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# MATH121, Chapter 2.3 Notes Math 121

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This 7 page Class Notes was uploaded by Mallory McClurg on Monday September 5, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 60 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 09/05/16

Math 121 Chapter 2 Notes Lesson 2.3 – Quadratic Equations In One Variable Example 1. 2 7y + 28y 2 0 (Since this is already in proper form “ax + bx + c = 0”, all we need to do is find the common factors in 7 and 28. Common factor here is 7y, since both can be divided by 7 and y, so factor out 7y.) 7y (y + 4) = 0 (By the Zero Factor Property, we know that for this product to equal zero, then at least one of the two factors must equal 0, as well. To determine the other solution of the equation, set (y + 4) equal to 0.) y = 0 y = -4 Example 2. 5y + 3y + 10 = 4y + 17y – 38 (Combine like terms 2 on same side to put2it in 2 ax + bx + c = 0 form. Subtract 4y from 5y . Subtract 17y from 3y. Add 38 to 10.) y – 14y + 48 = 0 (Now, we need to factor. 2 Since y has a coefficient of 1, then we know it will be (y +/- ?) (y +/- ?). So what numbers can we multiply together and get 48, but add together and get -14?) (y – 6) (y – 8) (Set each group equal to 0 and solve for y.) y = 6 y = 8 Example 3. 2 y – 6y – 7 = 0 (Since it’s already in its proper form, all we need to do is factor. 2 y has a coefficient of 1, so we know it’s going to be factored like (y +/- ?) (y +/- ?). So what numbers can we multiply together to get -7 and add together to get -6?) (y – 7) (y + 1) (Set both groups equal to 0 to find the solutions.) y = 7 y = -1 Example 4. y + 16y = -60 (To put this in proper form, add 602to the other side, so it looks like ax + bx + c = 0.) y + 16y + 60 = 0 (Factor it out. What two numbers can you multiply together to get 60, and add together to get 16?) (y + 10) (y + 6) (Set both groups equal to o and solve for y.) y = -10 y = -6 Example 5. A rectangular auditorium seats 2,009 people. The number of seats in each row exceeds the number of rows by 8. Find the number of seats in each row. (First, we need to set up a quadratic equation that represents the information we’re given. Now, this is confusing, but if x = the number of rows, and x+8 = the number of seats in each row, then we need to set it up like this.) 2 2 x + 8x = 2009 (To put this in ax + bx + c = 0 form, subtract 2009 from both sides.) x + 8x – 2009 = 0 (Now we must factor it. What two numbers can you multiply together to get -2009, but add together to get 8? This took me a while…) ( x – 41) ( x + 49) (Set both groups equal to zero to solve for x.) x = 41 x = -49 (Don’t forget, word problems can be tricky. Since we’re looking for the number of rows of seats right now, obviously there can’t be - 49 rows. So there’s one solution here. 41 rows.) Number of seats in each row: 49 (To find the number of seats in each row, I divided 2,009 seats by 41 rows and got 49.) Example 6. 28x + 5x = 12 (First, we need to put this in proper form.) 2 5x + 28x – 12 = 0 (Now, we need to factor this out. Since there is a coefficient in front of x , we know it will look a little something like this: (5x +/- ?) (x +/- ?). So, what numbers can you multiply together to get -12, and multiply one of them by 5 then added together to get 28?) (5x – 2) ( x + 6) (Since this is a little difficult, you can check yourself by FOILing. Mutiply the First numbers by eachother (5x * x = 5x ). Multiply the Outer numbers (5x * 6 = 30x). Multiply the Inner numbers ( -2 * x = -2x). Multiply the Last numbers (-2 * 6 = -12). Now 2 put all of those together and get 5x + 30x – 2x – 12, which when simplified, is the same as the original problem. So set each group equal to zero to solve for x.) x = (2/5) x = -6 Example 7. 2 36x – 25 = 0 (First of all, we need to recognize that even though this isn’t in the usual form. However, 2 36x and 25 are both perfect squares, which can be factored like this: A – B = (A – B) (A + B). So take the square root of each and put it in this form! Hint: A = 6x & B = 5) (6x – 5) (6x + 5) = 0 (Set each group equal to zero and solve for x.) x = (5/6) x = (-5/6) Example 8. 4x = 32 (First, we want to try and get x by itself, so divide both sides by 4.) x = 8 (Take the square root of each side to solve for x. Remember that square roots can be both the positive and negative of the root number.) x = +/- √8 (Don’t forget to simplify the square root…which is a weird process. Here, you can say that √8 = √(4*2). We can easily take the square root of 4, which is 2, so take out the 2 and leave the √2.) x = 2√2 x = -2√2 Example 9. (y – 15) = 100 (Using the square root method, we can easily take the square 2 root of (y – 15) , which is (y – 15). We also need to take the square root of 100, which is +/-√100….or…+/-10.) y – 15 = +/- 10 (Make two equations and solve for y.) y = 25 y = 5 Example 10. -3 (z +4) = -108 (First, divide both sides by -3 to get it in the same form as the last problem.) (z + 4) = 36 (Obviously the square root of the left side is (z + 4). The square root of 36 is +/- 6.) z + 4 = +/- 6 (Make two equations and solve for z.) z = 2 z = -10 Example 11. y + 7y – 2 = 0 (Sometimes, there are quadratic equations that are hard to factor. Often, we’re asked to solve with the quadratic formula, which is “x = (-b +/- √(b – 4ac))/ 2a”. So figure out first what a, b and c are in the proper form 2 ax + bx + c = 0. Hint: x is the variable in the problem, so technically our x here is y.) a = 1 (Since y has a coefficient of 1.) b = 7 (Since the coefficient of b is 7.) c = -2 (At this point, we can substitute these numbers into the quadratic formula.) 2 y = -7 +/- √((7 ) – 4 (1) (-2)) / 2 (1) (Write this equation out with the + in one and the – in the other.) y = (-7 + √57)/2 y = (-7 - √57)/2 Example 12. x – 8x + 5 = 0 (Instead of using the quadratic formula, we may be asked to solve by 2 completing the square. First, put this in “ax + bx = -c” form. Then, we will divide b by 2, which is -4 here, then square our answer, 2 which is 16. Now, we need to add 16 to both sides.) x – 8x + 16 = 11 (Now, looking at the left side of the equation, factor it out.) (x – 4) (x – 4) = 11 2 (x – 4) = 11 (Take the square root of each side!) x – 4 = +/- √11 (Now, set up two equations, and in both, get x on a side by itself.) x = √11 + 4 x = -√11 + 4 Example 13. 3x – 4x – 2 = -6x (First we need to put this in 2 ax 2bx+c=0 form.) 3x + 2x – 2 = 0 (The question asks to use the quadratic formula – see Example 11 – so we just need to substitute our a, b and c into the formula.) 2 x = (-2 +/- √(2 ) – (4)(3)(-2)) / (2)(3) (Once we compute this in two separate equations, we get…) x = (-2 – √28) / 6 x = (-2 + √28) / 10 (So this looks done, right? No. √28 can be simplified to “2√7” because √28 = √(7*2*2), so we can take out the (2*2) and put a “2” on the outside the square root symbol, since we found the square root of that part of 28. If that makes sense….) x = (-2 – 2√7) / 6 x = (-2 + 2√7) / 6 (If I could type this to make it look more like a fraction, you’d be able to see here that we can simplify even more. First, you can factor out a 2 from the numerators of both fractions.) x = (2 * (-1 - √7)) / 6 x = (2 * (-1 + √7)) / 6 (Aaaaaand, we can simplify again here too, because if this was in fraction form, you’d have a 2 over 6 that could be reduced to a “1/3”. You can also get rid of the parentheses.) x = (-1 - √7) / 3 x = (-1 + √7) / 3 (Remember, both of these are FRACTIONS and would NOT have parentheses in the numerator. Hawkes doesn’t like parenthesis apparently.)

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