4: Solving Quadratic Equations—Which Method?
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10.4 • Solving Quadratic Equations—Which Method?
437
Quadratic Formula Method
x2 ϩ 4x Ϫ 12 ϭ 0
xϭ
-4 Ϯ 242 Ϫ 41121 - 122
2112
-4 Ϯ 264
2
-4 Ϯ 8
xϭ
2
xϭ
-4 ϩ 8
2
xϭ2
or
xϭ
-4 Ϫ 8
2
x ϭ -6
xϭ
or
The solution set is 5- 6, 26.
We have also discussed the use of the property x2 ϭ a if and only if x ϭ Ϯ2a for certain types of quadratic equations. For example, we can solve x2 ϭ 4 easily by applying the
property and obtaining x ϭ 24 or x ϭ -24; thus, the solutions are 2 and Ϫ2.
Which method should you use to solve a particular quadratic equation? Let’s consider
some examples in which the different techniques are used. Keep in mind that this is a decision you must make as the need arises. So become as familiar as you can with the strengths
and weaknesses of each method.
Classroom Example
Solve 3x2 ϩ 18x Ϫ 120 ϭ 0.
EXAMPLE 1
Solve 2x2 ϩ 12x Ϫ 54 ϭ 0.
Solution
First, it is very helpful to recognize a factor of 2 in each of the terms on the left side.
2x2 ϩ 12x Ϫ 54 ϭ 0
1
Multiply both sides by
x2 ϩ 6x Ϫ 27 ϭ 0
2
Now you should recognize that the left side can be factored. Thus we can proceed as follows.
1x ϩ 921x Ϫ 32 ϭ 0
xϩ9ϭ0
or
x ϭ -9
or
The solution set is 5-9, 36.
Classroom Example
Solve (8n Ϫ 11)2 ϭ 49.
EXAMPLE 2
xϪ3ϭ0
xϭ3
Solve 14x ϩ 322 ϭ 16.
Solution
The form of this equation lends itself to the use of the property x2 ϭ a if and only if x ϭ Ϯ1a.
14x ϩ 322 ϭ 16
4x ϩ 3 ϭ Ϯ216
4x ϩ 3 ϭ 4
or
4x ϭ 1
or
1
or
xϭ
4
4x ϩ 3 ϭ - 4
4x ϭ -7
7
xϭ 4
7 1
The solution set is e- , f.
4 4
438
Chapter 10 • Quadratic Equations
Classroom Example
1
Solve x ϩ ϭ 3.
x
EXAMPLE 3
Solve n ϩ
1
ϭ 5.
n
Solution
First, we need to clear the equation of fractions by multiplying both sides by n.
nϩ
1
ϭ 5,
n
n Z 0
1
nan ϩ b ϭ 51n2
n
n2 ϩ 1 ϭ 5n
Now we can change the equation to standard form.
n2 Ϫ 5n ϩ 1 ϭ 0
Because the left side cannot be factored using integers, we must solve the equation by using
either the method of completing the square or the quadratic formula. Using the formula, we
obtain
nϭ
nϭ
-1- 52 Ϯ 21 -522 Ϫ 4112112
2112
5 Ϯ 221
2
The solution set is e
Classroom Example
Solve x2 ϭ 27x.
5 Ϫ 221 5 ϩ 221
,
f.
2
2
Solve t2 ϭ 22t.
EXAMPLE 4
Solution
A quadratic equation without a constant term can be solved easily by the factoring method.
t 2 ϭ 22t
t 2 Ϫ 22t ϭ 0
t1t Ϫ 222 ϭ 0
tϭ0
or
t Ϫ 22 ϭ 0
tϭ0
or
t ϭ 22
The solution set is 50, 226. (Check each of these solutions in the given equation.)
Classroom Example
Solve x2 ϩ 9x Ϫ 162 ϭ 0.
Solve x2 Ϫ 28x ϩ 192 ϭ 0.
EXAMPLE 5
Solution
Determining whether or not the left side is factorable presents a bit of a problem because of
the size of the constant term. Therefore, let’s not concern ourselves with trying to factor;
instead we will use the quadratic formula.
x2 Ϫ 28x ϩ 192 ϭ 0
-1-282 Ϯ 21- 2822 Ϫ 411211922
2112
28 Ϯ 2784 Ϫ 768
xϭ
2
xϭ
10.4 • Solving Quadratic Equations—Which Method?
xϭ
28 Ϯ 216
2
xϭ
28 ϩ 4
2
x ϭ 16
xϭ
or
or
The solution set is 512, 16}.
EXAMPLE 6
Classroom Example
Solve d 2 ϩ 16d ϭ 4.
439
28 Ϫ 4
2
x ϭ 12
Solve x2 ϩ 12x ϭ 17.
Solution
The form of this equation, and the fact that the coefficient of x is even, makes the method of
completing the square a reasonable approach.
x2 ϩ 12x ϭ 17
x ϩ 12x ϩ 36 ϭ 17 ϩ 36
1x ϩ 622 ϭ 53
x ϩ 6 ϭ Ϯ253
x ϭ - 6 Ϯ 253
2
The solution set is 5-6 Ϫ 253, - 6 ϩ 253}.
Concept Quiz 10.4
For Problems 1–7, choose the method that you think is most appropriate for solving the
given equation.
1.
2.
3.
4.
5.
6.
7.
2x2 ϩ 6x Ϫ 3 ϭ 0
(x ϩ 1)2 ϭ 36
x2 Ϫ 3x ϩ 2 ϭ 0
x2 ϩ 6x ϭ 19
4x2 ϩ 2x Ϫ 5 ϭ 0
4x2 ϭ 3
x2 Ϫ 4x Ϫ 12 ϭ 0
A.
B.
C.
D.
Factoring
Square root property (Property 10.1)
Completing the square
Quadratic formula
Problem Set 10.4
Solve each of the following quadratic equations using the
method that seems most appropriate to you. (Objective 1)
1. x ϩ 4x ϭ 45
2
3. 15n ϩ 62 ϭ 49
2
2. x ϩ 4x ϭ 60
2
4. (3n Ϫ 1)2 ϭ 25
5. t Ϫ t Ϫ 2 ϭ 0
6. t ϩ 2t Ϫ 3 ϭ 0
7. 8x ϭ 3x
8. 5x ϭ 7x
2
2
9. 9x Ϫ 6x ϩ 1 ϭ 0
2
2
2
10. 4x ϩ 36x ϩ 81 ϭ 0
2
11. 5n ϭ 28n
12. 23n ϭ 2n2
13. n2 Ϫ 14n ϭ 19
14. n2 Ϫ 10n ϭ 14
15. 5x2 Ϫ 2x Ϫ 7 ϭ 0
16. 3x2 Ϫ 4x Ϫ 2 ϭ 0
2
17. 15x2 ϩ 28x ϩ 5 ϭ 0
18. 20y2 Ϫ 7y Ϫ 6 ϭ 0
19. x2 Ϫ 28x Ϫ 7 ϭ 0
20. x2 ϩ 25x Ϫ 5 ϭ 0
21. y2 ϩ 5y ϭ 84
22. y2 ϩ 7y ϭ 60
3
n
25. 3x2 Ϫ 9x Ϫ 12 ϭ 0
24. n ϩ
23. 2n ϭ 3 ϩ
26. 2x2 ϩ 10x Ϫ 28 ϭ 0
27. 2x2 Ϫ 3x ϩ 7 ϭ 0
1
ϭ7
n
440
Chapter 10 • Quadratic Equations
28. 3x2 Ϫ 2x ϩ 5 ϭ 0
38. x2 Ϫ 33x ϩ 266 ϭ 0
29. n1n Ϫ 462 ϭ -480
39.
x2
1
Ϫxϭ 3
2
30. n1n ϩ 422 ϭ -432
31. n Ϫ
3
ϭ -1
n
40.
2x Ϫ 1
5
ϭ
3
xϩ2
32. n Ϫ
2
3
ϭ
n
4
41.
2
1
Ϫ ϭ3
x
xϩ2
1
25
33. x ϩ ϭ
x
12
42.
3
2
3
ϩ ϭ
x
xϪ1
2
1
65
34. x ϩ ϭ
x
8
43.
2
nϩ2
ϭ
3n Ϫ 1
6
35. t ϩ 12t ϩ 36 ϭ 49
44.
x2
1
ϭxϩ
2
4
36. t2 Ϫ 10t ϩ 25 ϭ 16
45. 1n Ϫ 221n ϩ 42 ϭ 7
2
46. 1n ϩ 321n Ϫ 82 ϭ - 30
37. x Ϫ 28x ϩ 187 ϭ 0
2
Thoughts Into Words
49. How can you tell by inspection that the equation
x2 ϩ x ϩ 4 ϭ 0 has no real number solutions?
47. Which method would you use to solve the equation
x2 ϩ 30x ϭ Ϫ216? Explain your reasons for making this
choice.
48. Explain how you
0 ϭ -x2 Ϫ x ϩ 6.
would
solve
Answers to the Concept Quiz
Answers for these questions may vary.
1. D
2. B
3. A
10.5
the
equation
4. C
5. D
6. B
7. A
Solving Problems Using Quadratic Equations
OBJECTIVE
1
Use quadratic equations to solve a variety of word problems
The following diagram indicates our approach in this text.
Develop
skills
Use skills
to solve
equations
Use equations
to solve word
problems
Now you should be ready to use your skills relative to solving systems of equations (Chapter
8) and quadratic equations to help with additional types of word problems. Before you consider such problems, let’s review and update the problem-solving suggestions we offered in
Chapter 3.
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