examples Example 1 Using transformations sketch the graph of the following functions a g x x2 3 Solution b fx x 5 1 Solution The first thing to do here is graph the function without the constant which by this point should be fairly simple for you Then shift accordingly a gx x2 3 In this case we first need to graph x2 the dotted line on the graph below and then pick this up and shift it upwards by 3 Coordinate wise this will mean adding 3 onto all the y coordinates of points on x2 Here is the sketch for this one IIll l 2 1 El 1 2 Return to Problems bfxx 5 2007 Paul Dawkins 229 httptutorialmathlamaredutermsaspx College Algebra Okay in this case we re going to be shifting the graph of J the dotted line on the graph below down by 5 Again from a coordinate standpoint this means that we subtract 5 from the y coordinates of points on Z Here is this graph IL i x 393939quotI I 1 2 3 4 5 E 4 II 4 IL 39I39I H5 42 IIIIIIIIIIIIIIIIIIIIIIIIII II JIIIIII 39T I Return to Problems So vertical shifts aren t all that bad if we can graph the base function first Note as well that if you re not sure that you believe the graphs in the previous set of examples all you need to do is plug a couple values of x into the function and verify that they are in fact the correct graphs Horizontal Shifts These are fairly simple as well although there is one bit where we need to be careful Given the graph of f x the graph of g x f x C will be the graph of f x shifted left by c units if c is positive and or right by c units if c is negative Now we need to be careful here A positive c shifts a graph in the negative direction and a negative c shifts a graph in the positive direction They are exactly opposite than vertical shifts and it s easy to ip these around and shift incorrectly if we aren t being careful Example 2 Using transformations sketch the graph of the following functions a hx x 23 E b g x xx 4 1 Solution a hx x 23 Okay with these we need to first identify the base function That is the function that s being shifted In this case it looks like we are shifting f x x3 We can then see that hxxIr 23 fa e 2 2007 Paul Dawkins 230 httptutorialmathlamaredutermsaspX College Algebra In this case c 2 and so we re going to shift the graph of f x x3 the dotted line on the graph below and move it 2 units to the left This will mean subtracting 2 from the x coordinates 3 of all the points on f x x Here is the graph for this problem 3 w 5quot U 3939I3Z39I39II uzmf nquot I IIIquotf39 LI 4 2 1 1 2 n39uu39r439r39m Return to Problems b gxxx 4 In this case it looks like the base function is J and it also looks like c 4 and so we will be shifting the graph of J the dotted line on the graph below to the right by 4 units In terms of coordinates this will mean that we re going to add 4 onto the x coordinate of all the points on J Here is the sketch for this function D1II El 1 2 3 1 5 E F E El Return to Problems Vertical and Horizontal Shifts Now we can also combine the two shifts we just got done looking at into a single problem If we know the graph of f x the graph of g x f x C1 kl39Wlll be the graph of f x shifted 2007 Paul Dawkins 231 httptutorialmathlamaredutermsaspX College Algebra left or right by c units depending on the sign of c and up or down by k units depending on the sign of k Let s take a look at a couple of examples Example 3 Use transformation to sketch the graph of each of the following a f X x 22 4 E b g x x 3 5 1 Solution a f x x 22 4 In this part it looks like the base function is x2 and it looks like will be shift this to the right by 2 since c 2 and up by 4 since k 4 Here is the sketch of this function fl I 3939L39I23939quotI139I39I39 IIIIII I IIIII 2 1 El 1 2 3 1 Return to Problems b g x x 3 5 For this part we will be shifting x to the left by 3 since c 3 and down 5 since k 5 Here is the sketch of this function o lxl 39 13quot 4 2 r quot 2 I I I I I I I I I I I I I o I I I I II 3 2 x35 E E3 4 is Return to Problems 2007 Paul Dawkins 232 httptutorialmathlamaredutermsaspX College Algebra Re ections The final set of transformations that we re going to be looking at in this section aren t shifts but instead they are called re ections and there are two of them Re ection about the xaxis Given the graph of f x then the graph of g x f x is the graph of f x reflected about the xaxis This means that the signs on the all the y coordinates are changed to the opposite sign Re ection about the yaxis Given the graph of f x then the graph of g x f x is the graph of f x reflected about the y aXis This means that the signs on the all the x coordinates are changed to the opposite sign Here is an example of each Example 4 Using transformation sketch the graph of each of the following a g x x2 E b hx x Solution Solution a Based on the placement of the minus sign i e it s outside the square and NOT inside the square or x2 it looks like we will be re ecting x2 about the xaxis So again the means that all we do is change the sign on all the y coordinates Here is the sketch of this graph quot39 39 IIIIIIIIIIIIIII I Return to Problems b Now with this one let s first address the minus sign under the square root in more general terms We know that we can t take the square roots of negative numbers however the presence of that minus sign doesn t necessarily cause problems We won t be able to plug positive values of x into the function since that would give square roots of negative numbers However if x were 2007 Paul Dawkins 233 httptutorialmathlamaredutermsaspX College Algebra negative then the negative of a negative number is positive and that is okay For instance h 4 4 JZ 2 So don t get all worried about that minus sign Now let s address the re ection here Since the minus sign is under the square root as opposed to in front of it we are doing a re ection about the yaxis This means that we ll need to change all the signs of points on J Note as well that this syncs up with our discussion on this minus sign at the start of this part Here is the graph for this function Return to Problems 2007 Paul Dawkins 234 httptutorialmathlamaredutermsaspX College Algebra Symmetry In this section we are going to take a look at something that we used back when we where graphing parabolas However we re going to take a more general view of it this section Many graphs have symmetry to them Symmetry can be useful in graphing an equation since it says that if we know one portion of the graph then we will also know the remaining and symmetric portion of the graph as well We used this fact when we were graphing parabolas to get an extra point of some of the graphs In this section we want to look at three types of symmetry 1 A graph is said to be symmetric about the xaxis if whenever a I9 is on the graph then so is a 19 Here is a sketch of a graph that is symmetric about the xaxis ah E3 bl 2 A graph is said to be symmetric about the yaxis if whenever a b is on the graph then so is ab Here is a sketch of a graph that is symmetric about the yaxis 3 A graph is said to be symmetric about the origin if whenever a I9 is on the graph then so is a 1 Here is a sketch of a graph that is symmetric about the origin 2007 Paul Dawkins 235 httptutorialmathlamaredutermsaspX College Algebra acJjI Note that most graphs don t have any kind of symmetry Also it is possible for a graph to have more than one kind of symmetry For example the graph of a circle centered at the origin exhibits all three symmetries Tests for Symmetry We ve some fairly simply tests for each of the different types of symmetry 1 A graph will have symmetry about the x axis if we get an equivalent equation when all the y s are replaced with y 2 A graph will have symmetry about the y axis if we get an equivalent equation when all the x s are replaced with x 3 A graph will have symmetry about the origin if we get an equivalent equation when all the y s are replaced with y and all the x s are replaced with x We will define just what we mean by an equivalent equation when we reach an example of that For the majority of the examples that we re liable to run across this will mean that it is exactly the same equation Let s test a few equations for symmetry Note that we aren t going to graph these since most of them would actually be fairly difficult to graph The point of this example is only to use the tests to determine the symmetry of each equation Example 1 Determine the symmetry of each of the following equations a y 252 6x 2 E b y 2x3 x5 E c y4 x3 5x O E d yx3 493 we 1 E e x2 y2 1 E 2007 Paul Dawkins 236 httptutorialmathlamaredutermsaspx College Algebra Solution a y x5 6x4 2 We ll first check for symmetry about the xaxis This means that we need to replace all the y s with y That s easy enough to do in this case since there is only one y yx2 6x4 9 Now this is not an equivalent equation since the terms on the right are identical to the original equation and the term on the left is the opposite sign So this equation doesn t have symmetry about the xaxis Next let s check symmetry about the yaxis Here we ll replace all x s with x 2 4 v X 6i X 2 y x5 6x4 2 After simplifying we got exactly the same equation back out which means that the two are equivalent Therefore this equation does have symmetry about the yaxis Finally we need to check for symmetry about the origin Here we replace both variables gtv 2amp2 6i 26 2 yx2 6x4 Q So as with the first test the left side is different from the original equation and the right side is identical to the original equation Therefore this isn t equivalent to the original equation and we don t have symmetry about the origin Return to Problems b y 2x3 x5 We ll not put in quite as much detail here First we ll check for symmetry about the xaxis y2x3 x 5 We don t have symmetry here since the one side is identical to the original equation and the other isn t So we don t have symmetry about the xaxis Next check for symmetry about the yaxis 3 5 Lv 2 36 6 36 y 2563 X5 Remember that if we take a negative to an odd power the minus sign can come out in front So upon simplifying we get the left side to be identical to the original equation but the right side is now the opposite sign from the original equation and so this isn t equivalent to the original equation and so we don t have symmetry about the yaxis Finally let s check symmetry about the origin 2007 Paul Dawkins 237 httptutorialmathlamaredutermsaspx College Algebra 3 5 v 2 36 6 36 y 2x 339r x5 Now this time notice that all the signs in this equation are exactly the opposite form the original equation This means that it IS equivalent to the original equation since all We would need to do is multiply the whole thing by 1 to get back to the original equation Therefore in this case we have symmetry about the origin Return to Problems c y4 x3 5x 0 First check for symmetry about the xaxis 4 y x3 5x O y4 x3 5x O This is identical to the original equation and so we have symmetry about the xaxis Now check for symmetry about the yaxis 3 y4 x 5 x O y4 x3 5x 0 So some terms have the same sign as the original equation and other don t so there isn t symmetry about the yaxis Finally check for symmetry about the origin 4 3 av lt xgt slt xgto y4 x3 5x 0 Again this is not the same as the original equation and isn t exactly the opposite sign from the original equation and so isn t symmetric about the origin Return to Problems d y x3 Jacz lac 4amp1 First symmetry about the xaxis y x3 Jacz Jet 461 It looks like no symmetry about the xaxis Next symmetry about the yaxis 3 2 y x x x 1 y xi x2 3 1 So no symmetry here either Finally symmetry about the origin 2007 Paul Dawkins 238 httptutorialmathlamaredutermsaspx College Algebra y x3 x2 x l y x 339r x2 35 1 And again no symmetry here either This function has no symmetry of any kind That s not unusual as most functions don t have any of these symmetries Return to Problems e x2 yz 1 Check xaxis symmetry first x2 y2 1 x2 yz 1 So it s got symmetry about the xaxis symmetry Next check for yaxis symmetry x2 y2 1 x2 yz 1 Looks like it s also got yaxis symmetry Finally symmetry about the origin So it s also got symmetry about the origin Note that this is a circle centered at the origin and as noted when we first started talking about symmetry it does have all three symmetries Return to Problems 2007 Paul Dawkins 239 httptutorialmathlamaredutermsaspx College Algebra Rational Functions In this final section we need to discuss graphing rational functions It s is probably best to start off with a fairly simple one that we can do without all that much knowledge on how these work 1 Let s sketch the graph of f x First since this is a rational function we are going to have x to be careful with division by zero issues So we can see from this equation that we ll have to avoid x 0 since that will give division by zero Now let s just plug in some values of x and see what we get 96 fx 4 025 2 05 1 1 01 10 001 100 001 100 01 10 1 1 2 05 4 025 So as x get large positively and negatively the function keeps the sign of x and gets smaller and smaller Likewise as we approach x O the function again keeps the same sign as x but starts getting quite large Here is a sketch of this graph 2 4 First notice that the graph is in two pieces Almost all rational functions will have graphs in multiple pieces like this Next notice that this graph does not have any intercepts of any kind That s easy enough to check for ourselves 2007 Paul Dawkins 240 httptutorialmathlamaredutermsaspX College Algebra Recall that a graph will have a yintercept at the point 0 f r However in this case we have to avoid x 0 and so this graph will never cross the yaxis It does get very close to the yaxis but it will never cross or touch it and so no yintercept Next recall that we can determine where a graph will have xintercepts by solving f x 0 For rational functions this may seem like a mess to deal with However there is a nice fact about rational functions that we can use here A rational function will be zero at a particular value of x only if the numerator is zero at that x and the denominator isn t zero at that x In other words to determine if a rational function is ever zero all that we need to do is set the numerator equal to zero and solve Once we have these solutions we just need to check that none of them make the denominator zero as well In our case the numerator is one and will never be zero and so this function will have no x intercepts Again the graph will get very close to the xaxis but it will never touch or cross it Finally we need to address the fact that graph gets very close to the x and yaxis but never crosses Since there isn t anything special about the axis themselves we ll use the fact that the x axis is really the line given by y O and the yaxis is really the line given by x O In our graph as the value of x approaches x 0 the graph starts gets very large on both sides of the line given by x 0 This line is called a vertical asymptote Also as x get very large both positive and negative the graph approaches the line given by y O This line is called a horizontal asymptote Here are the general definitions of the two asymptotes l The line x a is a vertical asymptote if the graph increases or decreases without bound on one or both sides of the line as x moves in closer and closer to x a 2 The line y b is a horizontal asymptote if the graph approaches y b as x increases or decreases without bound Note that it doesn t have to approach y b as x BOTH increases and decreases It only needs to approach it on one side in order for it to be a horizontal asymptote Determining asymptotes is actually a fairly simple process First let s start with the rational function axquot r x bx where n is the largest exponent in the numerator and m is the largest exponent in the denominator We then have the following facts about asymptotes 2007 Paul Dawkins 241 httptutorialmathlamaredutermsaspx College Algebra 1 The graph will have a vertical asymptote at x a if the denominator is zero at x a and the numerator isn t zero at x a 2 If n lt m then the x aXis is the horizontal asymptote 3 If n m then the line y is the horizontal asymptote 4 If n gt m there will be no horizontal asymptotes The process for graphing a rational function is fairly simple Here it is Process for Graphing a Rational Function 1 Find the intercepts if there are any Remember that the yintercept is given by O f xT and we find the x intercepts by setting the numerator equal to zero and solving 2 Find the vertical asymptotes by setting the denominator equal to zero and solving 3 Find the horizontal asymptote if it exits using the fact above 4 The vertical asymptotes will divide the number line into regions In each region graph at least one point in each region This point will tell us whether the graph will be above or below the horizontal asymptote and if we need to we should get several points to determine the general shape of the graph 5 Sketch the graph Note that the sketch that we ll get from the process is going to be a fairly rough sketch but that is okay That s all that we re really after is a basic idea of what the graph will look at Let s take a look at a couple of examples Example 1 Sketch the graph of the following function 3x6 X f x 1 Solution So we ll start off with the intercepts The yintercept is 6 f0 1 6 0 6 The xintercepts will be 3x 6 0 x 2 gt 2 0 Now we need to determine the asymptotes Let s first find the Vertical asymptotes x 1 O gt x 1 So we Ve got one Vertical asymptote This means that there are now two regions of x s They 2007 Paul Dawkins 242 httptutorialmathlamaredutermsaspX College Algebra are xltl and xgtl Now the largest exponent in the numerator and denominator is l and so by the fact there will be a horizontal asymptote at the line 3 yr Now we just need points in each region of x s Since the yintercept and xintercept are already in the left region we won t need to get any points there That means that we ll just need to get a point in the right region It doesn t really matter what Value of x we pick here we just need to keep it fairly small so it will fit onto our graph f2 1T212 3 212 Okay putting all this together gives the following graph H l 4342 245 Note that the asymptotes are shown as dotted lines Example 2 Sketch the graph of the following function 9 x f x2 9 Solution Okay we ll start with the intercepts The yintercept is 9 fO 9 1 gt 0 1 The numerator is a constant and so there won t be any xintercepts since the function can never be zero Next we ll have Vertical asymptotes at x2 9 0 gt x 3 So in this case we ll have three regions to our graph x lt 3 3 lt x lt 3 x gt 3 Also the largest exponent in the denominator is 2 and since there are no x s in the numerator the 2007 Paul Dawkins 243 httptutorialmathlamaredutermsaspx College Algebra largest exponent is 0 so by the fact the xaxis will be the horizontal asymptote Finally we need some points We ll use the following points here es Notice that along with the yintercept we actually have three points in the middle region This is because there are a couple of possible behaviors in this region and we ll need to determine the actual behavior We ll see the other main behaviors in the next examples and so this will make more sense at that point Here is the sketch of the graph I H I J 5 4 4 E Example 3 Sketch the graph of the following function x2 4 Solution This time notice that if we were to plug in x 0 into the denominator we would get division by zero This means there will not be a yintercept for this graph We have however managed to find a vertical asymptote already Now let s see if we ve got xintercepts x2 4 O gt x 2 So we ve got two of them 2007 Paul Dawkins 244 httptutorialmathlamaredutermsaspx College Algebra We ve got one vertical asymptote but there may be more so let s go through the process and see x2 4xxx 4 6 gt x Ox 4 So we ve got two again and the three regions that we ve got are x lt O O lt x lt 4 and x gt 4 Next the largest exponent in both the numerator and denominator is 2 so by the fact there will be a horizontal asymptote at the line l yl Now one of the xintercepts is in the far left region so we don t need any points there The other xintercept is in the middle region So we ll need a point in the far right region and as noted in the previous example we will want to get a couple more points in the middle region to completely determine its behavior 10 Notice that this time the middle region doesn t have the same behavior at the asymptotes as we saw in the previous example This can and will happen fairly often Sometimes the behavior at the two asymptotes will be the same as in the previous example and sometimes it will have the opposite behavior at each asymptote as we see in this example Because of this we will always need to get a couple of points in these types of regions to determine just what the behavior will be 2007 Paul Dawkins 245 httptutorialmathlamaredutermsaspx College Algebra Polynomial Functions Introduction In this chapter we are going to take a more in depth look at polynomials We ve already solved and graphed second degree polynomials i e quadratic equationsfunctions and we now want to extend things out to more general polynomials We will take a look at finding solutions to higher degree polynomials and how to get a rough sketch for a higher degree polynomial We will also be looking at Partial Fractions in this chapter It doesn t really have anything to do with graphing polynomials but needed to be put somewhere and this chapter seemed like as good a place as any Here is a brief listing of the material in this chapter Dividing Polynomials We ll review some of the basics of dividing polynomials in this section ZeroesRoots of Polynomials In this section we ll define just what zeroesroots of polynomials are and give some of the more important facts concerning them Graphing Polynomials Here we will give a process that will allow us to get a rough sketch of some polynomials Finding Zeroes of Polynomials We ll look at a process that will allow us to find some of the zeroes of a polynomial and in special cases all of the zeroes Partial Fractions In this section we will take a look at the process of partial fractions and finding the partial fraction decomposition of a rational expression 2007 Paul Dawkins 246 httptutorialmathlamaredutermsaspX College Algebra Dividing Polynomials In this section we re going to take a brief look at dividing polynomials This is something that we ll be doing off and on throughout the rest of this chapter and so we ll need to be able to do this Let s do a quick example to remind us how long division of polynomials works Example 1 Divide 5x3 x2 6 by x 4 Solution Let s first get the problem set up x 45x3 x2 0x6 Recall that we need to have the terms written down with the exponents in decreasing order and to make sure we don t make any mistakes we add in any missing terms with a zero coefficient Now we ask ourselves what we need to multiply x 4 to get the first term in first polynomial In this case that is 5x2 So multiply x 4 by 5x2 and subtract the results from the first polynomial 5x2 x 45x3 x20x6 5x3 20x2 19x2 Ox 6 The new polynomial is called the remainder We continue the process until the degree of the remainder is less than the degree of the divisor which is x 4 in this case So we need to continue until the degree of the remainder is less than 1 Recall that the degree of a polynomial is the highest exponent in the polynomial Also recall that a constant is thought of as a polynomial of degree zero Therefore we ll need to continue until we get a constant in this case Here is the rest of the work for this example 2007 Paul Dawkins 247 httptutorialmathlamaredutermsaspx College Algebra 5x2 l9x76 x 45x3 x2 0x6 5x3 20x2 19x2 0x6 l9x2 76x 76x6 76x 304 310 Okay now that we ve gotten this done let s remember how we write the actual answer down The answer is 3 2 1252 19 g6 x 4 x There is actually another way to write the answer from the previous example that we re going to find much more useful if for no other reason that it s easier to write down If we multiply both sides of the answer by x 4 we get 5x3 x26x 45x2 419x 76 310 In this example we divided the polynomial by a linear polynomial in the form of x r and we will be restricting ourselves to only these kinds of problems Long division works for much more general division but these are the kinds of problems we are going to seeing the later sections In fact we will be seeing these kinds of divisions so often that we d like a quicker and more efficient way of doing them Luckily there is something out there called synthetic division that works wonderfully for these kinds of problems In order to use synthetic division we must be dividing a polynomial by a linear term in the form x r If we aren t then it won t work Let s redo the previous problem with synthetic division to see how it works Example 2 Use synthetic division to divide 5x3 x2 6 by x 4 Solution Okay with synthetic division we pretty much ignore all the x s and just work with the numbers in the polynomials First let s notice that in this case r4 Now we need to set up the process There are many different notations for doing this We ll be using the following notation 5 106 2007 Paul Dawkins 248 httptutorialmathlamaredutermsaspX College Algebra The numbers to the right of the vertical bar are the coefficients of the terms in the polynomial written in order of decreasing exponent Also notice that any missing terms are acknowledged with a coefficient of zero Now it will probably be easier to write down the process and then explain it so here it is g5 1r1 5 V20 5 304i 5 quotRial 5731 1 The first thing we do is drop the first number in the top line straight down as shown Then along each diagonal we multiply the starting number by r which is 4 in this case and put this number in the second row Finally add the numbers in the first and second row putting the results in the third row We continue this until we get reach the final number in the first row Now notice that the numbers in the bottom row are the coefficients of the quadratic polynomial from our answer written in order of decreasing exponent and the final number in the third row is the remainder The answer is then the same as the first example 5x3 x26x 45x2 419x 76 310 We ll do some more examples of synthetic division in a bit However we really should generalize things out a little first with the following fact Division Algorithm Given a polynomial P x with degree at least 1 and any number r there is another polynomial Qx called the quotient with degree one less than the degree of Px and a number R called the remainder such that Px x r Qx R Note as well that Qx and R are unique or in other words there is only one Qx and R that will work for a given P x and r So with the one example we ve done to this point we can see that Qx5x2 l9x76 and R31O Now let s work a couple more synthetic division problems 2007 Paul Dawkins 249 httptutorialmathlamaredutermsaspx College Algebra Example 3 Use synthetic division to do each of the following divisions a Divide 2x3 3x 5 by x 2 E b Divide 4x4 10x2 1 by x 6 E Solution a Divide 2x3 3x 5 by x 2 Okay in this case we need to be a little careful here We MUST divide by a term in the form x r in order for this to work and that minus sign is absolutely required So we re first going to need to write x 2 as x2x 2 and in doing so we can see that r 2 We can now do synthetic division and this time we ll just put up the results and leave it to you to check all the actual numbers in 0 3 5 4 8 10 2 4 5 15 So in this case we have 2x3 3x 5 x 422x2 4x 45 15 Return to Problems b Divide 4x4 10x2 1 by x 6 In this case we ve got r6 Here is the work Q 4 0 l0 0 l 0 24 144 804 4824 4 24 134 804 4825 In this case we then have 4x4 l0x2lx 64x3 424 434x 804 4825 Return to Problems So just why are we doing this That s a natural question at this point One answer is that down the road in a later section we are going to want to get our hands on the Qx Just why we might want to do that will have to wait for an explanation until we get to that point There is also another reason for this that we are going to make heavy usage of later on Let s first start out with the division algorithm Px x rQx R Now let s evaluate the polynomial Px at r If we had an actual polynomial here we could evaluate Px directly of course but let s use the division algorithm and see what we get 2007 Paul Dawkins 250 httptutorialmathlamaredutermsaspX College Algebra I I N PrH new R ltogtQltrgt R R Now that s convenient The remainder of the division algorithm is also the value of the polynomial evaluated at r So from our previous examples we now know the following function evaluations If Px 5x3 acz 46 then P4 3lO If Px2 3ac 5 then P 2 15 If Px 4x4 l0x2 1 then P6 4825 This is a very quick method for evaluating polynomials For polynomials with only a few terms andor polynomials with small degree this may not be much quicker that evaluating them directly However if there are many terms in the polynomial and they have large degrees this can be much quicker and much less prone to mistakes than computing them directly As noted we will be using this fact in a later section to greatly reduce the amount of work we ll need to do in those problems 2007 Paul Dawkins 251 httptutorialmathlamaredutermsaspX College Algebra ZeroesRoots of Polynomials We ll start off this section by defining just what a root or zero of a polynomial is We say that x r is a root or zero of a polynomial Px if Pr O In other words x r is a root or zero of a polynomial if it is a solution to the equation P x O In the next couple of sections we will need to find all the zeroes for a given polynomial So before we get into that we need to get some ideas out of the way regarding zeroes of polynomials that will help us in that process The process of finding the zeros of P x really amount to nothing more than solving the equation P x O and we already know how to do that for second degree quadratic polynomials So to help illustrate some of the ideas were going to be looking at let s get the zeroes of a couple of second degree polynomials Let s first find the zeroes for P x x2 2x 15 To do this we simply solve the following equation x22x l5Irx 5c amp O gt x 5x 3 So this second degree polynomial has two zeroes or roots Now let s find the zeroes for Px x2 14 49 That will mean solving x2 l4x49x 72O gt x 7 So this second degree polynomial has a single zero or root Also recall that when we first looked at these we called a root like this a double root We solved each of these by first factoring the polynomial and then using the zero factor property on the factored form When we first looked at the zero factor property we saw that it said that if the product of two terms was zero then one of the terms had to be zero to start off with The zero factor property can be extended out to as many terms as we need In other words if we ve got a product of n terms that is equal to zero then at least one of them had to be zero to start off with So if we could factor higher degree polynomials we could then solve these as well Let s take a look at a couple of these Example 1 Find the zeroes of each of the following polynomials a Px 5x5 2Ox4 5x3 5Ox2 20x 40 5x 192 x 23 b Qx x8 4x7 l8x6 lO8x5 l35x4 x4 9 33 we 5 c Rx x7 lOx6 27x5 57x3 3Ox2 29x 20 x l3x l2x 5x 4 Solution In each of these the factoring has been done for us Do not worry about factoring anything like this You won t be asked to do any factoring of this kind anywhere in this material There are 2007 Paul Dawkins 252 httptutorialmathlamaredutermsaspX College Algebra only here to make the point that the zero factor property works here as well We will also use these in a later example a Px5x5 2Ox4 5x3 5Ox2 20x 40 5x 192 x 23 Okay in this case we do have a product of 3 terms however the first is a constant and will not make the polynomial zero So from the final two terms it looks like the polynomial will be zero for x l and x 2 Therefore the zeroes of this polynomial are x landx 2 b Qxx8 4x7 l8x6 lO8x5 l35x4 x4Ex 33kx 5 We ve also got a product of three terms in this polynomial However since the first is now an x this will introduce a third zero The zeroes for this polynomial are ac 5x 0andx 3 because each of these will make one of the terms and hence the whole polynomial zero c Rx x7 lOx6 27x5 57x3 3Ox2 29x 20 x l3x l2x 5x 4 With this polynomial we have four terms and the zeroes here are ac 5x l x l and x 4 Now we ve got some terminology to get out of the way If r is a zero of a polynomial and the exponent on the term that produced the root is k then we say that r has multiplicity k Zeroes with a multiplicity of l are often called simple zeroes For example the polynomial P x x2 l0x 25 x 5 2 will have one zero x 5 and its multiplicity is 2 In some way we can think of this zero as occurring twice in the list of all zeroes since we could write the polynomial as Px x2 l0x 25 x 5x 5 Written this way the term x 5 shows up twice and each term gives the same zero x 5 Saying that the multiplicity of a zero is k is just a shorthand to acknowledge that the zero will occur k times in the list of all zeroes Example 2 List the multiplicities of the zeroes of each of the following polynomials a Px x2 2x 15 x5x5 2Ox4 5x3 5Ox2 20x 40 5x 192 x 23 d Qxx8 4x7 l8x6 lO8x5 l35x4 x4 33x 5 2007 Paul Dawkins 253 httptutorialmathlamaredutermsaspx College Algebra Solution We ve already determined the zeroes of each of these in previous work or examples in this section so we won t redo that work In each case we will simply write down the previously found zeroes and then go back to the factored form of the polynomial look at the exponent on each term and give the multiplicity a In this case we ve got two simple zeroes 96 5 x 3 b Here x 7 is a zero of multiplicity 2 c There are two zeroes for this polynomial x l with multiplicity 2 and x 2 with multiplicity 3 d We have three zeroes in this case x 5 which is simple x 0 with multiplicity of 4 and x 3 with multiplicity 3 e In the final case we ve got four zeroes x 5 which is simple x l with multiplicity of 3 x 1 with multiplicity 2 and x 4 which is simple This example leads us to several nice facts about polynomials Here is the first and probably the most important Fundamental Theorem of Algebra If P x is a polynomial of degree n then P x will have exactly n zeroes some of which may repeat This fact says that if you list out all the zeroes and listing each one k times where k is its multiplicity you will have exactly n numbers in the list Another way to say this fact is that the multiplicity of all the zeroes must add to the degree of the polynomial We can go back to the previous example and verify that this fact is true for the polynomials listed there This will be a nice fact in a couple of sections when we go into detail about finding all the zeroes of a polynomial If we know an upper bound for the number of zeroes for a polynomial then we will know when we ve found all of them and so we can stop looking The next fact is also very useful at times The Factor Theorem For the polynomial P x 1 If r is a zero of Px then x r will be a factor of Px 2 If x r is a factor of Px then r will be a zero of Px Again if we go back to the previous example we can see that this is verified with the polynomials listed there 2007 Paul Dawkins 254 httptutorialmathlamaredutermsaspx College Algebra The factor theorem leads to the following fact Fact 1 If P x is a polynomial of degree n and r is a zero of P x then P x can be written in the Pltxgtltx rgtQltxgt where Q x is a polynomial with degree n 1 Q x can be found by dividing P x by following form C Iquot There is one more fact that we need to get out of the way Fact 2 If Px x rQx and xt is azero of Qx then xt will also be azero of Px This fact is easy enough to Verify directly First if x t is a zero of Q x then we know that Q t 0 since that is what it means to be a zero So if x t is to be a zero of P x then all we need to do is show that P t 0 and that s actually quite simple Here it is Prr west ltrrgtltogt o and so xt isazero of Px Let s work an example to see how these last few facts can be of use to us Example 3 Given that x 2 is a zero of P x x3 2x2 5x 6 find the other two zeroes Solution First notice that we really can say the other two since we know that this is a third degree polynomial and so by The Fundamental Theorem of Algebra we will have exactly 3 zeroes with some repeats possible So since we know that x 2 is a zero of Px x3 2x2 5x 6 the Fact 1 tells us that we can write Px as Pltxgt x 2gtQltxgt and Q x will be a quadratic polynomial Then we can find the zeroes of Q x by any of the methods that we Ve looked at to this point and by Fact 2 we know that the two zeroes we get from Q x will also by zeroes of P x At this point we ll have 3 zeroes and so we will be done So let s find Q x To do this all we need to do is a quick synthetic division as follows 2007 Paul Dawkins 255 httptutorialmathlamaredutermsaspX College Algebra Q1 25 6 2 8 6 14 3 0 Before writing down Q x recall that the final number in the third row is the remainder and that we know that P 2 must be equal to this number So in this case we have that P2 0 If you think about it we should already know this to be true We were given in the problem statement the fact that x 2 is a zero of P x and that means that we must have P 2 0 So why go on about this This is a great check of our synthetic division Since we know that x 2 is a zero of P x and we get any other number than zero in that last entry we will know that we ve done something wrong and we can go back and find the mistake Now let s get back to the problem From the synthetic division we have Px x 2x2 4x 3 So this means that Qx x2 4x 3 and we can find the zeroes of this Here they are Qxx3r 49 3 x 3fx 1 gt x x 1 So the three zeroes of Px are x 3 x 1 and x2 As an aside to the previous example notice that we can also now completely factor the polynomial P x x3 2x2 5x 6 Substituting the factored form of Q x into P x we Pxac 2Jac 3Jac 1 This is how the polynomials in the first set of examples were factored by the way Those require a little more work than this but they can be done in the same manner g t 2007 Paul Dawkins 256 httptutorialmathlamaredutermsaspX College Algebra Graphing Polynomials In this section we are going to look at a method for getting a rough sketch of a general polynomial The only real information that we re going to need is a complete list of all the zeroes including multiplicity for the polynomial In this section we are going to either be given the list of zeroes or they will be easy to find In the next section we will go into a method for determining a large portion of the list for most polynomials We are graphing first since the method for finding all the zeroes of a polynomial can be a little long and we don t want to obscure the details of this section in the mess of finding the zeroes of the polynomial Let s start off with the graph of couple of polynomials mm mu 3 an 3 203 3 ED 3 at W 2 39 I III IIIIIIII Do not worry about the equations for these polynomials We are giving these only so we can use them to illustrate some ideas about polynomials 4392 3 I First notice that the graphs are nice and smooth There are no holes or breaks in the graph and there are no sharp comers in the graph The graphs of polynomials will always be nice smooth curves Secondly the humps where the graph changes direction from increasing to decreasing or decreasing to increasing are often called turning points If we know that the polynomial has degree n then we will know that there will be at most 11 l turning points in the graph While this won t help much with the actual graphing process it will be a nice check If we have a fourth degree polynomial with 5 turning point then we will know that we ve done something wrong since a fourth degree polynomial will have no more than 3 turning points Next we need to explore the relationship between the xintercepts of a graph of a polynomial and the zeroes of the polynomial Recall that to find the xintercepts of a function we need to solve the equation P x 0 Also recall that x r is a zero of the polynomial P x provided Pr 0 But this means that x r is also a solution to Px 0 2007 Paul Dawkins 257 httptutorialmathlamaredutermsaspx College Algebra In other words the zeroes of a polynomial are also the xintercepts of the graph Also recall that xintercepts can either cross the xaxis or they can just touch the xaxis without actually crossing the axis Notice as well from the graphs above that the xintercepts can either atten out as they cross the xaxis or they can go through the xaxis at an angle The following fact will relate all of these ideas to the multiplicity of the zero Fact If x r is a zero of the polynomial P x with multiplicity k then 1 If k is odd then the x intercept corresponding to x r will cross the x axis 2 If k is even then the x intercept corresponding to x r will only touch the x axis and not actually cross it Furthermore if k gt 1 then the graph will atten out at x r Finally notice that as we let x get large in both the positive or negative sense ie at either end of the graph then the graph will either increase without bound or decrease without bound This will always happen with every polynomial and we can use the following test to determine just what will happen at the endpoints of the graph Leading Coefficient Test Suppose that P x is a polynomial with degree n So we know that the polynomial must look like P x axquot We don t know if there are any other terms in the polynomial but we do know that the first term will have to be the one listed since it has degree n We now have the following facts about the graph of Px at the ends of the graph 1 If a gt O and n is even then the graph of Px will increase without bound positively at both endpoints A good example of this is the graph of x2 2 If a gt 0 and n is odd then the graph of Px will increase without bound positively at the right end and decrease without bound at the left end A good example of this is the graph of xi 2007 Paul Dawkins 25 8 httptutorialmathlamaredutermsaspx College Algebra 3 If a lt O and n is even then the graph of Px will decrease without bound positively at both endpoints A good example of this is the graph of x2 4 If a lt 0 and n is odd then the graph of Px will decrease without bound positively at the right end and increase without bound at the left end A good example of this is the graph of xi Okay now that we ve got all that out of the way we can finally give a process for getting a rough sketch of the graph of a polynomial Process for Graphing a Polynomial 1 Determine all the zeroes of the polynomial and their multiplicity Use the fact above to determine the x intercept that corresponds to each zero will cross the x aXis or just touch it and if the x intercept will atten out or not 2 Determine the y intercept O P 3 Use the leading coefficient test to determine the behavior of the polynomial at the end of the graph 4 Plot a few more points This is left intentionally vague The more points that you plot the better the sketch At the least you should plot at least one at either end of the graph and at least one point between each pair of zeroes 2007 Paul Dawkins 259 httptutorialmathlamaredutermsaspx College Algebra We should give a quick warning about this process before we actually try to use it This process assumes that all the zeroes are real numbers If there are any complex zeroes then this process may miss some pretty important features of the graph Let s sketch a couple of polynomials Example 1 Sketch the graph of Px 5x5 20x4 5x3 5Ox2 20x 40 Solution We found the zeroes and multiplicities of this polynomial in the previous section so we ll just write them back down here for reference purposes x l multiplicity 2 x 2 multiplicity 3 So from the fact we know that x 1 will just touch the xaxis and not actually cross it and that x 2 will cross the xaxis and will be at as it does this since the multiplicity is greater than 1 Next the yintercept is 0 40 The coefficient of the 5 degree term is positive and since the degree is odd we know that this polynomial will increase without bound at the right end and decrease without bound at the left end Finally we just need to evaluate the polynomial at a couple of points The points that we pick aren t really all that important We just want to pick points according to the guidelines in the process outlined above and points that will be fairly easy to evaluate Here are some points We will leave it to you to verify the evaluations P 2a 320 191 20 P3 80 Now to actually sketch the graph we ll start on the left end and work our way across to the right end First we know that on the left end the graph decreases without bound as we make x more and more negative and this agrees with the point that we evaluated at x 2 So as we move to the right the function will actually be increasing at x 2 and we will continue to increase until we hit the first xintercept at x 0 At this point we know that the graph just touches the xaxis without actually crossing it This means that at x 0 the graph must be a turning point The graph is now decreasing as we move to the right Again this agrees with the next point that we ll run across the yintercept Now according to the next point that we ve got x l the graph must have another tuming point somewhere between x 0 and x 1 since the graph is higher at x 1 than at x 0 Just where this turning point will occur is very difficult to determine at this level so we won t worry 2007 Paul Dawkins 260 httptutorialmathlamaredutermsaspx College Algebra about trying to find it In fact determining this point usually requires some Calculus So we are moving to the right and the function is increasing The next point that we hit is the x intercept at x 2 and this one crosses the xaxis so we know that there won t be a tuming point here as there was at the first xintercept Therefore the graph will continue to increase through this point until we hit the final point that we evaluated the function at x 3 At this point we ve hit all the xintercepts and we know that the graph will increase without bound at the right end and so it looks like all we need to do is sketch in an increasing curve Here is a sketch of the polynomial 3392 II II I Note that one of the reasons for plotting points at the ends is to see just how fast the graph is increasing or decreasing We can see from the evaluations that the graph is decreasing on the left end much faster than it s increasing on the right end Okay let s take a look at another polynomial This time we ll go all the way through the process of finding the zeroes Example 2 Sketch the graph of P x x4 x3 6x2 Solution First we ll need to factor this polynomial as much as possible so we can identify the zeroes and get their multiplicities Pxx4 663 6x2 vc2x2 9e 6 A2x 3x 2 Here is a list of the zeroes and their multiplicities 2007 Paul Dawkins 261 httptutorialmathlamaredutermsaspX College Algebra x 2 multiplicity 1 x O multiplicity 2 3 x multiplicity 1 So the zeroes at x 2 and x 3 will correspond to xintercepts that cross the xaxis since their multiplicity is odd and will do so at an angle since their multiplicity is NOT at least 2 The zero at x 0 will not cross the xaxis since its multiplicity is even The yintercept is O O and notice that this is also an xintercept The coefficient of the 4 degree term is positive and so since the degree is even we know that the polynomial will increase without bound at both ends of the graph Finally here are some function evaluations P 354 141 4 Pa1 6 494 96 Now starting at the left end we know that as we make x more and more negative the function must increase without bound That means that as we move to the right the graph will actually be decreasing At x 3 the graph will be decreasing and will continue to decrease when we hit the first x intercept at x 2 since we know that this xintercept will cross the xaxis Next since the next xintercept is at x 0 we will have to have a tuming point somewhere so that the graph can increase back up to this xintercept Again we won t worry about where this turning point actually is Once we hit the xintercept at x O we know that we ve got to have a tuming point since this x intercept doesn t cross the xaxis Therefore to the right of x 0 the graph will now be decreasing It will continue to decrease until it hits another turning point at some unknown point so that the graph can get back up to the xaxis for the next xintercept at x 3 This is the final xintercept and since the graph is increasing at this point and must increase without bound at this end we are done Here is a sketch of the graph 2007 Paul Dawkins 262 httptutorialmathlamaredutermsaspx College Algebra lII DI Z39I39 III 1IIIIIIIIIIIIIIIIIIIIIIIIIIIII i IIIIIqII 2I 15 Example 3 Sketch the graph of P x E 969 4x3 Solution As with the previous example we ll first need to factor this as much as possible Pxa 6 49 x5 4 ae3 4 x3ae 2x 2 Notice that we first factored out a minus sign to make the rest of the factoring a little easier Here is a list of all the zeroes and their multiplicities x 2 multiplicity 1 x O multiplicity 3 x 2 multiplicity 1 So all three zeroes correspond to xintercepts that actually cross the xaxis since all their multiplicities are odd however only the xintercept at x 0 will cross the xaxis attened out The yintercept is O 0 and as with the previous example this is also an xintercept In this case the coefficient of the 5 degree term is negative and so since the degree is odd the graph will increase without bound on the left side and decrease without bound on the right side Here are some function evaluations p3 135 14 1 3 131 3 333 135 Alright this graph will start out much as the previous graph did At the left end the graph will be decreasing as we move to the right and will decrease through the first xintercept at x 2 since know that this xintercept crosses the xaxis Now at some point we ll get a turning point so the graph can get back up to the next xintercept at 2007 Paul Dawkins 263 httptutorialmathlamaredutermsaspx College Algebra x 0 and the graph will continue to increase through this point since it also crosses the xaxis Note as well that the graph should be at at this point as well since the multiplicity is greater than one Finally the graph will reach another turning point and start decreasing so it can get back down to the final xintercept at x 2 Since we know that the graph will decrease without bound at this end we are done Here is the sketch of this polynomial 23 13 I I I I I I I I H I I I I I 4 11 4 2 lIII 20 The process that we ve used in these examples can be a difficult process to leam It takes time to learn how to correctly interpret the results Also as pointed out at various spots there are several situations that we won t be able to deal with here To find the majority of the turning points we would need some Calculus which we clearly don t have Also the process does require that we have all the zeroes and that they all be real numbers Even with these drawbacks however the process can at least give us an idea of what the graph of a polynomial will look like 2007 Paul Dawkins 264 httptutorialmathlamaredutermsaspX College Algebra Finding Zeroes of Polynomials We ve been talking about zeroes of polynomial and why we need them for a couple of sections now We haven t however really talked about how to actually find them for polynomials of degree greater than two That is the topic of this section Well that s kind of the topic of this section In general finding all the zeroes of any polynomial is a fairly difficult process In this section we will give a process that will find all rational i e integer or fractional zeroes of a polynomial We will be able to use the process for finding all the zeroes of a polynomial provided all but at most two of the zeroes are rational If more than two of the zeroes are not rational then this process will not find all of the zeroes We will need the following theorem to get us started on this process Rational Root Theorem If the rational number x is a zero of the nu degree polynomial C Px sxquot 4c where all the coefficients are integers then I9 will be a factor of t and c will be a factor of 5 Note that in order for this theorem to work then the zero must be reduced to lowest terms In 4 2 other words it will work for E but not necessarily for Let s verify the results of this theorem with an example Example 1 Verify that the roots of the following polynomial satisfy the rational root theorem Px l2x3 4lx2 38x 40 x 43x 24x 5 Solution From the factored form we can see that the zeroes are 4 2 5 x 4 X X l 3 4 Notice that we wrote the integer as a fraction to fit it into the theorem Also with the negative zero we can put the negative onto the numerator or denominator It won t matter So according to the rational root theorem the numerators of these fractions with or without the minus sign on the third zero must all be factors of 40 and the denominators must all be factors of 12 Here are several ways to factor 40 and 12 40 410 4e 220 40 58 40 5 8 12 112 12 34 2 9 3 4 From these we can see that in fact the numerators are all factors of 40 and the denominators are all factors of 12 Also note that as shown we can put the minus sign on the third zero on either the numerator or the denominator and it will still be a factor of the appropriate number 2007 Paul Dawkins 265 httptutorialmathlamaredutermsaspX College Algebra So why is this theorem so useful Well for starters it will allow us to write down a list of possible rational zeroes for a polynomial and more importantly any rational zeroes of a polynomial WILL be in this list In other words we can quickly determine all the rational zeroes of a polynomial simply by checking all the numbers in our list Before getting into the process of finding the zeroes of a polynomial let s see how to come up with a list of possible rational zeroes for a polynomial Example 2 Find a list of all possible rational zeroes for each of the following polynomials a Px x4 7x3 l7x2 l7xIr 6 Solution b P x 2x4 963 43x2 43x 9 Solution Solution a Px x4 7x3 l7x2 l7xIr 6 b Now Just what does the rational root theorem say It says that if x is to be a zero of P x c then 9 must be a factor of 6 and c must be a factor of 1 Also as we saw in the previous example we can t forget negative factors So the first thing to do is actually to list all possible factors of l and 6 Here they are 6 l 1 i2 i3 i6 1 Now to get a list of possible rational zeroes of the polynomial all we need to do is write down all possible fractions that we can form from these numbers where the numerators must be factors of 6 and the denominators must be factors of 1 This is actually easier than it might at first appear to be There is a very simple shorthanded way of doing this Let s go through the first one in detail then we ll do the rest quicker First take the first factor from the numerator list including the i and divide this by the first factor okay only factor in this case from the denominator list again including the i Doing this gives This looks like a mess but it isn t too bad There are four fractions here They are 1 1 1 i 1 1 1 1 1 1 1 1 1 1 Notice however that the four fractions all reduce down to two possible numbers This will always happen with these kinds of fractions What we ll do from now on is form the fraction do any simplification of the numbers ignoring the i and then drop one of the i So the list possible rational zeroes for this polynomial is 2007 Paul Dawkins 266 httptutorialmathlamaredutermsaspX College Algebra J 3 J 6 il 1 39T39 I 39T ll i 3939 3 0 So it looks there are only 8 possible rational zeroes and in this case they are all integers Note as well that any rational zeroes of this polynomial WILL be somewhere in this list although we haven t found them yet Return to Problems b Px 2x4 963 3x2 3x 9 We ll not put quite as much detail into this one First get a list of all factors of 9 and 2 Note that the minus sign on the 9 isn t really all that important since we will still get a i on each of the factors 9 2 l 3 i9 1 2 Now the factors of 9 are all the possible numerators and the factors of 2 are all the possible denominators Here then is a list of all possible rational zeroes of this polynomial 1 1 3 3 i9 9 i1 i1 i1 i1 1 i3 2 i9 2 12quot 2 i2 2 i2 2 So we ve got a total of 12 possible rational zeroes half are integers and half are fractions Return to Problems The following fact will also be useful on occasion in finding the zeroes of a polynomial Fact If P x is a polynomial and we know that P a gt O and P 9 lt 0 then somewhere between a and bis a zero of Px What this fact is telling us is that if we evaluate the polynomial at two points and one of the evaluations gives a positive value ie the point is above the xaxis and the other evaluation gives a negative value i e the point is below the xaxis then the only way to get from one point to the other is to go through the xaxis Or in other words the polynomial must have a zero since we know that zeroes are where a graph touches or crosses the xaxis Note that this fact doesn t tell us what the zero is it only tells us that one will exist Also note that if both evaluations are positive or both evaluations are negative there may or may not be a zero between them Here is the process for determining all the rational zeroes of a polynomial 2007 Paul Dawkins 267 httptutorialmathlamaredutermsaspX College Algebra Process for Finding Rational Zeroes 1 Use the rational root theorem to list all possible rational zeroes of the polynomial P x 2 Evaluate the polynomial at the numbers from the first step until we find a zero Let s suppose the zero is x r then we will know that it s a zero because P r 0 Once this has been determined that it is in fact a zero write the original polynomial as Pltxgtltx rgtQltxgt 3 Repeat the process using Q x this time instead of P x This repeating will continue until we reach a second degree polynomial At this point we can solve this directly for the remaining zeroes To simplify the second step we will use synthetic division This will greatly simplify our life in several ways First recall that the last number in the final row is the polynomial evaluated at r and if we do get a zero the remaining numbers in the final row are the coefficients for Q x and so we won t have to go back and find that Also in the evaluation step it is usually easiest to evaluate at the possible integer zeroes first and then go back and deal with any fractions if we have to Let s see how this works Example 3 Determine all the zeroes of Px x4 7x3 l7x2 l7xk 6 Solution We found the list of all possible rational zeroes in the previous example Here they are i1 i2 i3 i6 We now need to start evaluating the polynomial at these numbers We can start anywhere in the list and will continue until we find zero To do the evaluations we will build a synthetic division table In a synthetic division table do the multiplications in our head and drop the middle row just writing down the third row and since we will be going through the process multiple times we put all the rows into a table Here is the first synthetic division table for this problem l 7 l7 l7 6 1 1 8 25 42 48P1 4 1 1611 6 0P1 0 So we found a zero Before getting into that let s recap the computations here to make sure you can do them 2007 Paul Dawkins 268 httptutorialmathlamaredutermsaspX College Algebra The top row is the coefficients from the polynomial and the first column is the numbers that we re evaluating the polynomial at Each row after the first is the third row from the synthetic division process Let s quickly look at the first couple of numbers in the second row The number in the second column is the first coefficient dropped down The number in the third column is then found by multiplying the 1 by l and adding to the 7 This gives the 8 For the fourth number is then 1 times 8 added onto 17 This is 25 etc You can do regular synthetic division if you need to but it s a good idea to be able to do these tables as it can help with the process Okay back to the problem We now know that x 1 is a zero and so we can write the polynomial as Pxx4 7x3 4l7x2 l7x 6 x Hx3 6x2 lklx 6 Now we need to repeat this process with the polynomial Q x x3 6x2 llx 6 So the first thing to do is to write down all possible rational roots of this polynomial and in this case we re lucky enough to have the first and last numbers in this polynomial be the same as the original polynomial that usually won t happen so don t always expect it Here is the list of all possible rational zeroes of this polynomial i1 i2 i3 i6 Now before doing a new synthetic division table let s recall that we are looking for zeroes to P x and from our first division table we determined that x l is NOT a zero of P x and so there is no reason to bother with that number again This is something that we should always do at this step Take a look at the list of new possible rational zeros and ask are there any that can t be rational zeroes of the original polynomial If there are some throw them out as we will already know that they won t work So a reduced list of numbers to try here is 1 i2 i3 i6 Note that we do need to include x l in the list since it is possible for a zero to occur more that once i e multiplicity greater than one Here is the synthetic division table for this polynomial 1 611 6 11 5 6 0P1 0 So x 1 is also a zero of Qx and we can now write Qx as Qxx3 6x2 lllx 6 x Hx2 5x 6 2007 Paul Dawkins 269 httptutorialmathlamaredutermsaspx College Algebra Now technically we could continue the process with x2 5x 6 but this is a quadratic equation and we know how to find zeroes of these without a complicated process like this so let s just solve this like we normally would x2 5x6x 2x 3 0 gt x 2x 3 Note that these two numbers are in the list of possible rational zeroes Finishing up this problem then gives the following list of zeroes for P x x 1 multiplicity 2 x 2 multiplicity 1 x 3 multiplicity 1 Note that x 1 has a multiplicity of 2 since it showed up twice in our work above Before moving onto the next example let s also note that we can now completely factor the polynomial P x x4 7x3 l7x2 l7xI 6 We know that each zero will give a factor in the factored form and that the exponent on the factor will be the multiplicity of that zero So the factored form is Pxx4 7x3 4l7x2 l7x 46 x H2x 2x 3 Let s take a look at another example Example 4 Find all the zeroes of P x 2x4 963 43x2 3x 9 Solution From the second example we know that the list of all possible rational zeroes is i11 1233 3 9 9 1 1 1 i11li3 E i9 3 2 2 2 2 2 2 The next step is to build up the synthetic division table When we ve got fractions it s usually best to start with the integers and do those first Also this time we ll start with doing all the negative integers first We are doing this to make a point on how we can use the fact given above to help us identify zeroes 2007 Paul Dawkins 270 httptutorialmathlamaredutermsaspx College Algebra 2 1 3 3 9 9 2 17 156 1401 12600P9 0 3 2 5 18 51 144P3 0 1 2 1 4 1 8P1 4 Now we haven t found a zero yet however let s notice that P 3 144 B and P 1 8lt 0 and so by the fact above we know that there must be a zero somewhere between 3 x 3 and x l Now we can also notice that x E 15 is 1nth1s range and is the only number in our list that is in this range and so there is a chance that this is a zero Let s run through synthetic division real quick to check and see if it s a zero and to get the coefficients for Q x if it is a zero 2l33 9 22660 I lIUJ So we got a zero in the final spot which tells us that this was a zero and Q x is Qx2x3 2x2 46x 6 We now need to repeat the whole process with this polynomial Also unlike the previous example we can t just reuse the original list since the last number is different this time So here are the factors of 6 and 2 6 ili2i3i6 2 il i 2 Here is a list of all possible rational zeroes for Q x 11 1 2 2 i3 3 i6 6 i1 i1 i1 i1 i1 1 2 1 i3 2 i6 3 2 2 2 2 2 2 Notice that some of the numbers appear in both rows and so we can shorten the list by only writing them down once Also remember that we are looking for zeroes of P x and so we can exclude any number in this list that isn t also in the original list we gave for P x So excluding previously checked numbers that were not zeros of P x as well as those that aren t in the original list gives the following list of possible number that we ll need to check 2007 Paul Dawkins 271 httptutorialmathlamaredutermsaspx College Algebra 1 3 ii i 2 lIUJ Again we Ve already checked x 3 and x l and know that they aren t zeroes so there is no reason to recheck them Let s again start with the integers and see what we get 2 2 6 6 12 0 6 0P1 0 So x 1 is a zero of Qx and we can now write Qx as Qx 2x3 2x2 6x 6 x H2x2 6 and as with the previous example we can solve the quadratic by other means 2x260 x2 3 xix3i So in this case we get a couple of complex zeroes That can happen Here is a complete list of all the zeroes for P x and note that they all have multiplicity of one 39 x x lx 3ix 3139 So as you can see this is a fairly lengthy process and we only did the work for two 4 degree polynomials The larger the degree the longer and more complicated the process With that being said however it is sometimes a process that we Ve got to go through to get zeroes of a polynomial 2007 Paul Dawkins 272 httptutorialmathlamaredutermsaspX College Algebra Partial Fractions This section doesn t really have a lot to do with the rest of this chapter but since the subject needs to be covered and this was a fairly short chapter it seemed like as good a place as any to put it So let s start with the following Let s suppose that we want to add the following two rational expressions 8 5 8x 4 5xl xlx 4xlx 4 xlx 4 8x 32 5x5 xlx 4 3x 37 x 1 x 4 What we want to do in this section is to start with rational expressions and ask what simpler rational expressions did we add andor subtract to get the original expression The process of doing this is called partial fractions and the result is often called the partial fraction decomposition The process can be a little long and on occasion messy but it is actually fairly simple We will start by trying to determine the partial fraction decomposition of P x Q x where both Px and Qx are polynomials and the degree of Px is smaller than the degree of Qx Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator That is important to remember So once we Ve determined that partial fractions can be done we factor the denominator as completely as possible Then for each factor in the denominator we can use the following table to determine the terms we pick up in the partial fraction decomposition Factor in Term in partial denominator fraction decomposition A ax b ax b axbk A1 A2 6196 axb2 axbk 2 b Ax B T ax X C axz bx c A1xB1 A2xB2 m AkxBk 2 k ax bxc 2 2 k ax bxC ax2bxc ax2bxc 2007 Paul Dawkins 273 httptutorialmathlamaredutermsaspx College Algebra Notice that the first and third cases are really special cases of the second and fourth cases respectively if we let k 1 Also it will always be possible to factor any polynomial down into a product of linear factors ax b and quadratic factors axz bx c some of which may be raised to a power There are several methods for determining the coefficients for each term and we will go over each of those as we work the examples Speaking of which let s get started on some examples Example 1 Determine the partial fraction decomposition of each of the following T u a 8x 42 Sol tion x2 3x 18 9 9x b 4x2 C T2 Solut1on x 1gtltx 2 d 9x 25 Solution x 3 Solution We ll go through the first one in great detail to show the complete partial fraction process and then we ll leave most of the explanation out of the remaining parts 8x 42 a 4 x2 3x 18 The first thing to do is factor the denominator as much as we can 8x 42 8x 42 x2 3x l8 x6x 3 So by comparing to the table above it looks like the partial fraction decomposition must look like 8x 42 A B I x2 3x l8 x6 x 339 Note that we ve got different coefficients for each term since there is no reason to think that they will be the same Now we need to determine the values of A and B The first step is to actually add the two terms back up This is usually simpler than it might appear to be Recall that we first need the least common denominator but we ve already got that from the original rational expression In this case it is LCDJac 6ac 3 Now just look at each term and compare the denominator to the LCD Multiply the numerator and denominator by whatever is missing then add In this case this gives 8x 42 Ax 3 Bx6 Ax 3Bx6 x23x l8 x6x 3 x6x 3 x6x 3 2007 Paul Dawkins 274 httptutorialmathlamaredutermsaspx College Algebra We need values of A and B so that the numerator of the expression on the left is the same as the numerator of the term on the right Or 8x 42Ax 3 Bx 6 This needs to be true regardless of the x that we plug into this equation As noted above there are several ways to do this One way will always work but can be messy and will often require knowledge that we don t have yet The other way will not always work but when it does it will greatly reduce the amount of work required In this set of examples the second and easier method will always work so we ll be using that here Here we are going to make use of the fact that this equation must be true regardless of the x that we plug in So let s pick an x plug it in and see what happens For no apparent reason let s try plugging in x 3 Doing this gives 83 42A3 3 B3 6 189B 2B Can you see why we choose this number By choosing x 3 we got the term involving A to drop out and we were left with a simple equation that we can solve for B Now we could also choose x 6 for exactly the same reason Here is what happens if we use this value of x 8 6 42A 6 3 B 6 6 90 9A 10A So by correctly picking x we were able to quickly and easily get the values of A and B So all that we need to do at this point is plug them in to finish the problem Here is the partial fraction decomposition for this part 8x 42 10 2 10 2 x23x l8 x6 x 3 x6 x 3 Notice that we moved the minus sign on the second term down to make the addition a subtraction We will always do that Return to Problems 9 9x 0 2x 7x 4 Okay in this case we won t put quite as much detail into the problem We ll first factor the denominator and then get the form of the partial fraction decomposition 9 9x 9 9x A B 2x27x 4 2x lx4 2x l x4 2007 Paul Dawkins 275 httptutorialmathlamaredutermsaspx College Algebra In this case the LCD is 2x 1 x 4 and so adding the two terms back up give 9 9x Ax4B2x l 2x27x 4 2x lx4 Next we need to set the two numerators equal 9 9xAx 4 B2x 9 Now all that we need to do is correctly pick values of x that will make one of the terms zero and solve for the constants Note that in this case we will need to make one of them a fraction This is fairly common so don t get excited about it Here is this work 4 16 9B B 5 xzlz 2 A 2 Agtl 2 2 2 The partial fraction decomposition for this expression is x 9 9x l 5 2x27x 4 2x l x4 Return to Problems 4x2 x 1gtltx 2f In this case the denominator has already been factored for us Notice as well that we ve now got a linear factor to a power So recall from our table that this means we will get 2 terms in the partial fraction decomposition from this factor Here is the form of the partial fraction decomposition for this expression c 4x2 2 A I B I C ltx 1gtltx 2gt2 H H ltx 2gt2 Now remember that the LCD is just the denominator of the original expression so in this case we ve got x lx 22 Adding the three terms back up gives us 4x2 Ax 22Bx lx 2Cx l ltx1gtltx2gt2 ltx1gtltx2gt2 Remember that we just need to add in the factors that are missing to each term Now set the numerators equal 4x2Ax 2 2 Bx l x 2 Gx H In this case we ve got a slightly different situation from the previous two parts Let s start by picking a couple of values of x and seeing what we get since there are two that should jump right 2007 Paul Dawkins 276 httptutorialmathlamaredutermsaspx College Algebra out at us as being particularly useful 3el 4gtA 12 A 4 x2 16C1 gt C16 So we can get A and C in the same manner that we ve been using to this point However there is no value of x that will allow us to eliminate the first and third term leaving only the middle term that we can use to solve for B While this may appear to be a problem it actually isn t At this point we know two of the three constants So all we need to do is chose any other value of x that would be easy to work with x 0 seems particularly useful here plug that in along with the values of A and C and we ll get a simple equation that we can solve for B Here is that work 4ltogt2lt4gtlt 2amp2 Blt 1x 2 16lt 1 0l6 2B 16 02B 0B In this case we got B 0 this will happen on occasion but do not expect it to happen in all cases Here is the partial fraction decomposition for this part 4x2 4 16 Return to Problems 9x 25 x 32 Again the denominator has already been factored for us In this case the form of the partial fraction decomposition is 1 9x 25 A Bl x32 X3 xl32 Adding the two terms together gives 9x25 Ax3B x 32 x 32 Notice that in this case the second term already had the LCD under it and so we didn t need to add anything in that time Setting the numerators equal gives 9x25Ax 3 B Now again we can get B for free by picking x 3 2007 Paul Dawkins 277 httptutorialmathlamaredutermsaspX College Algebra 9 325A 3 B 2B To find A we will do the same thing that we did in the previous part We ll use x O and the fact that we know what B is 25Ac9 2 2723A 9A In this case notice that the constant in the numerator of the first isn t zero as it was in the previous part Here is the partial fraction decomposition for this part 9x25 9 2 x32 X3 x32 Return to Problems Now we need to do a set of examples with quadratic factors Note however that this is where the work often gets fairly messy and in fact we haven t covered the material yet that will allow us to work many of these problems We can work some simple examples however so let s do that Example 2 Determine the partial fraction decomposition of each of the following 8x2 12 a E x x 2x 6 3 3 7 4 b x2 Solution x2 2 Solution 8x2 12 ltagt x x 2x 6 In this case the x that sits in the front is a linear term since we can write it as x x G and so the form of the partial fraction decomposition is 8x2 12 A BxCT xx22x 6 X 2966 Now we ll use the fact that the LCD is x x2 2x 6 and add the two terms together 8x2 12 Ax2 2x 6xBxC xx22x 6 xx22x 6 2007 Paul Dawkins 278 httptutorialmathlamaredutermsaspX College Algebra Next set the numerators equal 8x2 l2Ax2 235 6 xkBx C4 This is where the process changes from the previous set of examples We could choose x 0 to get the Value of A but that s the only constant that we could get using this method and so it just won t work all that well here What we need to do here is multiply the right side out and then collect all the like terms as follows 8x2 12 Axz 2Ax 6A Bxz ex 8x2 12 A Bx2 2A Cx 6A Now we need to choose A B and C so that these two are equal That means that the coefficient of the x2 term on the right side will have to be 8 since that is the coefficient of the x2 term on the left side Likewise the coefficient of the x term on the right side must be zero since there isn t an x term on the left side Finally the constant term on the right side must be 12 since that is the constant on the left side We generally call this setting coefficients equal and we ll write down the following equations AB8 2ACO 6Aim Now we haVen t talked about how to solve systems of equations yet but this is one that we can do without that knowledge We can solve the third equation directly for A to get that A 2 We can then plug this into the first two equations to get 2B8 33 6 22CO gt e 4 So the partial fraction decomposition for this expression is 8x2 12 2 6x 4 xx2 2x 6 2 X2 2966 Return to Problems 3x3 7x 4 x2 22 Here is the form of the partial fraction decomposition for this part 3x3 7x 4 AxB Cxl x222 quotx22 22 2 1 Adding the two terms up gives 2007 Paul Dawkins 279 httptutorialmathlamaredutermsaspx College Algebra 3x3 7x4 AxBx2 2CxD x2 22 x2 22 Now set the numerators equal and we might as well go ahead and multiply the right side out and collect up like terms while we re at it 3x37x 4Ax ax2 29 Gac 1 3x37x 4Ax3 2IAx Baez 21B Gac 1 3x37x 4Ax3 Baez QEA Gx 213 D Setting coefficients equal gives A3 BO 2AC7 2BD 4 In this case we got A and B for free and don t get excited about the fact that B 0 This is not a problem and in fact when this happens the remaining work is often a little easier So plugging the known Values of A and B into the remaining two equations gives 23 C 7 ampgt 1 20 D 4 gt 9 4 The partial fraction decomposition is then 3x37x 4 3x x 4 x222 x22 xu y Return to Problems 2007 Paul Dawkins 280 httptutorialmathlamaredutermsaspX College Algebra Exponential and Logarithm Functions Introduction In this chapter we are going to look at exponential and logarithm functions Both of these functions are very important and need to be understood by anyone who is going on to later math courses These functions also have applications in science engineering and business to name a few areas In fact these functions can show up in just about any field that uses even a small degree of mathematics Many students find these to be difficult functions to deal with especially logarithms This is probably because they are so different from any of the other functions that they ve looked at to this point and logarithms use a notation that will be new to almost everyone in an algebra class However you will find that once you get past the notation and start to understand some of their properties they really aren t too bad Here is a listing of the topics covered in this chapter Exponential Functions In this section we will introduce exponential functions We will be taking a look at some of the properties of exponential functions Logarithm Functions Here we will introduce logarithm functions We be looking at how to evaluate logarithms as well as the properties of logarithms Solving Exponential Equations We will be solving equations that contain exponentials in this section Solving Logarithm Equations Here we will solve equations that contain logarithms Application In this section we will look at a couple of applications of exponential functions and an application of logarithms 2007 Paul Dawkins 281 httptutorialmathlamaredutermsaspx College Algebra Exponential Functions Let s start off this section with the definition of an exponential function If I9 is any number such that b gt O and b 7amp1 then an exponential function is a function in the form fx 19quot where I9 is called the base and x can be any real number Notice that the x is now in the exponent and the base is a fixed number This is exactly the opposite from what we ve seen to this point To this point the base has been the variable x in most cases and the exponent was a fixed number However despite these differences these functions evaluate in exactly the same way as those that we are used to We will see some examples of exponential functions shortly Before we get too far into this section we should address the restrictions on b We avoid one and zero because in this case the function would be fx0quot 0 and fxlquot l and these are constant functions and won t have many of the same properties that general exponential functions have Next we avoid negative numbers so that we don t get any complex values out of the function evaluation For instance if we allowed I 4 the function would be rltxgt lt4gt r1 4 J7 and as you can see there are some function evaluations that will give complex numbers We only want real numbers to arise from function evaluation and so to make sure of this we require that 9 not be a negative number Now let s take a look at a couple of graphs We will be able to get most of the properties of exponential functions from these graphs 1 X Example 1 Sketch the graph of f x 2 and g x on the same axis system Solution Okay since we don t have any knowledge on what these graphs look like we re going to have to pick some values of x and do some function evaluations Function evaluation with exponential functions works in exactly the same manner that all function evaluation has worked to this point Whatever is in the parenthesis on the left we substitute into all the x s on the right side Here are some evaluations for these two functions 2007 Paul Dawkins 282 httptutorialmathlamaredutermsaspx College Algebra lI n 2339 E 1 I I 2 Note as well that we could have written g x in the following way gXGjx 3 2 Sometimes we ll see this kind of exponential function and so it s important to be able to go between these two forms Now let s talk about some of the properties of exponential functions Properties of f x b 2007 Paul Dawkins 283 httptutorialmathlamaredutermsaspx College Algebra 1 The graph of f x will always contain the point 01 Or put another way f 0 1 regardless of the value of I9 2 For every possible I9 bx gt O Note that this implies that bx 75 O 3 If 0 lt 9 lt1 then the graph of bx will decrease as we move from left to right Check out 1 X the graph of above for verification of this property 4 If 7 gt1 then the graph of bx will increase as we move from left to right Check out the graph of 2x above for verification of this property 5 If bxbx then xy All of these properties except the final one can be Verified easily from the graphs in the first example We will hold off discussing the final property for a couple of sections where we will actually be using it As a final topic in this section we need to discuss a special exponential function In fact this is so special that for many people this is THE exponential function Here it is r x e Where e 2718281828 Note the difference between f x bx and f x ex In the first case bis any number that is meets the restrictions given above while e is a Very specific number Also note that e is not a terminating decimal This special exponential function is Very important and arises naturally in many areas As noted above this function arises so often that many people will think of this function if you talk about exponential functions We will see some of the applications of this function in the final section of this chapter Let s get a quick graph of this function Example 2 Sketch the graph of f x ex Solution Let s first build up a table of Values for this function x 2 1 0 1 2 fx I 01353 03679 1 2718 7389 To get these evaluation with the exception of x 0 you will need to use a calculator In fact that is part of the point of this example Make sure that you can run your calculator and Verify these numbers Here is a sketch of this graph 2007 Paul Dawkins 284 httptutorialmathlamaredutermsaspx College Algebra 4 39 4 IIIII III IIIIIIIIIIIIIIIIIIIIIIIIIIII IT 43 39 H Ll 2 1 u 1 2 Notice that this is an increasing graph as we should expect since e 27l828l827 gt 1 There is one final example that we need to work before moving onto the next section This example is more about the evaluation process for exponential functions than the graphing process We need to be very careful with the evaluation of exponential functions Example 3 Sketch the graph of g x Se 4 Solution Here is a quick table of values for this function x 1 0 1 2 3 gx 32945 9591 1 2161 3323 Now as we stated above this example was more about the evaluation process than the graph so let s go through the first one to make sure that you can do these g 1 5e1 4 5e2 4 5 7389 4 Notice that when evaluating exponential functions we first need to actually do the exponentiation before we multiply by any coefficients 5 in this case Also we used only 3 decimal places here since we are only graphing In many applications we will want to use far more decimal places in these computations Here is a sketch of the graph 2007 Paul Dawkins 285 httptutorialmathlamaredutermsaspx College Algebra 4 E quot2 39A39 III Iquot39 III I I I I I I I I I I I I I I ICI I39 Ll 1 1 1 Notice that this graph violates all the properties we listed above That is okay Those properties are only valid for functions in the form f x bx or f x e We ve got a lot more going on in this function and so the properties as Written above won t hold for this function 2007 Paul Dawkins 286 httptutorialmathlamaredutermsaspx College Algebra Logarithm Functions In this section we now need to move into logarithm functions This can be a tricky function to graph right away There is going to be some different notation that you aren t used to and some of the properties may not be all that intuitive Do not get discouraged however Once you figure these out you will find that they really aren t that bad and it usually just takes a little working with them to get them figured out Here is the definition of the logarithm function If b is any number such that b gt O and b 7amp1 and x gt 0 then y log x is equivalent to by x We usually read this as log base b of x In this definition y log b x is called the logarithm form and by x is called the exponential form Note that the requirement that x gt O is really a result of the fact that we are also requiring b gt 0 If you think about it it will make sense We are raising a positive number to an exponent and so there is no way that the result can possibly be anything other than another positive number It is very important to remember that we can t take the logarithm of zero or a negative number Now let s address the notation used here as that is usually the biggest hurdle that students need to overcome before starting to understand logarithms First the log part of the function is simply three letters that are used to denote the fact that we are dealing with a logarithm They are not variables and they aren t signifying multiplication They are just there to tell us we are dealing with a logarithm Next the b that is subscripted on the log part is there to tell us what the base is as this is an important piece of information Also despite what it might look like there is no exponentiation in the logarithm form above It might look like we ve got bx in that form but it isn t It just looks like that might be what s happening It is important to keep the notation with logarithms straight if you don t you will find it very difficult to understand them and to work with them Now let s take a quick look at how we evaluate logarithms Example 1 Evaluate each of the following logarithms a log 4 16 Solution b lOg2 16 Solution C log 6 216 Solution 1 d 10g5E E e log18l Solution 5 2007 Paul Dawkins 287 httptutorialmathlamaredutermsaspx College Algebra f log 3 2 7 J 5 8 Solution Now the reality is that evaluating logarithms directly can be a very difficult process even for those who really understand them It is usually much easier to first convert the logarithm form into exponential form In that form we can usually get the answer pretty quickly a log 4 16 Okay what we are really asking here is the following log 4 16 As suggested above let s convert this to exponential form log 4 16 gt 4 16 Most people cannot evaluate the logarithm log 4 16 right off the top of their head However most people can determine the exponent that we need on 4 to get 16 once we do the exponentiation So since 42 16 we must have the following value of the logarithm log 4 16 2 Return to Problems b log 16 This one is similar to the previous part Let s first convert to exponential form logz 16 gt 2 16 If you don t know this answer right off the top of your head start trying numbers In other words compute 22 23 24 etc until you get 16 In this case we need an exponent of 4 Therefore the value of this logarithm is log 16 4 Before moving on to the next part notice that the base on these is a very important piece of notation Changing the base will change the answer and so we always need to keep track of the base Return to Problems c log6 216 We ll do this one without any real explanation to see how well you ve got the evaluation of logarithms down log6 216 3 because 63 216 Return to Problems 1 d 10g5 E Now this one looks different from the previous parts but it really isn t any different As always let s first convert to exponential form 2007 Paul Dawkins 288 httptutorialmathlamaredutermsaspx College Algebra gt 5quot 1 g5125 125 First notice that the only way that we can raise an integer to an integer power and get a fraction as an answer is for the exponent to be negative So we know that the exponent has to be negative Now let s ignore the fraction for a second and ask 5 125 In this case if we cube 5 we will get 125 So it looks like we have the following 1 1 logs 3 because 5393 3 125 5 125 Return to Problems e logl 81 3 Converting this logarithm to exponential form gives 1 2 log181 gt E 81 5 3 Now just like the previous part the only way that this is going to work out is if the exponent is negative Then all we need to do is recognize that 34 81 and we can see that 1 quot4 3 4 4 log181 4 because 3 81 3 3 1 Return to Problems 27 D 1Og3 5 8 Here is the answer to this one 3 3 2 log3 E 3 because Y 3 3 7 5 8 2 2 8 Return to Problems Hopefully you now have an idea on how to evaluate logarithms and are starting to get a grasp on the notation There are a few more evaluations that we want to do however we need to introduce some special logarithms that occur on a very regular basis They are the common logarithm and the natural logarithm Here are the definitions and notations that we will be using for these two logarithms common logarithm log x log10 x natural logarithm ln x loge x So the common logarithm is simply the log base 10 except we drop the base 10 part of the notation Similarly the natural logarithm is simply the log base e with a different notation and 2007 Paul Dawkins 289 httptutorialmathlamaredutermsaspx College Algebra where e is the same number that we saw in the previous section and is defined to be e2718281827 Let s take a look at a couple more evaluations Example 2 Evaluate each of the following logarithms a log 1000 1 bl 0g10O c lnl B d In E e log34 34 f logs 1 Solution To do the first four evaluations we just need to remember what the notation for these are and what base is implied by the notation The final two evaluations are to illustrate some of the properties of all logarithms that we ll be looking at eventually a log1000 3 because 103 1000 1 1 1 bl ab 102 Og1OO muse 102 100 1 1 c ln 1 because e391 e e 1 1 d ln 5 because e2 E Notice that with this one we are really just acknowledging a change of notation from fractional exponent into radical form e log34 34 1 because 341 34 Notice that this one will work regardless of the base that we re using f logs 1 0 because 80 1 Again note that the base that we re using here won t change the 3IlSWCI39 So when evaluating logarithms all that we re really asking is what exponent did we put onto the base to get the number in the logarithm Now before we get into some of the properties of logarithms let s first do a couple of quick graphs 2007 Paul Dawkins 290 httptutorialmathlamaredutermsaspx College Algebra Example 3 Sketch the graph of the common logarithm and the natural logarithm on the same axis system Solution This example has two points First it will familiarize us with the graphs of the two logarithms that we are most likely to see in other classes Also it will give us some practice using our calculator to evaluate these logarithms because the reality is that is how we will need to do most of these evaluations Here is a table of values for the two logarithms log x 111 x 03010 06931 0 0 03010 06931 04771 10986 06021 13863 Igtugtto N gtlt Here is a sketch of the graphs of these two functions Now let s start looking at some properties of logarithms We ll start off with some basic evaluation properties Properties of Logarithms 1 logb 1 0 This follows from the fact that 90 1 2 log 9 1 This follows from the fact that 91 b 3 log bx x This can be generalized out to log bf f 4 bbg x This can be generalized out to blogb x f 2007 Paul Dawkins 291 httptutorialmathlamaredutermsaspx College Algebra Properties 3 and 4 leads to a nice relationship between the logarithm and exponential function Let s first compute the following function compositions for f x bx and g x log b x f08xf8xl f10bx X 8 fx8fxl SW 1 gb quot X Recall from the section on inverse functions that this means that the exponential and logarithm functions are inverses of each other This is a nice fact to remember on occasion We should also give the generalized Version of Properties 3 and 4 in terms of both the natural and common logarithm as we ll be seeing those in the next couple of sections on occasion lnem fx log10fx fx elnfx d 10logfx f Now let s take a look at some manipulation properties of the logarithm More Properties of Logarithms For these properties we will assume that x gt O and y gt O 5 log xylogbx logb 3 x 6 log P log x logb 3 7 logb x rlogb x 8 If logbxlogb y then x y We won t be doing anything with the final property in this section it is here only for the sake of completeness We will be looking at this property in detail in a couple of sections The first two properties listed here can be a little confusing at first since on one side we Ve got a product or a quotient inside the logarithm and on the other side we Ve got a sum or difference of two logarithms We will just need to be careful with these properties and make sure to use them correctly Also note that there are no rules on how to break up the logarithm of the sum or difference of two terms To be clear about this let s note the following logb x y logb xlogb y logb x y logb x logb y Be careful with these and do not try to use these as they simply aren t true 2007 Paul Dawkins 292 httptutorialmathlamaredutermsaspx College Algebra Note that all of the properties given to this point are valid for both the common and natural logarithms We just didn t write them out explicitly using the notation for these two logarithms the properties do hold for them nonetheless Now let s see some examples of how to use these properties Example 4 Simplify each of the following logarithms a log x3y5 1 9 5 b 1ogx f J E Z c lnxE 1 x y2 d 1033 2 E x y Solution The instructions here may be a little misleading When we say simplify we really mean to say that we want to use as many of the logarithm properties as we can a 1ogx3y5 Note that we can t use Property 7 to bring the 3 and the 5 down into the front of the logarithm at this point In order to use Property 7 the whole term in the logarithm needs to be raised to the power In this case the two exponents are only on individual terms in the logarithm and so Property 7 can t be used here We do however have a product inside the logarithm so we can use Property 5 on this logarithm log4 x3y5 log4 x3 log4 325 Now that we ve done this we can use Property 7 on each of these individual logarithms to get the final simplified answer log4 x3y5 3log4x 5log4 3 Return to Problems 9 5 x b 1og l J Z In this case we ve got a product and a quotient in the logarithm In these cases it is almost always best to deal with the quotient before dealing with the product Here is the first step in this part 9 5 logvczg jlogx9y5 log z3 Now we ll break up the product in the first term and once we ve done that we ll take care of the exponents on the terms 2007 Paul Dawkins 293 httptutorialmathlamaredutermsaspx College Algebra 9 5 logvczg jlogx9y5 logz3 log x9 log y5 log z3 9logx5logy 3logz Return to Problems c 111 J For this part let s first rewrite the logarithm a little so that we can see the first step 1 ln 5 lnxy5 Written in this form we can see that there is a single exponent on the whole term and so we ll take care of that first 1 ln xy lnxy Now we will take care of the product 1 lnxy lnx ln y 1 Notice the parenthesis in this the answer The E multiplies the original logarithm and so it will also need to multiply the whole simplified logarithm Therefore we need to have a set of parenthesis there to make sure that this is taken care of correctly Return to Problems x y2 d10g3 T x y We ll first take care of the quotient in this logarithm 2 x y log3 log x y2 log3 x2 ygr We now reach the real point to this problem The second logarithm is as simplified as we can make it Remember that we can t break up a log of a sum or difference and so this can t be broken up any farther Also we can only deal with exponents if the term as a whole is raised to the exponent The fact that both pieces of this term are squared doesn t matter It needs to be the whole term squared as in the first logarithm So we can further simplify the first logarithm but the second logarithm can t be simplified any more Here is the final answer for this problem 2 log3 2log3x y loggfxz yz Return to Problems 2007 Paul Dawkins 294 httptutorialmathlamaredutermsaspx College Algebra Now we need to work some examples that go the other way This next set of examples is probably more important than the previous set We will be doing this kind of logarithm work in a couple of sections Example 5 Write each of the following as a single logarithm with a coefficient of l a 7log12 x 2log12 y E b 310g x 610g y E c 5lnx y 2ln y 8lnx J Solution The instruction requiring a coefficient of 1 means that the when we get down to a final logarithm there shouldn t be any number in front of the logarithm Note as well that these examples are going to be using Properties 5 7 only we ll be using them in reverse We will have expressions that look like the right side of the property and use the property to write it so it looks like the left side of the property a The first step here is to get rid of the coefficients on the logarithms This will use Property 7 in reverse In this direction Property 7 says that we can move the coefficient of a logarithm up to become a power on the term inside the logarithm Here is that step for this part 7log12 x 2log12 y log12 x7 loglz y2 We ve now got a sum of two logarithms both with coefficients of l and both with the same base This means that we can use Property 5 in reverse Here is the answer for this part 7log12 x 2log12 y log12 x7y2 Return to Problems b Again we will first take care of the coefficients on the logarithms 3log x 6log y log x3 log y6 We now have a difference of two logarithms and so we can use Property 6 in reverse When using Property 6 in reverse remember that the term from the logarithm that is subtracted off goes in the denominator of the quotient Here is the answer to this part 3 3logx 6log y logx 6j 3 Return to Problems c In this case we ve got three terms to deal with and none of the properties have three terms in them That isn t a problem Let s first take care of the coefficients and at the same time we ll factor a minus sign out of the last two terms The reason for this will be apparent in the next step 5lnxy 2lny 8lnxlnx 35 lny2 lnxs Now notice that the quantity in the parenthesis is a sum of two logarithms and so can be combined into a single logarithm with a product as follows 2007 Paul Dawkins 295 httptutorialmathlamaredutermsaspx College Algebra 5lnxy 2lny 8lnxlnx 5 lny2x8 Now we are down to two logarithms and they are a difference of logarithms and so we can write it as a single logarithm with a quotient 5 5lnxy 2lny 8lnxln j 2 8 yx Return to Problems The final topic that we need to discuss in this section is the change of base formula Most calculators these days are capable of evaluating common logarithms and natural logarithms However that is about it so what do we do if we need to evaluate another logarithm that can t be done easily as we did in the first set of examples that we looked at To do this we have the change of base formula Here is the change of base formula loga x jlogb X log b a where we can choose 9 to be anything we want it to be In order to use this to help us evaluate logarithms this is usually the common or natural logarithm Here is the change of base formula using both the common logarithm and the natural logarithm l l logaxz Ogx logax log a 1n a Let s see how this works with an example Example 6 Evaluate logs 7 Solution First notice that we can t use the same method to do this evaluation that we did in the first set of examples This would require us to look at the following exponential form 5 7 and that s just not something that anyone can answer off the top of their head If the 7 had been a 5 or a 25 or a 125 etc we could do this but it s not Therefore we have to use the change of base formula Now we can use either one and we ll get the same answer So let s use both and verify that We ll start with the common logarithm form of the change of base 1 7 4 4 l4 10g 7 0g 0 8 30980 00 120906195512 log 5 0698970004336 Now let s try the natural logarithm form of the change of base formula ln7 19i5910149062120906195512 ln5 160943791243 logs So we got the same answer despite the fact that the fractions involved different answers 2007 Paul Dawkins 296 httptutorialmathlamaredutermsaspx College Algebra Solving Exponential Equations Now that we ve seen the definitions of exponential and logarithm functions we need to start thinking about how to solve equations involving them In this section we will look at solving exponential equations and we will look at solving logarithm equations in the next section There are two methods for solving exponential equations One method is fairly simple but requires a very special form of the exponential equation The other will work on more complicated exponential equations but can be a little messy at times Let s start off by looking at the simpler method This method will use the following fact about exponential functions If bquotby then x y Note that this fact does require that the base in both exponentials to be the same If it isn t then this fact will do us no good Let s take a look at a couple of examples Example 1 Solve each of the following a 53x 57x2 Solution b 4t2 464 Solution C 3Z 9Z5 Solution d 45399quot Solution 87 Solution a 53x 57x 2 In this first part we have the same base on both exponentials so there really isn t much to do other than to set the two exponents equal to each other and solve for x 3x7x 2 24x 1 jx 2 1 So if we were to plug x E into the equation then we would get the same number on both sides of the equal sign Return to Problems b 4392 46 Again there really isn t much to do here other than set the exponents equal since the base is the same in both exponentials 2007 Paul Dawkins 297 httptutorialmathlamaredutermsaspx College Algebra 6 F t2t 60 t3t 20 gt t 3t 2 In this case we get two solutions to the equation That is perfectly acceptable so don t worry about it when it happens Return to Problems c 3Z 9 5 Now in this case we don t have the same base so we can t just set exponents equal However with a little manipulation of the right side we can get the same base on both exponents To do this all we need to notice is that 9 32 Here s what we get when we use this fact 3Z 32z5 Now we still can t just set exponents equal since the right side now has two exponents If we recall our exponent properties we can fix this however 3g 32z5 We now have the same base and a single exponent on each base so we can use the property and set the exponents equal Doing this gives Z 2z 5 z 2z l0I 10z So after all that work we get a solution of z 10 Return to Problems 1 d 45916 j 8C 2 In this part we ve got some issues with both sides First the right side is a fraction and the left side isn t That is not the problem that it might appear to be however so for a second let s ignore that The real issue here is that we can t write 8 as a power of 4 and we can t write 4 as a power of 8 as we did in the previous part The first thing to do in this problem is to get the same base on both sides and to so that we ll have to note that we can write both 4 and 8 as a power of 2 So let s do that 22 5 9x lt23 It s now time to take care of the fraction on the right side To do this we simply need to remember the following exponent property 2007 Paul Dawkins 298 httptutorialmathlamaredutermsaspx College Algebra Using this gives 225 9x 2 3x 2 So we now have the same base and each base has a single exponent on it so we can set the exponents equal 25 9x 3x 2 l0 l8x 396 6 4l5x 4 x 15 And there is the answer to this part Return to Problems Now the equations in the previous set of examples all relied upon the fact that we were able to get the same base on both exponentials but that just isn t always possible Consider the following equation 7 9 This is a fairly simple equation however the method we used in the previous examples just won t work because we don t know how to write 9 as a power of 7 In fact if you think about it that is exactly what this equation is asking us to find So the method we used in the first set of examples won t work The problem here is that the x is in the exponent Because of that all our knowledge about solving equations won t do us any good We need a way to get the x out of the exponent and luckily for us we have a way to do that Recall the following logarithm property from the last section logb a rlogb a Note that to avoid confusion with x s we replaced the x in this property with an a The important part of this property is that we can take an exponent and move it into the front of the term So if we had log b 7 we could use this property as follows xlogb 7 The x in now out of the exponent Of course we are now stuck with a logarithm in the problem and not only that but we haven t specified the base of the logarithm The reality is that we can use any logarithm to do this so we should pick one that we can deal with This usually means that we ll work with the common logarithm or the natural logarithm So let s work a set of examples to see how we actually use this idea to solve these equations 2007 Paul Dawkins 299 httptutorialmathlamaredutermsaspx College Algebra Example 2 Solve each of the following equations 3 7x 9 J b 24y1 3y 0 Solution C CH6 2 Solution d l05x 8 Solution e 5e2 4 8 0 Solution Solution a 7 9 Okay so we say above that if we had a logarithm in front the left side we could get the x out of the exponent That s easy enough to do We ll just put a logarithm in front of the left side However if we put a logarithm there we also must put a logarithm in front of the right side This is commonly referred to as taking the logarithm of both sides We can use any logarithm that we d like to so let s try the natural logarithm ln 7 ln 9 x ln 7 ln 9 Now we need to solve for x This is easier than it looks If we had 7x 9 then we could all solve for x simply by dividing both sides by 7 It works in exactly the same manner here Both ln7 and ln9 are just numbers Admittedly it would take a calculator to determine just what those numbers are but they are numbers and so we can do the same thing here xln7 li ln7 ln7 ln9 x ln7 Now that is technically the exact answer However in this case it s usually best to get a decimal answer so let s go one step further ln9 219722458 x ll29l5007 ln7 l9459l0l5 Note that the answers to these are decimal answers more often than not Also be careful here to not make the following mistake ll29l5OO7L 2 ln2j 025l3l44283 ln7 7 The two are clearly different numbers Finally let s also use the common logarithm to make sure that we get the same answer 2007 Paul Dawkins 300 httptutorialmathlamaredutermsaspx College Algebra log 7 log9 xlog7 log9 x log9 O95l242509 log7 O845098040 ll29l5007 So sure enough the same answer We can use either logarithm although there are times when it is more convenient to use one over the other Return to Problems b 2 3y 0 In this case we can t just put a logarithm in front of both sides There are two reasons for this First on the right side we ve got a zero and we know from the previous section that we can t take the logarithm of zero Next in order to move the exponent down it has to be on the whole term inside the logarithm and that just won t be the case with this equation in its present form So the first step is to move on of the terms to the other side of the equal sign then we will take the logarithm of both sides using the natural logarithm 24yl 3y ln24y1 ln3y 4ylln2 yln3 Okay this looks messy but again it s really not that bad Let s look at the following equation first 24yl3y 8y23y 5y 2 2 V 5 We can all solve this equation and so that means that we can solve the one that we ve got Again the ln2 and ln3 are just numbers and so the process is exactly the same The answer will be messier than this equation but the process is identical Here is the work for this one 4ylln2 yln3 4yln2ln2 yln3 4yln2 yln3 ln2 y41n2 1n3 ln2 ln2 y 2 T1113 So we get all the terms with y in them on one side and all the other terms on the other side Once this is done we then factor out a y and divide by the coefficient Again we would prefer a decimal answer so let s get that 2007 Paul Dawkins 301 httptutorialmathlamaredutermsaspx College Algebra ln2 0693l47l8l 4 0414072245 4ln2 ln3 40693147181 1098612289 y Return to Problems P0b el6 2 Now this one is a little easier than the previous one Again we ll take the natural logarithm of both sides ln e 6 ln 2 Notice that we didn t take the exponent out of this one That is because we want to use the following property with this one lne x fx We saw this in the previous section in more general form and by using this here we will make our life significantly easier Using this property gives t 6 ln 2 tln2 6 O69314748 6 530685202 Notice the parenthesis around the 2 in the logarithm this time They are there to make sure that We don t make the following mistake 530685202ln2 6 l n2 6 ln 4 can39tbedone Be very careful with this mistake It is easy to make when you aren t paying attention to what you re doing or are in a hurry Return to Problems d 105quot 8 The equation in this part is similar to the previous part except this time We ve got a base of 10 and so recalling the fact that 1og10 quot fx it makes more sense to use common logarithms this time around Here is the work for this equation log 105quot log 8 5 x log 8 5 log 8 x gt Ex 5 0903089987 40969l00l3 This could have been done with natural logarithms but the work would have been messier Return to Problems e 5e2 4 8 0 With this final equation we ve got a couple of issues First we ll need to move the number over to the other side In order to take the logarithm of both sides we need to have the exponential on one side by itself Doing this gives 2007 Paul Dawkins 302 httptutorialmathlamaredutermsaspx College Algebra 5e2z4 8 Next we Ve got to get a coefficient of 1 on the exponential We can only use the facts to simplify this if there isn t a coefficient on the exponential So divide both sides by 5 to get 2z4 E 5 B At this point we will take the logarithm of both sides using the natural logarithm since there is an e in the equation 2z 4 ln G 5 2z ln H 4 5 1 8 1 z ln E 4 0470003629 4 176499819 Return to Problems Note that we could have used this second method on the first set of examples as well if we d Wanted to although the work would have been more complicated and prone to mistakes if we d done that 2007 Paul Dawkins 303 httptutorialmathlamaredutermsaspx College Algebra Solving Logarithm Equations In this section we will now take a look as solving logarithmic equations or equations with logarithms in them We will be looking at two specific types of equations here In particular we will look at equations in which every term is a logarithm and we also look at equations in which all but one term in the equation is a logarithm and the term without the logarithm will be a constant Also we will be assuming that the logarithms in each equation will have the same base If there is more than one base in the logarithms in the equation the solution process becomes much more difficult Before we get into the solution process we will need to remember that we can only plug positive numbers into a logarithm This will be important down the road and so we can t forget that Now let s start off by looking at equations in which each term is a logarithm and all the bases on the logarithms are the same In this case we will use the fact that If logbxzlogb y then x y In other words if we ve got two logs in the problem one on either side of an equal sign and both with a coefficient of one then we can just drop the logarithms Let s take a look at a couple of examples Example 1 Solve each of the following equations a 2log9 log9 6x 1 O J b log x logx 1 log3x 12 J c ln10 ln7 x lnx Solution a 2log9 log9 6x 1 0 With this equation there are only two logarithms in the equation so it s easy to get on one either side of the equal sign We will also need to deal with the coefficient in front of the first term 2 logg L logg 6x 1 log9 x logg 6x 1 Now that we ve got two logarithms with the same base and coefficients of 1 on either side of the equal sign we can drop the logs and solve x 6x 1 15x gt x 1 5 Now we do need to worry if this solution will produce any negative numbers or zeroes in the logarithms so the next step is to plug this into the original equation and see if it does 2007 Paul Dawkins 304 httptutorialmathlamaredutermsaspx College Algebra l lrggl l LE1 1 g9 Note that we don t need to go all the way out with the check here We just need to make sure that once we plug in the x we don t have any negative numbers or zeroes in the logarithms Since we don t in this case we have the solution it is x F Return to Problems b logxlogx 1log3x 12 Okay in this equation we ve got three logarithms and we can only have two So we saw how to do this kind of work in a set of examples in the previous section so we just need to do the same thing here It doesn t really matter how we do this but since one side already has one logarithm on it we might as well combine the logs on the other side log xx 1 log 3x 12 Now we ve got one logarithm on either side of the equal sign they are the same base and have coefficients of one so we can drop the logarithms and solve xx l3x 12 x2 x 3x l20 x2 4x 120 x 6x2O gt c 2x 6 Now before we declare these to be solutions we MUST check them in the original equation x 6 log6log6 l log36 12 log6log5 log3O No logarithms of negative numbers and no logarithms of zero so this is a solution x 2 log 2log 2 llog3 2 42 We don t need to go any farther there is a logarithm of a negative number in the first term the others are also negative and that s all we need in order to exclude this as a solution Be careful here We are not excluding x 2 because it is negative that s not the problem We are excluding it because once we plug it into the original equation we end up with logarithms of negative numbers It is possible to have negative values of x be solutions to these problems so don t mistake the reason for excluding this value 2007 Paul Dawkins 305 httptutorialmathlamaredutermsaspx College Algebra Also along those lines we didn t take x 6 as a solution because it was positive but because it didn t produce any negative numbers or zero in the logarithms upon substitution It is possible for positive numbers to not be solutions So with all that out of the way we ve got a single solution to this equation x 6 Return to Problems c lnlO ln7 x lnx We will work this equation in the same manner that we worked the previous one We ve got two logarithms on one side so we ll combine those drop the logarithms and then solve ln 10 jlnx 7 x 10 x 7 x l0x7 x l07x x2 x2 7xl00 x 5x 20 gt x x 5 We ve got two possible solutions to check here x2 lnlO ln7 2 ln2 lnlO ln5ln2 This one is okay x5 lnlO ln7 5 ln5 lnlO ln2ln5 This one is also okay In this case both possible solutions x 2 and x 5 end up actually being solutions There is no reason to expect to always have to throw one of the two out as a solution Return to Problems Now we need to take a look at the second kind of logarithmic equation that we ll be solving here This equation will have all the terms but one be a logarithm and the one term that doesn t have a logarithm will be a constant In order to solve these kinds of equations we will need to remember the exponential form of the logarithm Here it is if you don t remember ylogbx gt by x 2007 Paul Dawkins 306 httptutorialmathlamaredutermsaspx College Algebra We will be using this conversion to exponential form in all of these equations so it s important that you can do it Let s work some examples so we can see how these kinds of equations can be solved Example 2 Solve each of the following equations a logs 2x 4 2 b log 351 logx 3 J c log x2 6x 3 log2 l x J Solution a logs 2x 4 2 To solve these we need to get the equation into exactly the form that this one is in We need a single log in the equation with a coefficient of one and a constant on the other side of the equal sign Once we have the equation in this form we simply convert to exponential form So let s do that with this equation The exponential form of this equation is 2x 4 52 25 Notice that this is an equation that we can easily solve 21 2x 21 gt x Now just as with the first set of examples we need to plug this back into the original equation and see if it will produce negative numbers or zeroes in the logarithms If it does it can t be a solution and if it doesn t then it is a solution em logs 25 2 2 21 Only positive numbers in the logarithm and so x Y is in fact a solution Return to Problems b logaezl logx 3 In this case we ve got two logarithms in the problem so we are going to have to combine them into a single logarithm as we did in the first set of examples Doing this for this equation gives logxlogx 3 1 logxx 3 1 Now that we ve got the equation into the proper form we convert to exponential form Recall as well that we re dealing with the common logarithm here and so the base is 10 Here is the exponential form of this equation 2007 Paul Dawkins 307 httptutorialmathlamaredutermsaspx College Algebra xx 3l01 x2 3x lOO x 5x2O gt c 2x 5 So we ve got two potential solutions Let s check them both x 2 log 2 l log 2 3 We ve got negative numbers in the logarithms and so this can t be a solution x5 log5l log5 3 log 5 1 log 2 No negative numbers or zeroes in the logarithms and so this is a solution Therefore we have a single solution to this equation x 5 Again remember that we don t exclude a potential solution because it s negative or include a potential solution because it s positive We exclude a potential solution if it produces negative numbers or zeroes in the logarithms upon substituting it into the equation and we include a potential solution if it doesn t Return to Problems c log x2 6x 3 log2 l x Again let s get the logarithms onto one side and combined into a single logarithm log x2 6x log 1 x 3 2 1Og2x 6xj 3 l x Now convert it to exponential form x2 6x 2 23 8 l x Now let s solve this equation x2 6x 8l x x2 6x 8 8x x2 2x 8 0 x4x 2O gt c 4x 2 Now let s check both of these solutions in the original equation x 4 2007 Paul Dawkins 308 httptutorialmathlamaredutermsaspx College Algebra log 42 6 4 3 log2 1 4 4 log 16 24 3 log 5 So upon substituting this solution in we see that all the numbers in the logarithms are positive and so this IS a solution Note again that it doesn t matter that the solution is negative it just can t produce negative numbers or zeroes in the logarithms x 2 log 22 62 3 log 1 2 log 4 12 43 log5 1 In this case despite the fact that the potential solution is positive we get negative numbers in the logarithms and so it can t possibly be a solution Therefore we get a single solution for this equation x 4 Return to Problems 2007 Paul Dawkins 309 httptutorialmathlamaredutermsaspx College Algebra Applications In this final section of this chapter we need to look at some applications of exponential and logarithm functions Compound Interest This first application is compounding interest and there are actually two separate formulas that we ll be looking at here Let s first get those out of the way If we were to put P dollars into an account that earns interest at a rate of r written as a decimal for t years yes it must be years then 1 if interest is compounded m times per year we will have APl 1 m 2 if interest is compounded continuously then we will have A Pe dollars after t years dollars after t years Let s take a look at a couple of examples Example 1 We are going to invest 100000 in an account that earns interest at a rate of 75 for 54 months Determine how much money will be in the account if a interest is compounded quarterly Solution b interest is compounded monthly Solution c interest is compounded continuously Solution Solution Before getting into each part let s identify the quantities that we will need in all the parts and won t change 75 54 P 100000 r 0075 t 45 100 12 Remember that interest rates must be decimals for these computations and t must be in years Now let s work the problems a Interest is compounded quarterly In this part the interest is compounded quarterly and that means it is compounded 4 times a year After 54 months we then have lt4gtlt4sgt A1000O01 L175 100000101875 100000139706686207 139706686207 13970669 Notice the amount of decimal places used here We didn t do any rounding until the Very last 2007 Paul Dawkins 310 httptutorialmathlamaredutermsaspx College Algebra step It is important to not do too much rounding in intermediate steps with these problems Return to Problems b Interest is compounded monthly Here we are compounding monthly and so that means we are compounding 12 times a year Here is how much we ll have after 54 months A 1000001 L5j12l4 5 12 10O0O010062554 1000O0139996843023 139996843023 13999684 So compounding more times per year will yield more money Return to Problems c Interest is compounded continuously Finally if we compound continuously then after 54 months we will have A 100000e 39 754395 100000140143960839 140143960839 14014396 Return to Problems Now as pointed out in the first part of this example it is important to not round too much before the final answer Let s go back and work the first part again and this time let s round to three decimal places at each step 0075 quot 5 T 18 A l00000l l00000l0l9 l00000l403 140 30000 QQ This answer is off from the correct answer by 59331 and that s a fairly large difference So how many decimal places should we keep in these Well unfortunately the answer is that it depends The larger the initial amount the more decimal places we will need to keep around As a general rule of thumb set your calculator to the maximum number of decimal places it can handle and take all of them until the final answer and then round at that point Let s now look at a different kind of example with compounding interest 2007 Paul Dawkins 311 httptutorialmathlamaredutermsaspx College Algebra Example 2 We are going to put 2500 into an account that earns interest at a rate of 12 If we want to have 4000 in the account when we close it how long should we keep the money in the account if a we compound interest continuously Solution b we compound interest 6 times a year Solution Solution Again 1et s identify the quantities that won t change with each part A4000 P2500 r 01 100 Notice that this time we ve been given A and are asking to find t This means that we are going to have to solve an exponential equation to get at the answer a Compound interest continuously Let s first set up the equation that we ll need to solve 4000 25006012 Now we saw how to solve these kinds of equations a couple of sections ago In that section we saw that We need to get the exponential on one side by itself with a coefficient of 1 and then take the natural logarithm of both sides Let s do that 4000 Z 2500 e0l2t ln16 ln eon ln16 ln16012t gt t 3917 We need to keep the amount in the account for 3917 years to get 4000 Return to Problems b Ccompound interest 6 times a year Again 1et s first set up the equation that we need to solve 6t 4000 25001 4000 25001026 We will solve this the same way that we solved the previous part The work will be a little messier but for the most part it will be the same 2007 Paul Dawkins 312 httptutorialmathlamaredutermsaspx College Algebra 1026 2500 16 1026 lnl6 1n1026 lnl6 6r ln 102 lnl6 O470003629246 t 3956 6lnl02 60019802627296 In this case we need to keep the amount slightly longer to reach 4000 Return to Problems Exponential Growth and Decay There are many quantities out there in the world that are governed at least for a short time period by the equation Q Qoekt where Q0 is positive and is the amount initially present at t 0 and k is a nonzero constant If k is positive then the equation will grow without bound and is called the exponential growth equation Likewise if k is negative the equation will die down to zero and is called the exponential decay equation Short term population growth is often modeled by the exponential growth equation and the decay of a radioactive element is governed the exponential decay equation Example 3 The growth of a colony of bacteria is given by the equation Q Q0eo195 1 If there are initially 500 bacteria present and t is given in hours determine each of the following a How many bacteria are there after a half of a day Solution b How long will it take before there are 10000 bacteria in the colony Solution Solution Here is the equation for this starting amount of bacteria Q e0l95 I a How many bacteria are there after a half of a day In this case if we want the number of bacteria after half of a day we will need to use t 12 since t is in hours So to get the answer to this part we just need to plug t into the equation Q 500 e 3919512 500103812365627 5190618 So since a fractional population doesn t make much sense we ll say that after half of a day there are 5190 of the bacteria present Return to Problems b How long will it take before there are 10000 bacteria in the colony Do NOT make the mistake of assuming that it will be approximately 1 day for this answer based on the answer to the previous part With exponential growth things just don t work that way as we ll see In order to answer this part we will need to solve the following exponential equation 10000 500 e 195 2007 Paul Dawkins 313 httptutorialmathlamaredutermsaspx College Algebra Let s do that 10000 eo9 500 20 e0l95t ln 20 1ne 195 ln 20 ln200l95t gt t 153627 0195 So it only takes approximately 154 hours to reach 10000 bacteria and NOT 24 hours if we just double the time from the first part In other words be careful Return to Problems Example 4 Carbon 14 dating works by measuring the amount of Carbon 14 a radioactive element that is in a fossil All living things have a constant level of Carbon 14 in them and once they die it starts to decay according to the formula Q Q e00001241 0 where t is in years and Q0 is the amount of Carbon 14 present at death and for this example let s assume that there will be 100 milligrams present at death a How much Carbon 14 will there be after 1000 years Solution b How long will it take for half of the Carbon 14 to decay Solution Solution a How much Carbon 14 will there be after 1000 years In this case all we need to do is plug in t1000 into the equation Q 100e 4 1 100O883379840883 88338 milligrams So it looks like we will have around 88338 milligrams left after 1000 years Return to Problems b How long will it take for half of the Carbon 14 to decay So we want to know how long it will take until there is 50 milligrams of the Carbon 14 left That means we will have to solve the following equation 50 100e 0000l24t Here is that work 2007 Paul Dawkins 314 httptutorialmathlamaredutermsaspx College Algebra 5i 000o124 100 A e 0ooo124 2 In A In e 0000l24t 2 111 41000124 ln 1 1 2 03969314718056 558989661742 000o124 000o124 So it looks like it will take about 5589897 years for half of the Carbon 14 to decay This number is called the halflife of Carbon 14 Return to Problems We Ve now looked at a couple of applications of exponential equations and we should now look at a quick application of a logarithm Earthquake Intensity The Richter scale is commonly used to measure the intensity of an earthquake There are many different ways of computing this based on a Variety of different quantities We are going to take a quick look at the formula that uses the energy released during an earthquake If E is the energy released measured in joules during an earthquake then the magnitude of the earthquake is given by 2 E M log 3 E0 where E0 l04394 joules Example 5 If 8 X1014 joules of energy is released during an earthquake what was the magnitude of the earthquake Solution There really isn t much to do here other than to plug into the formula 2 8x10 2 2 M 1 E1 8 10 1050308999 7002 3 g1044 J 3 g 3 So it looks like we ll have a magnitude of about 7 2007 Paul Dawkins 315 httptutorialmathlamaredutermsaspx College Algebra Example 6 How much energy will be released in an earthquake with a magnitude of 59 Solution In this case we will need to solve the following equation 2 E 59 1 3 g1o44 We saw how solve these kinds of equations in the previous section First we need the logarithm on one side by itself with a coefficient of one Once we have it in that form we convert to exponential form and solve E 2 10885 1044 2 101325 So it looks like there would be a release of 101325 joules of energy in an earthquake with a magnitude of 59 2007 Paul Dawkins 316 httptutorialmathlamaredutermsaspx College Algebra Systems of Equations Introduction This is a fairly short chapter devoted to solving systems of equations A system of equations is a set of equations each containing one or more variable We will focus exclusively on systems of two equations with two unknowns and three equations with three unknowns although the methods looked at here can be easily extended to more equations Also with the exception of the last section we will be dealing only with systems of linear equations Here is a list of the topics in this section Linear Systems with Two Variables In this section we will use systems of two equations and two variables to introduce two of the main methods for solving systems of equations Linear Systems with Three Variables Here we will work a quick example to show how to use the methods to solve systems of three equations with three variables Augmented Matrices We will look at the third main method for solving systems in this section We will look at systems of two equations and systems of three equations More on the Augmented Matrix In this section we will take a look at some special cases to the solutions to systems and how to identify them using the augmented matrix method Nonlinear Systems We will take a quick look at solving nonlinear systems of equations in this section 2007 Paul Dawkins 317 httptutorialmathlamaredutermsaspx College Algebra Linear Systems with Two Variables A linear system of two equations with two Variables is any system that can be written in the form ax by p we dy q where any of the constants can be zero with the exception that each equation must have at least one Variable in it Also the system is called linear if the Variables are only to the first power are only in the numerator and there are no products of Variables in any of the equations Here is an example of a system with numbers 3x y 7 2x3y1 Before we discuss how to solve systems we should first talk about just what a solution to a system of equations is A solution to a system of equations is a Value of x and a Value of y that when substituted into the equations satisfies both equations at the same time For the example above x 2 and y 1 is a solution to the system This is easy enough to check slt2gt1gt 223 1 7 1 So sure enough that pair of numbers is a solution to the system Do not worry about how we got these Values This will be the Very first system that we solve when we get into examples Note that it is important that the pair of numbers satisfy both equations For instance x 1 and y 4 will satisfy the first equation but not the second and so isn t a solution to the system Likewise x 1 and y 1 will satisfy the second equation but not the first and so can t be a solution to the system Now just what does a solution to a system of two equations represent Well if you think about it both of the equations in the system are lines So let s graph them and see what we get 2007 Paul Dawkins 318 httptutorialmathlamaredutermsaspx College Algebra 1 2 3 1 As you can see the solution to the system is the coordinates of the point where the two lines intersect So when solving linear systems with two variables we are really asking where the two lines will intersect We will be looking at two methods for solving systems in this section The first method is called the method of substitution In this method we will solve one of the equations for one of the variables and substitute this into the other equation This will yield one equation with one variable that we can solve Once this is solved we substitute this value back into one of the equations to find the value of the remaining variable In words this method is not always very clear Let s work a couple of examples to see how this method works Example 1 Solve each of the following systems 3x y7 a Solution 2x3yl 5x4yl b Solution 3x 6y2 Solution 3x y7 a 2x3yl So this was the first system that we looked at above We already know the solution but this will give us a chance to verify the values that we wrote down for the solution Now the method says that we need to solve one of the equations for one of the variables Which equation we choose and which variable that we choose is up to you but it s usually best to pick an equation and variable that will be easy to deal with This means we should try to avoid fractions if at all possible 2007 Paul Dawkins 319 httptutorialmathlamaredutermsaspx College Algebra In this case it looks like it will be really easy to solve the first equation for y so let s do that 3x 7 y Now substitute this into the second equation 2x33x 7 1 This is an equation in x that we can solve so let s do that 2x9x 2ll llx22 x2 So there is the x portion of the solution Finally do NOT forget to go back and find the y portion of the solution This is one of the more common mistakes students make in solving systems To so this we can either plug the x value into one of the original equations and solve for y or we can just plug it into our substitution that we found in the first step That will be easier so let s do that y3ae 1 39 7 1 So the solution is x 2 and y l as we noted above Return to Problems 5x4yl 3x 6y2 With this system we aren t going to be able to completely avoid fractions However it looks like if we solve the second equation for x we can minimize them Here is that work 3x6y 2 2 X22 Now substitute this into the first equation and solve the resulting equation for y 52yj4y 1 1Oy104y 1 10 7 14yE13 E 7 1 rlglal 1 y6 2007 Paul Dawkins 320 httptutorialmathlamaredutermsaspx College Algebra Finally substitute this into the original substitution to find x l l 2 6 3 3 3 1 So the solution to this system is x 5 and y 0 Lp 1 6 Return to Problems As with single equations we could always go back and check this solution by plugging it into both equations and making sure that it does satisfy both equations Note as well that we really would need to plug into both equations It is quite possible that a mistake could result in a pair of numbers that would satisfy one of the equations but not the other one Let s now move into the next method for solving systems of equations As we saw in the last part of the previous example the method of substitution will often force us to deal with fractions which adds to the likelihood of mistakes This second method will not have this problem Well that s not completely true If fractions are going to show up they will only show up in the final step and they will only show up if the solution contains fractions This second method is called the method of elimination In this method we multiply one or both of the equations by appropriate numbers ie multiply every term in the equation by the number so that one of the variables will have the same coefficient with opposite signs Then next step is to add the two equations together Because one of the variables had the same coefficient with opposite signs it will be eliminated when we add the two equations The result will be a single equation that we can solve for one of the variables Once this is done substitute this answer back into one of the original equations As with the first method it s much easier to see what s going on here with a couple of examples Example 2 Solve each of the following systems of equations 5x4yl 3 3 6 2 J x y 2x4y 10 6x3y 6 ml Solution 5x4yl a 3x 6y 2 This is the system in the previous set of examples that made us work with fractions Working it here will show the differences between the two methods and it will also show that either method can be used to get the solution to a system So we need to multiply one or both equations by constants so that one of the variables has the same coefficient with opposite signs So since the y terms already have opposite signs let s work with these terms It looks like if we multiply the first equation by 3 and the second equation by 2 2007 Paul Dawkins 321 httptutorialmathlamaredutermsaspx College Algebra the y terms will have coefficients of 12 and 12 which is what we need for this method Here is the work for this step 5x4yl gt3 l5xl2y3 3x 6y2 2 6x l2y4 21x 7 So as the description of the method promised we have an equation that can be solved for x Doing this gives x E which is exactly what we found in the previous example Notice however that the only fraction that we had to deal with to this point is the answer itself which is different from the method of substitution Now again don t forget to find y In this case it will be a little more work than the method of substitution To find y we need to substitute the value of x into either of the original equations and solve for y Since x is a fraction let s notice that in this case if we plug this value into the second equation we will lose the fractions at least temporarily Note that often this won t happen and we ll be forced to deal with fractions whether we want to or not 1 3 6 2 3 y l 6y2 6yl 1 6 Again this is the same value we found in the previous example Return to Problems 2x 4 y 10 6x 3 y 6 In this part all the variables are positive so we re going to have to force an opposite sign by multiplying by a negative number somewhere Let s also notice that in this case if we just multiply the first equation by 3 then the coefficients of the x will be 6 and 6 Sometimes we only need to multiply one of the equations and can leave the other one alone Here is this work for this part 2x4y lO gtlt 3 6x l2y30 6x3y6 Lne 6x3y6 9y36 y 4 Finally plug this into either of the equations and solve for x We will use the first equation this 2007 Paul Dawkins 322 httptutorialmathlamaredutermsaspx College Algebra time 2x4 4E 10 2x l6 10 2x6 x3 So the solution to this system is x 3 and y 4 Return to Problems There is a third method that we ll be looking at to solve systems of two equations but it s a little more complicated and is probably more useful for systems with at least three equations so we ll look at it in a later section Before leaving this section we should address a couple of special case in solving systems Example 3 Solve the following systems of equations x y6 2x 2y 1 Solution We can use either method here but it looks like substitution would probably be slightly easier We ll solve the first equation for x and substitute that into the second equation x6 yr 26y2yl l2 2y2yl 1217 So this is clearly not true and there doesn t appear to be a mistake anywhere in our work So what s the problem To see let s graph these two lines and see what we get 3 25 3 2 1 II 2 3 d1 I 253 2392u39l 503 x y5 It appears that these two lines are parallel can you Verify that with the slopes and we know that 2007 Paul Dawkins 323 httptutorialmathlamaredutermsaspx College Algebra two parallel lines with different yintercepts that s important will never cross As we saw in the opening discussion of this section solutions represent the point where two lines intersect If two lines don t intersect we can t have a solution So when we get this kind of nonsensical answer from our work we have two parallel lines and there is no solution to this system of equations The system in the previous example is called inconsistent Note as well that if we d used elimination on this system we would have ended up with a similar nonsensical answer Example 4 Solve the following system of equations 2x5y l lOx 25y 5 Solution In this example it looks like elimination would be the easiest method 2x5y l x 5 lOx25y 5 lOx 25y5 same l0x 25y5 O 0 On first glance this might appear to be the same problem as the previous example However in that case we ended up with an equality that simply wasn t true In this case we have 00 and that is a true equality and so in that sense there is nothing wrong with this However this is clearly not what we were expecting for an answer here and so we need to determine just what is going on We ll leave it to you to verify this but if you find the slope and yintercepts for these two lines you will find that both lines have exactly the same slope and both lines have exactly the same y intercept So what does this mean for us Well if two lines have the same slope and the same y intercept then the graphs of the two lines are the same graph In other words the graphs of these two lines are the same graph In these cases any set of points that satisfies one of the equations will also satisfy the other equation Also recall that the graph of an equation is nothing more than the set of all points that satisfies the equation In other words there is an infinite set of points that will satisfy this set of equations In these cases we do want to write down something for a solution So what we ll do is solve one of the equations for one of the variables it doesn t matter which you choose We ll solve the first for y 2x5y l 5y 266 l 2 1 V 3 5 Then given any x we can find a y and these two numbers will form a solution to the system of equations We usually denote this by writing the solution as follows 2007 Paul Dawkins 324 httptutorialmathlamaredutermsaspx College Algebra 1 wheret is any real number 5 So show that these give solutions let s work through a couple of values of t t0 1 xO y 5 To show that this is a solution we need to plug it into both equations in the system 2o5 1 100 g 5 5 1 1 5 5 So x 0 and y is a solution to the system Let s do another one real quick t3 Ullllox 1 1 1 5 5 Again we need to plug it into both equations in the system to show that it s a solution 2 351 1 10 3 251 5 1 1 Sure enough x 3 and y 1 is a solution 55 xt So since there are an infinite number of possible t s there must be an infinite number of solutions to this system and they are given by 1 wheret is any real number 5 Systems such as those in the previous examples are called dependent We ve now seen all three possibilities for the solution to a system of equations A system of equation will have either no solution exactly one solution or infinitely many solutions 2007 Paul Dawkins 325 httptutorialmathlamaredutermsaspx College Algebra Linear Systems with Three Variables This is going to be a fairly short section in the sense that it s really only going to consist of a couple of examples to illustrate how to take the methods from the previous section and use them to solve a linear system with three equations and three Variables So let s get started with an example Example 1 Solve the following system of equations x 2y 3z 7 2xyz4 3x 2y 2z 10 Solution We are going to try and find Values of x y and a z that will satisfy all three equations at the same time We are going to use elimination to eliminate one of the Variables from one of the equations and two of the Variables from another of the equations The reason for doing this will be apparent once we Ve actually done it The elimination method in this case will work a little differently than with two equations As with two equations we will multiply as many equations as we need to so that if we start adding pairs of equations we can eliminate one of the Variables In this case it looks like if we multiply the second equation by 2 it will be fairly simple to eliminate the y term from the second and third equation by adding the first equation to both of them So let s first multiply the second equation by two x 2y3z7 same x 2y3z7 2xyz4 2 X 4x2y2z8 3x2y 2z 10 same 3x2y 2z 10 Now with this new system we will replace the second equation with the sum of the first and second equations and we will replace the third equation with the sum of the first and third equations Here is the resulting system of equations x 2y3Z7 5x 5zl5 2x z 3 So we Ve eliminated one of the Variables from two of the equations We now need to eliminate either x or z from either the second or third equations Again we will use elimination to do this In this case we will multiply the third equation by 5 since this will allow us to eliminate z from this equation by adding the second onto is 2007 Paul Dawkins 326 httptutorialmathlamaredutermsaspx College Algebra x 2y3z7 same x 2y3z7 5x 5z 15 same 5x 5z 15 2x z 3 x 5 10x 5z15 Now replace the third equation with the sum of the second and third equation x 2y3Z7 5x 5z15 15x 30 Now at this point notice that the third equation can be quickly solved to find that x 2 Once we know this we can plug this into the second equation and that will give us an equation that we can solve for z as follows 525z15 105z15 5z5 z1 Finally we can substitute both x and z into the first equation which we can use to solve for y Here is that work 2 2y37 2y57 2y2 y 1 So the solution to this system is x 2 y 1 and z 1 That was a fair amount of work and in this case there was even less work than normal because in each case we only had to multiply a single equation to allow us to eliminate variables In the next section we ll be looking at a third method for solving systems that is basically a shorthand method for what we did in the previous example The work using that method will be messy as well but it will be slightly easier to do once you get the hang of it In the previous example all we did was use the method of elimination until we could start solving for the variables and then just back substitute known values of variables into previous equations to find the remaining unknown variables Not every linear system with three equations and three variables uses the elimination method exclusively so let s take a look at another example where the substitution method is used at least partially I Example 2 Solve the following system of equations 2007 Paul Dawkins 327 httptutorialmathlamaredutermsaspx College Algebra 2x 4y5 33 4x y E 5 2x2y 3z 19 Solution Before we get started on the solution process do not get excited about the fact that the second equation only has two variables in it That is a fairly common occurrence when we have more than two equations in the system In fact we re going to take advantage of the fact that it only has two variables and one of them the y has a coefficient of 1 This equation is easily solved for y to get y 4x 5 We can then substitute this into the first and third equation as follows 2x 44x55z 33 2x24x5 3zl9 Now if you think about it this is just a system of two linear equations with two variables x and z and we know how to solve these kinds of systems from our work in the previous section First we ll need to do a little simplification of the system 2x 16x 205z 33 l4x5Z 13 gt 2x8xlO 3zl9 6x 3z9 The simplified version looks just like the systems we were solving in the previous section Well it s almost the same The variables this time are x and z instead of x and y but that really isn t a difference The work of solving this will be the same We can use either the method of substitution or the method of elimination to solve this new system of two linear equations If we wanted to use the method of substitution we could easily solve the second equation for z you do see why it would be easiest to solve the second equation for z right and substitute that into the first equation This would allow us to find x and we could then find both z and y However to make the point that often we use both methods in solving systems of three linear equations let s use the method of elimination to solve the system of two equations We ll just need to multiply the first equation by 3 and the second by 5 Doing this gives l4x5z 13 x 3 4x l 5z 39 6x 3z 9 5 x 30x l5Z 45 l2x 6 2007 Paul Dawkins 328 httptutorialmathlamaredutermsaspx College Algebra 1 We can now easily solve for x to get x 2 The coefficients on the second equation are smaller so let s plug this into that equation and solve for z Here is that work 6 l 3z 9 2 3 3z9 3zl2 z 4 Finally we need to determine the value of y This is very easy to do Recall in the first step we used substitution and in that step we used the following equation y4x 5 Since we know the value of x all we need to do is plug that into this equation and get the value of y 4 1 5 3 y 2 Note that in many cases where we used substitution on the very first step the equation you ll have at this step will contain both x s and z s and so you will need both values to get the third variable 1 Okay to finish this example up here is the solution x 2 y 3 and z 4 As we ve seen with the two examples above there are a variety of paths that we could choose to take when solving a system of three linear equations with three variables That will always be the case There is no one true path for solving these However having said that there is often a path that will allow you to avoid some of the mess that can arise in solving these types of systems Once you work enough of these types of problems you ll start to get a feel for a good path through the solution process that will hopefully avoid some of the mess Interpretation of solutions in these cases is a little harder in some senses All three of these equations in the examples above are equations of planes in three dimensional space and solution to this systems in the examples above is the one point that all three of the planes have in common Note as well that it is completely possible to have no solutions to these systems or infinitely many systems as we saw in the previous section with systems of two equations We will look at these cases once we have the next section out of the way 2007 Paul Dawkins 329 httptutorialmathlamaredutermsaspx College Algebra Augmented Matrices In this section we need to take a look at the third method for solving systems of equations For systems of two equations it is probably a little more complicated than the methods we looked at in the first section However for systems with more equations it is probably easier than using the method we saw in the previous section Before we get into the method we first need to get some definitions out of the way An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation both the coefficients and the constant on the other side of the equal sign and each column represents all the coefficients for a single variable Let s take a look at an example Here is the system of equations that we looked at in the previous section x 2y3Z7 2xyz4 3x2y2z 10 Here is the augmented matrix for this system 12 35 7 2 1 1i 4 3 2 2i 10 The first row consists of all the constants from the first equation with the coefficient of the x in the first column the coefficient of the y in the second column the coefficient of the z in the third column and the constant in the final column The second row is the constants from the second equation with the same placement and likewise for the third row The dashed line represents where the equal sign was in the original system of equations and is not always included This is mostly dependent on the instructor andor textbook being used Next we need to discuss elementary row operations There are three of them and we will give both the notation used for each one as well as an example using the augmented matrix given above 1 Interchange Two Rows With this operation we will interchange all the entries in row i and row j The notation we ll use here is Rt lt gt Rj Here is an example 1 2 3 E 7 3 2 2 E 10 i R1lt gt R3 2 1 1 4 2 1 1 5 4 i gt i 3 2 2 10 1 2 3 i 7 So we do exactly what the operation says Every entry in the third row moves up to the first row and every entry in the first row moves down to the third row Make sure that you move all the entries One of the more common mistakes is to forget to move one or more entries 2007 Paul Dawkins 330 httptutorialmathlamaredutermsaspx College Algebra Once we have the augmented matrix in this form we are done The solution to the system will be x h and y k This method is called GaussJordan Elimination Example 1 Solve each of the following systems of equations 3x 2y14 a x3y1 E 2xyE 3 b x4y 2 J 3x 6y 9 C2x2y12 J Solution 3x 2y14 a x3yl The first step here is to write down the augmented matrix for this system 3 214 1 331 To convert it into the final form we will start in the upper left corner and the work in a counter clockwise direction until the first two columns appear as they should be So the first step is to make the red three in the augmented matrix above into a 1 We can use any of the row operations that we d like to We should always try to minimize the work as much as possible however So since there is a one in the first column already it just isn t in the correct row let s use the first row operation and interchange the two rows 3 214 R1lt gtR21 31 1 351 gt 3 2514 The next step is to get a zero below the 1 that we just got in the upper left hand corner This means that we need to change the red three into a zero This will almost always require us to use third row operation If we add 3 times row 1 onto row 2 we can convert that 3 into a 0 Here is that operation 1 3i1R2 3R1 gtR21 31 3 214 gt 0 11511 Next we need to get a 1 into the lower right corner of the first two columns This means changing the red 11 into a 1 This is usually accomplished with the second row operation If we divide the second row by 11 we will get the 1 in that spot that we need 2007 Paul Dawkins 332 httptutorialmathlamaredutermsaspx College Algebra 1 3i1 1 11R2l3i 1 0 11211 0121 Okay we re almost done The final step is to turn the red three into a zero Again this almost always requires the third row operation Here is the operation for this final step 13 1R1 3R2 gtR1 1 0 4 0131 gt 0151 We have the augmented matrix in the required form and so we re done The solution to this systemis x4 and y 1 Return to Problems 2xyE 3 b x 4y 2 In this part we won t put in as much explanation for each step We will mark the next number that we need to change in red as we did in the previous part We ll first write down the augmented matrix and then get started with the row operations 2 153 R1lt gtR2 l 4 2 R22R1 gtR2 l 4 2 1432 2 153 0737 Before proceeding with the next step let s notice that in the second matrix we had one s in both spots that we needed them However the only way to change the 2 into a zero that we had to have as well was to also change the l in the lower right corner as well This is okay Sometimes it will happen and trying to keep both ones will only cause problems Let s finish the problem 142 R2 1 42 R14R2 gtR1 1 02 0 7 7 0 131 gt 0121 The solution to this system is then x 2 and y 1 Return to Problems 3x 6y 9 c 2x 2 y 12 Let s first write down the augmented matrix for this system 3 6 9 2 2512 Now in this case there isn t a l in the first column and so we can t just interchange two rows as the first step However notice that since all the entries in the first row have 3 as a factor we can divide the first row by 3 which will get a l in that spot and we won t put any fractions into the problem 2007 Paul Dawkins 333 httptutorialmathlamaredutermsaspx College Algebra Here is the work for this system 3 69R1 1 23R22R1 gtR1 23 2 2212 2 2 12 gt 1 23R2 1 2 3 R12R2 gtR1 1 055 0656 0 151 gt 011 The solution to this system is x 5 and y 1 Return to Problems It is important to note that the path we took to get the augmented matrices in this example into the final form is not the only path that we could have used There are many different paths that we could have gone down All the paths would have arrived at the same final augmented matrix however so we should always choose the path that we feel is the easiest path Note as well that different people may well feel that different paths are easier and so may well solve the systems differently They will get the same solution however For two equations and two unknowns this process is probably a little more complicated than just the straight forward solution process we used in the first section of this chapter This process does start becoming useful when we start looking at larger systems So let s take a look at a couple of systems with three equations in them In this case the process is basically identical except that there s going to be more to do As with two equations we will first set up the augmented matrix and then use row operations to put it into the form Once the augmented matrix is in this form the solution is x p y q and z r As with the two equations case there really isn t any set path to take in getting the augmented matrix into this form The usual path is to get the l s in the correct places and 0 s below them Once this is done we then try to get zeroes above the l s Let s work a couple of examples to see how this works 2007 Paul Dawkins 334 httptutorialmathlamaredutermsaspx College Algebra Example 2 Solve each of the following systems of equations 3xy 2z2 a x2yz3 1 2x y 3z3 3xy 2zE 7 b2x2yz9 J x y3Z6 Solution 3xy 2z2 a x 2yz3 2x y 3z3 Let s first write down the augmented matrix for this system 3 l 252 1 2 153 2 1 3E3 As with the previous examples we will mark the numbers that we want to change in a given step in red The first step here is to get a l in the upper left hand corner and again we have many ways to do this In this case we ll notice that if we interchange the first and second row we can get a l in that spot with relatively little work 3 12 2 12 1E3 R1lt gtR2 12 153 3 122 21 33 21 33 The next step is to get the two numbers below this 1 to be 0 s Note as well that this will almost always require the third row operation to do Also we can do both of these in one step as follows 12 1j3R2 3R1 gtR2 1 2 1 3 3 1 2 2 R3 2R1 gtR3 0 7 57 21 33 gt 0 353 Next we want to turn the 7 into a 1 We can do this by dividing the second row by 7 141311213 i R O 0 7537720 151 0 3 5l 3 0 3 553 So we got a fraction showing up here That will happen on occasion so don t get all that excited about it The next step is to change the 3 below this new 1 into a 0 Note that we aren t going to bother with the 2 above it quite yet Sometimes it is just as easy to turn this into a 0 in the same step In this case however it s probably just as easy to do it later as we ll see 2007 Paul Dawkins 335 httptutorialmathlamaredutermsaspx College Algebra So using the third row operation we get 12 13 12 13 5 R 3R R 5 0 15513 gt30 1 1 7 gt 7 0 3 5 3 O OEO Next we need to get the number in the bottom right comer into a 1 We can do that with the second row operation 12 1 3 7 12 1 3 E R E 0 1 1 2030 1 1 7 71 205 0 0 150 0 0750 39 Now we need zeroes above this new 1 So using the third row operation twice as follows will do what we need done 5 1 2 1 3R27R3 gtR2 1 2 O 3 0 1 1 R1 R3 gtR1 0 10 1 0 0 15 0 quot 0 0110 Notice that in this case the final column didn t change in this step That was only because the final entry in that column was zero In general this won t happen The final step is then to make the 2 above the 1 in the second column into a zero This can easily be done with the third row operation 1 2 Of 3 10 OE1 R12R2 gtR1 0 1051 0 1051 gt O O 1 O O O 1 0 So we have the augmented matrix in the final form and the solution will be x l y 1 z 0 This can be Verified by plugging these into all three equations and making sure that they are all satisfied Return to Problems 2007 Paul Dawkins 336 httptutorialmathlamaredutermsaspx College Algebra 3x y 2z 2 7 b 2x2yz9 x y 3Z 6 Again the first step is to write down the augmented matrix 3 1 2 7 2 2 1 l 9 1 1 3 PKS 6 We can t get a 1 in the upper left corner simply by interchanging rows this time We could interchange the first and last row but that would also require another operation to turn the 1 into a 1 While this isn t difficult it s two operations Note that we could use the third row operation to get a 1 in that spot as follows 3 1 2 E7 1 1 3E 16 RfR1 2 2 159 2 2 15 9 2 1 1 3 6 1 1 3 6 Now we can use the third row operation to turn the two red numbers into zeroes 1 1 3516 R22R1 gtR2 1 1 3516 2 2 1E 9R3R1 gtR3 0 4 7E 41 11 31 6 gt 02 oi10 The next step is to get a 1 in the spot occupied by the red 4 We could do that by dividing the whole row by 4 but that would put in a couple of somewhat unpleasant fractions So instead of doing that we are going to interchange the second and third row The reason for this will be apparent soon enough 1 1 3 E16 1 1 3 E16 Ig egtEg O 4 7 41 O 2 0 110 i 2 i O 2 O i 10 0 4 7 41 Now if we divide the second row by 2 we get the 1 in that spot that we want 1 1 3E 16 1 1 1 3i 16 5 41 I R I 02 010 22010 047i41 gt04739 Before moving onto the next step let s think notice a couple of things here First we managed to avoid fractions which is always a good thing and second this row is now done We would have eventually needed a zero in that third spot and we ve got it there for free Not only that but it won t change in any of the later operations This doesn t always happen but if it does that will make our life easier Now let s use the third row operation to change the red 4 into a zero 2007 Paul Dawkins 337 httptutorialmathlamaredutermsaspx College Algebra 11 3E 16R 4R R 11 3E 16 i i 0 1 0 5 3 2 3 0 5 i gt i 0 4 7 41 0 0 7 21 We now can divide the third row by 7 to get that the number in the lower right corner into a one 1 1 3 P 16 1 1 1 3 0 16 O 1 0 5 0 0 7 3 R3 57 010 21gt001 Next we can use the third row operation to get the 3 changed into a zero 1 1 3E 16 1 1 Of 7 R13R3 gtR O 1 05 5 O 1 05 5 i 9 i O O 1 3 O O 1 3 The final step is to then make the 1 into a 0 using the third row operation again 1 1 O E 7 1 0 0 E 2 39 R1 R2 gt R g 0 1 O 0 1 0 g 5 1 i 3 O The solution to this system is then ac 27 5z 3 Return to Problems Using GaussJordan elimination to solve a system of three equations can be a lot of work but it is often no more work than solving directly and is many cases less work If we were to do a system of four equations which we aren t going to do at that point GaussJordan elimination would be less work in all likelihood that if we solved directly Also as we saw in the final example worked in this section there really is no one set path to take through these problems Each system is different and may require a different path and set of operations to make Also the path that one person finds to be the easiest may not by the path that another person finds to be the easiest Regardless of the path however the final answer will be the same 2007 Paul Dawkins 338 httptutorialmathlamaredutermsaspx College Algebra More on the Augmented Matrix In the first section in this chapter we saw that there were some special cases in the solution to systems of two equations We saw that there didn t have to be a solution at all and that we could in fact have infinitely many solutions In this section we are going to generalize this out to general systems of equations and we re going to look at how to deal with these cases when using augmented matrices to solve a system Let s first give the following fact Fact Given any system of equations there are exactly three possibilities for the solution 1 There will not be a solution 2 There will be exactly one solution 3 There will be infinitely many solutions This is exactly what we found the possibilities to be when we were looking at two equations It just turns out that it doesn t matter how many equations we ve got There are still only these three possibilities Now let s see how we can identify the first and last possibility when we are using the augmented matrix method for solving In the previous section we stated that we wanted to use the row operations to convert the augmented matrix into the following form I 10 Oip 10h 5 or O l Ogq 0 15k 39 0 0 lir depending upon the number of equations present in the system It turns out that we should have added the qualifier if possible to this instruction because it isn t always possible to do this In fact if it isn t possible to put it into one of these forms then we will know that we are in either the first or last possibility for the solution to the system Before getting into some examples let s first address how we knew what the solution was based on these forms of the augmented matrix Let s work with the two equation case 10 0 13k is an augmented matrix we can always convert back to equations Each row represents an equation and the first column is the coefficient of x in the equation while the second column is the coefficient of the y in the equation The final column is the constant that will be on the right side of the equation Since So if we do that for this case we get lxOyh xgth 0xlyk yk and this is exactly what we said the solution was in the previous section 2007 Paul Dawkins 339 httptutorialmathlamaredutermsaspx College Algebra This idea of tuming an augmented matrix back into equations will be important in the following examples Speaking of which let s go ahead and work a couple of examples We will start out with the two systems of equations that we looked at in the first section that gave the special cases of the solutions Example 1 Use augmented matrices to solve each of the following systems x y6 a Solution 2x 2 y l 2 5 1 b 1Oxx5L 5 1 Solution x y6 a 2x 2 y 1 Now we ve already worked this one out so we know that there is no solution to this system Knowing that let s see what the augmented matrix method gives us when we try to use it 1156 2 251 Notice that we ve already got a 1 in the upper left corner so we don t need to do anything with that So we next need to make the 2 into a 0 1156 R22R1 gtR2 1 156 2 251 gt 0 0513 We ll start with the augmented matrix Now the next step should be to get a 1 in the lower right corner but there is no way to do that without changing the zero in the lower left corner That s a problem because we must have a zero in that spot as well as a one in the lower right corner What this tells us is that it isn t possible to put this augmented matrix form Now go back to equations and see what we ve got in this case x y6 0 13 The first row just converts back into the first equation The second row however converts back to nonsense We know this isn t true so that means that there is no solution Remember if we reach a point where we have an equation that just doesn t make sense we have no solution 1156 0 150 we would have been okay since the last row would return the equation y 0 so don t get confused between this case and what we actually got for this system Note that if we d gotten Return to Problems 2007 Paul Dawkins 340 httptutorialmathlamaredutermsaspx College Algebra 2x5y 1 b 10x 25y 5 In this case we know from the first section that there are infinitely many solutions to this system Let s see what we get when we use the augmented matrix method for the solution Here is the augmented matrix for this system 2 551 10 255 5 In this case we ll need to first get a 1 in the upper left corner and there isn t going to be any easy way to do this that will avoid fractions so we ll just divide the first row by 2 2 51R 1 10 25 5 10 255 5 Now we can get a zero in the lower left corner 1 551 R21OR1 gtR2 1 51 2 2 2 2 10 25 5 quot 0 0 0 Now as with the first part we are never going to be able to get a 1 in place of the red zero without changing the first zero in that row However this isn t the nonsense that the first part got Let s convert back to equations x E 3 2 2 O 0 That last equation is a true equation and so there isn t anything wrong with this In this case we have infinitely many solutions Recall that we still need to do a little work to get the solution We solve one of the equations for one of the variables Note however that if we use the equation from the augmented matrix this is very easy to do 2 2 1 2 2 We then write the solution as x it 1 2 2 where t is any real number y t We get solutions by picking t and plugging this into the equation for x Note that this is NOT the same set of equations we got in the first section That is okay When there are infinitely many solutions there are more than one way to write the equations that will describe all the solutions Return to Problems 2007 Paul Dawkins 341 httptutorialmathlamaredutermsaspx College Algebra Example 3 Solve the following system of equations using augmented matrices 3x 3y 6 3 2x 2y 4amp 2 2x3yz7 Solution Notice that this system is almost identical to the system in the previous example The only difference is the number to the right of the equal sign in the second equation In this system it is 2 and in the previous example it was 10 Changing that one number completely changes the type of solution that we re going to get Often this kind of simple change won t affect the type of solution that we get but in some rare cases it can Since the first two steps of the process are identical to the previous part we won t discuss them Here they are 3 3 6 31 1 1 239 1R2 2R1 gtR2 1 1 211 E R E E 2 2 4 2 31 2 2 4 2 R32R1 gtR3 O O O 0 I I 5 2317 gt 2317 gt 01 We ve got a row of all zeroes so we instantly know that we ve got infinitely many solutions Unlike the two equation case we aren t going to stop however It looks like with a couple of row operations we can make the second column look like it is supposed to in the final form so let s do that 1 1 2 DV 1 1 1 2 j 1 1 O 5 U R2 gtR3 R1R2 gtR1 O O 05 O O 1 3 5 O 1 35 gt gt 0 1 3 5 0 O 0 0 O O 0amp0 In this case we were able to make the second column look like it s supposed to and the third column will never look correct However it is possible that the situation could be reversed and it would be the third column that we can make look correct and the second wouldn t look correct Every system is different Once we reach this point we go back to equations x 5z4 y 3z5 Now both of these equations contain a z and so we ll move that to the other side in each equation x 5 z 4 y3z 5 This means that we get to pick the value of z for free and we ll write the solution as 2007 Paul Dawkins 343 httptutorialmathlamaredutermsaspx College Algebra x5t 4 y3t 5 Zt Where t is any real number Since there are an infinite number of ways to choose t there are an infinite number of solutions to this system 2007 Paul Dawkins 344 httptutorialmathlamaredutermsaspx College Algebra NonLinear Systems In this section we are going to be looking at nonlinear systems of equations A nonlinear system of equations is a system in which at least one of the variables has an exponent other than 1 andor there is a product of variables in one of the equations To solve these systems we will use either the substitution method or elimination method that we first looked at when we solved systems of linear equations The main difference is that we may end up getting complex solutions in addition to real solutions Just as we saw in solving systems of two equations the real solutions will represent the coordinates of the points where the graphs of the two functions intersect Let s work some examples Example 1 Solve the following system of equations x2y2lO 2x y 1 Solution In linear systems we had the choice of using either method on any given system With nonlinear systems that will not always be the case In the first equation both of the variables are squared and in the second equation both of the variables are to the first power In other words there is no way that we can use elimination here and so we are must use substitution Luckily that isn t too bad to do for this system since we can easily solve the second equation for y and substitute this into the first equation y 1 2x x2 l 2x2 10 This is a quadratic equation that we can solve x2 l 4x4x2 10 5x2 4x 9 O 9 xl5x 9O gt 4 lx g So we have two values of x Now we need to determine the values of y and we are going to have to be careful to not make a common mistake here We determine the values of y by plugging x into our substitution x2 1 gt 3 1 21 3 4 Now we only have two solutions here Do not just start mixing and matching all possible values of x and y into solutions We get y 3 as a solution ONLY if x l and so the first solution is aely3 2007 Paul Dawkins 345 httptutorialmathlamaredutermsaspx College Algebra 13 9 Likewise we only get y 5 ONLY if x g and so the second solution is 9 13 x y T 5 5 So we have two solutions Now as noted at the start of this section these two solutions will represent the points of intersection of these two curves Since the first equation is a circle and the second equation is a line have two intersection points is definitely possible Here is a sketch of the two equations as a verification of this 2xy1 Note that when the two equations are a line and a circle as in the previous example we know that we will have at most two real solutions since it is only possible for a line to intersect a circle zero one or two times Example 2 Solve the following system of equations x2 2 y2 2 xy 2 Solution Okay in this case we have a hyperbola the first equation although it isn t in standard form and a rational function the second equation if we solved for y As with the first example we can t use elimination on this system so we will have to use substitution The best way is to solve the second equation for either x or y Either one will give us pretty much the same work so we ll solve for y since that is probably the one that will make the equation look more like those that we ve looked at in the past In other words the new equation will be in terms of x and that is the variable that we are used to seeing in equations 2 y X 2007 Paul Dawkins 346 httptutorialmathlamaredutermsaspx College Algebra 4 x2 2 22 X 8 2 2 6 x2 The first step towards solving this equation will be to multiply the whole thing by x2 to clear out the denominators x4 8 2x2 x4 2x2 8 0 Now this is quadratic in form and we know how to solve those kinds of equations If we define u x2gt uz x2 2 x4 and the equation can be written as uz 2u 8 O u 4u20 gt u Q u 4 In terms of x this means that we have the following x2 4 gt i x 2 x2 2gt i x E i So we have four possible Values of x and two of them are complex To determine the Values of y we can plug these into our substitution 2 2 1 X 3 y 2 2 2 2 1 ac y 2 8 y x2 xiii x2 5 2 2i 2i 2i Z 3 D 3D WE For the complex solutions notice that we made sure the i was in the numerator The for solutions are then x2y 1 and ye 2 y 1 and 2139 2139 x 2139 y and x 95139 y E 2007 Paul Dawkins 347 httptutorialmathlamaredutermsaspx College Algebra Two of the solutions are real and so represent intersection points of the graphs of these two equations The other two are complex solutions and while solutions will not represent intersection points of the curves For reference purposes here is a sketch of the two curves Note that there are only two intersection points of these two graphs as suggested by the two real solutions Complex solutions never represent intersections of two curves Example 3 Solve the following system of equations 2x2 y2 24 x2 y2 12 Solution This time we have an ellipse and a hyperbola Neither one are in standard form however In the first two examples we ve used the substitution method to solve the system and we can use that here as well Let s notice however that if we just add the two equations we will eliminate the y s from the system so we ll do it that way 2x2 y2 24 x2 y2 12 3x2 12 This is easy enough to solve for x 3x2 12 x2 4 x i 2 2007 Paul Dawkins 348 httptutorialmathlamaredutermsaspx College Algebra To determine the Values of the y s we can substitute these into either of the equations We will use the first since there won t be any minus signs to worry about x2 222y2 24 8y2 24 y2l6gt ll quotlt P 2 22 y2 24 8y2 24 y2l6gt i y 4 Note that for this system unlike the previous examples each Value of x actually gave two possible Values of y That means that there are in fact four solutions They are 24 14 14 24 This also means that there should be four intersection points to the two curves Here is a sketch for Verification Z39I39 I 24 23 333 24 Fl J E IIIIIIIIII I u39n I 2007 Paul Dawkins 349 httptutorialmathlamaredutermsaspx

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