9.3 Solving Quadratic Equations by Using the … Quadratic Equations by Using the Quadratic Formula 9.3 9.3 OBJECTIVES 1. Solve a quadratic equation by using the quadratic formula

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Solving Quadratic Equations by Using the Quadratic Formula

9.3

9.3 OBJECTIVES

1. Solve a quadratic equation by using the quadraticformula

2. Determine the nature of the solutions of a quadratic equation by using the discriminant

3. Use the Pythagorean theorem to solve a geometric application

4. Solve applications involving quadratic equations

Every quadratic equation can be solved by using the quadratic formula. In this section, wewill first describe how the quadratic formula is derived, then we will examine its use. Recallthat a quadratic equation is any equation that can be written in the form

ax2 � bx � c � 0 when a � 0

We now use the result derived above to state the quadratic formula, a formula thatallows us to find the solutions for any quadratic equation.

Step 1 Isolate the constanton the right side ofthe equation.

Step 2 Divide both sides bythe coefficientof the x2 term.

Step 3 Add the square ofone-half thex coefficient toboth sides.

Step 4 Factor the left sideas a perfect-square binomial. Then applythe square rootproperty.

Step 5 Solve the resultinglinear equations.

Step 6 Simplify.

Step by Step: Deriving the Quadratic Formula

ax2 � bx � �c

��b � 2b2 � 4ac

2a

x � �b2a

� 2b2 � 4ac

2a

x �b2a

� �Bb2 � 4ac4a2

�x �b2a�

2

��4ac � b2

4a2

x2 �ba

x �b2

4a2 � �ca

�b2

4a2

x2 �ba

x � �ca

Given any quadratic equation in the form

ax2 � bx � c � 0 when a � 0

the two solutions to the equation are found using the formula

x ��b � 2b2 � 4ac

2a

Rules and Properties: The Quadratic Formula

694 CHAPTER 9 QUADRATIC EQUATIONS, FUNCTIONS, AND INEQUALITIES

Using the Quadratic Formula

Solve, using the quadratic formula.

6x2 � 7x � 3 � 0

First, we determine the values for a, b, and c. Here,

a � 6 b � �7 c � �3

Substituting those values into the quadratic formula, we have

Simplifying inside the radical gives us

This gives us the solutions

Note that because the solutions for the equation of this example are rational, the origi-nal equation could have been solved by our earlier method of factoring.

x �3

2 or x � �

1

3 or �3

2, �

1

3�

�7 � 11

12

x �7 � 1121

12

x ��(�7) � 2(�7)2 � 4(6)(�3)

2(6)

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C H E C K Y O U R S E L F 1

Solve, using the quadratic formula.

3x2 � 2x � 8 � 0

To use the quadratic formula, we often must write the equation in standard form.Example 2 illustrates this approach.

NOTE Notice that the equationis in standard form.

NOTE Because b2 � 4ac � 121is a perfect square, the twosolutions in this case arerational numbers.

NOTE The equation must be instandard form to determine a,b, and c.

NOTE Compare these solutionsto the graph of y � 6x2 � 7x � 3

Example 1

Example 2


Solve by using the quadratic formula.

9x2 � 12x � 4

Our first example uses an equation in standard form.

SOLVING QUADRATIC EQUATIONS BY USING THE QUADRATIC FORMULA SECTION 9.3 695©

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Use the quadratic formula to solve the equation.

4x2 � 4x � �1

Thus far our examples and exercises have led to rational solutions. That is not always thecase, as Example 3 illustrates.

First, we must write the equation in standard form.

9x2 � 12x � 4 � 0

Second, we find the values of a, b, and c. Here,

a � 9 b � �12 c � 4

Substituting these values into the quadratic formula, we find

and simplifying yields

x �2

3 or �2

3�

�12 � 10

18

x ��(�12) � 2(�12)2 � 4(9)(4)

2(9)

Example 3


Using the quadratic formula, solve

x2 � 3x � 5

Once again, to use the quadratic formula, we write the equation in standard form.

x2 � 3x � 5 � 0

We now determine values for a, b, and c and substitute.

Simplifying as before, we have

x �3 � 129

2 or �3 � 129

2 �

x ��(�3) � 2(�3)2 � 4(1)(�5)

2(1)

NOTE a � 1, b � �3, andc � �5

NOTE The graph of

y � 9x2 � 12x � 4

intercepts the x axis only at the

point .�23

, 0�


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Example 4


Using the quadratic formula, solve

3x2 � 6x � 2 � 0

Here, we have a � 3, b � �6, and c � 2. Substituting gives

We now look for the largest perfect-square factor of 12, the radicand.

Simplifying, we note that is equal to or . We can then write the solutionsas

x � 6 � 223

6�

2(3 � 13)

6�

3 � 13

3

21314 � 3,112

�6 � 112

6

x � �(�6) � 2(�6)2 � 4(3)(2)

2(3)



x2 � 4x � 6

Let’s examine a case in which the solutions are nonreal complex numbers.

Example 5



x2 � 2x � �2

Rewriting in standard form, we have

x2 � 2x � 2 � 0

Labeling the coefficients, we find that

a � 1 b � �2 c � 2

C A U T I O N

Students are sometimestempted to reduce this result to

� 1 �

This is not a valid step. We mustdivide each of the terms in thenumerator by 2 whensimplifying the expression.

213

6 � 2136

NOTE The solutions will havean imaginary part any time

is negative.2b2 � 4ac

Example 4 requires some special care in simplifying the solution.


Using the quadratic equation, solve 2x2 � x � 7.


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Example 6

NOTE Although not necessarilydistinct or real, every second-degree equation has twosolutions.

Analyzing the Discriminant

How many real solutions are there for each of the following quadratic equations?

(a) x2 � 7x � 15 � 0



x2 � 4x � 6 � 0

NOTE The graph ofy � x2 � 2x � 2

does not intercept the x axis, sothere are no real solutions.

NOTE Graphically, we can seethe number of real solutions asthe number of times therelated quadratic functionintercepts the x axis.

In attempting to solve a quadratic equation, you should first try the factoring method. Ifthis method does not work, you can apply the quadratic formula or the square root methodto find the solution. The following algorithm outlines the steps for solving equations usingthe quadratic formula.

Step 1 Write the equation in standard form (one side is equal to 0).

ax2 � bx � c � 0Step 2 Determine the values for a, b, and c.Step 3 Substitute those values into the quadratic formula.

Step 4 Simplify.

x ��b � 2b2 � 4ac

2a

Step by Step: Solving a Quadratic Equation byUsing the Quadratic Formula

Given a quadratic equation, the radicand b2 � 4ac determines the number of real solu-tions. Because of this, we call it the discriminant. Because the value of the discriminant isa real number, there are three possibilities.

� 0 there are no real solutions, but two imaginary solutionsIf b2 � 4ac � 0 there is one real solution (a double solution)

� 0 there are two distinct real solutions

Rules and Properties: The Discriminant

�

Applying the quadratic formula, we have

and noting that is 2i, we can simplify to

x � 1 � i or �1 � i�

1�4

x �2 � 1�4

2


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How many real solutions are there for each of the following quadratic equations?

(a) 2x2 � 3x � 2 � 0 (b) 3x2 � x � 11 � 0

(c) 4x2 � 4x � 1 � 0 (d) x2 � �5x � 7

Consider the following two applications involving thrown balls that can be solved byusing the quadratic formula.

Example 7

Solving a Thrown-Ball Application

If a ball is thrown upward from the ground, the equation to find the height h of such a ballthrown with an initial velocity of 80 ft/s is

h(t) � 80t � 16t 2

Find the time it takes the ball to reach a height of 48 ft.First we substitute 48 for h, and then we rewrite the equation in standard form.

16t2 � 80t � 48 � 0

To simplify the computation, we divide both sides of the equation by the common factor,16. This yields

t2 � 5t � 3 � 0

We solve for t as before, using the quadratic equation, with the result

t �5 � 113

2

NOTE Here h measures theheight above the ground, infeet, t seconds (s) after the ballis thrown upward.

NOTE Notice that the result ofdividing by 16

is 0 on the right.

016

� 0

NOTE We could find twoimaginary solutions by usingthe quadratic formula.

The value of the discriminant is [49 � 4(1)(�15)] or 109. This indicates that there aretwo real solutions.

(b) 3x2 � 5x � 7 � 0

The value of the discriminant is negative (�59). There are no real solutions.

(c) 9x2 � 12x � 4 � 0

The value of the discriminant is 0. There is exactly one real solution (a double solution).


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The equation to find the height h of a ball thrown upward with an initial velocityof 64 ft/s is

h(t) � 64t � 16t2

Find the time it takes the ball to reach a height of 32 ft.

Example 8

Solving a Thrown-Ball Application

The height, h, of a ball thrown downward from the top of a 240-ft building with an initialvelocity of 64 ft/s is given by

h(t) � 240 � 64t � 16t2

At what time will the ball reach a height of 176 ft?Let h(t) � 176, and write the equation in standard form.

176 � 240 � 64t � 16t2

0 � 64 � 64t � 16t2

16t2 � 64t � 64 � 0

or

t2 � 4t � 4 � 0

Applying the quadratic formula with a � 1, b � 4, and c � �4 yields

t � �2 � 2

Estimating these solutions, we have t � �4.8 and t � 0.8 s, but of these two values only thepositive value makes any sense. (To accept the negative solution would be to say that theball reached the specified height before it was thrown.)

12

NOTE The graph of

h(t) � 240 � 64t � 16t2

shows the height, h, at anytime t.

�2 2

100

200

NOTE Again, we divide bothsides of the equation by 16 tosimplify the computation.

NOTE The ball has a height of176 ft at approximately 0.8 s.


The height h of a ball thrown upward from the top of a 96-ft building with an initialvelocity of 16 ft/s is given by

h(t) � 96 � 16t � 16t2

When will the ball have a height of 32 ft above the ground? (Estimate your answerto the nearest tenth of a second.)

Another geometric result that generates quadratic equations in applications is thePythagorean theorem, which we discussed in Section 8.5.

This gives us two solutions, and . But, because we have specified units

of time, we generally estimate the answer to the nearest tenth or hundredth of a second.In this case, estimating to the nearest tenth of a second gives solutions of 0.7 and 4.3 s.

5 � 113

2

5 � 113

2NOTE There are two solutionsbecause the ball reaches theheight twice, once on the wayup and once on the way down.


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A Triangular Application

One leg of a right triangle is 4 cm longer than the other leg. The length of the hypotenuseof the triangle is 12 cm. Find the length of the two legs.

As in any geometric problem, a sketch of the information will help us visualize.

Now we apply the Pythagorean theorem to write an equation.

x2 � (x � 4)2 � (12)2

x2 � x2 � 8x � 16 � 144

or

2x2 � 8x � 128 � 0

Dividing both sides by 2, we have the equivalent equation

x2 � 4x � 64 � 0

Using the quadratic formula, we get

x � �2 � 2 or x � �2 � 2

Now, we check our answers for reasonableness. We can reject (do you seewhy?), but we should still check the reasonableness of the value . We couldsubstitute into the original equation, but it seems more prudent to simplycheck that it “makes sense” as a solution. Remembering that we estimate

as

�2 � 2(4) � 6

Our equation,

x2 � (x � 4)2 ? (12)2

when x equals 6, becomes

36 � 100 144

This indicates that our answer is at least reasonable.

�2 � 2117116 � 4,

�2 � 2117�2 � 2117

�2 � 2117

117117

x � 4

x12We assign variable x to the shorter legand x � 4 to the other leg.

NOTE The sum of the squaresof the legs of the triangle isequal to the square of thehypotenuse.

NOTE Dividing both sides of aquadratic equation by acommon factor is always aprudent step. It simplifies yourwork with the quadraticformula.

NOTE is just slightly morethan or 4.116

117

Example 9

In Example 9, the solution of the quadratic equation contains radicals. Substituting a

pair of solutions such as is a very difficult process. As in our thrown-ball appli-

cations, always check the “reasonableness” of the answer.

3 � 15

2


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One leg of a right triangle is 2 cm longer than the other. The hypotenuse is 1 cm lessthan twice the length of the shorter leg. Find the length of each side of the triangle.

An important economic application involves supply and demand. Our last example illus-trates that application.

Example 10

An Economic Application

The number of intermediate algebra workbooks that a publisher is willing to produce isdetermined by the supply curve

S(p) � �p2 � 30p � 180 in which p is the unit price in dollars

The demand for these workbooks is determined by the equation

D(p) � �10p � 130

Find the equilibrium price (the price at which supply and demand are equal).Because supply equals demand (S � D at equilibrium), we can write

�p2 � 30p � 180 � �10p � 130

Rewriting this statement as a quadratic equation in standard form yields

p2 � 40p � 310 � 0

When we apply the quadratic formula, we find the solutions

p � 20 �

p 10.51 or p 29.49

Although you might assume that the publisher will choose the higher price, it will, in fact,choose $10.51. If you want to discover why, try substituting the two solutions into the orig-inal demand equation.

3110

C H E C K Y O U R S E L F 1 0

The demand equation for CDs that accompany a text is predicted to be

D � �6p � 30 in which p is the unit price in dollars

The supply equation is predicted to be

S � �p2 � 12p � 20

Find the equilibrium price.


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C H E C K Y O U R S E L F A N S W E R S

1. 2. 3. 4. � �

5. � � 6. (a) None; (b) two; (c) one; (d) none 7. 0.6 and 3.4 s

8. 2.6 s 9. Approximately 4.3, 6.3, and 7.7 cm 10. $3.43

2 � i12

2 � 110�1 � 157

4 ��1

2��2, 4

3�

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Section Date

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

703

Exercises

In exercises 1 to 8, solve each quadratic equation by first factoring and then using thequadratic formula.

1. x2 � 5x � 14 � 0 2. x2 � 7x � 18 � 0

3. t2 � 8t � 65 � 0 4. q2 � 3q � 130 � 0

5. 5x2 � 4x � 1 � 0 6. 3x2 � 2x � 1 � 0

7. 16t2 � 24t � 9 � 0 8. 6m2 � 23m � 10 � 0

In exercises 9 to 20, solve each quadratic equation by (a) completing the square, and(b) using the quadratic formula.

9. x2 � 2x � 5 � 0 10. x2 � 6x � 1 � 0

11. x2 � 3x � 27 � 0 12. t2 � 4t � 7 � 0

13. 2x2 � 6x � 3 � 0 14. 2x2 � 6x � 1 � 0

15. 2q2 � 4q � 1 � 0 16. 4r2 � 2r � 1 � 0

17. 3x2 � x � 2 � 0 18. 2x2 � 8x � 3 � 0

19. 2y2 � y � 5 � 0 20. 3m2 � 2m � 1 � 0

9.3

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ANSWERS

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

704

In exercises 21 to 42, solve each equation by using the quadratic formula.

21. x2 � 4x � 3 � 0 22. x2 � 7x � 3 � 0

23. p2 � 8p � 16 � 0 24. u2 � 7u � 30 � 0

25. 2x2 � 2x � 3 � 0 26. 2x2 � 3x � 7 � 0

27. �3s2 � 2s � 1 � 0 28. 5t2 � 2t � 2 � 0

Hint: Clear each of the following equations of fractions or parentheses first.

29. 2x2 � x � 5 � 0 30. 3x2 � x � 3 � 0

31. 5t2 � 2t � � 0 32. 3y2 � 2y � � 0

33. (x � 2)(x � 3) � 4 34. (x � 1)(x � 8) � 3

35. (t � 1)(2t � 4) � 7 � 0 36. (2w � 1)(3w � 2) � 1

37. 3x � 5 � 38. x � 3 �

39. 2t � � 3 40. 4p � � 6

41. 42.6

x2 �2

x� 1

5

y2 �2

y� 1 � 0

1

p

3

t

1

x

1

x

3

4

2

3

1

3

1

2

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In exercises 43 to 50, for each quadratic equation, find the value of the discriminant andgive the number of real solutions.

43. 2x2 � 5x � 0 44. 3x2 � 8x � 0

45. m2 � 8m � 16 � 0 46. 4p2 � 12p � 9 � 0

47. 3x2 � 7x � 1 � 0 48. 2x2 � x � 5 � 0

49. 2w2 � 5w � 11 � 0 50. 7q2 � 3q � 1 � 0

In exercises 51 to 62, find all the solutions of each quadratic equation. Use any applicablemethod.

51. x2 � 8x � 16 � 0 52. 4x2 � 12x � 9 � 0

53. 3t2 � 7t � 1 � 0 54. 2z2 � z � 5 � 0

55. 5y2 � 2y � 0 56. 7z2 � 6z � 2 � 0

57. (x � 1)(2x � 7) � �6 58. 4x2 � 3 � 0

59. x2 � 9 � 0 60. (4x � 5)(x � 2) � 1

61. x � 3 � 62. 1 � � 02

x�

2

x2

10

x� 0

ANSWERS

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

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The equation

h(t) � 112t � 16t2

is the equation for the height of an arrow, shot upward from the ground with an initialvelocity of 112 ft/s, in which t is the time, in seconds, after the arrow leaves the ground.Use this information to solve exercises 63 and 64. Your answers should be expressed tothe nearest tenth of a second.

63. Find the time it takes for the arrow to reach a height of 112 ft.

64. Find the time it takes for the arrow to reach a height of 144 ft.

The equation

h(t) � 320 � 32t � 16t2

is the equation for the height of a ball, thrown downward from the top of a 320-ft buildingwith an initial velocity of 32 ft/s, in which t is the time after the ball is thrown down fromthe top of the building. Use this information to solve exercises 65 and 66. Express yourresults to the nearest tenth of a second.

65. Find the time it takes for the ball to reach a height of 240 ft.

66. Find the time it takes for the ball to reach a height of 96 ft.

67. Number problem. The product of two consecutive integers is 72. What are the twointegers?

68. Number problem. The sum of the squares of two consecutive whole numbers is 61.Find the two whole numbers.

69. Rectangles. The width of a rectangle is 3 ft less than its length. If the area of therectangle is 70 ft2, what are the dimensions of the rectangle?

70. Rectangles. The length of a rectangle is 5 cm more than its width. If the area of therectangle is 84 cm2, find the dimensions.

71. Rectangles. The length of a rectangle is 2 cm more than 3 times its width. If the areaof the rectangle is 85 cm2, find the dimensions of the rectangle.

x � 3

x

ANSWERS

63.

64.

65.

66.

67.

68.

69.

70.

71.

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72. Rectangles. If the length of a rectangle is 3 ft less than twice its width, and the areaof the rectangle is 54 ft2, what are the dimensions of the rectangle?

73. Triangles. One leg of a right triangle is twice the length of the other. The hypotenuseis 6 m long. Find the length of each leg.

74. Triangles. One leg of a right triangle is 2 ft longer than the shorter side. If the lengthof the hypotenuse is 14 ft, how long is each leg?

75. Triangles. One leg of a right triangle is 1 in. shorter than the other leg. Thehypotenuse is 3 in. longer than the shorter side. Find the length of each side.

76. Triangles. The hypotenuse of a given right triangle is 5 cm longer than the shorterleg. The length of the shorter leg is 2 cm less than that of the longer leg. Find thelength of the three sides.

77. Triangles. The sum of the lengths of the two legs of a right triangle is 25 m. Thehypotenuse is 22 m long. Find the length of the two legs.

78. Triangles. The sum of the lengths of one side of a right triangle and the hypotenuse is15 cm. The other leg is 5 cm shorter than the hypotenuse. Find the length of each side.

79. Thrown ball. If a ball is thrown vertically upward from the ground, its height, h,after t seconds is given by

h(t) � 64t � 16t2

(a) How long does it take the ball to return to the ground? [Hint: Let h(t) � 0.]

(b) How long does it take the ball to reach a height of 48 ft on the way up?

80. Thrown ball. If a ball is thrown vertically upward from the ground, its height, h,after t seconds is given by

h(t) � 96t � 16t2

(a) How long does it take the ball to return to the ground?

(b) How long does it take the ball to pass through a height of 128 ft on the way backdown to the ground?

x

x � 2

14 ft

ANSWERS

72.

73.

74.

75.

76.

77.

78.

79.

80.

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81. Cost. Suppose that the cost C(x), in dollars, of producing x chairs is given by

C(x) � 2400 � 40x � 2x2

How many chairs can be produced for $5400?

82. Profit. Suppose that the profit T(x), in dollars, of producing and selling x appliancesis given by

T(x) � �3x2 � 240x � 1800

How many appliances must be produced and sold to achieve a profit of $3000?

If a ball is thrown upward from the roof of a building 70 m tall with an initial velocity of15 m/s, its approximate height, h, after t seconds is given by

h(t) � 70 � 15t � 5t2

Note: The difference between this equation and the one we used in Example 8 has to dowith the units used. When we used feet, the t2 coefficient was �16 (from the fact that theacceleration due to gravity is approximately 32 ft/s2). When we use meters as the height,the t2 coefficient is �5 (that same acceleration becomes approximately 10 m/s2). Use thisinformation to solve exercises 83 and 84.

83. Thrown ball. How long does it take the ball to fall back to the ground?

84. Thrown ball. When will the ball reach a height of 80 m?

Changing the initial velocity to 25 m/s will only change the t coefficient. Our new equationbecomes

h(t) � 70 � 25t � 5t2

85. Thrown ball. How long will it take the ball to return to the thrower?


The only part of the height equation that we have not discussed is the constant. You haveprobably noticed that the constant is always equal to the initial height of the ball (70 m inour previous problems). Now, let’s have you develop an equation.

ANSWERS

81.

82.

83.

84.

85.

86.

A ball is thrown upward from the roof of a 100-m building with an initial velocity of 20 m/s.Use this information to solve exercises 87 to 90.

87. Thrown ball. Find the equation for the height, h, of the ball after t seconds.

88. Thrown ball. How long will it take the ball to fall back to the ground?


90. Thrown ball. Will the ball ever reach a height of 125 m? (Hint: Check thediscriminant.)

A ball is thrown upward from the roof of a 100-ft building with an initial velocity of 20 ft/s.Use this information to solve exercises 91 to 94.

91. Thrown ball. Find the equation for the height, h, of the ball after t seconds.

92. Thrown ball. How long will it take the ball to fall back to the ground?

93. Thrown ball. When will the ball reach a height of 80 ft?

94. Thrown ball. Will the ball ever reach a height of 120 ft? Explain.

95. Profit. A small manufacturer’s weekly profit in dollars is given by

P(x) � �3x2 � 270x

Find the number of items x that must be produced to realize a profit of $5100.

96. Profit. Suppose the profit in dollars is given by

P(x) � �2x2 � 240x

Now how many items must be sold to realize a profit of $5100?

97. Equilibrium price. The demand equation for a certain computer chip is given by

D � �2p � 14


S � �p2 � 16p � 2

Find the equilibrium price.© 2

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87.

88.

89.

90.

91.

92.

93.

94.

95.

96.

97.

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98. Equilibrium price. The demand equation for a certain type of print is predictedto be

D � �200p � 36,000


S � �p2 � 400p � 24,000

Find the equilibrium price.

99. Can the solution of a quadratic equation with integer coefficients include one realand one imaginary number? Justify your answer.

100. Explain how the discriminant is used to predict the nature of the solutions of aquadratic equation.

In exercises 101 to 108, solve each equation for x.

101. x2 � y2 � z2 102. 2x2y2z2 � 1

103. x2 � 36a2 � 0 104. ax2 � 9b2 � 0

105. 2x2 � 5ax � 3a2 � 0 106. 3x2 � 16bx � 5b2 � 0

107. 2x2 � ax � 2a2 � 0 108. 3x2 � 2bx � 2b2 � 0

109. Given that the polynomial x3 � 3x2 � 15x � 25 � 0 has as one of its solutionsx � 5, find the other two solutions. (Hint: If you divide the given polynomial byx � 5 the quotient will be a quadratic equation. The remaining solutions will be thesolutions for that equation.)

110. Given that 2x3 � 2x2 � 5x � 2 � 0 has as one of its solutions x � �2, find theother two solutions. (Hint: In this case, divide the original polynomial by x � 2.)

111. Find all the zeros of the function f(x) � x3 � 1.

ANSWERS

98.

99.

100.

101.

102.

103.

104.

105.

106.

107.

108.

109.

110.

111.

710

112. Find the zeros of the function f(x) � x2 � x � 1.

113. Find all six solutions to the equation x6 � 1 � 0. (Hint: Factor the left-hand side ofthe equation first as the difference of squares, then as the sum and difference ofcubes.)

114. Find all six solutions to x6 � 64.

115. (a) Use the quadratic formula to solve x2 � 3x � 5 � 0. For each solution give adecimal approximation to the nearest tenth.

(b) Graph the function f(x) � x2 � 3x � 5 on your graphing calculator. Use a zoomutility and estimate the x intercepts to the nearest tenth.

(c) Describe the connection between parts (a) and (b).

116. (a) Solve the following equation using any appropriate method:

x2 � 2x � 3

(b) Graph the following functions on your graphing calculator:

f(x) � x2 � 2x and g(x) � 3

Estimate the points of intersection of the graphs of f and g. In particular notethe x coordinates of these points.

(c) Describe the connection between parts (a) and (b).

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112.

113.

114.

115. (a)(b)(c)

116. (a)(b)(c)

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Answers

1. ��2, 7� 3. ��13, 5� 5. 7. 9.

11. 13. 15. 17.

19. 21. �1, 3� 23. �4� 25.

27. 29. 31.

33. 35. 37.

39. 41. 43. 25, two 45. 0, one

47. 37, two 49. �63, none 51. �4� 53. 55.

57. 59. ��3i, 3i� 61. ��2, 5� 63. �1.2 or 5.8 s�

65. 1.4 s 67. �9, �8, or 8, 9 69. 7 by 10 ft 71. 5 by 17 cm

73. 2.7 m, 5.4 m 75. 5.5, 6.5, 8.5 in. 77. 3.2 m, 21.8 m

79. (a) 4 s; (b) 1 s 81. 50 chairs 83. 5.5 s 85. 5 s

87. h(t) � 100 � 20t � 5t 2 89. 5 s 91. h(t) � 100 � 20t � 16t 2

93. 1.9 s 95. 63 or 27 97. $0.94 99. 101. ��

103. ��6a, 6a� 105. 107. 109. ��1 � �

111. 113.

115. (a) ��1.2, 4.2�; (b) �1.2, 4.2; (c) Solutions to quadratic equation are the x interceptsof the graph

��1, 1, 1 � i13

2,

�1 � i13

2 ��1, 1 � i13

2 �16��a � a117

4 ��3a, a

2�2z2 � y2

��5 � 133

4 ��0,

2

5��7 � 137

6 ��1 � 16��3 � 133

4 ��5 � 137

6 ��1 � 123

2 ��1 � 141

2 ��3 � 139

15 ��1 � 1161

8 �� 1 � i12

3 ��1 � 17

2 ��1 � 141

4 ��

2

3, 1��

2 � 12

2 �� 3 � 115

2 ��3 � 3113

2 ��1 � 16��3

4��1, 1

5�

712

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9.3 Solving Quadratic Equations by Using the … Quadratic Equations by Using the Quadratic Formula 9.3 9.3 OBJECTIVES 1. Solve a quadratic equation by using the quadratic formula