5: Solving Problems Using Quadratic Equations
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10.5 • Solving Problems Using Quadratic Equations
441
Suggestions for Solving Word Problems
1. Read the problem carefully and make certain that you understand the meanings of all the
words. Be especially alert for any technical terms used in the statement of the problem.
2. Read the problem a second time (perhaps even a third time) to get an overview of the
situation being described and to determine the known facts as well as what is to be found.
3. Sketch any figure, diagram, or chart that might be helpful in analyzing the problem.
*4. Choose meaningful variables to represent the unknown quantities. Use one or two
variables, whichever seems easiest. The term “meaningful” refers to the choice of
letters to use as variables. Choose letters that have some significance for the problem
under consideration. For example, if the problem deals with the length and width of
a rectangle, then l and w are natural choices for the variables.
*5. Look for guidelines that you can use to help set up equations. A guideline might be
a formula such as area of a rectangular region equals length times width, or a
statement of a relationship such as the product of the two numbers is 98.
*6. (a) Form an equation containing the variable, which translates the conditions of the
guideline from English into algebra; or
(b) Form two equations containing the two variables, which translate the guidelines
from English into algebra.
*7. Solve the equation (system of equations) and use the solution (solutions) to
determine all facts requested in the problem.
8. Check all answers back in the original statement of the problem.
The asterisks indicate those suggestions that have been revised to include using systems of
equations to solve problems. Keep these suggestions in mind as you study the examples and
work the problems in this section.
Classroom Example
The length of a rectangular region is
8 inches more than its width. The
area of the region is 48 square inches.
Find the length and width of the
rectangle.
EXAMPLE 1
The length of a rectangular region is 2 centimeters more than its width. The area of the
region is 35 square centimeters. Find the length and width of the rectangle.
Solution
We let l represent the length, and we let w represent the width (see Figure 10.10). We can
use the area formula for a rectangle, A ϭ lw, and the statement “the length of a rectangular
region is 2 centimeters greater than its width” as guidelines to form a system of equations.
Area is 35 cm2.
l
Figure 10.10
w
a
lw ϭ 35
b
lϭwϩ2
The second equation indicates that we can substitute w ϩ 2 for l. Making this substitution in
the first equation yields
1w ϩ 221w2 ϭ 35
Solving this quadratic equation by factoring, we get
w2 ϩ 2w ϭ 35
w2 ϩ 2w Ϫ 35 ϭ 0
1w ϩ 721w Ϫ 52 ϭ 0
wϩ7ϭ0
or
wϪ5ϭ0
w ϭ -7
or
wϭ5
442
Chapter 10 • Quadratic Equations
The width of a rectangle cannot be a negative number, so we discard the solution -7. Thus
the width of the rectangle is 5 centimeters and the length (w ϩ 2) is 7 centimeters.
Classroom Example
Find two consecutive whole numbers
whose product is 702.
EXAMPLE 2
Find two consecutive whole numbers whose product is 506.
Solution
We let n represent the smaller whole number. Then n ϩ 1 represents the next larger whole
number. The phrase “whose product is 506” translates into the equation
n1n ϩ 12 ϭ 506
Changing this quadratic equation into standard form produces
n2 ϩ n ϭ 506
n2 ϩ n Ϫ 506 ϭ 0
Because of the size of the constant term, let’s not try to factor; instead, we can use the
quadratic formula.
nϭ
-1 Ϯ 212 Ϫ 41121 -5062
2112
-1 Ϯ 22025
2
-1 Ϯ 45
22025 ϭ 45
nϭ
2
-1 ϩ 45
-1 Ϫ 45
nϭ
or
nϭ
2
2
or
n ϭ -23
n ϭ 22
nϭ
Since we are looking for whole numbers, we discard the solution Ϫ23. Therefore, the whole
numbers are 22 and 23.
Classroom Example
The perimeter of a rectangular lot is
122 feet, and its area is 888 square
feet. Find the length and width of
the lot.
EXAMPLE 3
The perimeter of a rectangular lot is 100 meters, and its area is 616 square meters. Find the
length and width of the lot.
Solution
We let l represent the length, and we let w represent
the width (see Figure 10.11).
Then
a
lw ϭ 616
b
2l ϩ 2w ϭ 100
2
Area is 616 m
Perimeter is 100 m
Multiplying the second equation by
Area is
616 m 2.
Perimeter is
100 m.
1
produces
2
l ϩ w ϭ 50, which can be changed to l ϭ 50 Ϫ w.
Substituting 50 Ϫ w for l in the first equation produces
the quadratic equation
l
Figure 10.11
Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
w
10.5 • Solving Problems Using Quadratic Equations
443
150 Ϫ w21w2 ϭ 616
50w Ϫ w2 ϭ 616
w2 Ϫ 50w ϭ - 616
Using the method of completing the square, we have
w2 Ϫ 50w ϩ 625 ϭ -616 ϩ 625
1w Ϫ 2522 ϭ 9
w Ϫ 25 ϭ Ϯ3
w Ϫ 25 ϭ 3
w ϭ 28
or
w Ϫ 25 ϭ -3
w ϭ 22
or
If w ϭ 28, then l ϭ 50 Ϫ w ϭ 22. If w ϭ 22, then l ϭ 50 Ϫ w ϭ 28. The rectangle is 28 meters
by 22 meters or 22 meters by 28 meters.
Classroom Example
Find two numbers such that their sum
is 4 and their product is 2.
Find two numbers such that their sum is 2, and their product is Ϫ1.
EXAMPLE 4
Solution
We let n represent one of the numbers, and we let m represent the other number.
a
nϩmϭ2
b
nm ϭ -1
Their sum is 2
Their product is Ϫ1
We can change the first equation to m ϭ 2 Ϫ n; then we can substitute 2 Ϫ n for m in the second equation.
n 12 Ϫ n2 ϭ - 1
2n Ϫ n2 ϭ - 1
2
-n ϩ 2n ϩ 1 ϭ 0
n2 Ϫ 2n Ϫ 1 ϭ 0
Multiply both sides by Ϫ1
-1 -22 Ϯ 21- 22 Ϫ 41121 -12
2
nϭ
nϭ
2112
2 Ϯ 28
2 Ϯ 222
ϭ
ϭ 1 Ϯ 22
2
2
If n ϭ 1 ϩ 22, then m ϭ 2 Ϫ 11 ϩ 222
ϭ 2 Ϫ 1 Ϫ 22
ϭ 1 Ϫ 22
If n ϭ 1 Ϫ 22, then m ϭ 2 Ϫ 11 Ϫ 222
ϭ 2 Ϫ 1 ϩ 22
ϭ 1 ϩ 22
The numbers are 1 ϩ 22 and 1 Ϫ 22 .
Classroom Example
Lynn drove 201 miles in 1 hour less
time than it took Michelle to drive
256 miles. Lynn drove at an average
rate of 3 miles per hour faster than
Michelle. How fast did each one
drive?
Perhaps you should check these numbers in the
original statement of the problem!
Finally, let’s consider a uniform motion problem similar to those we solved in Chapter 7.
Now we have the flexibility of using two equations in two variables.
EXAMPLE 5
Larry drove 156 miles in 1 hour more than it took Mike to drive 108 miles. Mike drove at an
average rate of 2 miles per hour faster than Larry. How fast did each one travel?
444
Chapter 10 • Quadratic Equations
Solution
We can represent the unknown rates and times like this:
let r represent Larry’s rate
let t represent Larry’s time
then r ϩ 2 represents Mike’s rate
and t Ϫ 1 represents Mike’s time
Because distance equals rate times time, we can set up the following system:
a
rt ϭ 156
b
(r ϩ 2)(t Ϫ 1) ϭ 108
Solving the first equation for r produces r ϭ
equation and simplifying, we obtain
156
ϩ 2b 1t Ϫ 12 ϭ 108
t
156
156 Ϫ
ϩ 2t Ϫ 2 ϭ 108
t
156
2t Ϫ
ϩ 154 ϭ 108
t
156
2t Ϫ
ϩ 46 ϭ 0
t
2t2 Ϫ 156 ϩ 46t ϭ 0
156
156
. Substituting
for r in the second
t
t
a
Multiply both sides by t, t
0
2t ϩ 46t Ϫ 156 ϭ 0
2
t2 ϩ 23t Ϫ 78 ϭ 0
We can solve this quadratic equation by factoring.
1t ϩ 2621t Ϫ 32 ϭ 0
t ϩ 26 ϭ 0
or
t ϭ -26
or
tϪ3ϭ0
tϭ3
We must disregard the negative solution. So Larry’s time is 3 hours, and Mike’s time is 3 Ϫ 1 ϭ
156
2 hours. Larry’s rate is
ϭ 52 miles per hour, and Mike’s rate is 52 ϩ 2 ϭ 54 miles per hour.
3
Problem Set 10.5
Solve each of the following problems. (Objective 1)
1. Find two consecutive whole numbers whose product
is 306.
2. Find two consecutive whole numbers whose product
is 702.
3. Suppose that the sum of two positive integers is 44 and
their product is 475. Find the integers.
4. Two positive integers differ by 6. Their product is 616.
Find the integers.
5. Find two numbers such that their sum is 6 and their
product is 4.
6. Find two numbers such that their sum is 4 and their
product is 1.
7. The sum of a number and its reciprocal is
Find the number.
322
.
2
73
8. The sum of a number and its reciprocal is . Find
24
the number.
9. Each of three consecutive even whole numbers is
squared. The three results are added and the sum
is 596. Find the numbers.
10. Each of three consecutive whole numbers is squared. The
three results are added, and the sum is 245. Find the
three whole numbers.
10.5 • Solving Problems Using Quadratic Equations
11. The sum of the square of a number and the square of
one-half of the number is 80. Find the number.
12. The difference between the square of a positive number,
and the square of one-half the number is 243. Find the
number.
13. Find the length and width of a rectangle if its length is 4
meters less than twice the width, and the area of the rectangle is 96 square meters.
14. Suppose that the length of a rectangular region is
4 centimeters greater than its width. The area of the
region is 45 square centimeters. Find the length and
width of the rectangle.
445
22. The area of a circle is numerically equal to twice the circumference of the circle. Find the length of a radius of
the circle.
23. The sum of the lengths of the two legs of a right triangle is
14 inches. If the length of the hypotenuse is 10 inches, find
the length of each leg.
24. A page for a magazine contains 70 square inches of type.
The height of a page is twice the width. If the margin
around the type is to be 2 inches uniformly, what are the
dimensions of the page?
25. A 5-by-7-inch picture is surrounded by a frame of uniform width (see Figure 10.13). The area of the picture
15. The perimeter of a rectangle is 80 centimeters, and its
area is 375 square centimeters. Find the length and width
of the rectangle.
and frame together is 80 square inches. Find the width
of the frame.
16. The perimeter of a rectangle is 132 yards and its area is
1080 square yards. Find the length and width of the rectangle.
17. The area of a tennis court is 2106 square feet (see
26
times the
9
7 inches
Figure 10.12). The length of the court is
width. Find the length and width of a tennis court.
5 inches
Figure 10.13
Figure 10.12
18. The area of a badminton court is 880 square feet. The
length of the court is 2.2 times the width. Find the length
and width of the court.
19. An auditorium in a local high school contains 300 seats.
There are 5 fewer rows than the number of seats per row.
Find the number of rows and the number of seats per
row.
20. Three hundred seventy-five trees were planted in rows
in an orchard. The number of trees per row was 10 more
than the number of rows. How many rows of trees are in
the orchard?
21. The area of a rectangular region is 63 square feet. If the
length and width are each increased by 3 feet, the area is
increased by 57 square feet. Find the length and width of
the original rectangle.
26. A rectangular piece of cardboard is 3 inches longer
than it is wide. From each corner, a square piece 2
inches on a side is cut out. The flaps are then turned up
to form an open box that has a volume of 140 cubic
inches. Find the length and width of the original piece
of cardboard.
27. A class trip was to cost $3000. If there had been ten more
students, it would have cost each student $25 less. How
many students took the trip?
28. Simon mowed some lawns and earned $40. It took him 3
hours longer than he anticipated, and thus he earned $3 per
hour less than he anticipated. How long did he expect the
mowing to take?
29. A piece of wire 56 inches long is cut into two pieces and
each piece is bent into the shape of a square. If the sum of
the areas of the two squares is 100 square inches, find the
length of each piece of wire.
30. Suppose that by increasing the speed of a car by
10 miles per hour, it is possible to make a trip of
200 miles in 1 hour less time. What was the original
speed for the trip?
446
Chapter 10 • Quadratic Equations
31. On a 50-mile bicycle ride, Irene averaged 4 miles per
hour faster for the first 36 miles than she did for the last
14 miles. The entire trip of 50 miles took 3 hours. Find
her rate for the first 36 miles.
Additional word problems can be found in Appendix B.
All of the problems in the Appendix marked as (10.5)
are appropriate for this section.
32. One side of a triangle is 1 foot more than twice
the length of the altitude to that side. If the area of the
triangle is 18 square feet, find the length of a side and
the length of the altitude to that side.
Thoughts Into Words
33. Return to Example 1 of this section and explain how the
problem could be solved using one variable and one equation.
34. Write a page or two on the topic “using algebra to solve
problems.”
Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
Chapter 10 Summary
OBJECTIVE
SUMMARY
EXAMPLE
Solve quadratic equations
by factoring.
A quadratic equation in the variable x is
any equation that can be written in the
form ax2 ϩ bx ϩ c ϭ 0, when a, b, and c
are real numbers and a Z 0 .We can solve
quadratic equations that are factorable
using integers by factoring and applying
the property ab ϭ 0 if and only if a ϭ 0 or
b ϭ 0.
Solve x2 ϩ 4x ϭ 21 by factoring.
(Section 10.1/Objective 1)
Solve quadratic equations of
the form x2 ϭ a.
(Section 10.1/Objective 2)
The property x2 ϭ a if and only if
x ϭ Ϯ1a can be used to solve certain
types of quadratic equations.
Solution
First set the equation equal to 0 and then
factor.
x2 ϩ 4x Ϫ 21 ϭ 0
(x ϩ 7)(x Ϫ 3) ϭ 0
or
xϩ7ϭ0
xϪ3ϭ0
or
x ϭ -7
xϭ3
The solution set is {Ϫ7, 3}.
Solve (2x ϩ 3)2 ϭ 15.
Solution
(2x ϩ 3)2 ϭ 15
Applying the property gives
2x ϩ 3 ϭ Ϯ 215
2x ϭ Ϫ 3 Ϯ 215
Ϫ 3 Ϯ 215
xϭ
2
The solution set is
e
Solve quadratic equations by
completing the square.
(Section 10.2/Objective 1)
You should be able to solve quadratic
equations by the method of completing the
square. To review this method, look back
over the examples in Section 10.2.
Ϫ 3 Ϫ 215 ϩ 3 ϩ 215
,
f.
2
2
Solve x2 ϩ 10x Ϫ 6 ϭ 0 by completing
the square.
Solution
First, add 6 to both sides of the equation.
x2 ϩ 10x ϭ 6
1
Now take of the coefficient of the x
2
term, 10, and square the result.
1
(10) ϭ 5
and
52 ϭ 25.
2
Add 25 to both sides of the equation.
x2 ϩ 10x ϩ 25 ϭ 6 ϩ 25
x2 ϩ 10x ϩ 25 ϭ 31
Now factor and then apply the
square-root property.
(x ϩ 5)2 ϭ 31
x ϩ 5 ϭ Ϯ231
x ϭ -5Ϯ231
The solution set is
{-5 Ϫ 231,-5 ϩ 231}.
(continued)
447
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448
Chapter 10 • Quadratic Equations
OBJECTIVE
SUMMARY
EXAMPLE
Solve quadratic equations by
using the quadratic formula.
We usually state the quadratic formula as
Solve x2 ϩ 6x ϩ 2 ϭ 0 by the quadratic
formula.
(Section 10.3/Objective 1)
Choose the most appropriate
method for solving a quadratic equation.
(Section 10.4/Objective 1)
Ϫ b Ϯ 2b2 Ϫ 4ac
. We can use it to
2a
solve any quadratic equation that is written
in the form ax2 ϩ bx ϩ c ϭ 0. The final
answer should be reduced and have the
radical in simplest terms. Be careful when
reducing the final answer; errors are often
made at that step.
xϭ
The three basic methods for solving
quadratic equations are factoring,
completing the square, and the quadratic
formula. Factoring only works if the
expression is factorable over the integers.
Completing the square can be very efficient
in situations where you can complete the
square without working with fractions. The
quadratic formula will work for any
quadratic equation.
Solution
For this problem, a ϭ 1, b ϭ 6, and
c ϭ 2.
xϭ
Ϫ (6) Ϯ 2(6)2 Ϫ 4(1)(2)
2(1)
Ϫ 6 Ϯ 236 Ϫ 8
2
Ϫ 6 Ϯ 228
xϭ
2
2(- 3 Ϯ 27)
Ϫ 6 Ϯ 227
xϭ
ϭ
2
2
x ϭ -3 Ϯ 27
The solution set is {- 3 -27, -3 +27}.
xϭ
Solve m2 ϩ 6m Ϫ 8 ϭ 0 using the
method that seems most appropriate.
Solution
The expression does not factor, so let’s
solve by completing the square.
m2 ϩ 6m Ϫ 8 ϭ 0
m2 ϩ 6m ϭ 8
m2 ϩ 6m ϩ 9 ϭ 8 ϩ 9
(m ϩ 3)2 ϭ 17
m ϩ 3 ϭ Ϯ 117
m ϭ -3 Ϯ117
The solution set is
{- 3 Ϫ 117, -3 ϩ 117}
Solve word problems
involving the Pythagorean
theorem and 30Њ– 60Њ right
triangles.
(Section 10.4/Objective 3)
The property x2 ϭ a if and only if
x ϭ Ϯ 2a can be used when working
with the Pythagorean theorem if the resulting equation is of the form, x2 ϭ a
Don’t forget:
1. In an isosceles right triangle, the lengths
of the two legs are equal.
2. In a 30°– 60° right triangle, the length of
the leg opposite the 30° angle is one-half
the length of the hypotenuse.
A rectangular football field measures
approximately 50 yards by 100 yards.
Find the length of a diagonal of the football field to the nearest tenth of a yard.
Solution
A diagonal of the rectangle divides the
rectangle into two right triangles. Use
the Pythagorean theorem to find the
hypotenuse of the triangle knowing that
the legs measure 50 yards and 100 yards.
a2 ϩ b2 ϭ c2
502 ϩ 1002 ϭ c2
12,500 ϭ c2
c ϭ 212500 L 111.8
The length of a diagonal to the nearest
tenth of a yard is 111.8 yards.
Chapter 10 • Review Problem Set
449
OBJECTIVE
SUMMARY
EXAMPLE
Solve word problems
involving quadratic equations.
Our knowledge of systems of equations
and quadratic equations provides us with a
stronger basis for solving word problems.
Find two numbers such that their sum is 8
and their product is 6.
(Section. 10.5/Objective 1)
Solution
Let n represent one number, then the
other number will be represented by
8 Ϫ n. Now we write an equation
showing the product.
n(8 Ϫ n) ϭ 6
8n Ϫ n2 ϭ 6
0 ϭ n2 Ϫ 8n ϩ 6
Using the quadratic formula we find
that the numbers are 4 Ϫ 210 and
4 ϩ 210.
Chapter 10 Review Problem Set
For Problems 1–22, solve each quadratic equation.
1. 12x ϩ 72 ϭ 25
2. x ϩ 8x ϭ -3
4. x2 ϭ 17x
5. n Ϫ
2
2
3. 21x2 Ϫ 13x ϩ 2 ϭ 0
4
ϭ -3
n
6. n2 Ϫ 26n ϩ 165 ϭ 0
7. 3a2 ϩ 7a Ϫ 1 ϭ 0
8. 4x2 Ϫ 4x ϩ 1 ϭ 0
9. 5x2 ϩ 6x ϩ 7 ϭ 0
10. 3x2 ϩ 18x ϩ 15 ϭ 0
11. 31x Ϫ 222 Ϫ 2 ϭ 4
12. x2 ϩ 4x Ϫ 14 ϭ 0
13. y2 ϭ 45
14. x1x Ϫ 62 ϭ 27
15. x2 ϭ x
}
17. n2 Ϫ 44n ϩ 480 ϭ 0
5x Ϫ 2
2
ϭ
3
xϩ1
4
5
ϩ ϭ6
21.
x
xϪ3
19.
16. n2 Ϫ 4n Ϫ 3 ϭ 6
18.
x2
ϭxϩ1
4
-1
2x ϩ 1
ϭ
3x Ϫ 1
-2
2
1
Ϫ ϭ3
22.
x
xϩ2
20.
For Problems 23–32, set up an equation or a system of
equations to help solve each problem.
23. The perimeter of a rectangle is 42 inches, and its area
is 108 square inches. Find the length and width of the
rectangle.
24. Find two consecutive whole numbers whose product
is 342.
25. Each of three consecutive odd whole numbers is
squared. The three results are added and the sum is 251.
Find the numbers.
26. The combined area of two squares is 50 square meters.
Each side of the larger square is three times as long as
a side of the smaller square. Find the lengths of the
sides of each square.
27. The difference in the lengths of the two legs of a right
triangle is 2 yards. If the length of the hypotenuse is
2113 yards, find the length of each leg.
28. Tony bought a number of shares of stock for a total of
$720. A month later the value of the stock increased by
$8 per share, and he sold all but 20 shares and regained
his original investment plus a profit of $80. How many
shares did Tony sell and at what price per share?
29. A company has a rectangular parking lot 40 meters wide
and 60 meters long. They plan to increase the area of the
lot by 1100 square meters by adding a strip of equal
width to one side and one end. Find the width of the strip
to be added.
30. Jay traveled 225 miles in 2 hours less time than it took
Jean to travel 336 miles. If Jay’s rate was 3 miles per hour
slower than Jean’s rate, find each rate.
31. The length of the hypotenuse of an isosceles right triangle is 12 inches. Find the length of each leg.
32. In a 30Њ– 60Њ right triangle, the side opposite the 60Њ angle
is 8 centimeters long. Find the length of the hypotenuse.
For more practice with word problems, consult
Appendix B. All Appendix problems that have a Chapter
10 reference would be appropriate for you to work on.
Chapter 10 Test
1. The two legs of a right triangle are 4 inches and 6 inches
long. Find the length of the hypotenuse. Express your
answer in simplest radical form.
15. n1n Ϫ 282 ϭ - 195
2. A diagonal of a rectangular plot of ground measures 14
meters. If the width of the rectangle is 5 meters, find the
length to the nearest meter.
17. 12x ϩ 1213x Ϫ 22 ϭ -2
3. A diagonal of a square piece of paper measures
10 inches. Find, to the nearest inch, the length of a side
of the square.
19. 14x Ϫ 122 ϭ 27
16. n ϩ
3
19
ϭ
n
4
18. 17x ϩ 222 Ϫ 4 ϭ 21
20. n2 Ϫ 5n ϩ 7 ϭ 0
4. In a 30Њ– 60Њ right triangle, the side opposite the 30Њ
angle is 4 centimeters long. Find the length of the side
opposite the 60Њ angle. Express your answer in simplest
radical form.
For Problems 5–20, solve each equation.
5. 13x ϩ 222 ϭ 49
6. 4x2 ϭ 64
For Problems 21– 25, set up an equation or a system of
equations to help solve each problem.
21. A room contains 120 seats. The number of seats per row
is 1 less than twice the number of rows. Find the number
of seats per row.
22. Abu rode his bicycle 56 miles in 2 hours less time than
it took Stan to ride his bicycle 72 miles. If Abu’s rate
was 2 miles per hour faster than Stan’s rate, find Abu’s
rate.
7. 8x2 Ϫ 10x ϩ 3 ϭ 0
8. x2 Ϫ 3x Ϫ 5 ϭ 0
9. n2 ϩ 2n ϭ 9
10. 12x Ϫ 122 ϭ -16
23. Find two consecutive odd whole numbers whose product is 255.
24. The combined area of two squares is 97 square feet.
Each side of the larger square is 1 foot more than twice
the length of a side of the smaller square. Find the
length of a side of the larger square.
11. y2 ϩ 10y ϭ 24
12. 2x2 Ϫ 3x Ϫ 4 ϭ 0
13.
4
xϪ2
ϭ
3
xϩ1
14.
2
1
5
ϩ ϭ
x
xϪ1
2
450
25. Dee bought a number of shares of stock for a total of
$160. Two weeks later, the value of the stock had
increased $2 per share, and she sold all but 4 shares and
regained her initial investment of $160. How many
shares did Dee originally buy?
11
Additional Topics
11.1 Equations and
Inequalities Involving
Absolute Value
11.2 3 ϫ 3 Systems of
Equations
11.3 Fractional Exponents
11.4 Complex Numbers
11.5 Quadratic Equations:
Complex Solutions
11.6 Pie, Bar, and Line
Graphs
11.7 Relations and
Functions
© Stephen Aaron Rees
11.8 Applications of
Functions
Be sure that you understand the
meaning of the markings on both
the horizontal and vertical axes.
We include this chapter to give you the opportunity to expand your knowledge of topics presented in earlier chapters. From the list of section titles, the
topics may appear disconnected; however, each section is a continuation of a
topic presented in a previous chapter.
Section 11.1 continues the development of techniques for solving equations
and inequalities, which was the focus of Chapter 3. Section 11.2 uses the method
of elimination by addition from Section 8.6 to solve systems containing three
linear equations in three variables. Section 11.3 is an extension of the work we
did with exponents in Section 5.6 and with radicals in Chapter 9. Sections 11.4
and 11.5 enhance the study of quadratic equations from Chapter 10. Sections
11.6 – 11.8 extend our work with coordinate geometry from Chapter 8.
Video tutorials based on section learning objectives are available in a variety of
delivery modes.
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