Hungerford T.W., Shaw D.J. Contemporary Precalculus: A Graphing Approach
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EXAMPLE 12
If an object is thrown upward, dropped, or thrown downward and travels in a
straight line subject only to gravity (with wind resistance ignored), the height h of
the object above the ground (in feet) after t seconds is given by
h 16t
2
v
0
t h
0
,
where h
0
is the height of the object when t 0 and v
0
is the initial velocity at time
t 0. The value of v
0
is taken as positive if the object moves upward and negative
if it moves downward. If a baseball is thrown down from the top of a
640-foot-high building with an initial velocity of 52 feet per second, how long
does it take to reach the ground?
SOLUTION In this case, v
0
is 52 and h
0
is 640, so that the height equation is
h 16t
2
52t 640.
The object is on the ground when h 0, so we must solve the equation
0 16t
2
52t 640.
Using the quadratic formula and a calculator, we see that
8.15
t
52
3
4
2
3,664
or
4.90.
Only the positive answer makes sense in this case. So it takes about 4.9 seconds
for the baseball to reach the ground. ■
HIGHER-DEGREE EQUATIONS
A polynomial equation of degree n is one that can be written in the form
a
n
x
n
a
3
x
3
a
2
x
2
a
1
x a
0
0,
where n is a positive integer, each a
i
is a constant, and a
n
0. For instance,
4x
6
3x
5
x
4
7x
3
8x
2
4x 9 0
is a polynomial equation of degree 6. As a general rule, polynomial equations of
degree 3 and above are best solved by the numerical or graphical methods pre-
sented in Section 2.2. However, some such equations can be solved algebraically
by making a suitable substitution, as we now see.
EXAMPLE 13
To solve 4x
4
13x
2
3 0, substitute u for x
2
and solve the resulting quadratic
equation:
4x
4
13x
2
3 0
4(x
2
)
2
13x
2
3 0
(52)
(52)
2
4(
16)(64
0)
2(16)
26 CHAPTER 1 Basics
TECHNOLOGY TIP
Most calculators have built-in
polynomial equation solvers that will
solve quadratic and other polynomial
equations. See Exercise 105.
Let u x
2
: 4u
2
13u 3 0
(u 3)(4u 1) 0
u 3 0or4u 1 0
u 34u 1
u
1
4
.
Since u x
2
, we see that
x
2
3orx
2
1
4
x 3
x
1
2
.
Hence, the original equation has four solutions: 3
, 3
, 1/2, 1/2. ■
EXAMPLE 14
To solve x
4
4x
2
1 0, let u x
2
:
x
4
4x
2
1 0
u
2
4u 1 0.
The quadratic formula shows that
u
4
2
12
4
2
4
3
4
2
4
3
4
2
23
2 3
.
Since u x
2
, we have the equivalent statements:
x
2
2 3
or x
2
2 3
x 2
3
x 2
3
.
Therefore the original equation has four solutions. ■
FRACTIONAL EQUATIONS
The first step in solving an equation that involves fractional expressions is to mul-
tiply both sides by a common denominator to eliminate the fractions. Then use the
methods presented earlier in this chapter to solve the resulting equation. However,
care must be used.
Multiplying both sides of an equation by a quantity involving the variable
(which may be zero for some values) may lead to an extraneous solution, a
number that does not satisfy the original equation.* To avoid errors in such situa-
tions, always check your solutions in the original equation.
(4)
(4)
2
4
1
1
2
1
SECTION 1.2 Solving Equations Algebraically 27
*The second basic principle for solving equations (page 19) applies only to nonzero quantities.
EXAMPLE 15
Solve
x
x
2
3
x
7
3
x
2
3
0
9
.
SOLUTION Note that (x 3)(x 3) x
2
9. Multiply both sides of the
equation by (x 3)(x 3) to eliminate the fractions.
x
x
2
3
(x 3)(x 3)
x
7
3
(x 3)(x 3)
x
2
3
0
9
(x 3)(x 3)
Cancel common factors: (x 2)(x 3) 7(x 3) 30
Multiply out left side: x
2
5x 6 7x 21 30
Simplify: x
2
2x 27 30
Subtract 30 from both sides: x
2
2x 3 0
Factor: (x 3)(x 1) 0
Zero Product Property: x 3 0orx 1 0
x 3 x 1
So the possible solutions are 3 and 1. When 3 is substituted for x in the original
equation, its first term becomes
3
3
3
3
6
0
, which is not defined. So 3 is not a
solution of the original equation. Next we substitute 1 for x in the original
equation and obtain:
Left side:
1
1
2
3
1
7
3
1
4
7
2
1
4
1
4
4
1
4
5
Right side:
(1
3
)
2
0
9
30
8
1
4
5
.
So the left and right sides are equal. Therefore, 1 is the only solution of the
original equation. ■
EXAMPLE 16
The relationship between the focal length F of a digital camera lens, the distance
u from the object being photographed to the lens, and the distance v from the lens
to the camera’s sensor is given by
F
1
1
v
1
u
.
Express u in terms of F and v.
SOLUTION We must solve the equation for u. First we eliminate fractions by
multiplying both sides by Fvu.
Fvu
F
1
Fvu
1
v
1
u
Distributive Law: Fvu
F
1
Fvu
1
v
Fvu
1
u
28 CHAPTER 1 Basics
Cancel: vu Fu Fv
Subtract Fu from both sides: vu Fu Fv
Distributive Law: (v F)u Fv
Divide both sides by v F: u
v
F
v
F
. ■
SECTION 1.2 Solving Equations Algebraically 29
In Exercises 1–6, solve the equation.
1. 3x 2 26 2.
5
y
3 14
3. 3x 2 9x 7 4. 7(t 2) 3(4t 1)
5.
3
4
y
6 y 2 6. 2(1 x) 3x 5
In Exercises 7–12, solve the equation for the indicated
variable.
7. x 3y 5 for y 8. 5x 2y 1 for x
9. A
h
2
(b c) for b 10. V pb
2
c for c
11. V
pd
4
2
h
for h 12.
1
r
1
s
1
t
for r
In Exercises 13–22, solve the equation by factoring.
13. x
2
8x 15 0 14. x
2
5x 6 0
15. x
2
5x 14 16. x
2
x 20
17. 2y
2
5y 3 0 18. 3t
2
t 2 0
19. 4t
2
9t 2 0 20. 9t
2
2 11t
21. 3u
2
4u 4 22. 5x
2
26x 5
In Exercises 23–26, solve the equation by completing the
square.
23. x
2
2x 12
24. x
2
4x 30 0
25. x
2
x 1 0
26. x
2
3x 2 0
In Exercises 27–32, find the number of real solutions of the
equation by computing the discriminant.
27. x
2
4x 1 0 28. 4x
2
4x 3 0
29. 9x
2
12x 1 30. 9t
2
15 30t
31. 25t
2
49 70t 32. 49t
2
5 42t
In Exercises 33–42, use the quadratic formula to solve the
equation.
33. x
2
4x 1 0 34. x
2
2x 1 0
35. x
2
6x 7 0 36. x
2
4x 3 0
37. x
2
6 2x 38. x
2
11 6x
39. 4x
2
4x 7 40. 4x
2
4x 11
41. 4x
2
8x 1 0 42. 2t
2
4t 1 0
In Exercises 43–52, solve the equation by any method.
43. x
2
9x 18 0 44. 3t
2
11t 20 0
45. 4x(x 1) 1 46. 25y
2
20y 1
47. 2x
2
7x 15 48. 2x
2
6x 3
49. t
2
4t 13 0 50. 5x
2
2x 2
51.
7
3
x
2
2
3
x
1 52. x
2
2
x 3 0
In Exercises 53–56, use a calculator to find approximate
solutions of the equation.
53. 4.42x
2
10.14x 3.79 0
54. 8.06x
2
25.8726x 25.047256 0
55. 3x
2
82.74x 570.4923 0
56. 7.63x
2
2.79x 5.32
In Exercises 57–64, find all real solutions of the equation exactly.
57. y
4
7y
2
6 0 58. x
4
2x
2
1 0
59. x
4
2x
2
35 0 60. x
4
2x
2
24 0
61. 2y
4
9y
2
4 0 62. 6z
4
7z
2
2 0
63. 10x
4
3x
2
1 64. 6x
4
7x
2
3
In Exercises 65–74, solve the equation and check your answers.
65.
2
1
t
5
2
t
1
1
0t
1 66.
1
2
2
y
1
3
3
y
67.
2
x
x
4
7
x
5
4
2 68.
z
z
4
5
z
1
5
EXERCISES 1.2
69. 25x
4
x
20 70. 1
3
x
4
x
0
2
71.
x
2
2
5
x
4 72.
x
x
1
x
x
2
3
73.
3x
2
4x
2
5
x
5
2
3x
4
1
x
3
2
74.
x
x
3
2
x
3
2
x
2
2
0
4
The gross federal debt y (in trillions of dollars) in year x is
approximated by
y .79x 3.93 (x 3),
where x is the number of years after 2000.* In Exercises 75–76,
find the year in which the approximate federal debt is:
75. $12.62 billion 76. $14.2 billion
The total health care expenditures E in the United States (in
billions of dollars) can be approximated by
E 73.04x 625.6,
where x is the number of years since 1990.
†
Determine the year
in which health care expenditures are at the given level.
77. $1794.25 billion 78. $1940.3 billion
In a simple model of the economy (by John Maynard Keynes),
equilibrium between national output and national expenditures
is given by the equilibrium equation
Y C I G (X M),
where Y is the national income, C is consumption (which
depends on national income), I is the amount of investment, G is
government spending, X is exports, and M is imports (which
also depend on national income). In Exercises 79 and 80, solve
the equilibrium equation for Y under the given conditions.
79. C 120 .9Y, M 20 .2Y, I 140, G 150, and
X 60
80. C 60 .85Y, M 35 .2Y, I 95, G 145, and
X 50
In Exercises 81–84, use the height equation in Example 12.
Note that an object that is dropped (rather than thrown
downward) has initial velocity v
0
0.
81. How long does it take a baseball to reach the ground if it is
dropped from the top of a 640-foot-high building? Compare
with Example 12.
30 CHAPTER 1 Basics
82. You are standing on a cliff that is 200 feet high. How long
will it take a rock to reach the ground if
(a) you drop it?
(b) you throw it downward at an initial velocity of 40 feet
per second?
(c) How far does the rock fall in 2 seconds if you throw it
downward with an initial velocity of 40 feet per second?
83. A rocket is fired straight up from ground level with an initial
velocity of 800 feet per second.
(a) How long does it take the rocket to rise 3200 feet?
(b) When will the rocket hit the ground?
84. A rocket loaded with fireworks is to be shot vertically up-
ward from ground level with an initial velocity of
200 feet per second. When the rocket reaches a height of
400 feet on its upward trip the fireworks will be detonated.
How many seconds after liftoff will this take place?
85. The atmospheric pressure a (in pounds per square foot) at
height h thousand feet above sea level is approximately
a .8315h
2
73.93h 2116.1.
(a) Find the atmospheric pressure at sea level and at the top
of Mount Everest, the tallest mountain in the world
(29,035 feet*). [Remember that h is measured in
thousands.]
(b) The atmospheric pressure at the top of Mount Rainier is
1223.43 pounds per square foot. How high is Mount
Rainier?
86. Data from the U.S. Department of Health and Human
Services indicates that the cumulative number N of reported
cases of AIDS in the United States in year x can be approx-
imated by the equation
N 3362.1x
2
17,270.3x 24,043,
where x 0 corresponds to 1980. In what year did the total
reach 550,000?
87. According to data from the U.S. Census Bureau, the popu-
lation P of Cleveland, Ohio (in thousands) in year x can
be approximated by P .08x
2
13.08x 927, where
x 0 corresponds to 1950. In what year in the past was
the population about 804,200?
88. The number N of AIDS cases diagnosed to date (in thou-
sands) is approximated by N .37x
2
59.5x 247.26,
where x is the number of years since 1990.
†
Assuming that
this equation remains valid through 2011, determine when
the number of diagnosed cases of AIDS was or will be
(a) 825,000
(b) 1.2 million
*Based on data and projections from the Congressional Budget Office in
2005.
†
Based on data from the U.S. Centers for Medicare and Medicaid
Services.
*Based on measurements in 1999 by climbers sponsored by the Boston
Museum of Science and the National Geographic Society, using satellite-
based technology.
†
Based on data from the U.S. Department of Health and Human Services.
SECTION 1.2 Solving Equations Algebraically 31
89. The number N of Walgreens drugstores in year x can be ap-
proximated by N 6.82x
2
1.55x 666.8, where x 0
corresponds to 1980.* Determine when the number of
stores was or will be
(a) 4240 (b) 5600 (c) 7000
90. The total resources T (in billions of dollars) of the Pension
Benefit Guaranty Corporation, the government agency
that insures pensions, can be approximated by the equation
T .26x
2
3.62x 30.18, where x is the number of
years after 2000.
†
Determine when the total resources are at
the given level.
(a) $42.5 billion
(b) $30 billion
(c) When will the Corporation be out of money (T 0)?
91. According to data from the National Highway Traffic
Safety Administration, the driver fatality rate D per 1000
licensed drivers every 100 million miles can be approxi-
mated by the equation D .0031x
2
.291x 7.1, where x
is the age of the driver.
(a) For what ages is the driver fatality rate about 1 death
per 1000?
(b) For what ages is the rate three times greater than in
part (a)?
92. The combined resistance of two resistors, with resistances
R
1
and R
2
respectively, connected in parallel is given by
R
1
R
1
R
2
R
2
. If the first resistor has 8 ohms resistance and the
combined resistance is 4.8 ohms, what is the resistance of
the second resistor?
93. The cost-benefit equation
10
1
0
8
x
x
D relates the cost D
(in thousands of dollars) needed to remove x percent of a pol-
lutant from the emissions of a factory. Find the percent of the
pollutant removed when the following amounts are spent.
(a) $50,000 [Here D 50]
(b) $100,000
(c) $200,000
94. (a) Let b be a real number. Multiply out the expression
x
b
2
2
.
(b) Explain why your computation in part (a) shows that
this statement is true: If you add
b
2
2
to the expression
x
2
bx, the resulting polynomial is a perfect square.
THINKERS
In Exercises 95–98, find a number k such that the given
equation has exactly one real solution.
95. x
2
kx 25 0 96. x
2
kx 49 0
97. kx
2
8x 1 0 98. kx
2
24x 16 0
In Exercises 99–101, the discriminant of the equation
ax
2
bx c 0 (with a, b, c integers) is given. Use it to
determine whether or not the solutions of the equation are
rational numbers.
99. b
2
4ac 25
100. b
2
4ac 0
101. b
2
4ac 72
102. Find the error in the following “proof” that 6 3.
x 3
Multiply both sides by x: x
2
3x
Subtract 9 from both sides: x
2
9 3x 9
Factor each side: (x 3)(x 3) 3(x 3)
Divide both sides by x 3: x 3 3
Since x 3: 3 3 3
6 3
103. Find a number k such that 4 and 1 are the solutions of
x
2
5x k 0.
104. Suppose a, b, c are fixed real numbers such that
b
2
4ac 0. Let r and s be the solutions of
ax
2
bx c 0.
(a) Use the quadratic formula to show that r s b/a
and rs c/a.
(b) Use part (a) to verify that ax
2
bx c
a(x r)(x s).
(c) Use part (b) to factor x
2
2x 1 and 5x
2
8x 2.
105. (a) Solve x
2
5x 2 0 (exact answer required).
(b) If you have one of the calculators listed below, use
its polynomial solver to solve the equation in part (a).
Does your answer agree with the one in part (a)?
Calculator Use this menu/choice
TI-84 APPS/PolySmlt*
TI-86 POLY
†
TI-89 ALGEBRA/Solve
‡
Casio 9850 EQUATION (Main Menu)
HP-39gs MATH/POLYNOM/Polyroot
§
(c) Use the solver to solve 3x
4
2x
3
5x
2
2x 1 0.
*
Based on data from the Walgreen Company.
†
Based on data from the Center on Federal Financial Institutions.
*If PolySmlt is not in the APPS menu, you can download it from TI.
†
When asked for “order”, enter the degree of the polynomial.
‡
Syntax: Solve(x
2
5x 2 0, x)
§
Syntax: Polyroot([1, 5, 2])
32 CHAPTER 1 Basics
1.2.A SPECIAL TOPICS Absolute Value Equations
Section Objective
■ Solve absolute value equations algebraically.
If c is a real number, then by the definition of absolute value, c is either c or c
(whichever one is positive). This fact can be used to solve absolute value equa-
tions algebraically.
EXAMPLE 1
To solve 3x 4 8, apply the fact stated above with c 3x 4. Then
3x 4 is either 3x 4 or (3x 4), so
3x 4 8or(3x 4) 8
3x 12 3x 4 8
x 4 3x 4
x 4/3.
So there are two possible solutions of the original equation 3x 4 8. You can
readily verify that both 4 and 4/3 actually are solutions. ■
EXAMPLE 2
Solve x 4 5x 2.
SOLUTION The left side of the equation is either x 4 or (x 4) (why?).
Hence,
x 4 5x 2or(x 4) 5x 2
4x 4 2 x 4 5x 2
4x 6 6x 2
x
6
4
3
2
x
2
6
1
3
.
We must check each of these possible solutions in the original equation,
x 4 5x 2.
We see that x 3/2 is a solution because
3
2
4
1
2
1
and 5
3
2
2
1
2
1
.
However, x 1/3 is not a solution, since
1
3
4
1
3
1
but 5
1
3
2
1
3
1
. ■
1.2.B SPECIAL TOPICS Variation
SPECIAL TOPICS 1.2.B Variation 33
EXAMPLE 3
Solve the equation x
2
4x 3 2.
SOLUTION The equation is equivalent to
x
2
4x 3 2or(x
2
4x 3) 2
x
2
4x 5 0 x
2
4x 3 2
x
2
4x 1 0
x
2
4x 1 0.
The first of these equations can be solved by factoring and the second by the quad-
ratic formula:
(x 5)(x 1) 0orx
x 5orx 1 x
4
2
20
4
2
25
x 2 5
or x 2 5
.
Verify that all four of these numbers are solutions of the original equation. ■
4
4
2
4
1
(1)
2
1
In Exercises 1–12, find all real solutions of each equation.
1. 2x 3 9 2. 3x 5 7
3. 6x 9 0 4. 4x 5 9
5. 2x 3 4x 1 6. 3x 2 5x 4
7. x 3 x 8. 2x 1 2x 1
9. x
2
4x 1 4 10. x
2
2x 9 6
11. x
2
5x 1 3 12. 12x
2
5x 7 4
13. In statistical quality control, one needs to find the propor-
tion of the product that is not acceptable. The upper and
lower control limits are found by solving the following
equation (in which p
is the mean percent defective and n is
the sample size) for CL.
CL p
3
p
(1
n
p
)
Find the control limits when p
.02 and n 200.
Section Objectives
■ Set up and solve a direct variation equation.
■ Set up and solve an indirect variation equation.
■ Set up and solve combined variation equations.
If a plane flies 400 mph for t hours, then it travels a distance of 400t miles. So the
distance d is related to the time t by the equation d 400t. When time t increases,
so does the distance d. We say that d varies directly as t and that the constant of
variation is 400. Direct variation also occurs in other settings and is formally
defined as follows.
EXERCISES 1.2.A
EXAMPLE 1
When you swim underwater, the pressure p varies directly with the depth d
at which you swim. At a depth of 20 feet, the pressure is 8.6 pounds per square
inch. What is the pressure at 65 feet deep?
SOLUTION The variation equation is p kd for some constant k. We first use
the given information to find k. Since p 8.6 when d 20, we have
p kd
8.6 k(20)
k
8
2
.
0
6
.43.
Therefore, the variation equation is p .43d. To find the pressure at 65 feet,
substitute 65 for d in the equation:
p .43d .43(65) 27.95 pounds per square inch. ■
The basic idea in direct variation is that the two quantities grow or shrink
together. For instance, when v 3u, then as u gets large, so does v. But two quan-
tities can also be related in such a way that one grows as the other shrinks or vice
versa. For example if v 5/u, then when u is large (say, u 500), v is small:
v 5/500 .01. This is an example of inverse variation.
EXAMPLE 2
According to one of Parkinson’s laws, the amount of time the Math Department
spends discussing an item in its budget is inversely proportional to the cost of the
item. If the department spent 40 minutes discussing the purchase of a copying
machine for $2100, how much time will it spend discussing a $280 appropriation
for the annual picnic?
34 CHAPTER 1 Basics
Direct
Variation
Suppose the quantities u and v are related by the equation
v ku,
where k is a nonzero constant. We say that v varies directly as u or that v
is directly proportional to u. In this case, k is called the constant of vari-
ation or the constant of proportionality.
Inverse
Variation
Suppose the quantities u and v are related by the equation
v
u
k
,
where k is a nonzero constant. We say that v varies inversely as u or that v
is inversely proportional to u. Once again, k is called the constant of
variation or the constant of proportionality.
SOLUTION Let t denote time (in minutes) and c the cost of an item. Then the
variation equation is
t
k
c
for some constant k. We know that t 40 when c 2100. Substituting these num-
bers in the equation, we see that
40
21
k
00
k 40
2100 84,000.
So the variation equation is
t
84,
c
000
.
To find the time spent discussing the picnic, let c 280. Hence,
t
84
2
,
8
0
0
00
300 minutes ( 5 hours!). ■
The terminology of variation also applies to situations involving powers of
variables, as illustrated in the following table.
EXAMPLE 3
The density of light I falling on an object (measured in lumens per square foot) is
inversely proportional to the square of the distance d from the light source. If an
object 2 feet from a standard 100-watt light bulb receives 34.82 lumens per square
foot, how much light falls on an object 10 feet from this bulb?
SOLUTION The variation equation is
I
d
k
2
SPECIAL TOPICS 1.2.B Variation 35
General
Example Terminology Case*
y varies directly as the fourth power of x.
y 7x
4
y is directly proportional to the fourth power of x.
(Constant of variation is 7.)
A varies directly as the square of r.
A pr
2
A is directly proportional to the square of r.
(Constant of variation is p).
y varies inversely as the cube of x.
y
1
x
0
3
y is inversely proportional to the cube of x.
(Constant of variation is 10.)
W varies inversely as the square of d.
W
3.48
d
2
10
9
W is inversely proportional to the square of d.
(Constant of variation is 3.48 10
9
.)
v ku
n
v
u
k
n
*k is a nonzero constant; u and v are variables.