1. Chapter 1.4 – Quadratic Equations and Applications What you should learn 1. Solve quadratic equations by factoring 2. Solve quadratic equations by extracting square roots 3. Solve quadratic equations by completing the square 4. Use Quadratic Formula to solve quadratic equations 5. Use quadratic equations to model and solve real-life problems

2. Chapter 1.4 – Definition A Quadratic Equation in x is an equation that can be written in the general form ax 2 + bx + c = 0 where a, b, and c are real numbers with a = 0 is also called a second-degree polynomial equation

3. Chapter 1.4 – Methods of solving Quadratic Equations 1. Factoring 2. Extracting Square roots 3. Completing the square 4. Quadratic formula

4. Chapter 1.4 – Factoring Factoring is based on Zero-Factor Property If ab = 0, then a=0 or b = 0 To use this property, write the left side of the general form of a quadratic equation as a product of tow linear factors. Then find the solutions by setting each factor equal to 0.

5. Factoring All terms must be collected on one side of the equation before factoring.

6. Example 1: Solve the quadratic equation by factoring : p. 117 13. 6x 2 + 3x = 0 18. 4x 2 + 12x + 9 = 0 20. 2x 2 = 19x + 33

7. Solution : 13. 6x 2 + 3x = 0 3x ( 2x + 1) = 0 3x = 0 or 2x + 1 = 0 x = 0 or 2x = -1 x = - ½

8. Solution : 18. 4x 2 + 12x + 9 = 0 (2x +3 ) (2x +3 ) = 0 2x + 3 = 0 2x = -3 x = - 3/2 20. 2x 2 = 19x + 33 2x 2 – 19x – 33 = 0 (2x +3 ) ( x – 11) = 0 2x + 3 = 0 x – 11 = 0 x = -3/2 x = 11

9. Extracting the square root The equation u 2 = d, where d > 0, has exactly two solutions: u = √d and u = - √d These solutions can also be written as u = ± √d Watch video

10. Example Solve the equation by extracting the square roots. P. 117 28. 3x 2 = 81 34. (x + 9) 2 = 24 37. ( x – 7) 2 = ( x + 3) 2

11. Solution 28. 3x 2 = 81 divide both sides by 3 x 2 = 27 take the square root of both sides x = ± √27 x = ± 3√3 34. (x + 9) 2 = 24 take the square root of both sides (x + 9) = ± √24 x + 9 = ± 2√6 solve for x, subtract 9 from both sides x = 9 ± 2√6

12. Solution 37. ( x – 7) 2 = ( x + 3) 2 take the square root of both sides ( x – 7) = ± ( x + 3) x – 7 = x + 3 ( not possible) x – 7 = - ( x + 3) x – 7 = -x – 3, then solve for x 2x = 4, x = 2, therefore the solution is x = 2

13. Completing the square If the left side of the equation is not factorable, and then you have to rewrite the equation by completing the square so it can be solved by extracting the square roots.

14. Completing the square To complete the square for the expression x 2 + bx, add (b/2) 2, which is the square of half of the coefficient of x. Consequently, x 2 + bx + (b/2) 2 = ( x + b/2) 2 Watch videovideo

15. Example Solve the quadratic equation by completing the square. p. 117 42. x 2 + 8x + 14 = 0 43. 9x 2 - 18x = - 3 46. - x 2 +x – 1 = 0

16. Solution 42. x 2 + 8x + 14 = 0 x 2 + 8x = -14 divide (8/2), then square, add to both sides x 2 + 8x + 16 = -14 + 16 Subtract 14 from both sides

17. Solution x 2 + 8x + 16 = -14 + 16 then factor the left side (x + 4) 2 = 2, then solve using extracting the square root, take the square root of both sides x + 4 = ± √2, then solve for x x = - 4 ± √2

18. Solution 43. 9x 2 - 18x = - 3 divide each side by 9 x 2 - 2x = - 1/3 divide (-2/2), then square, add to both sides x 2 - 2x +1 = - 1/3 +1 then factor the left side, simplify the right side (x – 1) 2 = 2/3 x = 1 ± √2/3

19. Solution 46. - x 2 +x – 1 = 0 multiply by -1 x 2 - x + 1 = 0 then follow the steps for completing the square x 2 - x = -1 divide -1 by 2, then square, then add to both sides x 2 - x + ¼ = -1 + ¼ (x + ½) 2 = - ¾ no real solution, since the square root of - ¾ will yield an imaginary number

20. The Quadratic Formula The solutions of a quadratic equation in the general form ax 2 + bx + c = 0, a = 0 are given by the Quadratic Formula x = -b ± √b 2 – 4ac 2a Watch videoCengage Learning Mathematics VideosCengage Learning Mathematics Videos

21. Solutions of a Quadratic Equation The quantity under the square root is called the discriminant. It can be used to determine the nature of the solutions of a quadratic equation.

22. If the discriminant b 2 – 4ac is PositiveEquation has 2 distinct real solutions graph has 2 x- intercepts ZeroEquation has one repeated real solutions graph has one x- intercepts NegativeEquation has no real solution graph has no x- intercepts

23. Example: Use the discriminant to determine the number of real solutions of the quadratic equation p.117, 118 67. 2x 2 – x – 1 = 0 68. x 2 – 4x + 4 = 0 94. (x + 6) 2 = - 2x

24. Application - p. 118 #118. A rectangular classroom seats 72 students. If the seats were rearranged with three more seats in each row, the classroom would have two fewer rows. Find the original number of seats in each row.

25. Solution Let x = number of rows y = number of seats xy = 72 Original y = 72 /x (x – 2) ( y + 3) = 72 replace y by 72/x (x – 2 ) (72/x + 3) = 72 Find the LCD, combine like terms, solve. x = 8 original # rows, 72/8 = 9 seats per row

26. Practice: Answer the following questions on pp. 117-119 # 22,19, 32, 36, 40, 41, 65, 91, 93, 113, 120 For the Position equation mentioned in #120, use S = -16t 2 + v 0 t + S 0, where v 0 is the initial velocity, S is the height, S 0 is the initial height, and t is the time