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# Extensive Study Guide for Exam 1 Math 1302

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This 11 page Study Guide was uploaded by Leighton Tidwell on Tuesday September 27, 2016. The Study Guide belongs to Math 1302 at University of Texas at Arlington taught by Dr. Glass in Fall 2016. Since its upload, it has received 103 views. For similar materials see College Algebra in Math at University of Texas at Arlington.

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Date Created: 09/27/16

9/29/16 Exam 1 Review – Created by Leighton Tidwell 1.1 Linear Equations The most important thing to remember is the properties of equality. o Addition Property A = B and A + C = B + C o Multiplication Property A = B and AC = BC Linear Equation follows structure: o ax+b = 0 Example of Solving for X in linear equations o 3(2x-4)=7-(x+5) o 6x-12 = 7-x-5 o 6x-12 = 12-x o 7x-12=12 o 7x=14 o X=2 Special Cases: Conditional One X=a Contradiction None 2=3 Identity Infinite 0=0 If the equation comes down to false, no solution. (Contradiction) If it comes down to true no matter what is plugged in, all real numbers. (Identity) Example Problem o 5(2-x)+8x = 3(x-6) o 10-5x+8x = 3x-18 o 10+3x = 3x-18 o 10 DOES NOT EQUAL 18 o No solution. Contradictory Also remember Simple Interest Formula o I = Prt 1.2 Application with Linear Equations How to solve an applied problem: o Read question and assign appropriate variables so that you may create an equation that can then be solved for. o Example: A train traveled at 100 MPH for 15 minutes, how far did it go? D = rt D = (100)(.25) D = 25 Miles Solving Geometry problems: o Example: The second side of a triangular deck is 3 feet longer than the shortest side, the third side of the deck is 3 ft shorter than twice the length of the shortest side. If the perimeter is 80 feet, what are the lengths? P = x+x+3+2x-3 80= 2x + 2x 80=4x (divide each side by 4) 20=x Solving an Investment Problem o Remember I = Prt P R T I X 11 1 X(.11) 140,00-x 9 1 (140,000-x)(.09) (1) o X(.11)+(140,000-x)(.09)=14100 o .02x+12600 = 14100 o .02x=1500 o X=75000 Solving a Mixture Problem o Example Problem How much pure acid is in 78 ML of a 14% solution? A = 780(.14) A = 109.2 ML Solving a Motion Problem o Example Problem Two Ferries leave at the same time. The first ferry travels at 28 MPH, the second at 22 MPH. In how many hours will the ferries be 12 miles apart? D = RT 28x = 22x – 12 6x = -12 X = -2 Hours (But time cannot be negative IN THIS CASE, so it is 2 hours.) 1.3 Complex Numbers Remember that if you take the square root of a negative, it always turns out like: o √−x=i √ Some more things to remember when regarding i are: i= −1 o √ 2 o i =−1 o Complex Number Standard form a+bi A = real number B = imaginary number a=0∧b≠0 Pure imaginary a ≠0∧b=0 Real numbers a≠0∧b≠0 Nonreal complex o Example Problem √−70=i √0 Noticethatits∈a+bi form(ais just0∈thiscase) o Multiplying and Dividing Square Roots with Negative Numbers Dividing Example: −8+ −128 √ 4 −8+i √28 4 −2+8i √ 4 −2+2i √ Multiplying Example: −4+3i − 6−7i )=−4+3i−6+7i −¿ 10+3i+7i −¿ 10+10i Multiplying (FOIL) example 4i(5−i)=4i 5−i )5−i ) 4i(25−5i−5i+i ) 4i(25−1−10i ) (24-10i) 4i(24-10i) 96i−40i 2 96i+40 1.4 Quadratic Equations Remember the standard form of a quadratic equation: a x +bx+c=0 o There will typically be two answers in a ‘2 nd degree’ equation (quadratic). Basically need to remember four methods of solving a quadratic equation o Zero-Factor Example: (8a+3 )a+2 =0 8a+3=0∧a+2=0 8a=−3∧a=−2 a= −3 ∧a=−2 8 o Factoring Example: 10x +x−2=0 Need ¿ find anumber that multiplies¿2,but addsup¿1 (2x+1 )5x−2 )=0 You can multiply/foil this to check your answer. Now use zero factor property 2 x+1=0∧5 x−2=0 2x=−1∧5x=2 x= −1 ∧x= 2 2 5 {−¿ 1/2, 2/5} o Completing the Square Example: 9x −12x+9=0 Divide 9 so that a = 1 2 4 x − x+1=0 3 Subtract 1¿each side x = −4 x=−1 3 2 1 −4 4 4 ( ) = ;Add ¿eachside [2] 3 9 9 x − 4x+ =−1+ 4 3 9 9 Factor . 2 2 −5 ( ) 3 = 9 Squareroot property. x− =± −5 3 √ 9 2 √5 x− 3± 3 i 2 √ 5 x= ± i 3 3 o Remember the Quadratic Formula 2 −b± b√−4ac x= 2a o Finding the Discriminant The discriminant tells you how many solutions there will be To find the discriminant, solve for the 2 −b± √ −4ac Portionof theQuadratic Formula Key pointers of discriminant value If it is positive and the square of an integer, you have 2 rational solutions If it is positive, but not the square, you have 2 irrational solutions If it is zero, you have 1 rational solution If it is negative, you have 2 imaginary solutions 1.5 Applications with Quadratic Equations For this section, remember: PythagoreanTheorem:a +b =c 2 o Heightof aProjectedObjectst)=−16t +v t+s o 0 0 The applications will usually just involve word problems. Same situation as above, however you will interpret the equation via the information in the problem 1.6 Other Types of Equations and Applications To solve a rational equation, multiply both sides by a common denominator and then solve regularly. Example: −3 2 5 + = o 20 x 4x 20 x o Multiplybothsides by1 −3 2 5 o The20∈ cancelsout ,thex∈ cancelsout∧the canbesimplified 20 x 4x o −3 x+40=25 o −3x=−15 o x=5 Solving a rate of work problem: Name Hours to Complete Rate to Complete Al 8 1/8 Mario 10 1/10 Both X 1/x 1 + 1 = 1Becausebothof their ratestogether should equalboth. o 8 10 x Multiplybothsidesby40x commondenominator ) o 5 x+4 x=40 o 9x=40 o 40 o x= 9 40 ' ' o But 9 doesn tmakesense¿say∈termsof time..So≤t ssimplifythat¿: 4 o 4+ Hours 9 Remember the power rule o “If both sides of the equation are raised to the same power, all solutions of the original equation are also solutions of the new equation.” Solving equations in Quadratic Form using Substitution 3 3 o (x+1 ) x(1 −)=0 ' 1 o ¿t slet u( x+1) o 1 2 1 3 3 2 1 ((+1 ) )− (+1 −)=0Wechangedthe exponent¿ squared sothatwecanusesubstitutioneasily 3 3 2 o u −u−2=0 o (u−2 )(u+1 =0 o Now we can use zero factor property and solve for x. Not U: 3 3 o (x+1 )2=0∧ x+1 (1=0) o Add two¿theother side∨¿Subtract1¿theother side 3 3 o (x+1 )2∧ x+( =−1) o ¿get ridof theexponent ,weneed¿cubeeachside. 3 3 33 3 o (x+1 )2 ∧ x+1(=−1) o x+1=8∧x+1=−1 o x=7∧x=−2 1.7Inequalities Always remember these four interval notations: a<x<b o Open Interval (a,b) or a≤x≤b o Closed Interval [a,b] or (−∞,a )∪ b,∞ ∨x<a∨x>b o Disjoint Interval o Infinite Interval (a ,∞)representsx>a (−∞,b represents x≤b o If you have a ≤∨≥ then you will use [ or ] brackets. Otherwise if it’s just < or >, you will use ( and ) brackets. Also remember that when solving inequalities if you multiply or divide by a negative, the sign will change. Solving a Linear Inequality o −3 x+5>−7 o −3x>−12 o x<4−−Noticethesignchangeas wedivided by−3 o Theanswer∈interval notationlookslike:(−∞,4 ) o Graph: o Solving a Quadratic Inequality 2x +5x−12≥0 o Setitequal¿zerosothatwecansolveitsimilarly¿aregular quadraticequation o o 2 o 2x +5x−12=0 Factor likenormal o o 2x−3 )(x+4)=0 o ZeroFactor property −3 X= ∧−4 o 2 o So the answer looks like this: o −∞,−4 ∪] o Graph: o Solving a Rational Inequality x+3 o <0 x−4 o Immedietlyyoucanseperatethetop∧bottomof the fractionlikeso: o o x+3=0∧x−4=0 o x=−3,4 o Now wecan plot these points ona numberline: o ¿−−−−−−¿−−−−−−−−−¿−−−−−¿ -3 4 o So now we will plot points in between each of the segments like so: o ¿−−¿−−−¿−−−−¿−−−−¿−−¿−−¿ -4 -3 0 4 5 o Now take those numbers and plug them into the original equation to test if they are true. −4+3 ) 1 o <0= whichisnotlessthan0sothisis false. −4−4 8 0+3 = 3 which islessthan0 sothisistrue. o 0−4 −4 5+3 8 o 5−4 = 1hichis notlessthan0,sothisis false. Theonlysegment of numebersthat workis−3¿ 4. o o Theanswer ,∈set notation: (-3,4) o Graph: o 1.8 Absolute Value Equations and Inequalities Things to remember: o An absolute value can never be negative o Absolute values are always greater than 0 Example Problem |4 x−1|=6 o o Splitthe problemintotwo problems,onewrittenasitappears,∧theother withthe ' o ¿of theequation ssignsswitched o 4 x−1=6∧−4 x+1=6 4 x=7∧−4 x=5 o 7 −5 o x= ∧x= 4 4 Solving an equation with both sides Absolute Value o Example |5a+7 = 6a−5 | Justrewriteoneside∧solvethesameas∈the previousquestion. 5a+7=6a−5∧−5a−7=6a−5 −1a=−12∧−11a=2 −2 a=12∧a= 11 If the equation is equal to the exact opposite absolute value, the set is all real numbers. This looks like: o |4y−7 = 7−4 y| Absolute value inequalities o Example |2x−7|≥12 2 x−7≥12∧−2 x+7≥12 2x≥19∧−2x≥5 x ≥19∧x≤− 5 2 2 −∞,− 5 ∪ 19,∞ ( 2] [ 2 ) Example of a problem with no solution o |6x+4|<−8 o This problem is asking which value will make this absolute value less than 8.. That cannot happen as discussed above. Absolute value is always positive.

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