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10.1

10.2

10.3

10.4

10.5

Factoring and Completing the Square

The Quadratic Formula

More on Quadratic Equations

Graphing Quadratic Functions

Quadratic Inequalities

Quadratic Equations, Functions, and Inequalities Is it possible to measure beauty? For thousands of years artists and philosophers

have been challenged to answer this question. The seventeenth-century philoso

pher John Locke said, “Beauty consists of a certain composition of color and figure

causing delight in the beholder.” Over the centuries many architects, sculptors,

and painters have searched for beauty in their work by exploring numerical

patterns in various art forms.

Today many artists and architects still use the concepts of beauty given to us

by the ancient Greeks. One principle,

called the Golden Rectangle, concerns

the most pleasing proportions of a rec

tangle. The Golden Rectangle appears

in nature as well as in many cultures.

Examples of it can be seen in Leonardo

da Vinci’s Proportions of the Human

Figure as well as in Indonesian temples

and Chinese pagodas. Perhaps one

of the best-known examples of the W

Golden Rectangle is in the façade and

floor plan of the Parthenon, built in

Athens in the fifth century B.C. W

L _ W L

In Exercise 89 of Section 10.3 we will see that the principle of

the Golden Rectangle is based on a proportion that

we can solve using the quadratic formula.

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628 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-2

10.1 Factoring and Completing the Square

Factoring and the even-root property were used to solve quadratic equations in Chapters 5, 6, and 9. In this section we first review those methods. Then you will learn the method of completing the square, which can be used to solve any quadratic equation.

In This Section

U1V Review of Factoring

U2V Review of the Even-Root Property

U3V Completing the Square

U4V Radicals and Rational Expressions

U5V Imaginary Solutions U1V Review of Factoring A quadratic equation has the form ax2 + bx + c = 0, where a, b, and c are real num bers with a * 0. In Section 5.6 we solved quadratic equations by factoring and then applying the zero factor property.

Zero Factor Property

The equation ab = 0 is equivalent to the compound equation

a = 0 or b = 0.

Of course we can only use the factoring method when we can factor the quadratic poly nomial. To solve a quadratic equation by factoring we use the following strategy.

E X A M P L E 1

Strategy for Solving Quadratic Equations by Factoring

1. Write the equation with 0 on one side.

2. Factor the other side.

3. Use the zero factor property to set each factor equal to zero.

4. Solve the simpler equations.

5. Check the answers in the original equation.

U Helpful Hint V

After you have factored the quadratic polynomial, use FOIL to check that you have factored correctly before proceeding to the next step.

Solving a quadratic equation by factoring Solve 3x2 - 4x = 15 by factoring.

Solution Subtract 15 from each side to get 0 on the right-hand side:

3x2 - 4x - 15 = 0

(3x + 5)(x - 3) = 0 Factor the left-hand side.

3x + 5 = 0 or x - 3 = 0 Zero factor property

3x = -5 or x = 3

x = -- 5 3

The solution set is {--5 3 -, 3}. Check the solutions in the original equation. Now do Exercises 1–10

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10-3 10.1 Factoring and Completing the Square 629

U2V Review of the Even-Root Property In Chapter 9 we solved some simple quadratic equations by using the even-root prop erty, which we restate as follows:

Even-Root Property

Suppose n is a positive even integer. nIf k > 0, then xn = k is equivalent to x = ±Vkk.

If k = 0, then xn = k is equivalent to x = 0. If k < 0, then xn = k has no real solution.

By the even-root property x2 = 4 is equivalent to x = ±2, x2 = 0 is equivalent to x = 0, and x2 = -4 has no real solutions.

E X A M P L E 2 Solving a quadratic equation by the even-root property Solve (a - 1)2 = 9.

Solution By the even-root property x2 = k is equivalent to x = ±Vkk.

(a - 1)2 = 9

a - 1 = ±V9k Even-root property a - 1 = 3 or a - 1 = -3

a = 4 or a = -2

Check these solutions in the original equation. The solution set is {-2, 4}. Now do Exercises 11–20

U Helpful Hint V

The area of an x by x square and two x by 3 rectangles is x2 + 6x. The area needed to “complete the square” in this figure is 9:

3 3 3

x x2 3x

93x

U3V Completing the Square We cannot solve every quadratic by factoring because not all quadratic polynomials can be factored. However, we can write any quadratic equation in the form of Example 2 and then apply the even-root property to solve it. This method is called completing the square.

The essential part of completing the square is to recognize a perfect square trinomial when given its first two terms. For example, if we are given x2 + 6x, how do we recognize that these are the first two terms of the perfect square trinomial x2 + 6x + 9? To answer this question, recall that x2 + 6x + 9 is a perfect square trinomial because it is the square of the binomial x + 3:

2 2(x + 3)2 = x + 2 · 3x + 32 = x + 6x + 9

Notice that the 6 comes from multiplying 3 by 2 and the 9 comes from squaring the 3. So to find the missing 9 in x2 + 6x, divide 6 by 2 to get 3, and then square 3 to get 9. This procedure can be used to find the last term in any perfect square trinomial in which the coefficient of x2 is 1.

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630 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-4

Rule for Finding the Last Term

The last term of a perfect square trinomial is the square of one-half of the coeffi cient of the middle term. In symbols, the perfect square trinomial whose first two

bterms are x2 + bx is x2 + bx + (--)2. 2

E X A M P L E 3 Finding the last term Find the perfect square trinomial whose first two terms are given.

a) x2 + 8x b) x2 - 5x c) x2 + - 4

7 - x d) x2 - -

2 - x

Solution a) One-half of 8 is 4, and 4 squared is 16. So the perfect square trinomial is

x2 + 8x + 16.

b) One-half of -5 is --5 2

-, and --5 2

- squared is -2 4 5 -. So the perfect square trinomial is

x2 - 5x + - 2 4 5 -.

c) Since -1 2

- · -4 7

- = - 2 7

- and -2 7

- squared is - 4 4 9 -, the perfect square trinomial is

x2 + - 4 7

- x + - 4 4 9 -.

d) Since -1 2

-(--3 2 -) = --3 4 - and (--3 4 -) 2

= - 1 9 6 -, the perfect square trinomial is

x2 - - 3

2 - x + -

6 -.

Now do Exercises 21–28

CAUTION The rule for finding the last term applies only to perfect square trinomials with a = 1. A trinomial such as 9x2 + 6x + 1 is a perfect square trinomial because it is (3x + 1)2, but the last term is certainly not the square of one-half the coefficient of the middle term.

Another essential step in completing the square is to write the perfect square trinomial as the square of a binomial. Recall that

2a + 2ab + b2 = (a + b)2

and 2a - 2ab + b2 = (a - b)2.

E X A M P L E 4 Factoring perfect square trinomials Factor each trinomial.

49 a) x2 + 12x + 36 b) y2 - 7y + --

4 4 4

c) z2 - -- z + -- 3 9

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10-5 10.1 Factoring and Completing the Square 631

Solution a) The trinomial x2 + 12x + 36 is of the form a2 + 2ab + b2 with a = x and

b = 6. So,

x2 + 12x + 36 = (x + 6)2 .

Check by squaring x + 6.

b) The trinomial y2 - 7y + -4 4 9 - is of the form a2 - 2ab + b2 with a = y and b = -7

2 -. So,

y2 - 7y + - 4

9 - = (y - -7 2 -)

2 .

Check by squaring y - -7 2 -.

c) The trinomial z2 - -4 3 - z + - 4 9

- is of the form a2 - 2ab + b2 with a = z and b = --2 3 -. So,

z2 - - 4 3

- z + - 4 9

- = (z - -2 3 -) 2 .

U Helpful Hint V

To square a binomial use the follow ing rule (not FOIL): • Square the first term. • Add twice the product of the terms. • Add the square of the last term.

Now do Exercises 29–36

In Example 5, we use the skills that we learned in Examples 2, 3, and 4 to solve the quadratic equation ax2 + bx + c = 0 with a = 1 by the method of completing the square. This method works only if a = 1 because the method for completing the square developed in Examples 2, 3, and 4 works only for a = 1.

E X A M P L E 5 Completing the square with a = 1 Solve x2 + 6x + 5 = 0 by completing the square.

Solution The perfect square trinomial whose first two terms are x2 + 6x is

x2 + 6x + 9.

So we move 5 to the right-hand side of the equation, and then add 9 to each side to create a perfect square on the left side:

x2 + 6x = -5 Subtract 5 from each side.

x2 + 6x + 9 = -5 + 9 Add 9 to each side to get a perfect square trinomial.

(x + 3)2 = 4 Factor the left-hand side.

x + 3 = ±V4k Even-root property x + 3 = 2 or x + 3 = -2

x = -1 or x = -5

Check in the original equation:

(-1)2 + 6(-1) + 5 = 0

and

(-5)2 + 6(-5) + 5 = 0

The solution set is {-1, -5}.

U Calculator Close-Up V

The solutions to

x2 + 6x + 5 = 0

correspond to the x-intercepts for the graph of

y = x2 + 6x + 5.

So we can check our solutions by graphing and using the TRACE fea ture as shown here.

-8 2

-6

Now do Exercises 37–44

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CAUTION

632 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-6

All of the perfect square trinomials that we have used so far had a leading coefficient of 1. If a * 1, then we must divide each side of the equation by a to get an equation with a leading coefficient of 1.

The strategy for solving a quadratic equation by completing the square is stated in the following box.

Strategy for Solving Quadratic Equations by Completing the Square

1. If a * 1, then divide each side of the equation by a.

2. Get only the x2- and the x-terms on the left-hand side.

3. Add to each side the square of -1 2

- the coefficient of x.

4. Factor the left-hand side as the square of a binomial.

5. Apply the even-root property.

6. Solve for x.

7. Simplify.

E X A M P L E 6 Completing the square with a = 1 Solve 2x2 + 3x - 2 = 0 by completing the square.

Solution For completing the square, the coefficient of x2 must be 1. So we first divide each side of the equation by 2:

- 2x2 +

2 3x - 2 - = -

0 2

- Divide each side by 2.

x2 + - 3 2

- x - 1 = 0 Simplify.

x2 + - 3 2

- x = 1 Get only x2- and x-terms on the left-hand side.

x2 + - 3 2

- x + - 1 9 6 - = 1 + -

1 9 6 - One-half of -3

2 - is -3

4 -, and (-3 4 -)

2 = -1

9 6 -.

(x + -3 4 -) 2

= - 2 1 5 6 - Factor the left-hand side.

x + - 3 4

- = ±�-2 1 5 6 - Even-root property x + -

3 4

- = - 5 4

- or x + - 3 4

- = -- 5 4

x = - 2 4

- = - 1 2

- or x = -- 8 4

- = -2

Check these values in the original equation. The solution set is {-2, -1 2 -}.

U Calculator Close-Up V

Note that the x-intercepts for the graph of

y = 2x2 + 3x - 2

are (-2, 0) and (-1 2 -, 0):

-4 2

Now do Exercises 45–46 -6

In Examples 5 and 6, the solutions were rational numbers, and the equations could have been solved by factoring. In Example 7, the solutions are irrational numbers, and factoring will not work.

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�

10-7 10.1 Factoring and Completing the Square 633

E X A M P L E 7 A quadratic equation with irrational solutions Solve x2 - 3x - 6 = 0 by completing the square.

Solution Because a = 1, we first get the x2- and x-terms on the left-hand side:

2x - 3x - 6 = 0 2x - 3x = 6 Add 6 to each side.

9 92 3 3 9x - 3x + -- = 6 + -- One-half of -3 is ---, and (---)2 = --. 2 2 44 4 23 33 9 24 9 33(x - -- = -- 6 + -- = -- + -- = --4 4 4 42) 4 3 33

x - -- = ± -- Even-root property 2 4

3 V33 3k x = -- ± -- Add -- to each side. 22 2

3 ± V33k x = --

The solution set is {-3 + 2 V33k -, -3 -2 V33k -}. Now do Exercises 47–56

U4V Radicals and Rational Expressions Examples 8 and 9 show equations that are not originally in the form of quadratic equa tions. However, after simplifying these equations, we get quadratic equations. Even though completing the square can be used on any quadratic equation, factoring and the square root property are usually easier and we can use them when applicable. In Examples 8 and 9, we will use the most appropriate method.

E X A M P L E 8 An equation containing a radical Solve x + 3 = V153k-kx.

Solution Square both sides of the equation to eliminate the radical:

x + 3 = V153 - x The original equationkk (x + 3)2 = (V153 -kx)2k Square each side.

x2 + 6x + 9 = 153 - x Simplify.

x2 + 7x - 144 = 0

(x - 9)(x + 16) = 0 Factor.

x - 9 = 0 or x + 16 = 0 Zero factor property

x = 9 or x = -16

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634 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-8

Because we squared each side of the original equation, we must check for extraneous roots. Let x = 9 in the original equation:

9 + 3 = V153 -k 9k

12 = V144k Correct

Let x = -16 in the original equation:

-16 + 3 = V153 -k (-16)k

-13 = V169k Incorrect because V169k = 13

Because -16 is an extraneous root, the solution set is {9}.

U Calculator Close-Up V

You can provide graphical support for the solution to Example 8 by graphing

y1 = x + 3

and

y2 = V153 -k xk. It appears that the only point of inter section occurs when x = 9.

-150 200

Now do Exercises 57–60

-50

E X A M P L E 9 An equation containing rational expressions Solve -1

x - + -

x - 3

2 - = -

5 8

Solution The least common denominator (LCD) for x, x - 2, and 8 is 8x(x - 2).

- 1 x

- + - x -

3 2

- = - 5 8

8x(x - 2) - 1 x

- + 8x(x - 2)- x -

3 2

- = 8x(x - 2) - 5 8

- Multiply each side by the LCD.

8x - 16 + 24x = 5x2 - 10x

32x - 16 = 5x2 - 10x

-5x2 + 42x - 16 = 0

5x2 - 42x + 16 = 0 Multiply each side by -1

(5x - 2)(x - 8) = 0 for easier factoring.

5x - 2 = 0 or x - 8 = 0

Factor.

x = - 2 5

- or x = 8

Check these values in the original equation. The solution set is {-2 5 -, 8}. Now do Exercises 61–64

U5V Imaginary Solutions In Chapter 9, we found imaginary solutions to quadratic equations using the even-root property. We can get imaginary solutions also by completing the square.

E X A M P L E 10 An equation with imaginary solutions Find the complex solutions to x2 - 4x + 12 = 0.

Solution Because the quadratic polynomial cannot be factored, we solve the equation by completing the square.

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10-9 10.1 Factoring and Completing the Square 635

Warm-Ups ▼

Fill in the blank. 1. In this section quadratic equations are solved by ,

the property, and the square.

2. If b = 0 in ax2 + bx + c = 0, then the equation can be solved by the .

3. The last term of a perfect square trinomial is the square of one-half the coefficient of the term.

4. If the leading coefficient is not 1, then the first step in completing the square is to divide both sides of the equation by the .

True or false? 5. Every quadratic equation can be solved by factoring.

6. All quadratic equations have two distinct complex solutions.

7. The trinomial x2 + 2 3 2

2x + 2 1 9 6 2 is a perfect square trinomial.

8. Every quadratic equation can be solved by completing the square.

9. (x - 3)2 = 12 is equivalent to x - 3 = 2V3l.

10. (2x - 3)(3x + 5) = 0 is equivalent to x = 2 3 2

2 or x = 2 5 3

11. x2 = 8 is equivalent to x = ±2V2l. 12. To complete the square for x2 - 3x = 4, add 2

9 4

2 to each side.

1 0

x2 - 4x + 12 = 0 The original equation

x2 - 4x = -12 Subtract 12 from each side.

x2 - 4x + 4 = -12 + 4 One-half of -4 is -2, and (-2)2 = 4.

(x - 2)2 = -8

x - 2 = ±V-8l Even-root property x = 2 ± iV8l

= 2 ± 2iV2l

Check these values in the original equation. The solution set is {2 ± 2iV2l }.

U Calculator Close-Up V

The answer key (ANS) can be used to check imaginary answers as shown here.

Now do Exercises 65–74

Exercises

U Study Tips V • Stay calm and confident.Take breaks when you study. Get 6 to 8 hours of sleep every night. • Keep reminding yourself that working hard throughout the semester will really pay off in the end.

U1V Review of Factoring 3. a2 + 2a = 15 4. w2 - 2w = 15 Solve by factoring. See Example 1. See the Strategy for Solving 5. 2x2 - x - 3 = 0 6. 6x2 - x - 15 = 0 Quadratic Equations by Factoring box on page 628.

1. x2 - x - 6 = 0 2. x2 + 6x + 8 = 0

7. y2 + 14y + 49 = 0 8. a2 - 6a + 9 = 0

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636 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-10

2 2 9. a - 16 = 0 10. 4w - 25 = 0

U2V Review of the Even-Root Property

Use the even-root property to solve each equation. See Example 2.

9 2 2 11. x = 81 12. x = -- 4

16 2 2 13. x = -- 14. a = 32 9

15. (x - 3)2 = 16 16. (x + 5)2 = 4

17. (z + 1)2 = 5 18. (a - 2)2 = 8

3 7 2 5 19. (w - --)2 = -- 20. (w + --)2 = --2 4 3 9

U3V Completing the Square

Find the perfect square trinomial whose first two terms are given. See Example 3.

21. x2 + 2x 22. m2 + 14m

2 2 23. x - 3x 24. w - 5w

1 3 25. y2 + -- y 26. z2 + -- z

4 2

2 6 27. x2 + -- x 28. p2 + -- p

3 5

Factor each perfect square trinomial. See Example 4.

29. x2 + 8x + 16 30. x2 - 10x + 25

25 1 2 31. y - 5y + -- 32. w2 + w + -- 4 4

4 4 6 9 2 2 33. z - -- z + -- 34. m - -- m + -- 7 49 5 25

3 9 3 9 35. t2 + -- t + -- 36. h2 + -- h + --

5 100 2 16

Solve by completing the square. See Examples 5–7. See the Strategy for Solving Quadratic Equations by Completing the Square box on page 632. Use your calculator to check.

2 37. x - 2x - 15 = 0

2 38. x - 6x - 7 = 0

2 39. 2x - 4x = 70

2 40. 3x - 6x = 24

2 41. w - w - 20 = 0

42. y2 - 3y - 10 = 0

43. q2 + 5q = 14

44. z2 + z = 2

45. 2h2 - h - 3 = 0

2 46. 2m - m - 15 = 0

47. x2 + 4x = 6

2 48. x + 6x - 8 = 0

49. x2 + 8x - 4 = 0

2 50. x + 10x - 3 = 0

51. x2 + 5x + 5 = 0

2 52. x - 7x + 4 = 0

2 53. 4x - 4x - 1 = 0

54. 4x2 + 4x - 2 = 0

55. 2x2 + 3x - 4 = 0

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10-11 10.1 Factoring and Completing the Square 637

56. 2x2 + 5x - 1 = 0

U4V Radicals and Rational Expressions

Solve each equation by an appropriate method. See Examples 8 and 9.

57. V2kx + 1 = x - 1 k58. V2x - 4 = x - 14

Vwk+ 1 Vy + 1k 59. w = -- 60. y - 1 = --

2 2

t 2t - 3 z 3z 61. -- = -- 62. -- = --

t - 2 t z + 3 5z - 1

2 4 63. -- + -- + 1 = 02x x

1 3 64. -- + -- + 1 = 02x x

U5V Imaginary Solutions

Use completing the square to find the imaginary solutions to each equation. See Example 10.

65. x2 + 2x + 5 = 0 66. x2 + 4x + 5 = 0

2 267. x - 6x + 11 = 0 68. x - 8x + 19 = 0

1 12 269. x = --- 70. x = --- 2 8

2 271. x + 12 = 0 72. -3x - 21 = 0

2 273. 5z - 4z + 1 = 0 74. 2w - 3w + 2 = 0

Miscellaneous

Find all real or imaginary solutions to each equation. Use the method of your choice.

275. x = -121 276. w = -225

77. 4x2 + 25 = 0

278. 5w - 3 = 0

1 9 79. (p + --)

= -- 2 4

2 4 80. (y - --)

= -- 3 9

81. 5t2 + 4t - 3 = 0

82. 3v2 + 4v - 1 = 0

83. m2 + 2m - 24 = 0

84. q2 + 6q - 7 = 0

85. (x - 2)2 = -9

86. (2x - 1)2 = -4

87. -x2 + x + 6 = 0

88. -x2 + x + 12 = 0

289. x - 6x + 10 = 0

290. x - 8x + 17 = 0

91. 2x - 5 = V7x + 7k

92. V7kx + 2k9 = x + 3

1 1 1 93. -- + -- = --

x x - 1 4

1 2 1 94. -- - -- = --

x 1 - x 2

Find the real solutions to each equation by examining the graphs on page 638.

95. x2 + 2x - 15 = 0

96. 100x2 + 20x - 3 = 0

97. x2 + 4x + 15 = 0

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638 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-12

298. 100x - 60x + 9 = 0

20 20

_8 6 _0.8 0.5

_20 _20

40 40

_10 5 _1 1

_40 _40

Applications

Solve each problem.

99. Approach speed. The formula 1211.1L = CA2S is used to determine the approach speed for landing an aircraft, where L is the gross weight of the aircraft in pounds, C is the coefficient of lift, S is the surface area of the wings in square feet (ft2), and A is approach speed in feet per second. Find A for the Piper Cheyenne, which has a gross weight of 8700 lb, a coefficient of lift of 2.81, and a wing surface area of 200 ft2.

100. Time to swing. The period T (time in seconds for one com plete cycle) of a simple pendulum is related to the length L (in feet) of the pendulum by the formula 8T2 = �2L . If a child is on a swing with a 10-foot chain, then how long does it take to complete one cycle of the swing?

101. Time for a swim. Tropical Pools figures that its monthly revenue in dollars on the sale of x aboveground pools is given by R = 1500x - 3x2, where x is less than 25. What number of pools sold would provide a revenue of $17,568?

102. Pole vaulting. In 1981 Vladimir Poliakov (USSR) 3set a world record of 19 ft -- in. for the pole vault 4

(www.polevault.com). To reach that height, Poliakov obtained a speed of approximately 36 feet per second on the runway. The formula h = -16t2 + 36t gives his height t seconds after leaving the ground.

a) Use the formula to find the exact values of t for which his height was 18 feet.

b) Use the accompanying graph to estimate the value of t for which he was at his maximum height.

c) Approximately how long was he in the air?

H ei

gh t (

ft )

0 0 1 2

Time (sec)

Figure for Exercise 102

Getting More Involved

103. Discussion

Which of the following equations is not a quadratic equation? Explain your answer.

a) �x2 - Vk5x - 1 = 0 b) 3x2 - 1 = 0 c) 4x + 5 = 0 d) 0.009x2 = 0

104. Exploration

Solve x2 - 4x + k = 0 for k = 0, 4, 5, and 10.

a) When does the equation have only one solution? b) For what values of k are the solutions real? c) For what values of k are the solutions imaginary?

105. Cooperative learning

Write a quadratic equation of each of the following types, and then trade your equations with those of a classmate. Solve the equations and verify that they are of the required types.

a) a single rational solution b) two rational solutions c) two irrational solutions d) two imaginary solutions

106. Exploration 2In Section 10.2 we will solve ax + bx + c = 0 for

x by completing the square. Try it now without looking ahead.

Graphing Calculator Exercises

For each equation, find approximate solutions rounded to two decimal places.

2107. x - 7.3x + 12.5 = 0

108. 1.2x2 - �x + V2k = 0 109. 2x - 3 = V20 - xk

2 2110. x - 1.3x = 22.3 - x

http:www.polevault.com

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10-13 10.2 The Quadratic Formula 639

Math at Work Financial Matters

In the United States, over 1 million new homes are sold annually, with a median price of about $200,000. Over 17 million new cars are sold each year with a median price over $20,000. Americans are constantly saving and borrowing. Nearly everyone will need to know a monthly payment or what their savings will total over time. The answers to these questions are in the following table.

In each case, n is the number of periods per year, r is the annual percentage rate (APR), t is the number of years, and i is the interest rate per period (i = -n

r - ). For periodic payments or

deposits these expressions apply only if the compounding period equals the payment period. So let’s see what these expressions do.

A person inherits $10,000 and lets it grow at 4% APR compounded daily for 20 years.

Use the first expression with n = 365, i = - 0 3 . 6 0 5 4

-, and t = 20 to get 10,000(1 + -0 3 .0 6 4 5

-) 365·20

$22,254.43, which is the amount after 20 years. More often, people save money with periodic

deposits. Suppose you deposit $100 per month at 4% compounded monthly for 20 years. Use the second

expression with R = 100, i = - 0 1 .0 2 4

-, n = 12, and t = 20 to

get 100 or $36,677.46, which is the

amount after 20 years. Suppose that you get a 20-year $200,000 mortgage at

7% APR compounded monthly to buy an average house. Try using the third expression to calculate the monthly payment of $1550.60. See the accompanying figure.

(1 + 0.04�12)12·20 - 1 ---

0.04�12

What $P Left at Compound What $R Deposited Periodic Payment That Will Interest Will Grow to Periodically Will Grow to Pay off a Loan of $P

P(1 + i)nt R - (1 + i)

nt - 1 - P -

1 - (1 i + i)-nt -

M on

th ly

p ay

m en

t ( $) 2000

1000

1500

500

2 4 10 86 APR (percent)

20-year $200,000 mortgage

In This Section

U1V Developing the Formula

U2V Using the Formula

U3V Number of Solutions

U4V Applications

10.2 The Quadratic Formula

Completing the square from Section 10.1 can be used to solve any quadratic equation. Here we apply this method to the general quadratic equation to get a formula for the solutions to any quadratic equation.

U1V Developing the Formula Start with the general form of the quadratic equation,

2ax + bx + c = 0.

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�

640 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-14

Assume a is positive for now, and divide each side by a:

ax2 + bx + c 0 -- = --

a a b c

x2 + -- x + -- = 0 a a

b c c2x + -- x = --- Subtract -- from each side. a a a

2b b b bOne-half of -- is --, and -- squared is --2: a 2a 2a 4a

b b2 c b2 2x + -- x + -- = - -- + --2 2a 4a a 4a

Factor the left-hand side and get a common denominator for the right-hand side:

2b b 4ac c(4a) 4ac(x + --) 2

= -- - -- -- = --22a 4a2 4a2 a(4a) 4a

b b2 - 4ac(x + --) 2

= --22a 4a

b b2 - 4ac x + -- = ± - Even-root property 2a 4a2

-b Vb2 - 4kkac x = -- ± -- Because a > 0, V4ka2 = 2a.

2a 2a

-b ± Vb2 - 4kkac x =--

We assumed a was positive so that V4a = 2a would be correct. If a is negative, thenk2 V4a = -2a, and we getk2

-b Vb2 - 4k kac x = -- ± -- .

2a -2a

However, the negative sign can be omitted in -2a because of the ± symbol preceding it. For example, the results of 5 ± (-3) and 5 ± 3 are the same. So when a is negative, we get the same formula as when a is positive. It is called the quadratic formula.

The Quadratic Formula

The solution to ax2 + bx + c = 0, with a * 0, is given by the formula

-b ± Vkacb2 - 4k x =--. 2a

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10-15 10.2 The Quadratic Formula 641

U2V Using the Formula The quadratic formula solves any quadratic equation. Simply identify a, b, and c and insert those numbers into the formula. Note that if b is positive then -b (the opposite of b) is a negative number. If b is negative, then -b is a positive number.

E X A M P L E 1 Two rational solutions Solve x2 + 2x - 15 = 0 using the quadratic formula.

Solution To use the formula, we first identify the values of a, b, and c:

1x2 + 2x - 15 = 0 ↑ ↑ ↑ a b c

The coefficient of x2 is 1, so a = 1. The coefficient of 2x is 2, so b = 2. The constant term is -15, so c = -15. Substitute these values into the quadratic formula:

x =

= - -2 ± V

2 4 + 60k -

= - -2 ±

2 V64k -

= - -2

2 ± 8 -

x = - -2

2 + 8 - = 3 or x = -

-2 2 - 8 - = -5

Check 3 and -5 in the original equation. The solution set is {-5, 3}.

-2 ± V22 - 4k(1)(-1k5)k ---

2(1) U Calculator Close-Up V

Note that the two solutions to

x2 + 2x - 15 = 0

correspond to the two x-intercepts for the graph of

y = x2 + 2x - 15.

-8 6

Now do Exercises 1–8 -20

CAUTION To identify a, b, and c for the quadratic formula, the equation must be in the standard form ax2 + bx + c = 0. If it is not in that form, then you must first rewrite the equation.

E X A M P L E 2 One rational solution Solve 4x2 = 12x - 9 by using the quadratic formula.

Solution Rewrite the equation in the form ax2 + bx + c = 0 before identifying a, b, and c:

24x - 12x + 9 = 0

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642 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-16

In this form we get a = 4, b = -12, and c = 9.

x = Because b = -12, -b = 12.

= - 12 ± V1

8 44 -k 144k -

= - 12

8 ± 0 -

= - 1 8 2 -

= - 3 2

Check -3 2

- in the original equation. The solution set is {-3 2 -}.

12 ± V(-12)2k - 4(4k)(9)k ---

2(4)

U Calculator Close-Up V

Note that the single solution to

4x2 - 12x + 9 = 0

corresponds to the single x-intercept for the graph of

y = 4x2 - 12x + 9.

-2

-2 4

Now do Exercises 9–14

Because the solutions to the equations in Examples 1 and 2 were rational num bers, these equations could have been solved by factoring. In Example 3, the solutions are irrational.

E X A M P L E 3 Two irrational solutions Solve -1

3 -x2 + x + -1

2 - = 0.

Solution We could use a = -1

3 -, b = 1, and c = -1 2 - in the quadratic formula, but it is easier to use the

formula with integers. So we first multiply each side of the equation by 6, the least com mon denominator. Multiplying by 6 yields

2x2 + 6x + 3 = 0.

Now let a = 2, b = 6, and c = 3 in the quadratic formula:

x =

= - -6 ± V

4 36 - 2k4k -

= - -6 ±

4 V12k -

= - -6 ±

4 2V3k -

= - 2(-3

2 ±

· 2 V3k)

= - -3 ±

2 V3k

Check these values in the original equation. The solution set is {--3 ± 2 V3k -}.

-6 ± V(6)2 -k 4(2)(3k)k ---

2(2)

U Calculator Close-Up V

The two irrational solutions to

2x2 + 6x + 3 = 0

correspond to the two x-intercepts for the graph of

y = 2x2 + 6x + 3.

-5 5

Now do Exercises 15–20 -3

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10-17 10.2 The Quadratic Formula 643

E X A M P L E 4

U Calculator Close-Up V

Because x2 + x + 5 = 0 has no real solutions, the graph of

y = x2 + x + 5

has no x-intercepts.

-6 6

Two imaginary solutions, no real solutions Find the complex solutions to x2 + x + 5 = 0.

Solution Let a = 1, b = 1, and c = 5 in the quadratic formula:

x =

= - -1 ±

2 V-19k -

= - -1 ±

2 iV19k -

Check these values in the original equation. The solution set is {--1 ± 2 iV19k -}. There are no real solutions to the equation.

-1 ± V(1)2 -k 4(1)(5k)k ---

2(1)

Now do Exercises 21–26 -2

You have learned to solve quadratic equations by four different methods: the even-root property, factoring, completing the square, and the quadratic formula. The even-root property and factoring are limited to certain special equations, but you should use those methods when possible. Any quadratic equation can be solved by completing the square or using the quadratic formula. Because the quadratic formula is usually faster, it is used more often than completing the square. However, complet ing the square is an important skill to learn. It will be used in the study of conic sec tions later in this text.

Summary of Methods for Solving ax2 + bx + c 0

Method Comments Examples

Even-root Use when b = 0. (x - 2)2 = 8 property x - 2 = ±V8k

Factoring Use when the polynomial x2 + 5x + 6 = 0 can be factored. (x + 2)(x + 3) = 0

Quadratic Solves any quadratic equation x2 + 5x + 3 = 0 formula

x =

Completing Solves any quadratic equation, x2 - 6x + 7 = 0 the square but quadratic formula is faster x2 - 6x + 9 = -7 + 9

(x - 3)2 = 2

-5 ± V25 - 4k(3)k --

U3V Number of Solutions The quadratic equations in Examples 1 and 3 had two real solutions each. In each of those examples, the value of b2 - 4ac was positive. In Example 2, the quadratic equa tion had only one solution because the value of b2 - 4ac was zero. In Example 4, the

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644 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-18

quadratic equation had no real solutions because b2 - 4ac was negative. Because b2 - 4ac determines the kind and number of solutions to a quadratic equation, it is called the discriminant.

Number of Solutions to a Quadratic Equation

The quadratic equation ax2 + bx + c = 0 with a * 0 has two real solutions if b2 - 4ac > 0, one real solution if b2 - 4ac = 0, and no real solutions (two imaginary solutions) if b2 - 4ac < 0.

E X A M P L E 5 Using the discriminant Use the discriminant to determine the number of real solutions to each quadratic equation.

a) x2 - 3x - 5 = 0 b) x2 = 3x - 9 c) 4x2 - 12x + 9 = 0

Solution a) For x2 - 3x - 5 = 0, use a = 1, b = -3, and c = -5 in b2 - 4ac:

b2 - 4ac = (-3)2 - 4(1)(-5) = 9 + 20 = 29

Because the discriminant is positive, there are two real solutions to this quadratic equation.

b) Rewrite x2 = 3x - 9 as x2 - 3x + 9 = 0. Then use a = 1, b = -3, and c = 9 in b2 - 4ac:

b2 - 4ac = (-3)2 - 4(1)(9) = 9 - 36 = -27

Because the discriminant is negative, the equation has no real solutions. It has two imaginary solutions.

c) For 4x2 - 12x + 9 = 0, use a = 4, b = -12, and c = 9 in b2 - 4ac:

b2 - 4ac = (-12)2 - 4(4)(9) = 144 - 144 = 0

Because the discriminant is zero, there is only one real solution to this quadratic equation.

Now do Exercises 27–42

U4V Applications With the quadratic formula we can easily solve problems whose solutions are irrational numbers. When the solutions are irrational numbers, we usually use a calculator to find rational approximations and to check.

E X A M P L E 6 Area of a tabletop The area of a rectangular tabletop is 6 square feet. If the width is 2 feet shorter than the length, then what are the dimensions?

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10-19 10.2 The Quadratic Formula 645

Solution Let x be the length and x - 2 be the width, as shown in Fig. 10.1. Because the area is 6 square feet and A = LW, we can write the equation

x(x - 2) = 6

x2 - 2x - 6 = 0.

Because this equation cannot be factored, we use the quadratic formula with a = 1, b = -2, and c = -6:

x =

= - 2 ±

2 V28k - = -

2 ± 2 2V7k - = 1 ± V7k

Because 1 - V7k is a negative number, it cannot be the length of a tabletop. If x = 1 + V7k, then x - 2 = 1 + V7k - 2 = V7k - 1. Checking the product of V7k + 1 and V7k - 1, we get

(V7k + 1)(V7k - 1) = 7 - 1 = 6.

The exact length is V7k + 1 feet, and the width is V7k - 1 feet. Using a calculator, we find that the approximate length is 3.65 feet and the approximate width is 1.65 feet.

2 ± V(-2)2k- 4(1)k(-6)k ---

2(1)

Figure 10.1

x - 2 ft

x ft

Now do Exercises 71–90

Warm-Ups ▼

Fill in the blank. 1. The formula can be used to solve any quadratic

equation.

2. The is b2 - 4ac.

3. In the number system every quadratic equation has at least one solution.

4. If b2 - 4ac = 0, then the quadratic equation has real solution.

5. If b2 - 4ac > 0, then the quadratic equation has real solutions.

6. If b2 - 4ac < 0, then the quadratic equation has imaginary solutions.

True or false? 7. Completing the square is used to develop the quadratic

formula. 8. The quadratic formula will not work on x2 - 3 = 0.

9. If a = 2, b = -3, and c = -4, then b2 - 4ac = 41.

10. If x2 + 4x - 5 = 0, then x = .

11. If 3x2 - 5x + 9 = 0, then x = .

12. If mx2 + nx + p = 0 and m � 0, then

x = . -n ± Vn2 - 4kmpk --

5 ± V25 - 4k(3)(9)k ---

-4 ± V16 - 4k(-5)k ---

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Exercises 1

0 .2

U Study Tips V • The last couple of weeks of the semester is not the time to slack off. This is the time to double your efforts. • Make a schedule and plan every hour of your time.

1 1 17 U2V Using the Formula 25. -- x2 + 13 = 5x 26. -- x2 + -- = 2x

2 4 4Solve each equation by using the quadratic formula. See Example 1.

1. x2 - 3x + 2 = 0 2. x2 - 7x + 12 = 0 U3V Number of Solutions3. x2 + 5x + 6 = 0 4. x2 + 4x + 3 = 0 Find b2 - 4ac and the number of real solutions to each

5. y2 + y = 6 6. m2 + 2m = 8 equation. See Example 5.

7. -6z2 + 7z + 3 = 0 8. -8q2 - 2q + 1 = 0 27. x 2 - 6x + 2 = 0 28. x2 + 6x + 9 = 0

29. -2x2 + 5x - 6 = 0 30. -x2 + 3x - 4 = 0

31. 4m2 + 25 = 20m 32. v2 = 3v + 5 Solve each equation by using the quadratic formula.

1 1 1 1 1 See Example 2. 33. y2 - - y + - 34. - w - - = 0- - = 0 - 2 - - w + -

2 4 2 3 4 9. 4x2 - 4x + 1 = 0 10. 4x2 - 12x + 9 = 0

11. -9x2 + 6x - 1 = 0 12. -9x2 + 24x - 16 = 0 35. -3t2 + 5t + 6 = 0 36. 9m2 + 16 = 24m

37. 9 - 24z + 16z2 = 0 38. 12 - 7x + x2 = 0 2 = 0 213. 9 + 24x + 16x 14. 4 + 20x = -25x 39. 5x2 - 7 = 0 40. -6x2 - 5 = 0

41. x2 = x 42. -3x2 + 7x = 0 Solve each equation by using the quadratic formula. See Example 3.

Miscellaneous 15. v2 + 8v + 6 = 0 16. p2 + 6p + 4 = 0

Solve by the method of your choice. See the Summary of Methods for Solving ax2 + bx + c = 0 on page 643.

17. -x2 - 5x + 1 = 0 18. -x2 - 3x + 5 = 0 43. -- y 44. - x2 + x = 1

1 2 + y = 1 1

- 4 2

1 1 3 1 19. - t2 - t + - 20. --x - = 0- - = 0 2 - 2x + - 1 1 1 4 53 6 4 2 45. -- x - x = - 46. -- w -- w2 + - - 2 + 1 =

3 2 3 9 3

Solve each equation by using the quadratic formula. 47. 3y2 + 2y - 4 = 0 48. 2y2 - 3y - 6 = 0 See Example 4.

21. 2t2 - 6t + 5 = 0 22. 2y2 + 1 = 2y w w y 2

49. -- = -- 50. -- = -- w - 2 w - 3 3y - 4 y + 4

23. -2x2 + 3x = 6 24. -3x2 - 2x - 5 = 0

9(3x - 5)2 25(2x + 1)2 51. -- = 1 52. -- = 0

4 9

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10-21

1 49 12 253. 25 - -- x = 0 54. -- - -- x = 0 3 2 4

20 8 34 6 55. 1 + -- = -- 56. -- = -- - 12 2x x x x

57. (x - 8)(x + 4) = -42 58. (x - 10)(x - 2) = -20

3(2y + 5) 7z - 4 59. y = -- 60. z = --

8(y - 1) 12(z - 1)

Use the quadratic formula and a calculator to solve each equation. Round answers to three decimal places and check your answers.

61. x2 + 3.2x - 5.7 = 0

262. x + 7.15x + 3.24 = 0

263. x - 7.4x + 13.69 = 0

64. 1.44x2 + 5.52x + 5.29 = 0

65. 1.85x2 + 6.72x + 3.6 = 0

66. 3.67x2 + 4.35x - 2.13 = 0

67. 3x2 + 14,379x + 243 = 0

68. x2 + 12,347x + 6741 = 0

69. x2 + 0.00075x - 0.0062 = 0

270. 4.3x - 9.86x - 3.75 = 0

U4V Applications

Find the exact solution(s) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. See Example 6.

71. Missing numbers. Find two positive real numbers that differ by 1 and have a product of 16.

72. Missing numbers. Find two positive real numbers that differ by 2 and have a product of 10.

73. More missing numbers. Find two real numbers that have a sum of 6 and a product of 4.

74. More missing numbers. Find two real numbers that have a sum of 8 and a product of 2.

10.2 The Quadratic Formula 647

75. Bulletin board. The length of a bulletin board is 1 foot more than the width. The diagonal has a length of V3k feet (ft). Find the length and width of the bulletin board.

76. Diagonal brace. The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures V6k m. Find the width and height.

V6 m

x � 2

Figure for Exercise 76

77. Area of a rectangle. The length of a rectangle is 4 ft longer than the width, and its area is 10 square feet (ft2). Find the length and width.

78. Diagonal of a square. The diagonal of a square is 2 m longer than a side. Find the length of a side.

If an object is given an initial velocity of v0 feet per second from a height of s0 feet, then its height S after t seconds is given by the formula S = -16t2 + v0t + s0.

79. Projected pine cone. If a pine cone is projected upward at a velocity of 16 ft/sec from the top of a 96-foot pine tree, then how long does it take to reach the earth?

80. Falling pine cone. If a pine cone falls from the top of a 96-foot pine tree, then how long does it take to reach the earth?

81. Tossing a ball. A ball is tossed into the air at 10 ft/sec from a height of 5 feet. How long does it take to reach the earth?

82. Time in the air. A ball is tossed into the air from a height of 12 feet at 16 ft/sec. How long does it take to reach the earth?

83. Penny tossing. If a penny is thrown downward at 30 ft/sec from the bridge at Royal Gorge, Colorado, how long does it take to reach the Arkansas River 1000 ft below?

84. Foul ball. Suppose Charlie O’Brian of the Braves hits a baseball straight upward at 150 ft/sec from a height of 5 ft.

a) Use the formula to determine how long it takes the ball to return to the earth.

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648 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-22

b) Use the accompanying graph to estimate the maximum height reached by the ball.

H ei

gh t (

ft )

400

300

200

100

0 0 2 4 6 8 10

Time (sec)

Figure for Exercise 84

Solve each problem.

85. Kitchen countertop. A 30 in. by 40 in. countertop for a work island is to be covered with green ceramic tiles, except for a border of uniform width as shown in the figure. If the area covered by the green tiles is 704 square inches (in.2), then how wide is the border?

40 in.30 in.

Figure for Exercise 85

86. Recovering an investment. The manager at Cream of the Crop bought a load of watermelons for $200. She priced the melons so that she would make $1.50 profit on each melon. When all but 30 had been sold, the manager

person will decrease by $2000. How many members are currently in the club?

89. Farmer’s delight. The manager of Farmer’s Delight bought a load of watermelons for $750 and priced the watermelons so that he would make a profit of $2 on each melon. When all but 100 of the melons had been sold, he broke even. How many did he buy originally?

90. Traveling club. The members of a traveling club plan to share equally the cost of a $150,000 motorhome. If they can find 10 more people to join and share the cost, then the cost per person will decrease by $1250. How many members are there originally in the club?

Getting More Involved

91. Discussion 2Find the solutions to 6x + 5x - 4 = 0. Is the sum of

byour solutions equal to ---? Explain why the sum of a

bthe solutions to any quadratic equation is ---. a

(Hint: Use the quadratic formula.)

92. Discussion 2 1Use the result of Exercise 91 to check whether {--, --}3 32is the solution set to 9x - 3x - 2 = 0. If this

solution set is not correct, then what is the correct solution set?

93. Discussion

What is the product of the two solutions to 6x2 + 5x - 4 = 0? Explain why the product of the

csolutions to any quadratic equation is --. a

94. Discussion

Use the result of Exercise 93 to check whether 9 2{--, -2} is the solution set to 2x - 13x + 18 = 0. 2

If this solution set is not correct, then what is the correct solution set?

had recovered her initial investment. How many did she Graphing Calculator Exercises buy originally?

87. Baby shower. A group of office workers plans to share equally the $100 cost of giving a baby shower for a coworker. If they can get six more people to share the cost, then the cost per person will decrease by $15. How many people are in the original group?

88. Sharing cost. The members of a flying club plan to share equally the cost of a $200,000 airplane. The members want to find five more people to join the club so that the cost per

Determine the number of real solutions to each equation by examining the calculator graph of y = ax2 + bx + c. Use the discriminant to check your conclusions.

295. x - 6.33x + 3.7 = 0 296. 1.8x + 2.4x - 895 = 0

297. 4x - 67.1x + 344 = 0 298. -2x - 403 = 0

99. -x2 + 30x - 226 = 0 2100. 16x - 648x + 6562 = 0

http:4x-67.1x

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10-23 10.3 More on Quadratic Equations 649

In This Section

U1V Writing a Quadratic Equation with Given Solutions

U2V Using the Discriminant in Factoring

10.3 More on Quadratic Equations

In this section, we use the ideas and methods of the previous sections to explore additional topics involving quadratic equations.

U1V Writing a Quadratic Equation with Given Solutions U3V Equations Quadratic in Form Not every quadratic equation can be solved by factoring, but the factoring method can U4V Applications be used (in reverse) to write a quadratic equation with given solutions.

E X A M P L E 1 Writing a quadratic given the solutions Write a quadratic equation that has each given pair of solutions.

a) 4, -6 b) -V2k, V2k c) -3i, 3i

Solution a) Reverse the factoring method using solutions 4 and -6:

x = 4 or x = -6 x - 4 = 0 or x + 6 = 0

(x - 4)(x + 6) = 0 Zero factor property x2 + 2x - 24 = 0 Multiply the factors.

b) Reverse the factoring method using solutions -V2k and V2k:

x = -V2k or x = V2k x + V2k = 0 or x - V2k = 0

(x + V2k)(x - V2k) = 0 Zero factor property x2 - 2 = 0 Multiply the factors.

c) Reverse the factoring method using solutions -3i and 3i:

x = -3i or x = 3i x + 3i = 0 or x - 3i = 0

(x + 3i)(x - 3i) = 0 Zero factor property x2 - 9i2 = 0 Multiply the factors.

x2 + 9 = 0 Note: i2 = -1

U Calculator Close-Up V

The graph of y = x2 + 2x - 24 sup ports the conclusion in Example 1(a) because the graph crosses the x-axis at (4, 0) and (-6, 0).

-30

-8 6

Now do Exercises 1–12

The process in Example 1 can be shortened somewhat if we observe the corre spondence between the solutions to the equation and the factors.

Correspondence Between Solutions and Factors

If a and b are solutions to a quadratic equation, then the equation is equivalent to

(x - a)(x - b) = 0.

So if 2 and -3 are solutions to a quadratic equation, then the quadratic equation is (x - 2)(x + 3) = 0 or x2 + x - 6 = 0. If the solutions are fractions, it is not necessary

2to use fractions in the factors. For example, if -- is a solution, then 3x - 2 is a factor 3

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650 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-24

2 1because 3x - 2 = 0 is equivalent to x = --. If --- is a solution, then 5x + 1 is a fac 3 51 2 1tor because 5x + 1 = 0 is equivalent to x = ---. So if -3 - and --5 - are solutions to a qua5

dratic equation, then the equation is (3x - 2)(5x + 1) = 0 or 15x2 - 7x - 2 = 0.

U2V Using the Discriminant in Factoring b2 - 4k-b ± VkacThe quadratic formula x = -- gives the solutions to the quadratic equation

2a ax2 + bx + c = 0. If a, b, and c are integers and b2 - 4ac is a perfect square, then

Vb2 - 4k is a whole number and the quadratic formula produces solutions that arekac rational. The quadratic equations with rational solutions are precisely the ones that we solve by factoring. So we can use the discriminant b2 - 4ac to determine whether a quadratic polynomial is prime.

Identifying Prime Quadratic Polynomials Using b2 � 4ac

Let ax2 + bx + c be a quadratic polynomial with integral coefficients having a greatest common factor of 1. The quadratic polynomial is prime if and only if the discriminant b2 - 4ac is not a perfect square.

E X A M P L E 2 Using the discriminant Use the discriminant to determine whether each polynomial can be factored.

a) 6x2 + x - 15 b) 5x2 - 3x + 2

Solution a) Use a = 6, b = 1, and c = -15 to find b2 - 4ac:

b2 - 4ac = 12 - 4(6)(-15) = 361

Because V361k = 19, 6x2 + x - 15 can be factored. Using the ac method, we get

6x2 + x - 15 = (2x - 3)(3x + 5).

b) Use a = 5, b = -3, and c = 2 to find b2 - 4ac:

b2 - 4ac = (-3)2 - 4(5)(2) = -31

Because the discriminant is not a perfect square, 5x2 - 3x + 2 is prime.

Now do Exercises 13–24

U3V Equations Quadratic in Form An equation in which an expression appears in place of x in ax2 + bx + c = 0 is called quadratic in form. So,

3(x - 7)2 + (x - 7) + 8 = 0,

-2(x2 + 3)2 - (x2 + 3) + 1 = 0, and -7x4 + 5x2 - 6 = 0

are quadratic in form. Note that the last equation is quadratic in form because it could be written as -7(x2)2 + 5(x2) - 6, where x2 is used in place of x. To solve an equation

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10-25 10.3 More on Quadratic Equations 651

that is quadratic in form we replace the expression with a single variable and then solve the resulting quadratic equation, as shown in Example 3.

E X A M P L E 3 An equation quadratic in form Solve (x + 15)2 - 3(x + 15) - 18 = 0.

Solution Note that x + 15 and (x + 15)2 both appear in the equation. Let a = x + 15 and substitute a for x + 15 in the equation:

(x + 15)2 - 3(x + 15) - 18 = 0

a2 - 3a - 18 = 0

(a - 6)(a + 3) = 0 Factor.

a - 6 = 0 or a + 3 = 0

a = 6 or a = -3

x + 15 = 6 or x + 15 = -3 Replace a by x + 15.

x = -9 or x = -18

Check in the original equation. The solution set is {-18, -9}. Now do Exercises 25–30

In Example 4, we have a fourth-degree equation that is quadratic in form. Note that the fourth-degree equation has four solutions.

E X A M P L E 4

U Helpful Hint V

The fundamental theorem of algebra says that the number of solutions to a polynomial equation is less than or equal to the degree of the polynomial. This famous theorem was proved by Carl Friedrich Gauss when he was a young man.

A fourth-degree equation Solve x4 - 6x2 + 8 = 0.

Solution Note that x4 is the square of x2. If we let w = x2, then w2 = x4. Substitute these expressions into the original equation.

x4 - 6x2 + 8 = 0

w2 - 6w + 8 = 0 Replace x4 by w2 and x2 by w.

(w - 2)(w - 4) = 0 Factor.

w - 2 = 0 or w - 4 = 0

w = 2 or w = 4

x2 = 2 or x2 = 4 Substitute x2 for w.

x = ±V2k or x = ±2 Even-root property

Check. The solution set is {-2, -V2k, V2k, 2}. Now do Exercises 31–38

CAUTION If you replace x2 by w, do not quit when you find the values of w. If the variable in the original equation is x, then you must solve for x.

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652 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-26

E X A M P L E 5 A quadratic within a quadratic Solve (x2 + 2x)2 - 11(x2 + 2x) + 24 = 0.

Solution Note that x2 + 2x and (x2 + 2x)2 appear in the equation. Let a = x2 + 2x and substitute.

a2 - 11a + 24 = 0

(a - 8)(a - 3) = 0 Factor.

a - 8 = 0 or a - 3 = 0

a = 8 or a = 3

x2 + 2x = 8 or x2 + 2x = 3 Replace a by x2 + 2x.

x2 + 2x - 8 = 0 or x2 + 2x - 3 = 0

(x - 2)(x + 4) = 0 or (x + 3)(x - 1) = 0

x - 2 = 0 or x + 4 = 0 or x + 3 = 0 or x - 1 = 0

x = 2 or x = -4 or x = -3 or x = 1

Check. The solution set is {-4, -3, 1, 2}.

U Calculator Close-Up V

The four x-intercepts on the graph of

y = (x2 + 2x)2 -11(x2 + 2x) + 24

support the conclusion in Example 5.

-20

-6 6

Now do Exercises 39–44

Example 6 involves a fractional exponent. To identify this type of equation as quadratic in form, recall how to square an expression with a fractional exponent. For example, (x1�2)2 = x, (x1�4)2 = x1�2, and (x1�3)2 = x2�3.

E X A M P L E 6 A fractional exponent Solve x - 9x1�2 + 14 = 0.

Solution Note that the square of x1�2 is x. Let w = x1�2; then w2 = (x1�2)2 = x. Now substitute w and w2 into the original equation:

w2 - 9w + 14 = 0 (w - 7)(w - 2) = 0

w - 7 = 0 or w - 2 = 0 w = 7 or w = 2

x1�2 = 7 or x1�2 = 2 Replace w by x1�2 . x = 49 or x = 4 Square each side.

Because we squared each side, we must check for extraneous roots. First evaluate x - 9x1�2 + 14 for x = 49:

49 - 9 · 491�2 + 14 = 49 - 9 · 7 + 14 = 0

Now evaluate x - 9x1�2 + 14 for x = 4:

4 - 9 · 41�2 + 14 = 4 - 9 · 2 + 14 = 0

Because each solution checks, the solution set is {4, 49}.

An equation of quadratic form with variable x must have a power of x and 4 1�2 1�3its square. Equations such as x - 5x3 + 6 = 0 or x - 3x - 18 = 0

are not quadratic in form and cannot be solved by substitution.

CAUTION

Now do Exercises 45–52

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10-27 10.3 More on Quadratic Equations 653

U4V Applications Applied problems often result in quadratic equations that cannot be factored. For such equations we use the quadratic formula to find exact solutions and a calculator to find decimal approximations for the exact solutions.

E X A M P L E 7 Changing area Marvin’s flower bed is rectangular in shape with a length of 10 feet and a width of 5 feet (ft). He wants to increase the length and width by the same amount to obtain a flower bed with an area of 75 square feet (ft2). What should the amount of increase be?

Solution Let x be the amount of increase. The length and width of the new flower bed are x + 10 ft

x ft and x + 5 ft, respectively, as shown in Fig. 10.2. Because the area is to be 75 ft2, we have

(x + 10)(x + 5) = 75.

Write this equation in the form ax2 + bx + c = 0:

x2 + 15x + 50 = 75

x2 + 15x - 25 = 0 Get 0 on the right. 10 ft

-15 ± V225 -k25)k 4(1)(-k x =---

2(1)

-15 ± V325 -15 ± kk 5V13 = -- = --

2 2

Because the value of x must be positive, the exact increase isx ft 5 ft

-15 + 5V13 kFigure 10.2 -- feet. 2

Using a calculator, we can find that x is approximately 1.51 ft. If x = 1.51 ft, then the new length is 11.51 ft, and the new width is 6.51 ft. The area of a rectangle with these dimensions is 74.93 ft2. Of course, the approximate dimensions do not give an area of exactly 75 ft2.

Now do Exercises 79–86

E X A M P L E 8 Mowing the lawn It takes Carla 1 hour longer to mow the lawn than it takes Sharon to mow the lawn. If they can mow the lawn in 5 hours working together, then how long would it take each girl by herself?

Solution 1If Sharon can mow the lawn by herself in x hours, then she works at the rate of -- of the x

lawn per hour. If Carla can mow the lawn by herself in x + 1 hours, then she works at the rate of -1 - of the lawn per hour. We can use a table to list all of the important quantities. x + 1

Rate Time Work

1 lawn 5 Sharon - - 5 hr - lawn xx hr

51 lawn Carla - - 5 hr - lawn x + 1 hr x + 1

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654 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-28

Because they complete the lawn in 5 hours, the portion of the lawn done by Sharon and the portion done by Carla have a sum of 1:

- 5 x

- + - x +

5 1

- = 1

x(x + 1) - 5 x

- + x(x + 1) - x +

5 1

- = x(x + 1)1 Multiply by the LCD.

5x + 5 + 5x = x2 + x 10x + 5 = x2 + x

-x2 + 9x + 5 = 0 x2 - 9x - 5 = 0

x =

= - 9 ± V

2 101k

Using a calculator, we find that -9 - V 2

101k - is negative. So Sharon’s time alone is

- 9 + V

2 101k

- hours.

To find Carla’s time alone, we add 1 hour to Sharon’s time:

- 9 + V

2 101k

- + 1 = - 9 + V

2 101k

- + - 2 2

- = - 11 +

2 V101k - hours

Sharon’s time alone is approximately 9.525 hours, and Carla’s time alone is approximately 10.525 hours.

9 ± V(-9)2k- 4(1)k(-5)k ---

2(1)

U Helpful Hint V

Note that the equation concerns the portion of the job done by each girl. We could have written an equation about the rates at which the two girls work. Because they can finish the lawn together in 5 hours, they are mowing together at the rate of -1

5 - lawn

per hour. So,

- 1 x

- + - x +

1 1

- = - 1 5

Now do Exercises 87–90

Warm-Ups ▼

Fill in the blank. 1. If d is a solution to a quadratic equation, then x - d is a

of the quadratic polynomial.

2. If b2 - 4ac is not a perfect square, then ax2 + bx + c is a polynomial.

3. An equation that is quadratic after a substitution is called in form.

4. If m and n are to a quadratic equation, then (x - m)(x - n) = 0 is a quadratic equation with those solutions.

True or false? 5. The equation (x - 4)(x + 5) = 0 is a quadratic equation

with solutions 4 and -5. 6. If w = x1�6, then w2 = x1�3 .

7. If y = 21�2, then y2 = 21�4 . 8. To solve x4 - 5x2 + 6 = 0 by substitution, let w = x2 .

9. To solve x5 - 3x3 - 10 = 0 by substitution, let w = x3 .

10. If Ann’s boat goes 10 mph in still water, then against a 5-mph current it will go 2 mph.

11. If Elvia drives 300 miles in x hours, then her rate is

- 30

x 0

- mph.

12. If John paints a 100-foot fence in x hours, then his rate is

- 10

x 0

- of the fence per hour.

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- - - -

Exercises

1 0

U Study Tips V • Establish a regular routine of eating, sleeping, and exercise. • The ability to concentrate depends on adequate sleep, decent nutrition, and the physical well-being that comes with exercise.

U1V Writing a Quadratic Equation with Given Solutions

For each given pair of numbers find a quadratic equation with integral coefficients that has the numbers as its solutions. See Example 1.

1. 3, -7 2. -8, 2 3. 4, 1 4. 3, 2 5. Vs5, -Vs5 6. -Vs7, Vs7 7. 4i, -4i 8. -3i, 3i 9. iVs2, -iVs2 10. 3iVs2, -3iVs2

1 1 1 1 11. --, -- 12. - --, - --

2 3 5 2

U2V Using the Discriminant in Factoring

Use the discriminant to determine whether each quadratic polynomial can be factored, and then factor the ones that are not prime. See Example 2.

13. x2 + 9 14. x2 - 9 15. 2x2 - x + 4 16. 2x2 + 3x - 5

17. 2x2 + 6x - 5 18. 3x2 + 5x - 1

19. 6x2 + 19x - 36 20. 8x2 + 6x - 27

21. 4x2 - 5x - 12 22. 4x2 - 27x + 45

23. 8x2 - 18x - 45 24. 6x2 + 9x - 16

U3V Equations Quadratic in Form

Find all real solutions to each equation. See Example 3.

25. (x - 1)2 - 2(x - 1) - 8 = 0 26. (m + 3)2 + 5(m + 3) - 14 = 0

27. (2a - 1)2 + 2(2a - 1) - 8 = 0

28. (3a + 2)2 - 3(3a + 2) = 10

29. (w - 1)2 + 5(w - 1) + 5 = 0

30. (2x - 1)2 - 4(2x - 1) + 2 = 0

Find all real solutions to each equation. See Example 4.

31. x4 - 13x2 + 36 = 0

32. x4 - 20x2 + 64 = 0

33. x6 - 28x3 + 27 = 0

34. x6 - 3x3 - 4 = 0

35. x4 - 14x2 + 45 = 0

36. x4 + 2x2 = 15

37. x6 + 7x3 = 8 38. a6 + 6a3 = 16

Find all real solutions to each equation. See Example 5.

39. (x2 + 1)2 - 11(x2 + 1) = -10 40. (x2 + 2)2 - 11(x2 + 2) = -30 41. (x2 + 2x)2 - 7(x2 + 2x) + 12 = 0 42. (x2 + 3x)2 + (x2 + 3x) - 20 = 0 43. (y2 + y)2 - 8(y2 + y) + 12 = 0 44. (w2 - 2w)2 + 24 = 11(w2 - 2w)

Find all real solutions to each equation. See Example 6.

45. x - 3x112 + 2 = 0

46. x112 - 3x114 + 2 = 0

47. x213 + 4x113 + 3 = 0

48. x213 - 3x113 - 10 = 0

49. x112 - 5x114 + 6 = 0

50. 2x - 5Vsx + 2 = 0

112 - 3 = 0 114 + 2 = 11251. 2x - 5x 52. x x

Find all real solutions to each equation.

53. x 2 + x 1 - 6 = 0 54. x 2 - 2x 1 = 8

116 -55. x x113 + 2 = 0 56. x213 - x113 - 20 = 0

57. --) 2

+ - = 6-(y - 1

1 (y - 1

1) 1 1

58. -- - 2(--) - 24 = 0(w + 1) 2

w + 1

59. 2x2 - 3 - 6V2sx2 - s3 + 8 = 0

60. x x + V 2 + x2 + xs - 2 = 0

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656 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-30

x -2 - 2x -161. - 1 = 0

x -2 - 6x -1 + 6 = 062.

Miscellaneous

Find all real and imaginary solutions to each equation. 2 263. w + 4 = 0 64. w + 9 = 0

65. a4 + 6a2 + 8 = 0

66. b4 + 13b2 + 36 = 0 467. m - 16 = 0

68. t4 - 4 = 0

69. 16b4 - 1 = 0 70. b4 - 81 = 0

371. x + 1 = 0

372. x - 1 = 0

373. x + 8 = 0

374. x - 27 = 0

a -275. - 2a -1 + 5 = 0

76. b-2 - 4b-1 + 6 = 0

77. (2x - 1)2 - 2(2x - 1) + 5 = 0

78. (4x - 1)2 - 6(4x - 1) + 25 = 0

U4V Applications

Find the exact solution to each problem. If the exact solution is an irrational number, then also find an approximate decimal solution. See Examples 7 and 8.

79. Country singers. Harry and Gary are traveling to Nashville to make their fortunes. Harry leaves on the train at 8:00 A.M. and Gary travels by car, starting at 9:00 A.M. To complete the 300-mile trip and arrive at the same time as Harry, Gary travels 10 miles per hour (mph) faster than the train. At what time will they both arrive in Nashville?

80. Gone fishing. Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4-mph current, it took her 20 minutes longer to get there than to return. How fast will her boat go in still water?

81. Cross-country cycling. Erin was traveling across the desert on her bicycle. Before lunch she traveled 60 miles (mi); after lunch she traveled 46 mi. She put in 1 hour more after lunch than before lunch, but her speed was

4 mph slower than before. What was her speed before lunch and after lunch?

Photo for Exercise 81

82. Extreme hardship. Kim starts to walk 3 mi to school at 7:30 A.M. with a temperature of 0°F. Her brother Bryan starts at 7:45 A.M. on his bicycle, traveling 10 mph faster than Kim. If they get to school at the same time, then how fast is each one traveling?

83. American pie. John takes 3 hours longer than Andrew to peel 500 pounds (lb) of apples. If together they can peel 500 lb of apples in 8 hours, then how long would it take each one working alone?

84. On the half shell. It takes Brent 1 hour longer than Calvin to shuck a sack of oysters. If together they shuck a sack of oysters in 45 minutes, then how long would it take each one working alone?

85. The growing garden. Eric’s garden is 20 ft by 30 ft. He wants to increase the length and width by the same amount to have a 1000-ft2 garden. What should be the new dimensions of the garden?

86. Open-top box. Thomas is going to make an open-top box by cutting equal squares from the four corners of an 11 inch by 14 inch sheet of cardboard and folding up the

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10-31 10.3 More on Quadratic Equations 657

sides. If the area of the base is to be 80 square inches, then what size square should be cut from each corner?

14 in. x x

11 in.

Figure for Exercise 86

87. Pumping the pool. It takes pump A 2 hours less time than pump B to empty a certain swimming pool. Pump A is started at 8:00 A.M., and pump B is started at 11:00 A.M. If the pool is still half full at 5:00 P.M., then how long would it take pump A working alone?

88. Time off for lunch. It usually takes Eva 3 hours longer to do the monthly payroll than it takes Cicely. They start working on it together at 9:00 A.M. and at 5:00 P.M. they have 90% of it done. If Eva took a 2-hour lunch break while Cicely had none, then how much longer will it take for them to finish the payroll working together?

89. Golden Rectangle. One principle used by the ancient Greeks to get shapes that are pleasing to the eye in art and architecture was the Golden Rectangle. If a square is removed from one end of a Golden Rectangle, as shown in the figure, the sides of the remaining rectangle are proportional to the original rectangle. So the length and width of the original rectangle satisfy

L W -- = --. W L - W

If the length of a Golden Rectangle is 10 meters, then what is its width?

L - W L

Figure for Exercise 89

90. Golden painting. An artist wants her painting to be in the shape of a Golden Rectangle. If the length of the painting is 36 inches, then what should be the width? See Exercise 89.

Getting More Involved 91. Exploration

2a) Given that P(x) = x4 + 6x - 27, find P(3i), P(-3i), P(V3k), and P(-V3k ).

b) What can you conclude about the values 3i, -3i, V3k, and -V3k and their relationship to each other?

92. Cooperative learning

Work with a group to write a quadratic equation that has each given pair of solutions.

a) 3 + V5k, 3 - V5k b) 4 - 2i, 4 + 2i 1 + iVk3 1 - iV3k

c) --, -- 2 2

Graphing Calculator Exercises

Solve each equation by locating the x-intercepts on a calculator graph. Round approximate answers to two decimal places.

93. (5x - 7)2 - (5x - 7) - 6 = 0

4 294. x - 116x + 1600 = 0

2 295. (x + 3x)2 - 7(x + 3x) + 9 = 0

2 1�296. x - 3x - 12 = 0

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658 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-32

Mid-Chapter Quiz Sections 10.1 through 10.3 Chapter 10

Solve each equation by factoring. Solve by any method.

1. x2 - 4x - 32 = 0 9. (x + 7) 2 - 5(x + 7) + 6 = 0

2. 6x2 - 5x + 1 = 0 10. x 4 - 6x2 + 5 = 0

Solve using the even-root property.

3. x2 = 4 1 2 6 5 4 4. (w + 3)2 = 6

11. x - 2x112 - 8 = 0

12. �2x - 3 = 4x - 2

4 4

Solve by completing the square. Miscellaneous.

5. x2 - 4x = 1 6. 2z2 + z = 1 13. Find the imaginary solutions to x2 - 10x + 26 = 0.

Solve by using the quadratic formula. 14. Find the discriminant for the equation 3x2 - x + 5 = 0.

7. 2x2 - 5x + 2 = 0 15. Find a quadratic equation that has -3 and 8 as its

8. 2h2 + 4h + 1 = 0 solutions.

In This Section

U1V Finding Ordered Pairs

U2V Graphing Quadratic Functions

U3V The Vertex and Intercepts

U4V Applications

10.4 Graphing Quadratic Functions

An equation of the form y = mx + b is a linear function. Its graph is a straight line. An equation of the form y = ax2 + bx + c (with a * 0) is a quadratic function. We will see in this section that all quadratic functions have similar graphs that are in the shape of a parabola. Note that a linear function is a first-degree polynomial function and a quadratic function is a second-degree polynomial function.

U1V Finding Ordered Pairs It is straightforward to calculate y when given x for an equation of the form y = ax2 + bx + c. However, if we are given y and want to find x, then we must use methods for solving quadratic equations.

E X A M P L E 1 Finding ordered pairs Complete each ordered pair so that it satisfies the given equation.

a) (2, ), ( , 0), y = x2 - x - 6

b) (0, ), ( , 20), y = -16x2 + 48x + 84

Solution a) If x = 2, then y = 22 - 2 - 6 = -4. So the ordered pair is (2, -4). To find x

when y = 0, replace y by 0 and solve the resulting quadratic equation:

x2 - x - 6 = 0 (x - 3)(x + 2) = 0

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10-33 10.4 Graphing Quadratic Functions 659

x - 3 = 0 or x + 2 = 0

x = 3 or x = -2

The ordered pairs are (-2, 0) and (3, 0).

b) If x = 0, then y = -16 � 02 + 48 � 0 + 84 = 84. The ordered pair is (0, 84). To find x when y = 20, replace y by 20 and solve the equation for x:

-16x2 + 48x + 84 = 20

-16x2 + 48x + 64 = 0 Subtract 20 from each side.

x2 - 3x - 4 = 0 Divide each side by -16.

(x - 4)(x + 1) = 0 Factor.

x - 4 = 0 or x + 1 = 0 Zero factor property

x = 4 or x = -1

The ordered pairs are (-1, 20) and (4, 20).

Now do Exercises 1–4

U2V Graphing Quadratic Functions All equations of the form y = ax2 + bx + c with a * 0 have graphs that are similar in shape. The graph of any equation of this form is called a parabola. Note that any real number can be used in place of x.

E X A M P L E 2 The simplest parabola Make a table of ordered pairs that satisfy y = x2, and then sketch the graph of y = x2 .

Solution Make a table of values for x and y :

x -2 -1 0 1 2

y = x2 4 1 0 1 4

Plot the ordered pairs from the table, and draw a parabola through the points as shown in Fig. 10.3.

Figure 10.3

y = x2

-4

1 -2

2 3

4-1-2-3

U Calculator Close-Up V

This close-up view of y = x2 shows how rounded the curve is at the bottom. When drawing a parabola by hand, be sure to draw it smoothly.

�1

�2

Now do Exercises 5–14

The parabola in Example 2 is said to open upward. In Example 3 we see a parabola that opens downward. If a � 0 in the equation y = ax2 + bx + c, then the parabola opens upward. If a � 0, then the parabola opens downward.

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�

660 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-34

Note the symmetry of the parabola in Fig. 10.3. If the paper was folded along the y-axis, the two sides of the parabola would come together. The point (-1, 1) would match up with (1, 1), the point (-2, 4) would match up with (2, 4), and so on.

E X A M P L E 3 A parabola opening downward Graph y = 4 - x2 .

Solution We plot enough points to get the correct shape of the graph:

y = 4 - x2

x-1

-3

1 -1

-2

Figure 10.4

x -2 -1 0 1 2

y = 4 - x2 0 3 4 3 0

See Fig. 10.4 for the graph.

Now do Exercises 15–20

U3V The Vertex and Intercepts The lowest point on a parabola that opens upward or the highest point on a parabola that opens downward is called the vertex. The y-coordinate of the vertex is the minimum value of y if the parabola opens upward, and it is the maximum value of y if the parabola opens downward. For y = x2 the vertex is (0, 0), and 0 is the minimum value of y. For y = 4 - x2 the vertex is (0, 4), and 4 is the maximum value of y.

If y = ax2 + bx + c has x-intercepts, they can be found by solving ax2 + bx + c = 0 by the quadratic formula. The vertex is midway between the x-intercepts as shown in Fig. 10.5. Note that in the quadratic formula

b2 x =44 ,

-b ± - 4ac 2a

-b -bb2 - 4ac is added and subtracted from the numerator of 44. So (44, 0) is the point2a 2a midway between the x-intercepts and the vertex has the same x-coordinate. Even if the

-bparabola has no x-intercepts, the x-coordinate of the vertex is still 44. 2a

�b � �b2 � 4ac 2a �b � �b2 � 4ac

y � ax

�b 2a

Vertex

2 � bx � c

Figure 10.5

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10-35 10.4 Graphing Quadratic Functions 661

Vertex of a Parabola -bThe x-coordinate of the vertex of y = ax2 + bx + c is 44, provided that a * 0. 2a

When you graph a parabola, you should always locate the vertex because it is the point at which the graph “turns around.” With the vertex and several nearby points you can see the correct shape of the parabola.

Using function notation we can write f(x) = ax2 + bx + c rather than y = ax2 + bx + c. With this notation, the coordinates of the vertex are

-b -b x = 44 and y = f (44). 2a 2a

Note that in this context we are thinking of f as the name of the function rather than as a variable. We are keeping x and y as the variables and using the function called f to find y for any given x.

E X A M P L E 4 Using the vertex in graphing a parabola Find the vertex and graph f(x) = -x2 - x + 2.

Solution First find the x-coordinate of the vertex:

x = 4 -

2a b 4 = 4

2( ( -

1 1 ) )

4 = 4 -

1 2 4 = -4

1 2

Now find f (-4 1 24):

f (-4 1 24) = -(-4 1 24) 2

- (-4 1 24) + 2 = -4 1 44 + 4 1 24 + 2 = 4 9 44 The vertex is (-4 1 24, 4 9 44). Now find a few points on either side of the vertex:

x -2 -1 -4 1 2

4 0 1

f (x) = -x2 - x + 2 0 2 4 9 4

4 2 0

Sketch a parabola through these points as in Fig. 10.6.

Now do Exercises 21–28

f (x) = -x2 - x + 2

x-1

-3

-1

-2

-3

-4

2 3

Figure 10.6

The y-intercept for the parabola y = ax2 + bx + c is the point that has 0 as its x-coordinate. If we let x = 0, we get y = a(0)2 + b(0) + c = c. So the y-intercept is (0, c). To find the x-intercepts let y = 0 and solve ax2 + bx + c = 0. A parabola may have 0, 1, or 2 x-intercepts depending on the number of solutions to this equation.

Finding Intercepts

The y-intercept for y = ax2 + bx + c is (c, 0). To find the x-intercepts solve ax2 + bx + c = 0.

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�

� � �

��

�

��

� � �

662 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-36

E X A M P L E 5 Using the intercepts in graphing a parabola Find the vertex and intercepts, and sketch the graph of each parabola.

a) f(x) = x2 - 2x - 8

b) s = -16t2 + 64t

Solution a) Use x = 4

2a b 4 to get x = 1 as the x-coordinate of the vertex. If x = 1, then

f(1) = 12 - 2 � 1 - 8

= -9.

So the vertex is (1, -9). If x = 0, then

f(0) = 02 - 2 � 0 - 8

= -8.

The y-intercept is (0, -8). To find the x-intercepts, replace f(x) by 0:

x2 - 2x - 8 = 0

(x - 4)(x + 2) = 0

x - 4 = 0 or x + 2 = 0

x = 4 or x = -2

The x-intercepts are (-2, 0) and (4, 0). The graph is shown in Fig. 10.7.

b) Because s is expressed in terms of t in the equation s = -16t2 + 64t, the inde pendent variable is t and the dependent variable is s. Since we always put the independent variable first in an ordered pair, the ordered pairs are written in

the form (t, s). To find the vertex use t = -4 2 b a 4 to get

t = 4 2(

6) 4 = 2.

If t = 2, then

s = -16 � 22 + 64 � 2

= 64.

So the vertex is (2, 64). If t = 0, then

s = -16 � 02 + 64 � 0

= 0.

So the s-intercept is (0, 0). To find the t-intercepts, replace s by 0:

-16t2 + 64t = 0

-16t(t - 4) = 0

-16t = 0 or t - 4 = 0

t = 0 or t = 4

The t-intercepts are (0, 0) and (4, 0). The graph is shown in Fig. 10.8.

21 3 5 6 7

s 16t2 64t

(2, 64)

Figure 10.8

21 3 5

(1, 9)10

f (x) x2 2x 8

Figure 10.7

Now do Exercises 29–44

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10-37 10.4 Graphing Quadratic Functions 663

U Calculator Close-Up V

U4V Applications In applications we are often interested in finding the maximum or minimum value of a variable. If the graph of a parabola opens downward, then the maximum value of the dependent variable is the second coordinate of the vertex. If the parabola opens upward, then the minimum value of the dependent variable is the second coordinate of the vertex.

E X A M P L E 6 Finding the maximum height If a projectile is launched with an initial velocity of v0 feet per second from an initial height of s0 feet, then its height s(t) in feet is determined by s(t) = -16t2 + v0t + s0, where t is the time in seconds. If a ball is tossed upward with velocity 64 feet per second from a height of 5 feet, then what is the maximum height reached by the ball?

Solution The height s(t) of the ball for any time t is given by s(t) = -16t2 + 64t + 5. Because the maximum height occurs at the vertex of the parabola, we use t = 4

2a b 4 to find the vertex:

t = 4 2(

6 1 4 6)

4 = 2

Now use t = 2 to find the second coordinate of the vertex:

s(2) = -16(2)2 + 64(2) + 5 = 69

The maximum height reached by the ball is 69 feet. See Fig. 10.9.

210 3 4 5 6 7

s(t) 16t2 64t 5 (2, 69)

Figure 10.9

Now do Exercises 53–61

You can find the vertex of a parabola with a calculator by using either the maximum or minimum feature. First graph the parabola as shown.

10 10

Because this parabola opens upward, the y-coordinate of the vertex is the minimum

y-coordinate on the graph. Press CALC and choose minimum.

The calculator will ask for a left bound, a right bound, and a guess. For the left bound choose a point to the left of the vertex by

moving the cursor to the point and pressing ENTER. For the right bound choose a point to the right of the vertex. For the guess choose a point close to the vertex.

10 10

Note that the graph in Fig. 10.9 shows the height of the ball as a function of time. It does not show the path of the ball. The ball in Example 6 is tossed straight upward and falls straight downward. The path of a projectile is discussed in trigonometry. It turns out that a ball thrown nonvertically travels through the air on a parabolic path, which depends on the velocity and the angle at which it is thrown.

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U1V Finding Ordered Pairs

Complete each ordered pair so that it satisfies the given equation. See Example 1.

1. y = x2 - x - 12 (3, ), ( , 0)

2. y = - 1 2

x2 - x + 1 (0, ), ( , -3)

3. y = -16x2 + 32x (4, ), ( , 0)

4. y = x2 + 4x + 5 (-2, ), ( , 2)

U2V Graphing Quadratic Functions

Determine whether the graph of each parabola opens upward or downward. See Examples 2 and 3.

5. y = x2 + 5 6. y = 2x2 + x - 1

7. y = -3x2 + 4x + 2 8. y = -x2 + 3

9. y = (-2x + 3)2 10. y = (5 - x)2

Graph each parabola. See Examples 2 and 3.

11. y = x2 + 2

664 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-38 1

0 .4

Warm-Ups ▼

Fill in the blank. 1. The graph of y = ax2 + bx + c with a � 0 is a . 2. The graph of y = ax2 + bx + c with a > 0

opens .

3. The graph of y = ax2 + bx + c with a < 0 opens .

4. The is the highest point on a parabola that opens downward or the lowest point on a parabola that opens upward.

5. The x-coordinate of the vertex for y = ax2 + bx + c is

6. To find the of the vertex, evaluate y = ax2 + bx + c with x = -b/(2a).

True or false? 7. The ordered pair (-2, -1) satisfies f(x) = x2 - 5.

8. The y-intercept for g(x) = x2 - 3x + 9 is (9, 0).

9. The x-intercepts for y = x2 - 5 are (V5s, 0) and (-V5s, 0).

10. The graph of f(x) = x2 - 12 opens upward. 11. The graph of y = 4 + x2 opens downward. 12. The parabola y = x2 + 1 has no x-intercepts. 13. The y-intercept for g(x) = ax2 + bx + c is (0, c). 14. If w = -2t2 + 9, then the maximum value of w is 9.

Exercises

U Study Tips V • Be sure to ask your instructor what to expect on the final exam. Will it be the same format as other tests? • If there are any sample final exams available, use them as a guide for your studying.

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10-39 10.4 Graphing Quadratic Functions 665

12. y = x2 - 4

1 13. y = 44 x2 - 4

1 14. y = 44 x2 - 6

15. y = -2x2 + 5

16. y = -x2 - 1

1 17. y = -44x2 + 5

1 18. y = -44x2 + 3

19. y = (x - 2)2

20. y = (x + 3)2

U3V The Vertex and Intercepts

Find the vertex for the graph of each parabola. See Example 4.

21. f (x) = x2 - 9 22. f (x) = x2 + 12 223. y = x - 4x + 1 24. y = x2 + 8x - 3

25. f (x) = -2x2 + 20x + 1 26. f (x) = -3x2 + 18x - 7

27. y = x2 - x + 1 28. y = 3x2 - 2x + 1

Find all intercepts for the graph of each parabola. See Example 5.

29. f (x) = 16 - x2 30. f (x) = x2 - 9

31. y = x2 - 2x - 15 32. y = x2 - x - 6

233. f (x) = -4x2 + 12x - 9 34. f (x) = -2x - x + 3

Find the vertex and intercepts for each parabola. Sketch the graph. See Examples 4 and 5.

35. f (x) = x2 - x - 2

36. f (x) = x2 + 2x - 3

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666 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-40

37. g(x) = x2 + 2x - 8

38. g(x) = x2 + x - 6

39. y = -x2 - 4x - 3

40. y = -x2 - 5x - 4

41. h(x) = -x2 + 3x + 4

242. h(x) = -x - 2x + 8 43. a = b2 - 6b - 16

244. v = -u - 8u + 9

Find the maximum or minimum value of y.

45. y = x2 - 8 46. y = 33 - x2

47. y = -3x2 + 14 48. y = 6 + 5x2

49. y = x2 + 2x + 3 50. y = x2 - 2x + 5

251. y = -2x - 4x 52. y = -3x2 + 24x

U4V Applications

Solve each problem. See Example 6.

53. Maximum height. If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height in feet is given by the function

s(t) = -16t2 + 64t

where t is time in seconds. Graph this parabola for 0 � t � 4. What is the maximum height reached by the ball?

54. Maximum height. If a soccer ball is kicked straight up from the ground with an initial velocity of 32 feet per second, then its height above the earth in feet is given by the function s(t) = -16t2 + 32t where t is time in seconds. Graph this parabola for 0 � t � 2. What is the maximum height reached by the ball?

55. Minimum cost. It costs Acme Manufacturing C dollars per hour to operate its golf ball division. An analyst has determined that C is related to the number of golf balls produced per hour, x, by the function C = 0.009x2 - 1.8x + 100. What number of balls per hour should Acme produce to minimize the cost per hour of manufacturing these golf balls?

56. Maximum profit. A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function

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10-41 10.4 Graphing Quadratic Functions 667

c) What was the maximum stabilization ratio from part (b)? P = -25x2 + 300x. What number of clerks will maximize d) What is the significance of a stabilization ratio of 1?the profit, and what is the maximum possible profit?

57. Maximum area. Jason plans to fence a rectangular area with 100 meters of fencing. He has written the function A = w(50 - w) to express the area in terms of the width w. What is the maximum possible area that he can enclose with his fencing?

St ab

ili za

tio n

ra tio

( b ir

th s/

de at

hs )

Stabilization ratio for South and Central America

50 x

0 10 20 30 40 Years after 1950

Photo for Exercise 57

Figure for Exercise 60

61. Suspension bridge. The cable of the suspension bridge shown in the figure hangs in the shape of a parabola with equation y = 0.0375x2, where x and y are in meters. What is the height of each tower above the roadway? What is the length z for the cable bracing the tower?

58. Minimizing cost. A company uses the function C(x) = 20.02x - 3.4x + 150 to model the unit cost in dollars for

producing x stabilizer bars. For what number of bars is the unit cost at its minimum? What is the unit cost at that level of production?

59. Air pollution. The amount of nitrogen dioxide A in parts per million (ppm) that was present in the air in the city of Homer on a certain day in June is modeled by the function

A(t) = -2t2 + 32t + 12,

0 5 10

15 25 30 35 40

Figure for Exercise 61 where t is the number of hours after 6:00 A.M. Use this function to find the time at which the nitrogen dioxide level was at its maximum.

60. Stabilization ratio. The stabilization ratio (births/deaths) for South and Central America can be modeled by the function

y = -0.0012x2 + 0.074x + 2.69,

where y is the number of births divided by the number of deaths in the year 1950 + x (World Resources Institute, www.wri.org).

a) Use the graph to estimate the year in which the stabi lization ratio was at its maximum.

b) Use the function to find the year in which the stabiliza tion ratio was at its maximum.

Getting More Involved

62. Exploration

a) Write the equation y = 3(x - 2)2 + 6 in the form y = ax2 + bx + c, and find the vertex of the parabola

-busing the formula x = 44. 2a

b) Repeat part (a) with the equations y = -4(x - 5)2 - 9 and y = 3(x + 2)2 - 6.

c) What is the vertex for a parabola that is written in the form y = a(x - h)2 + k? Explain your answer.

http:www.wri.org

http:0.02x-3.4x

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� � �

668 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-42

Graphing Calculator Exercises 2 163. Graph y = x , y = 44 x2, and y = 2x2 on the same coordinate

2 system. What can you say about the graph of y = ax2 for a � 0?

264. Graph y = x , y = (x - 3)2, and y = (x + 3)2 on the same coordinate system. How does the graph of y = (x - h)2

compare to the graph of y = x2?

65. The equation x = y2 is equivalent to y = ± x. Graph both y = x and y = - x on a graphing calculator. How does the graph of x = y2 compare to the graph of y = x2?

66. Graph each of the following equations by solving for y. a) x = y2 - 1 b) x = -y2

c) x2 + y2 = 4

67. Graph each parabola using a viewing window that contains the vertex and all intercepts. Answers may vary.

a) y = 100x2 - 30x + 2

b) y = x2 - 110x + 3000

c) y = 999x - 10 - 10x2

68. Determine the approximate vertex and x-intercepts for each parabola.

a) y = 3.2x2 - 5.4x + 1.6 b) y = -1.09x2 + 13x + 7.5

In This Section

U1V Solving Quadratic Inequalities Graphically

10.5 Quadratic Inequalities

In this section, we solve inequalities involving quadratic polynomials. We use two methods, which are based on the graphs of the corresponding quadratic functions.

U2V Solving Quadratic Inequalities with the Test-Point Method U1V Solving Quadratic Inequalities Graphically

U3V Applications An inequality involving a quadratic polynomial is called a quadratic inequality.

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10-43 10.5 Quadratic Inequalities 669

Quadratic Inequality

A quadratic inequality has one of the following forms:

ax2 + bx + c � 0, ax2 + bx + c � 0,

where a, b, and c are real numbers with a * 0.

To solve ax2 + bx + c � 0 we can examine the graph of the corresponding qua dratic function y = ax2 + bx + c. The values of x that satisfy the inequality are the same as the values of x for which y � 0 on the graph of y = ax2 + bx + c. We use the following strategy for the graphical method.

Strategy for the Graphical Method

1. Rewrite the inequality (if necessary) so that 0 is on the right side and a quadratic polynomial is on the left side.

2. Find the roots to the quadratic polynomial.

3. Plot the x-intercepts using the roots found in step 2, and graph the parabola passing through the x-intercepts.

4. Read the solution set to the inequality from the graph.

E X A M P L E 1 Solving quadratic inequalities graphically Solve each quadratic inequality. Write the solution set in interval notation and graph it.

a) x2 + 3x � 10 b) x2 - 2x - 1 � 0 c) -x2 + 3 � 0

Solution a) Rewrite the inequality as x2 + 3x - 10 � 0. Then find the roots to the quadratic

polynomial:

x2 + 3x - 10 = 0 (x + 5)(x - 2) = 0

x + 5 = 0 or x - 2 = 0 x = -5 or x = 2

The graph of y = x2 + 3x - 10 is a parabola that opens upward with x-intercepts at (-5, 0) and (2, 0) as shown in Fig. 10.10. The y-coordinates on the parabola are negative between the intercepts and positive to the left and right of the intercepts. Since y = x2 + 3x - 10, whenever y is positive x2 + 3x - 10 is positive and the inequality is satisfied. So the solution set to the inequality is (-�, -5) � (2, �). The graph of the solution set is shown in Fig. 10.11.

b) Find the roots to the quadratic polynomial using the quadratic formula:

x2 - 2x - 1 = 0

x =

= 4 2 ±

2 8

4 = 4 2 ±

2 2 2 4 = 1 ± 2

-(-2) ± (-2)2 - 4(1)(-1) 4444

2(1)

Figure 10.10

Figure 10.11

-4-5-6-7-8

x1 -2

-4

-6

-8

-10

-12

-14

2 3

4 5-1-2-3

y � 0

y = x2 + 3x - 10

5 4 3 2 1 0 21

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670 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-44

Figure 10.12

y x2 2x 1

x1 1

2 3

4 5 1 2 3

(1 v2, 0) (1 + v2, 0)

1 v2

Figure 10.13

x1 1

4 1 3

y = x2 + 3

( 03, 0) (03, 0)

The graph of y = x2 - 2x - 1 is a parabola that opens upward with x-intercepts at (1 - 032, 0) and (1 + 032, 0) as shown in Fig. 10.12. The y-coordinates on the parabola are negative between the intercepts, positive to the left and right of the intercepts, and zero at the intercepts. Because the inequality symbol is -, the solution set includes the roots to the polynomial. So the solution set to the inequality is [1 - 032, 1 + 023]. The graph of the solution set is shown in Fig. 10.13.

c) Find the roots to the quadratic polynomial:

-x2 + 3 = 0

-x2 = -3

x2 = 3

x = ±0331 + v2

The graph of y = -x2 + 3 is a parabola that opens downward with x-intercepts at (-033, 0) and (033, 0) as shown in Fig. 10.14. The y-coordinates on the parabola are greater than or equal to zero whenever x is in the interval [-033, 033], and that is the solution set to -x2 + 3 : 0. The graph of the solution set is shown in Fig. 10.15.

-03 03

Figure 10.15

Now do Exercises 1–18

Figure 10.14

E X A M P L E 2

Figure 10.16

y x2 4x + 4

(2, 0)

x1 1

2 3 4 5

1 2

The graphs of the corresponding quadratic polynomials in Example 1 all had two x-intercepts. In Example 2 the graphs have fewer than two x-intercepts.

Solving quadratic inequalities graphically Solve each quadratic inequality. Write the solution set in interval notation and graph it.

a) x2 - 4x + 4 - 0 b) x2 + 2x + 3 > 0 c) -x2 - 4 : 0

Solution a) Find the roots to the quadratic polynomial:

x2 - 4x + 4 = 0

(x - 2)2 = 0

x - 2 = 0

x = 2

The graph of y = x2 - 4x + 4 is a parabola that opens upward with an x-intercept at (2, 0) as shown in Fig. 10.16. The y-coordinates on the parabola are positive except when x = 2. At x = 2 the y-coordinate is zero. So there is only one value for x that satisfies x2 - 4x + 4 - 0, and that is x = 2. So the solution set to the inequality is {2}. The graph of the solution set is shown in Fig. 10.17.

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�

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�

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10-45 10.5 Quadratic Inequalities 671

b) Find the roots to the quadratic polynomial using the quadratic formula:

x2 + 2x + 3 = 0

x = = 4 -2 ±

2 -8

Since the radical contains a negative number, there are no real solutions to the equation and no x-intercepts. The graph of y = x2 + 2x + 3 is a parabola that opens upward from its vertex (-1, 2) as shown in Fig. 10.18. Since all y-coordinates on this graph are positive for any value of x, the solution set to x2 + 2x + 3 � 0 is the set of all real numbers, (-�, �). The graph of the solution set is shown in Fig. 10.19.

c) Find the roots to the quadratic polynomial:

-x2 - 4 = 0

-x2 = 4

x2 = -4

x = ± -4 = ±2i

Since there are no real solutions to this equa tion, there are no x-intercepts for the graph of y = -x2 - 4. The graph of y = -x2 - 4 opens downward from its vertex (0, -4) as shown in Fig. 10.20. The y-coordinates on the parabola are negative for every value of x. So there are no values of x that would make -x2 - 4 � 0 and the solution set for the inequality is the empty set, �.

-2 ± 22 - 4(1)(3) 444

2(1)

Figure 10.18

Figure 10.19

Figure 10.20

Figure 10.17

-1 10 32

x �1

�2

1 2 3

�1�2�3�4

y � x2 � 2x � 3

(�1, 2)

-2 -1 0 1 2

x 2

1 2123

(0, 4)

y x2 4

Now do Exercises 19–30

U2V Solving Quadratic Inequalities with the Test-Point Method The test-point method is a variation of the graphical method, but we don’t graph the parabola. We have seen that the y-coordinates on a parabola can change sign only at an x-intercept. So we find the x-intercepts (if there are any) and then test points in the intervals determined by the intercepts to see if they satisfy the inequality. Here is the strategy.

Strategy for the Test-Point Method

1. Rewrite the inequality (if necessary) with 0 on the right.

2. Solve the quadratic equation that results from replacing the inequality symbol with the equals symbol.

3. Locate the solutions to the quadratic equation on a number line.

4. Select one test point in each interval determined by the solutions.

5. Check to see whether each test point satisfies the original inequality.

6. Write the solution set using interval notation.

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672 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-46

E X A M P L E 3 Solving quadratic inequalities with the test-point method Solve each quadratic inequality. Write the solution set in interval notation and graph it.

a) x2 - x - 6 - 0 b) x2 - 4x > 6 c) x2 + 6x +10 : 0

Figure 10.21

Figure 10.22

-2 -1 0 1 2 3

Solution a) We can solve the quadratic equation x2 - x - 6 = 0 by factoring:

(x - 3)(x + 2) = 0 x - 3 = 0 or x + 2 = 0

x = 3 or x = -2

Locate -2 and 3 on the number line, and select three test points as shown in Fig. 10.21. We have chosen the points -5, 0, and 5. Now test -5, 0, and 5 in the original inequality x2 - x - 6 - 0:

(-5)2 - (-5) - 6 - 0 Incorrect 02 - 0 - 6 - 0 Correct 52 - 5 - 6 - 0 Incorrect

Of the three test points, only 0 satisfies the inequality. So the solution set is the inter val containing 0. Since the inequality symbol includes equality, -2 and 3 are included in the solution set [-2, 3]. The graph of the solution set is shown in Fig. 10.22.

b) First rewrite the inequality as x2 - 4x - 6 > 0. Then solve x2 - 4x - 6 = 0 using the quadratic formula:

x = = 4 ±

V404 = 2 ± V104

Now 2 - V104 = -1.2 and 2 + V104 = 5.2. Plot these points on a number line, and select three test points as shown in Fig. 10.23. We have chosen the points -2, 0, and 7. Now test -2, 0, and 7 in the original inequality x2 - 4x > 6:

(-2)2 - 4(-2) > 6 Correct 02 - 4(0) > 6 Incorrect 72 - 4(7) > 6 Correct

-(-4) ± V(-4)24- 4(1)4(-6)4

2(1)

2 - √ —– 10 2 + √

—– 10

10-1-2-3-4-5 2 3 4 5

Test points

= 5.2= -1.2Test point Test point Test point

-2 -1 0 1 2 3 4 5 6 7

Figure 10.23

So the inequality is satisfied on the intervals containing -2 and 7. The solution set is (-o, 2 - V10) U (2 + V10, o), and its graph is shown in Fig. 10.24.4 4

—– —– 2 - √10 2 + √10

-2 -1 0 1 2 3 4 5 6 7

Figure 10.24

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� �

10-47 10.5 Quadratic Inequalities 673

c) Solve x2 + 6x + 10 = 0 using the quadratic formula:

x = =

Since there is a negative number inside the radical, there are no real solutions to the equation. So there are no points to plot on the number line. There is just one interval to consider, and that is (-�, �). We can select any point in (-�, �) as a test point. Let’s try 2:

22 + 6(2) + 10 � 0 Correct

Since the inequality is satisfied for x = 2, it is satisfied for every real number and the solution set is (-�, �). The graph is shown in Fig. 10.25.

Now do Exercises 31–44

-6 ± -4 44

2 -6 ± 62 - 4(1)(10) 444

2(1)

Figure 10.25

-2 -1 0 1 2

If there are no solutions to the quadratic equation, then the quadratic polynomial does not change sign. The solution set is either all real numbers or no real numbers. If the inequality in Example 3(c) was not satisfied when x = 2, then the solution set would have been the empty set. In fact, the solution set to x2 + 6x + 10 � 0 is the empty set, �.

U3V Applications Example 4 shows how a quadratic inequality can be used to solve a problem.

E X A M P L E 4

Figure 10.26

Making a profit Charlene’s daily profit P (in dollars) for selling x magazine subscriptions is determined by the formula

P = -x2 + 80x - 1500.

For what values of x is her profit positive?

Solution The profit is positive whenever -x2 + 80x - 1500 � 0. Find the solutions to the corre sponding quadratic equation:

-x2 + 80x - 1500 = 0 x2 - 80x + 1500 = 0 (x - 30)(x - 50) = 0

x - 30 = 0 or x - 50 = 0 x = 30 or x = 50

Locate 30 and 50 on a number line as shown in Fig. 10.26, and select 0, 40, and 60 as test points. Check the test points in the original inequality -x2 + 80x - 1500 � 0:

-02 + 80(0) - 1500 � 0 Incorrect -402 + 80(40) - 1500 � 0 Correct -602 + 80(60) - 1500 � 0 Incorrect

Since 40 satisfies the inequality, every point between 30 and 50 also satisfies the inequal ity. So for a positive profit, she must sell between 30 and 50 magazine subscriptions.

20100 30 40 50 60

Test points

Now do Exercises 65–70

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674 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-48

Warm-Ups ▼

Fill in the blank. 1. A inequality has the form ax2 � bx � c � 0.

2. A quadratic inequality can be solved by the method or the method.

True or false? 3. The solution set to x2 � 4 is (2, �). 4. To solve x2 � x � 2 � 0 by graphing, we graph

y � x2 � x � 2.

5. In solving quadratic inequalities, we must get 0 on one side.

6. To solve (x � 3)(x � 5) � 0 using test points, the test points could be �5, 0, and 3.

7. We can’t solve quadratic inequalities that do not factor.

8. The parabola y � x2 � 4 has no x-intercepts.

1 0

.5 Exercises

U Study Tips V • Keep track of your time for one entire week. Account for every half hour. • You should be sleeping 50 to 60 hours per week and studying 1 to 2 hours for every credit hour you are taking. For a 3-credit-hour

class, you should be studying 3 to 6 hours per week.

7. 2u2 � 5u � 12

Use the graphical method to solve each inequality. State the solution set using interval notation and graph it. See Example 1. See the Strategy for the Graphical Method on page 669.

U1V Solving Quadratic Inequalities Graphically

8. 2v2 � 7v � 4 1. x2 � x � 6 � 0

2. x2 � 3x � 4 � 0

9. 4x2 � 8x � 0

3. z2 � 16 � 0 10. x2 � x � 0

4. y2 � 4 � 0

11. 5x � 10x2 � 0 5. x2 � 2x � 8 0

6. x2 � x � 12 0 12. 3x � x2 � 0

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10-49 10.5 Quadratic Inequalities 675

13. x2 - 5 > 0 25. 25x2 + 10x + 1 > 0

14. x2 - 3 < 0

15. x2 - 2x - 5 � 0

16. x2 - 2x - 4 > 0

17. 2x2 - 6x + 3 : 0

18. 2x2 - 8x + 3 < 0

Use the graphical method to solve each inequality. State the solution set using interval notation and graph it. See Example 2.

19. x2 + 6x + 9 : 0

20. x2 + 10x + 25 : 0

21. x2 + 4 < 4x

22. x2 < 8x - 16

23. 4x2 - 20x + 25 � 0

26. 16x2 - 16x + 4 > 0

27. x2 + 5x + 12 : 0

28. x2 + 3x + 9 > 0

29. 2x2 + 5x + 5 < 0

30. -3x2 + x - 6 : 0

U2V Solving Quadratic Inequalities with the Test-Point Method

Use the test-point method to solve each inequality. State the solution set using interval notation and graph it. See Example 3. See the Strategy for the Test-Point Method on page 671.

31. x2 - 4x - 12 > 0

32. x2 + 7x - 18 � 0

33. x2 + 3x < 40

34. x2 - 15x > 16

35. x2 + 8x + 17 : 0

36. x2 + 10x + 27 � 0

37. 9x - 4x2 > x

24. 9x2 + 12x + 4 � 0 38. 5x - x2 � x

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676 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-50

39. x2 � 4 � 6x

40. x2 � 2 � 4x

41. �5x2 � 2x � 4

242. 3x � 5 � 3x

43. y2 � 3y � 9 � 0

44. z2 � 5z � 7 � 0

Miscellaneous

Solve each inequality. State the solution set using interval notation when possible.

45. x2 � 0

46. x2 0

47. x2 � 4 0

48. x2 � 1 � 0

49. x2 � 9

50. x2 36

51. 16 � x2 � 0

52. 9 � x2 � 0

53. x2 � 4x 0

54. 4x2 � 9 � 0

55. 3(2w2 � 5) � w

56. 6(y2 � 2) � y � 0

57. z2 4(z � 3)

58. t2 � 3(2t � 3)

59. (q � 4)2 � 10q � 31

60. (2p � 4)( p � 1) � ( p � 2)2

1 61. �� x2 4 � x

1 62. �� x2 � x � 12

63. 0.23x2 � 6.5x � 4.3 � 0

64. 0.65x2 � 3.2x � 5.1 � 0

U3V Applications

Solve each problem by using a quadratic inequality. See Example 4.

65. Positive profit. The monthly profit P (in dollars) that Big Jim makes on the sale of x mobile homes is determined by the formula P � x2 � 5x � 50. For what values of x is his profit positive?

66. Profitable fruitcakes. Sharon’s revenue R (in dollars) on the sale of x fruitcakes is determined by the formula R � 50x � x2. Her cost C (in dollars) for producing x fruitcakes is given by the formula C � 2x � 40. For what values of x is Sharon’s profit positive? (Profit � revenue � cost.)

If an object is given an initial velocity straight upward of v0 feet per second from a height of s0 feet, then its altitude S after t seconds is given by the formula

S � �16t2 � v0 t � s0.

67. Flying high. An arrow is shot straight upward with a velocity of 96 feet per second (ft/sec) from an altitude of 6 feet. For how many seconds is this arrow more than 86 feet high?

68. Putting the shot. In 1978 Udo Beyer (East Germany) set a world record in the shot-put of 72 ft 8 in. If Beyer had pro jected the shot straight upward with a velocity of 30 ft/sec from a height of 5 ft, then for what values of t would the shot be under 15 ft high?

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�

10-51 10.5 Quadratic Inequalities 677

If a projectile is fired at a 45° angle from a height of s0 feet with initial velocity v0 ft/sec, then its altitude S in feet after t seconds is given by

v0 S = -16t2 + 4 t + s0.

2 69. Siege and garrison artillery. An 8-inch mortar used in

the Civil War fired a 44.5-lb projectile from ground level a distance of 3600 ft when aimed at a 45° angle (Harold R. Peterson, Notes on Ordinance of the American Civil War). The accompanying graph shows the altitude of the projectile when it is fired with a velocity of 240 2 ft/sec.

a) Use the graph to estimate the maximum altitude reached by the projectile.

b) Use the graph to estimate approximately how long the altitude of the projectile was greater than 864 ft.

c) Use the formula to determine the length of time for which the projectile had an altitude of more than 864 ft.

H ei

gh t (

ft )

800

600

400

200

0 0

4 8 12 Time (sec)

Figure for Exercise 69

70. Seacoast artillery. The 13-inch mortar used in the Civil War fired a 220-lb projectile a distance of 12,975 ft when aimed at a 45° angle. If the 13-inch mortar was fired from a hill 100 ft above sea level with an initial velocity of 644 ft /sec, then for how long was the projectile more than 800 ft above sea level?

45�

100 ft

Figure for Exercise 70

Getting More Involved 71. Cooperative learning

Work in a small group to solve each inequality for x, given that h and k are real numbers with h � k.

a) (x - h)(x - k) � 0

b) (x - h)(x - k) � 0

c) (x + h)(x + k) � 0

d) (x + h)(x + k) � 0

72. Cooperative learning

Work in a small group to solve ax2 + bx + c � 0 for x in each case.

a) b2 - 4ac = 0 and a � 0

b) b2 - 4ac = 0 and a � 0

c) b2 - 4ac � 0 and a � 0

d) b2 - 4ac � 0 and a � 0

e) b2 - 4ac � 0 and a � 0

f) b2 - 4ac � 0 and a � 0

C h

a p

t e

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� �

678 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-52

10 Wrap-Up Summary

Quadratic Equations Examples 2Quadratic equation An equation of the form ax2 + bx + c = 0, x = 11

where a, b, and c are real numbers, with a * 0 (x - 5)2 = 99 x2 + 3x - 20 = 0

Methods for solving Factoring: x2 + x - 6 = 0 quadratic equations Factor the quadratic polynomial, and then (x + 3)(x - 2) = 0

set each factor equal to 0. x + 3 = 0 or x - 2 = 0

The even-root property: (x - 5)2 = 10 If x2 = k (k � 0), then x = ± k. x - 5 = ± 10 If x2 = 0, then x = 0.

2There are no real solutions to x = k for k � 0.

Completing the square: x2 + 6x = -4 Take one-half of middle term, square it, and x2 + 6x + 9 = -4 + 9 then add it to each side. (x + 3)2 = 5

Quadratic formula: If ax2 + bx + c = 0 with a * 0, then 2x2 + 3x - 5 = 0

-b ± b2 - 4ac -3 ± 32 - 4(2)(-5)x =44. x = 444 2a 2(2)

Number of solutions Determined by the discriminant b2 - 4ac: b2 - 4ac � 0 2 real solutions

x2 + 6x - 12 = 0 62 - 4(1)(-12) � 0

b2 - 4ac = 0 1 real solution x2 + 10x + 25 = 0 102 - 4(1)(25) = 0

b2 - 4ac � 0 no real solutions, 2 imaginary solutions

x2 + 2x + 20 = 0 22 - 4(1)(20) � 0

Writing equations To write an equation with given solutions, reverse the steps in solving an equation by factoring.

x = 2 or x = -3 (x - 2)(x + 3) = 0

x2 + x - 6 = 0

Factoring The quadratic polynomial ax2 + bx + c (with integral coefficients) can be factored if and only if b2 - 4ac is a perfect square.

2x2 - 11x + 12 b2 - 4ac = 25 (2x - 3)(x - 4)

Equations quadratic in form

Use substitution to convert to a quadratic. x4 + 3x2 - 10 = 0 Let a = x2 . a2 + 3a - 10 = 0

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�

10-53 Chapter 10 Enriching Your Mathematical Word Power 679

Graphing Quadratic

Functions

Quadratic function A function of the form y = ax2 + bx + c where a ¥ 0

Parabola The graph of a quadratic function is a parabola.

Properties of parabolas If a � 0, then the parabola opens upward. If a � 0, then the parabola opens downward.

The first coordinate of the vertex is 4 -2a b 4.

The second coordinate of the vertex is the minimum y-value if a � 0 or the maximum y-value if a � 0.

The x-intercepts are found by solving ax2 + bx + c = 0. Let x = 0 to find the y-intercept.

Quadratic Inequalities

Quadratic inequality An inequality involving a quadratic polynomial

The Graphical Method Graph the corresponding parabola, and determine the solution from the graph.

The Test-Point Method Test points in the intervals on the number line that are determined by the roots to the quadratic polynomial.

Examples

x-1-3-2-5

-3

-4

-2

-1

-5

-6

-7

-8

1 3

y = x2 + 2x - 8

y = x2 + 2x - 8 Opens upward

-b -2x = 44 = 44 = -1 2a 2(1)

Vertex: (-1, -9) Minimum y-value: -9

x-intercepts: (-4, 0), (2, 0)

y-intercept: (0, -8)

Examples 22x - 7x + 6 � 0

2x - 4x - 5 � 0

2To solve x - 5x + 6 � 0 graph y = x2 - 5x + 6. Solution set: (2, 3)

To solve x2 - 4 � 0 plot -2 and 2 on a number line. Test -5, 0, and 5 in the inequality. Solution set: (-�, -2) � (2, �)

Enriching Your Mathematical Word Power

Fill in the blank. 4. Finding the third term of a perfect square trinomial is the square.1. A equation has the form ax2 + bx + c = 0

where a ¥ 0. b2 -b ± - 4ac 5. The equation x =44 is the2. A function has the form y = ax2 + bx +c where 2aformula.a ¥ 0.

3. The trinomial a2 + 2ab + b2 is a square trinomial. 6. The expression b2 - 4ac is the .

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680 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-54

7. The graph of y = ax2 + bx + c with a ¥ 0 is 9. The inequality ax2 + bx + c � 0 with a ¥ 0 is a a . inequality.

8. An equation that is quadratic after a substitution is quadratic 10. A number that is used to check if an inequality is satisfied in . is a point.

Review Exercises

10.1 Factoring and Completing the Square 2Solve by factoring. 28. 6x = x + 2

21. x - 2x - 15 = 0 29. x2 + 4x + 2 = 0

22. x - 2x - 24 = 0 30. x2 + 6x = 2

3. 2x2 + x = 15 31. 3x2 + 1 = 5x

24. 2x + 7x = 4 32. 2x2 + 3x - 1 = 0

2 5. w - 25 = 0 2 Find the value of the discriminant and the number of real 6. a - 121 = 0

solutions to each equation. 27. 4x - 12x + 9 = 0 233. 25x - 20x + 4 = 0

34. 16x2 + 1 = 8x 8. x2 - 12x + 36 = 0 235. x - 3x + 7 = 0

236. 3x - x + 8 = 0Solve by using the even-root property. 2 2 37. 2x2 + 1 = 5x9. x = 12 10. x = 20

38. -3x2 + 6x - 2 = 0 11. (x - 1)2 = 9 12. (x + 4)2 = 4

3 1 Find the complex solutions to the quadratic equations.13. (x - 2)2 = 44 14. (x - 3)2 = 44 4 4

239. 2x - 4x + 3 = 02 215. 4x = 9 16. 2x = 3

240. 2x - 6x + 5 = 0 Solve by completing the square.

217. x - 6x + 8 = 0 241. 2x + 3 = 3x 18. x2 + 4x + 3 = 0

219. x - 5x + 6 = 0 2

42. x2 + x + 1 = 0 20. x - x - 6 = 0

2 43. 3x2 + 2x + 2 = 021. 2x - 7x + 3 = 0

2 44. x2 + 2 = 2x

22. 2x - x = 6 1 245. 44 x + 3x + 8 = 0 2 223. x + 4x + 1 = 0

1 22 46. 44 x - 5x + 13 = 024. x + 2x - 2 = 0 2

10.2 The Quadratic Formula 10.3 More on Quadratic Equations

Solve by the quadratic formula. Use the discriminant to determine whether each quadratic polynomial can be factored, and then factor the ones that are 225. x - 3x - 10 = 0 not prime.226. x - 5x - 6 = 0

247. 8x - 10x - 3 27. 6x - 7x = 32 48. 18x2 + 9x - 2

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� � � �

�

10-55 Chapter 10 Review Exercises 681

49. 4x2 - 5x + 2 71. g(x) = x2 - 4x - 12 250. 6x - 7x - 4

51. 8y2 + 10y - 25 252. 25z - 15z - 18

Write a quadratic equation that has each given pair of solutions.

53. -3, -6 54. 4, -9

55. -5 2, 5 2 72. g(x) = x2 + 2x - 2456. -2i 3, 2i 3

Find all real solutions to each equation.

57. x6 + 7x3 - 8 = 0

58. 8x6 + 63x3 - 8 = 0

59. x4 - 13x2 + 36 = 0

60. x4 + 7x2 + 12 = 0

61. (x2 + 3x)2 - 28(x2 + 3x) + 180 = 0 73. h(x) = -2x2 + 8x62. (x2 + 1)2 - 8(x2 + 1) + 15 = 0

2 263. x - 6x + 6 x - 6x - 40 = 0

2 264. x - 3x - 3 x - 3x + 2 = 0

t-2 + 5t-165. - 36 = 0

166. a -2 + a - - 6 = 0 74. h(x) = -3x2 + 6x

67. w - 13 w + 36 = 0

68. 4a - 5 a + 1 = 0

10.4 Graphing Quadratic Functions Find the vertex and intercepts for each parabola, and sketch its graph.

69. f (x) = x2 - 6x 75. y = -x2 + 2x + 3

76. y = -x2 - 3x - 270. f (x) = x2 + 4x

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�

682 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-56

Determine whether each equation has a maximum or minimum y-value and find it. 77. f (x) = x2 + 4x + 1

78. f (x) = x2 - 6x + 2

79. y = -2x2 - x + 4

80. y = -3x2 + 2x + 7

10.5 Quadratic Inequalities Solve each inequality. State the solution set using interval notation and graph it.

281. a + a � 6

282. x - 5x + 6 � 0

283. x - x - 20 � 0

84. a2 + 2a � 15

285. w - w � 0

86. x - x2 � 0

287. 2x + 5x � 3

288. 3x - 4 � x

89. x2 + 2x + 4 � 0

90. 10x - x2 � 28

291. x - 10x + 25 � 0

92. 4x � 4x2 + 1

293. x - 2x + 10 � 0

94. -x2 + 4x - 5 � 0

Miscellaneous

Find all real or imaginary solutions to each equation. 295. 144x - 120x + 25 = 0

96. 49x2 + 9 = 42x

97. (2x + 3)2 + 7 = 12

19x + 25 98. 6x = -44

x + 1

20 8 99. 1 + 44 = 4429x 3x

x - 1 2x - 3 100. 44 = 44

x + 2 x + 4

101. 3x2 + 7x - 30 = x 4x 2102. 44 = x + 6

3 103. 2(2x + 1)2 + 5(2x + 1) = 3

2 2104. (w - 1)2 + 2(w - 1) = 15 105. x112 - 15x114 + 50 = 0

106. x -2 - 9x -1 + 18 = 0

Find exact and approximate solutions to each problem.

107. Missing numbers. Find two positive real numbers that differ by 4 and have a product of 4.

108. One on one. Find two positive real numbers that differ by 1 and have a product of 1.

109. Big screen TV. On a 19-inch diagonal measure television picture screen, the height is 4 inches less than the width. Find the height and width.

19 in. x � 4 in.

x in.

Figure for Exercise 109

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10-57 Chapter 10 Review Exercises 683

110. Boxing match. A boxing ring is in the shape of a square, 20 ft on each side. How far apart are the fighters when they are in opposite corners of the ring?

111. Students for a Clean Environment. A group of environ mentalists plans to print a message on an 8 inch by 10 inch paper. If the typed message requires 24 square inches of paper and the group wants an equal border on all sides, then how wide should the border be?

10 in.

8 in.

Figure for Exercise 111

112. Winston works faster. Winston can mow his dad’s lawn in 1 hour less than it takes his brother Willie. If they take 2 hours to mow it when working together, then how long would it take Winston working alone?

113. Ping Pong. The table used for table tennis is 4 ft longer than it is wide and has an area of 45 ft2. What are the dimensions of the table?

Figure for Exercise 113

114. Swimming pool design. An architect has designed a motel pool within a rectangular area that is fenced on three sides as shown in the figure. If she uses 60 yards of fencing to enclose an area of 352 square yards, then what are the dimensions marked L and W in the figure? Assume L is greater than W.

L W

Figure for Exercise 114

115. Minimizing cost. The unit cost in dollars for manufactur ing n starters is given by C(n) = 0.004n2 - 3.2n + 660. What is the unit cost when 390 starters are manufactured? For what number of starters is the unit cost at a minimum?

116. Maximizing profit. The total profit (in dollars) for sales of x rowing machines is given by P(x) = -0.2x2 + 300x - 200. What is the profit if 500 are sold? For what value of x will the profit be at a maximum?

117. Decathlon champion. For 1989 and 1990 Dave Johnson had the highest decathlon score in the world. When Johnson reached a speed of 32 ft/sec on the pole vault runway, his height above the ground t seconds after leaving the ground was given by h = -16t2 + 32t. (The elasticity of the pole converts the horizontal speed into vertical speed.) Find the value of t for which his height was 12 ft.

118. Time of flight. Use the information from Exercise 117 to determine how long Johnson was in the air. For how long was he more than 14 ft in the air?

119. Golden ratio. The ancient Greeks believed that a rectan gle had the most pleasing shape when the ratio of its

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684 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-58

length to width was the golden ratio. To find the golden ratio remove a 1 by 1 square from a 1 by x rectangle as shown in the diagram. The ratio of the length to width of the small rectangle that remains should be equal to the ratio of the length to width of the original rectangle. So,

x 1 �� . 1 x � 1

Find x (the golden ratio) to three decimal places.

Chapter 10 Test

Calculate the value of b2 � 4ac, and state how many real solutions each equation has.

1. 2x2 � 3x � 2 � 0

2. �3x2 � 5x � 1 � 0

3. 4x2 � 4x � 1 � 0

Solve by using the quadratic formula.

4. 2x2 � 5x � 3 � 0

5. x2 � 6x � 6 � 0

Solve by completing the square.

6. x2 � 10x � 25 � 0

7. 2x2 � x � 6 � 0

Solve by any method.

8. x(x � 1) � 12

9. a4 � 5a2 � 4 � 0

10. x � 2 � 8�x � 2� � 15 � 0

Find the complex solutions to the quadratic equations.

11. x2 � 36 � 0

12. x2 � 6x � 10 � 0

13. 3x2 � x � 1 � 0

Graph each parabola. Identify the vertex, intercepts, and the maximum or minimum y-value.

14. f (x) � 16 � x2

x – 11

Figure for Exercise 119

15. g(x) � x2 � 3x

Write a quadratic equation that has each given pair of solutions.

16. �4, 6 17. �5i, 5i

Solve each inequality. State and graph the solution set.

18. w2 � 3w � 18

19. x2 � 2x � 1

20. x2 � 6x � 13 0

21. x � x2 � 4

Find the exact solution to each problem.

22. The length of a rectangle is 2 ft longer than the width. If the area is 16 ft2, then what are the length and width?

23. A new computer can process a company’s monthly payroll in 1 hour less time than the old computer. To really save time, the manager used both computers and finished the payroll in 3 hours. How long would it take the new computer to do the payroll by itself?

24. The height in feet for a ball thrown upward at 48 feet per second is given by s(t) � �16t2 � 48t, where t is the time in seconds after the ball is tossed. What is the maximum height that the ball will reach?

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�

10-59 Chapter 10 Making Connections 685

MakingConnections A Review of Chapters 1–10

Evaluate each expression. Solve each equation for y.

16 8 29. 2x - 3y = 91. {44 2. {3 44I I81 27 y - 3 1

30. 44 = -44 3. (100 + 21)112 4. (4 + 4)213 x + 2 2

1110 31. 3y2 + cy + d = 032 5. 8116 � 8112 6. 44 -311032

32. my2 - ny = w 213 27. (4112 + 36112)- 8. (3-2 + 5 - - 9 -1)312

1 2 5 33. 44 x - 44 y = 44

3 5 6 Factor completely. 2

34. y - 3 = -44(x - 4) 9. y2 + 97y - 300 3

210. 20y - 7y - 3 y2 - y1 3 2 Let m = 44. Find the value of m for each of the following 11. 6a - 60a + 150a x2 - x1

12. b3 + 2b2 + 4b + 8 choices of x1, x2, y1, and y2.

13. ab - ay2 - by2 + y4 35. x1 = 2, x2 = 5, y1 = 3, y2 = 7 314. -2m - 16 36. x1 = -3, x2 = 4, y1 = 5, y2 = -6

37. x1 = 0.3, x2 = 0.5, y1 = 0.8, y2 = 0.4 Solve each equation.

1 1 3 4 38. x1 = 44 , x2 = 44 , y1 = 44 , y2 = -44

15. 2x - 15 = 0 2 3 5 3

2 Solve each problem. 16. 2x - 15 = 0 39. Ticket prices. If the price of a concert ticket goes up, then

the number sold will go down, as shown in the figure. If 17. 2x2 + x - 15 = 0 you use the formula n = 48,000 - 400p to predict the

number sold depending on the price p, then how many will 18. 2x2 + 4x - 15 = 0 be sold at $20 per ticket? How many will be sold at $25

per ticket? Use the bar graph to estimate the price if 35,000 19. I 4x + 11 I = 3 tickets were sold.

220. I 4x + 11x I = 3

21. x = x - 6 22. (2x - 5)213 = 4

Solve each inequality. State the solution set using interval notation.

23. 1 - 2x � 5 - x 24. (1 - 2x)(5 - x) � 0

T ic

ke ts

s ol

d (i

n th

ou sa

nd s)

0 10 15 20 25 30 35 405

25. x2 � x 26. 5x2 + 3 � 0 Price (in dollars)

27. 3x - 1 � 5 and -3 � x 28. x - 3 � 1 or 2x � 8 Figure for Exercise 39

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686 Chapter 10 Quadratic Equations, Functions, and Inequalities 10-60

40. Increasing revenue. Even though the number of tickets sold for a concert decreases with increasing price, the revenue generated does not necessarily decrease. Use the formula R = p(48,000 - 400p) to determine the revenue when the price is $20 and when the price is $25. What price would produce a revenue of $1.28 million? Use the graph to find the price that determines the maximum revenue.

R ev

en ue

( in

m ill

io ns

o f

do lla

rs ) 2

0 10 20 30 40 50 60 70 80 90100

Ticket price (in dollars)

Figure for Exercise 40

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10-61 Chapter 10 Critical Thinking 687

CriticalThinking For Individual or Group Work Chapter 10

These exercises can be solved by a variety of techniques, which may or may not require algebra. So be creative and think critically. Explain all answers. Answers are in the Instructor’s Edition of this text.

1. Ant parade. An ant marches from point A to point B on the cylindrical garbage can shown in the accompanying figure. The can is 1 foot in diameter and 2 feet high. If the ant makes two complete revolutions of the can in a perfect spiral, then exactly how far did he travel?

Figure for Exercise 1

2. Connecting points. Draw a circle and pick any three points on the circle.

a) How many line segments can be drawn connecting these points?

b) How many line segments can be drawn connecting four points on a circle? Five points? Six points?

c) How many line segments can be drawn connecting n points on a circle?

3. Summing the digits. Find the sum of the digits in the standard form of the number 22005 � 52007.

4. Consecutive odd numbers. Find three consecutive odd whole numbers such that the sum of their squares is a four-digit whole number whose digits are all the same.

5. Reversible prime numbers. The prime number 13 has an interesting property. When its digits are reversed, the new number 31 is also prime. Find the sum of all prime numbers greater than 10 yet less than 125 that have this property.

6. Circles and squares. Start with a square piece of paper. Draw the largest possible circle inside the square. Cut out the circle and keep it. Now draw the largest possible square inside the circle. Cut out the square and keep it. What is the ratio of the area of the original square to the area of the final square? If you repeat this process six more times, then what is the ratio of the area of the original square to the area of the final square?

7. Perpendicular hands. What are the first two times (to the nearest second) after 12 noon for which the minute hand and hour hand of a clock are perpendicular to each other?

Photo for Exercise 7

8. Going broke. Albert and Zelda agreed to play a game. If heads appeared on the toss of an ordinary coin, Zelda had to double the amount of money that Albert had. If the result was tails, then Albert had to pay Zelda $24. As it turned out, the coin came up heads, tails, heads, tails, heads, tails. Then Albert was broke. How much money did Albert start with?

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