X2 2x 15 0 quadratic formula

C h

a p

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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