2-6 Nonlinear Inequalities
29. PROM A group of friends decides to share a limo for prom. The cost of rental is $750 plus a $25 fee for each occupant. There is a minimum of two passengers, and the limo can hold up to 14 people. Write and solve an
inequality to determine how many people can share the limo for less than $120 per person.
SOLUTION: Let x be the total number of people that share the limo. The total cost of the limo is $750 + $25x. The cost per
person is the total cost divided by the total number of people or
. Write an inequality to determine how
many people can share the limo for less than $120 per person.
< 120
Solve for x.
When more than 8 people share the limo, the cost per person will be less than $120 per person. Thus, 8 to 14 people
are needed to be able to share the limo for less than $120 per person.
Find the domain of each expression.
31. SOLUTION: 2
For the expression to be defined, x − 3x − 40 ≥ 0.
2
Let f (x) = x − 3x −40. A factored form of f (x) is f (x) = (x − 8)(x + 5). f (x) has real zeros at x = 8 and x = −5. Set
up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive
or negative at that point.
2
The solutions of x − 3x −40 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can
see that the solution set is (– , –5] [8, ).
33. SOLUTION: 2
For the expression to be defined, x − 9 ≥ 0.
2
Let f (x) = x − 9. A factored form of f (x) is f (x) = (x − 3)(x + 3). f (x) has real zeros at x = 3 and x = −3. Set up a
sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or
negative at that point.
2
The solutions of x − 9 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that
the solution set is (– , –3] [3, ).
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35. Page 1
2
solutions ofInequalities
x − 9 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that
2-6 The
Nonlinear
the solution set is (–
, –3]
[3, ).
35. SOLUTION: 2
2
For the expression to be defined, the denominator of the radicand, 36 − x , cannot equal 0. Let f (x) = 36 − x . Set
this function equal to 0 and solve for x.
2
36 − x = 0 at x = 6 and x = −6. Thus, the domain of the expression is (–
, –6)
(−6, 6)
(6, ).
Solve each inequality.
41. −
≤ 4
SOLUTION: For the inequality to be defined, 4x + 4 ≥ 0 and x − 4 ≥ 0. Solve both inequalities for x.
The domain of x must be restricted to x ≥4. Solve the original inequality.
Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign.
Let f (x) = (9x − 40)(x − 8). f (x) has real zeros at x =
and x = 8. Set up a sign chart. Substitute an x-value in
each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.
The solutions of (9x − 40)(x − 8) ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you
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can see that the solution set is
Page 2
. Recall that x ≥ 4. Create a chart that includes this restriction. When solving inequalities that involve raising each side to a power to eliminate a radical, it is important to test every interval using
2-6 Nonlinear
Inequalities
2
36 − x = 0 at x = 6 and x = −6. Thus, the domain of the expression is (–
, –6)
(−6, 6)
(6, ).
Solve each inequality.
41. −
≤ 4
SOLUTION: For the inequality to be defined, 4x + 4 ≥ 0 and x − 4 ≥ 0. Solve both inequalities for x.
The domain of x must be restricted to x ≥4. Solve the original inequality.
Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign.
Let f (x) = (9x − 40)(x − 8). f (x) has real zeros at x =
and x = 8. Set up a sign chart. Substitute an x-value in
each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.
The solutions of (9x − 40)(x − 8) ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you
can see that the solution set is
. Recall that x ≥ 4. Create a chart that includes this restriction. When solving inequalities that involve raising each side to a power to eliminate a radical, it is important to test every interval using
the original inequality. Substitute an x-value in each test interval into the original inequality to determine if f (x) is a
solution.
Thus, the solution set is
43. −
.
< 5
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SOLUTION: For the inequality to be defined, 25 − 12x ≥ 0 and 16 − 4x ≥ 0. Solve both inequalities for x.
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Thus, the solution set is
2-6 Nonlinear Inequalities
43. −
.
< 5
SOLUTION: For the inequality to be defined, 25 − 12x ≥ 0 and 16 − 4x ≥ 0. Solve both inequalities for x.
The domain of x must be restricted to x ≤ . Solve the original inequality.
Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign.
Let f (x) = 4(4x − 7)(x + 12). f (x) has real zeros at x =
and x = −12. Set up a sign chart. Substitute an x-value in
each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.
The solutions of 4(4x − 7)(x + 12) < 0 are x-values such that f (x) is negative. From the sign chart, you can see that
the solution set is
. Recall that x ≤ . Create a chart that includes this restriction. When solving
inequalities that involve raising each side to a power to eliminate a radical, it is important to test every interval using
the original inequality. Substitute an x-value in each test interval into the original inequality to determine if f (x) is a
solution.
Notice that x-values that lie outside of the solution set found by solving the inequality for x are also solutions. Thus,
the solution set is
.
Solve each inequality.
47. 3a 4 + 7a 3 – 56a 2 – 80a < 0
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SOLUTION: 4
3
2
3
2
Let f (a) = 3a + 7a – 56a – 80a. f (a) can be written as f (a) = a(3a + 7a − 56a − 80). By using synthetic
Notice that x-values that lie outside of the solution set found by solving the inequality for x are also solutions. Thus,
2-6 the
Nonlinear
solution setInequalities
is
.
Solve each inequality.
47. 3a 4 + 7a 3 – 56a 2 – 80a < 0
SOLUTION: 4
3
2
3
2
Let f (a) = 3a + 7a – 56a – 80a. f (a) can be written as f (a) = a(3a + 7a − 56a − 80). By using synthetic
3
2
division, it can be determined that a = 4 is a rational zero of 3a + 7a − 56a − 80.
2
The remaining quadratic factor (3a + 19a + 20) can be written as (3a + 4)(a + 5). A factored form of f (a) is f (a) =
a(a − 4)(3a + 4)(a + 5). f (a) has real zeros at a = 0, a = 4,
, and a = −5. Set up a sign chart. Substitute an
a-value in each test interval into the factored polynomial to determine if f (a) is positive or negative at that point.
4
3
2
The solutions of 3a + 7a – 56a – 80a < 0 are a-values such that f (a) is negative. From the sign chart, you can
see that the solution set is
.
49. 3x5 + 13x4 – 137x3 – 353x2 + 330x + 144 ≤ 0
SOLUTION: 5
4
3
2
Let f (x) = 3x + 13x – 137x – 353x + 330x + 144. By using synthetic division, it can be determined that x = 1 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = −3 is a rational zero.
By using synthetic division on the new depressed polynomial, it can be determined that x = 6 is a rational zero.
2
The remaining quadratic factor (3x + 25x + 8) can be written as (3x + 1)(x + 8). A factored form of f (x) is f (x) =
(x − 1)(x + 3)(x − 6)(3x + 1)(x + 8). f (x) has real zeros at x = 1, x = −3, x = 6,
, and x = −8. Set up a sign
chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or
negative at that point.
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5
4
Page 5
3
2
The solutions of 3x + 13x – 137x – 353x + 330x + 144 ≤ 0 are x-values such that f (x) is negative or equal to 0.
4
3
2
The solutions of 3a + 7a – 56a – 80a < 0 are a-values such that f (a) is negative. From the sign chart, you can
see that the solution set is
.
2-6 Nonlinear Inequalities
49. 3x5 + 13x4 – 137x3 – 353x2 + 330x + 144 ≤ 0
SOLUTION: 5
4
3
2
Let f (x) = 3x + 13x – 137x – 353x + 330x + 144. By using synthetic division, it can be determined that x = 1 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = −3 is a rational zero.
By using synthetic division on the new depressed polynomial, it can be determined that x = 6 is a rational zero.
2
The remaining quadratic factor (3x + 25x + 8) can be written as (3x + 1)(x + 8). A factored form of f (x) is f (x) =
(x − 1)(x + 3)(x − 6)(3x + 1)(x + 8). f (x) has real zeros at x = 1, x = −3, x = 6,
, and x = −8. Set up a sign
chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or
negative at that point.
5
4
3
2
The solutions of 3x + 13x – 137x – 353x + 330x + 144 ≤ 0 are x-values such that f (x) is negative or equal to 0.
From the sign chart, you can see that the solution set is
.
Solve each inequality.
51. (x + 3)2(x – 4)3(2x + 1)2 < 0
SOLUTION: 2
3
2
Let f (x) = (x + 3) (x – 4) (2x + 1) . f (x) has real zeros at x = −3 (multiplicity: 2), x = 4 (multiplicity: 3), and
(multiplicity: 2). Set up a sign chart. Substitute an x-value in each test interval into the polynomial to
determine if f (x) is positive or negative at that point.
2
3
2
The solutions of (x + 3) (x – 4) (2x + 1) < 0 are x-values such that f (x) is negative. Since the inequality cannot
equal 0, the real zeros are not included in the solution set. From the sign chart, you can see that the solution set is
.
3
3
2
53. (a –Manual
3) (a -+Powered
2) (a –by6)
>0
eSolutions
Cognero
Page 6
SOLUTION: 3
3
2
2
3
2
The solutions of (x + 3) (x – 4) (2x + 1) < 0 are x-values such that f (x) is negative. Since the inequality cannot
equal 0, the real zeros are not included in the solution set. From the sign chart, you can see that the solution set is
.
2-6 Nonlinear Inequalities
53. (a – 3)3(a + 2)3(a – 6)2 > 0
SOLUTION: 3
3
2
Let f (a) = (a – 3) (a + 2) (a – 6) . f (a) has real zeros at a = 3 (multiplicity: 3), y = −2 (multiplicity: 3), and a = 6
(multiplicity: 2). Set up a sign chart. Substitute an a-value in each test interval into the polynomial to determine if f (a)
is positive or negative at that point.
3
3
2
The solutions of (a – 3) (a + 2) (a – 6) > 0 are a-values such that f (a) is positive. Since the inequality cannot equal
0, the real zeros are not included in the solution set. From the sign chart, you can see that the solution set is (– , –
2) (3, 6) (6, ).
55. STUDY TIME Jarrick determines that with the information that he currently knows, he can achieve a score of a 75% on his test. Jarrick believes that for every 5 complete minutes he spends studying, he will raise his score by 1%.
a. If Jarrick wants to obtain a score of at least 89.5%, write an inequality that could be used to find the time t that he
will have to spend studying.
b. Solve the inequality that you wrote in part a and interpret the solution.
SOLUTION: a. Without studying, Jarrick starts at a score of 75%. For every 5 complete minutes that he spends studying, 1% can
be added to his score. Since fractions of 5 minutes will not increase his score, the greatest integer function is used to
find out how many times he spends 5 complete minutes studying out of a total of t minutes. Thus, his score S after
studying for t minutes can be found using
If Jarrick wants to obtain a score of at least 89.5%, the equation becomes the following inequality:
b. Sample answer:
Since t ≥ 72.5, Jarrick will have to spend 75 minutes studying for the test.
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