1. College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

2. Equations and Inequalities 1

3. Basic Equations 1.3

4. Linear Equations vs. Quadratic Equations Linear equations are first-degree equations, such as: 2x + 1 = 5 or 4 – 3x = 2 Quadratic equations are second-degree equations, such as: x 2 + 2x – 3 = 0 or 2x 2 + 3 = 5x

5. Quadratic Equation—Definition A quadratic equation is an equation of the form ax 2 + bx + c = 0 where a, b, and c are real numbers with a ≠ 0.

6. Solving Quadratic Equations by Factoring

7. Solving Quadratic Equations Some quadratic equations can be solved by factoring and using the following basic property of real numbers.

8. Zero-Product Property AB = 0 if and only if A = 0 or B = 0 This means that, if we can factor the LHS of a quadratic (or other) equation, then we can solve it by setting each factor equal to 0 in turn. This method works only when the RHS is 0.

9. E.g. 1—Solving a Quadratic Equation by Factoring Solve the equation x 2 + 5x = 24 We must first rewrite the equation so that the RHS is 0.

10. E.g. 1—Solving a Quadratic Equation by Factoring x 2 + 5x = 24 x 2 + 5x – 24 = 0 (Subtract 24) (x – 3)(x + 8) = 0 (Factor) x – 3 = 0 or x + 8 = 0 (Zero-Product Property) x = 3x = –8 (Solve)

11. Solving a Quadratic Equation by Factoring Do you see why one side of the equation must be 0 in Example 1? Factoring the equation as x(x + 5) = 24 does not help us find the solutions. 24 can be factored in infinitely many ways, such as 6. 4, ½. 48, (–2/5). (–60), and so on.

12. Solving Quadratic Equations by Completing the Square

13. Solving Simple Quadratics As we saw in Section 1.1, Example 5(b), if a quadratic equation is of the form (x ± a) 2 = c, we can solve it by taking the square root of each side. In an equation of this form, the LHS is a perfect square: the square of a linear expression in x.

14. Completing the Square So, if a quadratic equation does not factor readily, we can solve it using the technique of completing the square.

15. Completing the Square To make x 2 + bx a perfect square, add, the square of half the coefficient of x. This gives the perfect square

16. Completing the Square To compete the square, we add a constant to an expression to make it a perfect square. For example, to make x 2 – 6x a perfect square, we must add (6/2) 2 = 9. Then x 2 – 6x + 9 = (x – 3) 2 is a perfect square.

17. Completing the Square The table gives some more examples of completing the square.

18. E.g. 2—Solving by Completing the Square Solve each equation. (a) x 2 – 8x + 13 = 0 (b) 3x 2 – 12x + 6 = 0

19. E.g. 2—Completing the Square x 2 – 8x + 13 = 0 x 2 – 8x = –13 (Subtract 13) x 2 – 8x + 16 = –13 + 16 (Complete the square) (x – 4) 2 = 3 (Perfect square) x – 4 = ± (Take square root) x = 4 ± (Add 4) Example (a)

20. E.g. 2—Completing the Square First, we subtract 6 from each side. Then, we factor the coefficient of x 2 (the 3) from the left side. This puts the equation in the correct form for completing the square. Example (b)

21. E.g. 2—Completing the Square 3x 2 – 12x + 6 = 0 3x 2 – 12x = –6 (Subtract 6) 3(x 2 – 4x) = –6 (Factor 3 from LHS)

22. E.g. 2—Completing the Square 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.

23. E.g. 2—Completing the Square 3(x 2 – 4x + 4) = –6 + 3. 4 (Complete the square) 3(x – 2) 2 = 6 (Perfect square) (x – 2) 2 = 2 (Divide by 3) (Take square root) (Add 2)

24. The Quadratic Formula

25. Deriving a Formula for the Roots 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

26. The Quadratic Formula The roots of the quadratic equation ax 2 + bx + c = 0 where a ≠ 0, are:

27. The Quadratic Formula—Proof First, we divide each side of the equation by a and move the constant to the right side, giving:

28. The Quadratic Formula—Proof We now complete the square by adding (b/2a) 2 to each side of the equation.

29. The Quadratic Formula The quadratic formula could be used to solve the equations in Examples 1 and 2. You should carry out the details of these calculations.

30. E.g. 3—Using the Quadratic Formula Find all solutions of each equation. (a) 3x 2 – 5x – 1 = 0 (b) 4x 2 + 12x + 9 = 0 (c) x 2 + 2x + 2 = 0

31. E.g. 3—Using Quadratic Formula In 3x 2 – 5x – 1 = 0, a = 3 b = –5 c = –1 By the quadratic formula, Example (a)

32. E.g. 3—Using Quadratic Formula If approximations are desired, we can use a calculator to obtain: Example (a)

33. E.g. 3—Using Quadratic Formula Using the quadratic formula with a = 4, b = 12, and c = 9 gives: This equation has only one solution, x = –3/2. Example (b)

34. E.g. 3—Using Quadratic Formula Using the quadratic formula with a = 1, b = 2, and c = 2 gives: Since the square of a real number is nonnegative, is undefined in the real number system. The equation has no real solution. Example (c)

35. Complex Number System In the next section, we study the complex number system, in which the square roots of negative numbers do exist. The equation in Example 3(c) does have solutions in the complex number system.

36. The Discriminant

37. 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. It is given the symbol D.

38. D < 0 If D < 0, then is undefined. The quadratic equation has no real solution—as in Example 3(c).

39. D = 0 and D > 0 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).

40. Discriminant The following box summarizes these observations.

41. E.g. 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) 1/3x 2 – 2x + 4 = 0

42. E.g. 4—Using the Discriminant x 2 + 4x – 1 = 0 The discriminant is: D = 4 2 – 4(1)(–1) = 20 > 0 So, the equation has two distinct real solutions. Example (a)

43. E.g. 4—Using the Discriminant 4x 2 – 12x + 9 = 0 The discriminant is: D = (–12) 2 – 4. 4. 9 = 0 So, the equation has exactly one real solution. Example (b)

44. E.g. 4—Using the Discriminant 1/3x 2 – 2x + 4 = 0 The discriminant is: D = (–2) 2 – 4(1/3)4 = –4/3 < 0 So, the equation has no real solution. Example (c)

45. Modeling with Quadratic Equations

46. Quadratic Equations in Real Life Let’s consider a real-life situation that can be modeled by a quadratic equation. The principles discussed in Section 1.2 for setting up equations as models are useful here as well.

47. E.g. 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.

48. E.g. 5—Dimensions of a Building Lot We are asked to find the width and length of the lot. So, let: w = width of lot

49. E.g. 5—Dimensions of a Building Lot Then, we translate the information in the problem into the language of algebra. In WordsIn Algebra Width of lotw Length of lotw + 8

50. E.g. 5—Dimensions of a Building Lot Now, we set up the model. Width of Lot. Length of Lot = Area of Lot

51. E.g. 5—Dimensions of a Building Lot w(w + 8) = 2900 w 2 + 8w = 2900 (Expand) w 2 + 8w – 2900 = 0 (w – 50)(w + 58) = 0 (Factor) w = 50 or w = –58 (Zero-Product Property)

52. E.g. 5—Dimensions of a Building Lot 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

53. E.g. 6—A Distance-Speed-Time Problem A jet flew from New York (NY) to Los Angeles (LA), a distance of 4,200 km. The speed for the return trip was 100 km/h faster than the outbound speed. If the total trip took 13 hours, what was the jet’s speed from NY to LA?

54. E.g. 6—A Distance-Speed-Time Problem We are asked for the speed of the jet from NY to LA. So, let: s = speed from NY to LA Then, s + 100 = speed from LA to NY

55. E.g. 6—A Distance-Speed-Time Problem Now, we organize the information in a table. First, we fill in the “Distance” column—as we know that the cities are 4,200 km apart.

56. E.g. 6—A Distance-Speed-Time Problem Then, we fill in the “Speed” column—as we have expressed both speeds (rates) in terms of the variable s.

57. E.g. 6—A Distance-Speed-Time Problem Finally, we calculate the entries for the “Time” column, using:

58. E.g. 6—A Distance-Speed-Time Problem The total trip took 13 hours. So, we have the model: Time from NY to LA + Time from LA to NY = Total time This gives:

59. E.g. 6—A Distance-Speed-Time Problem Multiplying by the common denominator, s(s + 100), we get: Although this equation does factor, with numbers this large, it is probably quicker to use the quadratic formula and a calculator.

60. E.g. 6—A Distance-Speed-Time Problem As s represents speed, we reject the negative answer and conclude that the jet’s speed from NY to LA was 600 km/h.

61. E.g. 7—The Path of a Projectile An object thrown or fired straight upward at an initial speed of v 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 + v 0 t

62. E.g. 7—The Path of a Projectile Suppose that a bullet is shot straight upward with an initial speed of 800 ft/s. a)When does the bullet fall back to ground level? b)When does it reach a height of 6,400 ft? c)When does it reach a height of 2 mi? d)How high is the highest point the bullet reaches?

63. E.g. 7—The Path of a Projectile The initial speed is v 0 = 800 ft/s. Thus, the formula is: h = –16t 2 + 800t

64. E.g. 7—The Path of a Projectile Ground level corresponds to h = 0. So, we must solve: 0 = –16t 2 + 800t (Set h = 0) 0 = –16t(t – 50) (Factor) Thus, t = 0 or t = 50. This means the bullet starts (t = 0) at ground level and returns to ground level after 50 s. Example (a)

65. E.g. 7—The Path of a Projectile Setting h = 6400 gives: 6400 = –16t 2 + 800t (Set h = 6400) 16t 2 – 800t + 6400 = 0 (All terms to LHS) t 2 – 50t + 400 = 0 (Divide by 16) Example (b)

66. E.g. 7—The Path of a Projectile (t – 10)(t – 40) = 0 (Factor) t = 10 or t = 40 (Solve) The bullet reaches 6400 ft after 10 s (on its ascent) and again after 40 s (on its descent to earth). Example (b)

67. E.g. 7—The Path of a Projectile Two miles is: 2 x 5,280 = 10,560 ft 10,560 = –16t 2 + 800t (Set h = 10,560) 16t 2 – 800t + 10,560 = 0 (All terms to LHS) t 2 – 50t + 660 = 0 (Divide by 16) Example (c)

68. E.g. 7—The Path of a Projectile The discriminant of this equation is: D = (–50) 2 – 4(660) = –140 It is negative. Thus, the equation has no real solution. The bullet never reaches a height of 2 mi. Example (c)

69. E.g. 7—The Path of a Projectile Each height 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. Example (d)

70. E.g. 7—The Path of a Projectile 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 This, in turn, means that the discriminant of the equation is 0. Example (d)

71. E.g. 7—The Path of a Projectile So, D = (–800) 2 – 4(16)h = 0 640,000 – 64h = 0 h = 10,000 The maximum height reached is 10,000 ft. Example (d)