1. Copyright © Cengage Learning. All rights reserved. 1.4 Solving Quadratic Equations

2. 2 Objectives ■ Solving Quadratic Equations by Factoring ■ Solving Quadratic Equations by Completing the Square ■ The Quadratic Formula ■ The Discriminant ■ Modeling with Quadratic Equations

3. 3 Solving Quadratic Equations In this section we learn how to solve quadratic equations, which are second-degree equations such as x 2 + 2x – 3 = 0 or 2x 2 + 3 = 5x. We will also see that many real-life problems can be modeled by using quadratic equations.

4. 4 Solving Quadratic Equations by Factoring

5. 5 Some quadratic equations can be solved by factoring and using the following basic property of real numbers.

6. 6 Example 1 – Solving a Quadratic Equation by Factoring Find all real solutions of the equation x 2 + 5x = 24. Solution: We must first rewrite the equation so that the right-hand side is 0: x 2 + 5x = 24 x 2 + 5x – 24 = 0 (x – 3)(x + 8) = 0 Given equation Subtract 24 Factor

7. 7 Example 1 – Solution x – 3 = 0 or x + 8 = 0 x = 3 x = –8 The solutions are x = 3 and x = –8. Check Your Answers x = 3: (3) 2 + 5(3) = 9 + 15 = 24 x = –8: (–8) 2 + 5(–8) = 64 – 40 = 24 Zero-Product Property Solve cont’d

8. 8 Solving Quadratic Equations by Completing the Square

9. 9 If a quadratic equation is of the form (x  a) 2 = c, then we can solve it by taking the square root of each side. In an equation of this form the left-hand side is a perfect square: the square of a linear expression in x. So if a quadratic equation does not factor readily, then we can solve it by completing the square.

10. 10 Example 2 – Solving Quadratic Equations by Completing the Square Find all real solutions of each equation. (a) x 2 – 8x + 13 = 0 (b) 3x 2 – 12x + 6 = 0 Solution: (a) x 2 – 8x + 13 = 0 x 2 – 8x = –13 x 2 – 8x + 16 = –13 + 16 (x – 4) 2 = 3 Given equation Subtract 13 Complete the square: add = 16 Perfect square

11. 11 Example 2 – Solution x – 4 =  x = 4  (b) After subtracting 6 from each side of the equation, we must factor the coefficient of x 2 (the 3) from the left side to put the equation in the correct form for completing the square: 3x 2 – 12x + 6 = 0 3x 2 – 12x = –6 Take square root Add 4 Given equation Subtract 6 cont’d

12. 12 Example 2 – Solution 3(x 2 – 4x) = –6 Now we complete the square by adding (–2) 2 = 4 inside the parentheses. Since everything inside the parentheses is multiplied by 3, this means that we are actually adding 3  4 = 12 to the left side of the equation. Thus we must add 12 to the right side as well. 3(x 2 – 4x + 4) = –6 + 3  4 Factor 3 from LHS Complete the square: add 4 cont’d

13. 13 Example 2 – Solution 3(x – 2) 2 = 6 (x – 2) 2 = 2 x – 2 =  x = 2  Perfect square Divide by 3 Take square root Add 2 cont’d

14. 14 The Quadratic Formula

15. 15 The Quadratic Formula We can use the technique of completing the square to derive a formula for the roots of the general quadratic equation ax 2 + bx + c = 0.

16. 16 Example 3 – Using the Quadratic Formula Find all real solutions of each equation. (a) 3x 2 – 5x – 1 = 0 (b) 4x 2 + 12x + 9 = 0 (c) x 2 + 2x + 2 = 0 Solution: (a) In this quadratic equation a = 3, b = –5, and c = –1: 3x 2 – 5x – 1 = 0

17. 17 Example 3 – Solution By the Quadratic Formula, If approximations are desired, we can use a calculator to obtain and cont’d

18. 18 Example 3 – Solution (b) Using the Quadratic Formula with a = 4, b = 12, and c = 9 gives This equation has only one solution, x = cont’d

19. 19 Example 3 – Solution (c) Using the Quadratic Formula with a = 1, b = 2, and c = 2 gives Since the square of any real number is nonnegative, is undefined in the real number system. The equation has no real solution. cont’d

20. 20 The Discriminant

21. 21 The Discriminant The quantity b 2 – 4ac that appears under the square root sign in the Quadratic Formula is called the discriminant of the equation ax 2 + bx + c = 0 and is given the symbol D. If D < 0, then is undefined, and the quadratic equation has no real solution, as in Example 3(c). If D = 0, then the equation has only one real solution, as in Example 3(b). Finally, if D > 0, then the equation has two distinct real solutions, as in Example 3(a).

22. 22 The Discriminant The following box summarizes these observations.

23. 23 Example 4 – Using the Discriminant Use the discriminant to determine how many real solutions each equation has. (a) x 2 + 4x – 1 = 0 (b) 4x 2 – 12x + 9 = 0 (c) x 2 – 2x + 4 = 0 Solution: (a) The discriminant is D = 4 2 – 4(1)(–1) = 20 > 0, so the equation has two distinct real solutions.

24. 24 Example 4 – Solution (b) The discriminant is D = (–12) 2 – 4  4  9 = 0, so the equation has exactly one real solution. (c) The discriminant is D = (–2) 2 – 4 4 = – < 0, so the equation has no real solution. cont’d

25. 25 Modeling with Quadratic Equations

26. 26 Example 5 – Dimensions of a Building Lot A rectangular building lot is 8 ft longer than it is wide and has an area of 2900 ft 2. Find the dimensions of the lot. Solution: Identify the variable. We are asked to find the width and length of the lot. So let w = width of lot

27. 27 Example 5 – Solution Translate from words to algebra. Then we translate the information given in the problem into the language of algebra (see Figure 1): Figure 1 cont’d

28. 28 Example 5 – Solution Set up the model. Now we set up the model: w (w + 8) = 2900 Solve. Now we solve for w. w 2 + 8w = 2900 w 2 + 8w – 2900 = 0 Expand Subtract 2900 cont’d

29. 29 Example 5 – Solution (w – 50)(w + 58) = 0 w = 50 or w = –58 Since the width of the lot must be a positive number, we conclude that w = 50 ft. The length of the lot is w + 8 = 50 + 8 = 58 ft. Factor Zero-Product Property cont’d

30. 30 Example 7 – The Path of a Projectile An object thrown or fired straight upward at an initial speed of 0 ft/s will reach a height of h feet after t seconds, where h and t are related by the formula h = –16t 2 + 0 t Suppose that a bullet is shot straight upward with an initial speed of 800 ft/s. Its path is shown in Figure 2. Figure 2

31. 31 Example 7 – The Path of a Projectile (a) When does the bullet fall back to ground level? (b) When does it reach a height of 6400 ft? (c) When does it reach a height of 2 mi? (d) How high is the highest point the bullet reaches? Solution: Since the initial speed in this case is 0 = 800 ft/s, the formula is h = –16t 2 + 800t

32. 32 Example 7 – Solution (a) Ground level corresponds to h = 0, so we must solve the equation 0 = –16t 2 + 800t 0 = –16t(t – 50) Thus t = 0 or t = 50. This means the bullet starts (t = 0) at ground level and returns to ground level after 50 s. Set h = 0 Factor cont’d

33. 33 Example 7 – Solution (b) Setting h = 6400 gives the equation 6400 = –16t 2 + 800t 16t 2 – 800t + 6400 = 0 t 2 – 50t + 400 = 0 (t – 10)(t – 40) = 0 t = 10 or t = 40 All terms to LHS Set h = 6400 cont’d Factor Solve Divide by 16

34. 34 Example 7 – Solution The bullet reaches 6400 ft after 10 s (on its ascent) and again after 40 s (on its descent to earth). (c) Two miles is 2  5280 = 10,560 ft: 10,560 = –16t 2 + 800t Set h = 10,560 cont’d

35. 35 Example 7 – Solution 16t 2 – 800t + 10,560 = 0 t 2 – 50t + 660 = 0 The discriminant of this equation is D = (–50) 2 – 4(660) = –140, which is negative. Thus the equation has no real solution. The bullet never reaches a height of 2 mi. All terms to LHS Divide by 16 cont’d

36. 36 Example 7 – Solution (d) Each height that the bullet reaches is attained twice: once on its ascent and once on its descent. The only exception is the highest point of its path, which is reached only once. This means that for the highest value of h, the following equation has only one solution for t: h = –16t 2 + 800t 16t 2 – 800t + h = 0 All terms to LHS cont’d

37. 37 Example 7 – Solution This in turn means that the discriminant D of the equation is 0, so D = (–800) 2 – 4(16)h = 0 640,000 – 64h = 0 h = 10,000 The maximum height reached is 10,000 ft. cont’d

38. 38 3.2 Height of a Propelled Object The coefficient of t ² (that is,  16), is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon or the other planets. Height of a Propelled Object If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height s 0 feet with initial velocity v 0 feet per second is where t is the number of seconds after the object is projected.

39. 39 A ball is thrown directly upward from an initial height of 100 feet with an initial velocity of 80 feet per second. (a)Give the function that describes height in terms of time t. (b)Graph this function. (c)The cursor in part (b) is at the point (4.8,115.36). What does this mean? 3.2 Solving a Problem Involving Projectile Motion After 4.8 seconds, the object will be at a height of 115.36 feet.

40. 40 3.2 Solving a Problem Involving Projectile Motion (d) After how many seconds does the projectile reach its maximum height? (e)For what interval of time is the height of the ball greater than 160 feet? Figure 19 pg 3-24 Using the graphs, t must be between.92 and 4.08 seconds.

41. 41 3.2 Solving a Problem Involving Projectile Motion (f) After how many seconds will the ball fall to the ground? When the ball hits the ground, its height will be 0, so we need to find the positive x-intercept. From the graph, the x-intercept is about 6.04, so the ball will reach the ground 6.04 seconds after it is projected. Figure 21 pg 3-25

42. Copyright © 2011 Pearson Education, Inc. Slide 3.2-42 EXAMPLE The coefficient of t ² (that is,  16), is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon or the other planets. Height of a Propelled Object If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height s 0 feet with initial velocity v 0 feet per second is where t is the number of seconds after the object is projected.

43. Copyright © 2011 Pearson Education, Inc. Slide 3.2-43 A ball is thrown directly upward from an initial height of 100 feet with an initial velocity of 80 feet per second. (a)Give the function that describes height in terms of time t. (b)Graph this function. (c)The cursor in part (b) is at the point (4.8,115.36). What does this mean? 3.2 Solving a Problem Involving Projectile Motion After 4.8 seconds, the object will be at a height of 115.36 feet.

44. Copyright © 2011 Pearson Education, Inc. Slide 3.2-44 3.2 Solving a Problem Involving Projectile Motion (d) After how many seconds does the projectile reach its maximum height? (e)For what interval of time is the height of the ball greater than 160 feet? Figure 19 pg 3-24 Using the graphs, t must be between.92 and 4.08 seconds.

45. Copyright © 2011 Pearson Education, Inc. Slide 3.2-45 3.2 Solving a Problem Involving Projectile Motion (f) After how many seconds will the ball fall to the ground? When the ball hits the ground, its height will be 0, so we need to find the positive x-intercept. From the graph, the x-intercept is about 6.04, so the ball will reach the ground 6.04 seconds after it is projected. Figure 21 pg 3-25