Find all the zeros of the equation. -4x4 - 44x2 + 3600 = 0

C h

a p

t e

r

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10

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

W

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

3

93x

x

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 - -

3

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 + -

1

9

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

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.

6

-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):

6

-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 = --

2

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.

50

-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

2.

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

.1

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

= -- 2 4

2 4 80. (y - --)

2

= -- 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 )

25

20

15

10

5

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

or

$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)

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

0

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 =--

2a

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.

10

-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.

10

-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 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.

10

-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 --

2

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

or

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 --

2m

5 ± V25 - 4k(3)(9)k ---

2

-4 ± V16 - 4k(-5)k ---

2

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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

x

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.

x

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).

10

-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.

50

-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

.3

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

x

11 in.

x

x

??

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?

W

W

W

W

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

y

x

6

-4

2

4

1 -2

2 3

8

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.

4

2

�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

y

x-1

3

5

-3

1

2

1 -1

-2

3

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

y

x

�b � �b2 � 4ac 2a �b � �b2 � 4ac

2a

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

4

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

y

x-1

3

4

-3

1

-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

1

4

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