1. 1 Copyright © Cengage Learning. All rights reserved. 2. Equations and Inequalities 2.3 Quadratic Equations

2. 2 Quadratic Equations A toy rocket is launched vertically upward from level ground, as illustrated in Figure 1. If its initial speed is 120 ft/sec and the only force acting on it is gravity, then the rocket’s height h (in feet) above the ground after t seconds is given by h = –16t 2 + 120t. Figure 1

3. 3 Quadratic Equations Some values of h for the first 7 seconds of flight are listed in the following table. We see from the table that, as it ascended, the rocket was 180 feet above the ground at some time between t = 2 and t = 3. As it descended, the rocket was 180 feet above the ground at some time between t = 5 and t = 6.

4. 4 Quadratic Equations To find the exact values of t for which h = 180 ft, we must solve the equation 180 = –16t 2 + 120t, or 16t 2 – 120t + 180 = 0 As indicated in the next chart, an equation of this type is called a quadratic equation in t.

5. 5 Quadratic Equations To enable us to solve many types of equations, we will make use of the next theorem. The zero factor theorem can be extended to any number of algebraic expressions—that is, pqr = 0 if and only if p = 0 or q = 0 or r = 0, and so on.

6. 6 Quadratic Equations It follows that if ax 2 + bx + c can be written as a product of two first-degree polynomials, then solutions can be found by setting each factor equal to 0, as illustrated in the next example. This technique is called the method of factoring.

7. 7 Example 2 – Solving an equation by factoring Solve the equation x 2 + 16 = 8x. Solution: We proceed as in Example 1: x 2 + 16 = 8x x 2 – 8x + 16 = 0 (x – 4)(x – 4) = 0 given subtract 8x factor

8. 8 Example 2 – Solving an equation by factoring x – 4 = 0, x – 4 = 0 x = 4, x = 4 Thus, the given quadratic equation has one solution, 4. zero factor theorem solve for x

9. 9 Quadratic Equations Since x – 4 appears as a factor twice in the previous solution, we call 4 a double root or root of multiplicity 2 of the equation x 2 + 16 = 8x. If a quadratic equation has the form x 2 = d for some number d > 0, then x 2 – d = 0 or, equivalently, (x + )(x – ) = 0. Setting each factor equal to zero gives us the solutions – and.

10. 10 Quadratic Equations We frequently use the symbol  (plus or minus ) to represent both and –. Thus, for d > 0, we have proved the following result. (The case d < 0 requires the system of complex numbers.) Note on Notation: It is common practice to allow one variable to represent more than one value, as in x =  3. A more descriptive notation is x 1,2 =  3, implying that x 1 = 3 and x 2 = –3.

11. 11 Quadratic Equations The process of solving x 2 = d as indicated in the preceding box is referred to as taking the square root of both sides of the equation. Note that if d > 0 we obtain both a positive square root and a negative square root, not just the principal square root.

12. 12 Example 3 – Solving equations of the form x 2 = d Solve the equations: (a) x 2 = 5 (b) (x + 3) 2 = 5 Solution: (a) x 2 = 5 x =  Thus, the solutions are and –. given take the square root

13. 13 Example 3 – Solution (b) (x + 3) 2 = 5 x + 3 =  x = –3  Thus, the solutions are –3 + and –3 –. given subtract 3 take the square root cont’d

14. 14 Quadratic Equations In the work that follows we will replace an expression of the form x 2 + kx by (x + d) 2, where k and d are real numbers. This procedure, called completing the square for x 2 + kx, calls for adding (k/2) 2, as described in the next box. (The same procedure is used for x 2 – kx.) In the next example we solve a quadratic equation by completing a square.

15. 15 Example 5 – Solving a quadratic equation by completing the square Solve the equation x 2 – 5x + 3 = 0. Solution: It is convenient to first rewrite the equation so that only terms involving x are on the left-hand side, as follows: x 2 – 5x + 3 = 0 x 2 – 5x = –3 x 2 – 5x + = –3 + given subtract 3 complete the square, adding to both sides

16. 16 Example 5 – Solution Thus, the solutions of the equation are (5 + )/2  4.3 and (5 – )/2  0.7. take the square root add cont’d equivalent equation

17. 17 Quadratic Equations In Example 5, we solved a quadratic equation of the form ax 2 + bx + c = 0 with a = 1. If a ≠ 1, we can solve the quadratic equation by adding a step to the procedure used in the preceding example. After rewriting the equation so that only terms involving x are on the left-hand side, ax 2 + bx = –c, we divide both sides by a, obtaining

18. 18 Quadratic Equations We then complete the square by adding to both sides. This technique is used in the proof of the following important formula. Note that if the quadratic formula is executed properly, it is unnecessary to check the solutions.

19. 19 Quadratic Equations The number b 2 – 4ac under the radical sign in the quadratic formula is called the discriminant of the quadratic equation. The discriminant can be used to determine the nature of the roots of the equation, as in the following chart.

20. 20 Example 7 – Using the quadratic formula Solve the equation 2x(3 – x) = 3. Solution: To use the quadratic formula, we must write the equation in the form ax 2 + bx + c = 0. The following equations are equivalent: 2x(3 – x) = 3 6x – 2x 2 = 3 –2x 2 + 6x – 3 = 0 2x 2 – 6x + 3 = 0 given multiply factors subtract 3 multiply by –1

21. 21 Example 7 – Solution We now let a = 2, b = –6, and c = 3 in the quadratic formula, obtaining cont’d

22. 22 Example 7 – Solution Since 2 is a factor of the numerator and denominator, we can simplify the last fraction as follows: Hence, the solutions are and cont’d

23. 23 Quadratic Equations The next example illustrates the case of a double root.

24. 24 Example 8 – Using the quadratic formula Solve the equation 9x 2 – 30x + 25 = 0. Solution: Let a = 9, b = –30, and c = 25 in the quadratic formula: Consequently, the equation has one (double) root,. simplify

25. 25 Example 9 – Clearing an equation of fractions Solve the equation Solution: For solving an equation containing rational expressions, we multiply by the lcd, (x + 3)(x – 3), remembering that, by guideline 2, the numbers (–3 and 3) that make the lcd zero cannot be solutions. Thus, we proceed as follows: given

26. 26 Example 9 – Solution 2x(x + 3) + 5(x – 3) = 36 2x 2 + 6x + 5x – 15 – 36 = 0 2x 2 + 11x – 51 = 0 (2x + 17)(x – 3) = 0 multiply by the lcd, (x + 3)(x – 3) multiply factors and subtract 36 simplify factor cont’d

27. 27 Example 9 – Solution 2x + 17 = 0, x – 3 = 0 x =, x = 3 Since x = 3 cannot be a solution, we see that x = is the only solution of the given equation. zero factor theorem solve for x cont’d

28. 28 Quadratic Equations The next example shows how the quadratic formula can be used to help factor trinomials.

29. 29 Example 10 – Factoring with the quadratic formula Factor the polynomial 21x 2 – 13x – 20. Solution: We solve the associated quadratic equation, 21x 2 – 13x – 20 = 0, by using the quadratic formula:

30. 30 Example 10 – Solution cont’d

31. 31 Example 10 – Solution We now write the equation as a product of linear factors, both of the form (x – solution): Eliminate the denominators by multiplying both sides by 3  7: 3  7 = 0  3  7 = 0 (3x – 4)(7x + 5) = 0 cont’d

32. 32 Example 10 – Solution The left side is the desired factoring—that is, 21x 2 – 13x – 20 = (3x – 4)(7x + 5). cont’d

33. 33 Quadratic Equations In the next example, we use the quadratic formula to solve an equation that contains more than one variable.

34. 34 Example 11 – Using the quadratic formula Solve y = x 2 – 6x – 5 for x, where x  3. Solution: The equation can be written in the form x 2 – 6x + 5 – y = 0, so it is a quadratic equation in x with coefficients a = 1, b = –6, and c = 5 – y. Notice that y is considered to be a constant since we are solving for the variable x.

35. 35 Example 11 – Solution Now we use the quadratic formula: simplify b 2 – 4ac factor out = 2 cont’d

36. 36 Example 11 – Solution Since is nonnegative, 3 + is greater than or equal to 3 and 3 – is less than or equal to 3. Because the given restriction is x  3, we have x = 3 –. cont’d divide 2 into both terms

37. 37 Quadratic Equations Many applied problems lead to quadratic equations. One is illustrated in the following example.

38. 38 Example 13 – Finding the height of a toy rocket The height above ground h (in feet) of a toy rocket, t seconds after it is launched, is given by h = –16t 2 + 120t. When will the rocket be 180 feet above the ground? Solution: Using h = –16t 2 + 120t, we obtain the following: 180 = –16t 2 + 120t 16t 2 – 120t + 180 = 0 4t 2 – 30t + 45 = 0 let h = 180 add 16t 2 – 120t divide by 4

39. 39 Example 13 – Solution Applying the quadratic formula with a = 4, b = –30, and c = 45 gives us cont’d

40. 40 Example 13 – Solution Hence, the rocket is 180 feet above the ground at the following times:  2.07 sec  5.43 sec cont’d

41. 41 Other Types of Equations (sec 2.5) In applications it is often necessary to consider powers x k with k > 2. Some equations involve absolute values or radicals. In this section we give examples of equations of these types that can be solved using elementary methods.

42. 42 Example 1 – Solving an equation containing an absolute value Solve the equation | x – 5 | = 3. Solution: If a and b are real numbers with b > 0, then | a | = b if and only if a = b or a = –b. Hence, if | x – 5 | = 3, then either x – 5 = 3 or x – 5 = –3.

43. 43 Example 1 – Solution Solving for x gives us x = 5 + 3 = 8 or x = 5 – 3 = 2. Thus, the given equation has two solutions, 8 and 2. cont’d

44. 44 Other Types of Equations If an equation is in factored form with zero on one side, then we may obtain solutions by setting each factor equal to zero. For example, if p, q, and r are expressions in x and if pqr = 0, then either p = 0, q = 0, or r = 0. In the next example we factor by grouping terms.

45. 45 Example 2 – Solving an equation using grouping Solve the equation x 3 + 2x 2 – x – 2 = 0. Solution: x 3 + 2x 2 – x – 2 = 0 x 2 (x + 2) – 1(x + 2) = 0 (x 2 – 1)(x + 2) = 0 (x + 1)(x – 1)(x + 2) = 0 Given group terms factor out x + 2 factor x 2 – 1

46. 46 Example 2 – Solution x + 1 = 0, x – 1 = 0, x + 2 = 0 x = – 1, x = 1, x = –2 Zero factor theorem solve for x cont’d

47. 47 Example 3 – Solving an equation containing rational exponents Solve the equation x 3/2 = x 1/2. Solution: x 3/2 = x 1/2 x 3/2 – x 1/2 = 0 x 1/2 (x – 1) = 0 x 1/2 = 0, x – 1 = 0 x = 0, x = 1 given subtract x 1/2 factor out x 1/2 solve for x Zero factor theorem

48. 48 Other Types of Equations In Example 3 it would have been incorrect to divide both sides of the equation x 3/2 = x 1/2 by x 1/2, obtaining x = 1, since the solution x = 0 would be lost. In general, avoid dividing both sides of an equation by an expression that contains variables—always factor instead.

49. 49 Example 4 – Solving an equation containing a radical Solve the equation Solution: x 2 – 1 = 8 x 2 = 9 given cube both sides property of add 1

50. 50 Example 4 – Solution x = Thus, the given equation has two solutions, 3 and –3. Except to detect algebraic errors, a check is unnecessary, since we raised both sides to an odd power. take the square root cont’d

51. 51 Other Types of Equations In the last solution we used the phrase cube both sides of In general, for the equation x m/n = a, where x is a real number, we raise both sides to the power n/m (the reciprocal of m/n) to solve for x. If m is odd, we obtain x = a n/m, but if m is even, we have

52. 52 Other Types of Equations If n is even, extraneous solutions may occur—for example, if x 3/2 = –8, then x = (–8) 2/3 = = (–2) 2 = 4. However, 4 is not a solution of x 3/2 = –8 since 4 3/2 = 8 not –8.

53. 53 Other Types of Equations Illustration: Solving x m/n = a, m odd, x real Equation Solution x 3/1 = 64 x = 64 1/3 = = 4 x 3/2 = 64 x = 64 2/3 = = 4 2 = 16

54. 54 Other Types of Equations Illustration: Solving x m/n = a, m even, x real Equation Solution x 4/1 = 16 x 2/3 = 16

55. 55 Other Types of Equations In the next example, before we raise both sides of the equation to a power, we isolate a radical—that is, we consider an equivalent equation in which only the radical appears on one side.

56. 56 Example 5 – Solving an equation containing a radical Solve the equation Solution: 3 + = x = x – 3 = (x – 3) 2 3x + 1 = x 2 – 6x + 9 given isolate the radical square both the sides simplify

57. 57 Example 5 – Solution x 2 – 9x + 8 = 0 (x – 1)(x – 8) = 0 x – 1 = 0, x – 8 = 0 x = 1, x = 8 subtract 3x + 1 solve for x Zero factor theorem factor cont’d

58. 58 Example 5 – Solution We raised both sides to an even power, so checks are required. Check: x = 1 LS: 3 + = 3 + = 3 + 2 = 5 RS: 1 Since 5 ≠ 1, x = 1 is not a solution. cont’d

59. 59 Example 5 – Solution Check: x = 8 LS: 3 + = 3 + = 3 + 5 = 8 RS: 8 Since 8 = 8 is a true statement, x = 8 is a solution. Hence, the given equation has one solution, x = 8. cont’d

60. 60 Other Types of Equations An equation is of quadratic type if it can be written in the form au 2 + bu + c = 0. where a ≠ 0 and u is an expression in some variable. If we find the solutions in terms of u, then the solutions of the given equation can be obtained by referring to the specific form of u.

61. 61 Example 7 – Solving an equation of quadratic type Solve the equation x 2/3 + x 1/3 – 6 = 0. Solution: Since x 2/3 = (x 1/3 ) 2, the form of the equation suggests that we let u = x 1/3, as in the second line below: x 2/3 + x 1/3 – 6 = 0 u 2 + u – 6 = 0 (u + 3)(u – 2) = 0 given let u = x 1/3 factor

62. 62 Example 7 – Solution u + 3 = 0, u – 2 = 0 u = –3, u = 2 x 1/3 = –3, x 1/3 = 2 x = –27, x = 8 A check is unnecessary, since we did not raise both sides to an even power. Hence, the given equation has two solutions, –27 and 8. Zero factor theorem cube both sides u = x 1/3 solve for u cont’d

63. 63 Example 7 – Solution An alternative method is to factor the left side of the given equation as follows: x 2/3 + x 1/3 – 6 = (x 1/3 + 3)(x 1/3 – 2) By setting each factor equal to 0, we obtain the solutions. cont’d