Polynomial Inequalities - College Algebra

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Question

When is the inequality true?

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Answer

We can rewrite this as . In order to solve the inequality we need to factor the polynomial. Taking a couple educated guesses we find that this polynomial factors as . So now we have our inequality written as

.

When is this true? When both linear factors are positive, and also when they are both negative.

$x-9 > 0 \rightarrow x > 9$

$x-9 < 0 \rightarrow x < 9$

$x+7 > 0 \rightarrow x > -7$

$x+7 < 0 \rightarrow x < -7$

If we match them up then we do get that the inequality is satisfied when and when .

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Question

Solve

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Answer

First we substract on each side

Now we can factor the left hand side

From this, we can see that the roots are at , and .

To make this inequality true, we will have two answers.

$5<x<\infty$

$-\infty<x<-2$

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Question

Which is correct graph of the following function?

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Answer

First graph

Since the inequality is $\geq$ , this means that the line is included in the solution, as indicated by a solid line.

Shade the area outside of the parabola since it is a negative parabola, but greater than or equal to.

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Question

Give the solution set of the inequality

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Answer

To solve the polynomial inequality

it is necessary to get it into standard form by writing all terms at the same side. Subtract from both sides:

The values of that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for in the equation:

Factor the polynomial by grouping and taking out the GCFs:

is the difference of squares and can be factored according to a pattern:

By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

,

in which case

,

or

in which case

.

, , and 2 are the boundaries of the intervals, which are $(-\infty, -7)$ , , , and $(2, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty, -7)$ : Choose

$-512 + 7 \cdot 64 < -32+ 28$

True - include $(-\infty, -7)$

: Choose

$-27 + 7 \cdot 9 < -12 + 28$

False - do not include

: Choose

True - include

$(2, \infty)$ : Choose

$27 + 7 \cdot 9 < 12+ 28$

False - do not include $(2, \infty)$

Also, the boundary values themselves should not be included, because the inequality symbol does not allow for equality.

The solution set is $\left \{ x| x < -7 \textrm{ or } -2 < x < 2 \right \}$

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Question

Give the solution set of the inequality

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

Tap to reveal answer

Answer

The boundaries of the intervals in the solution set of the polynomial inequality

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

are the zeroes of the equation

$(x- 5) (x+ 5 $)^{2}$ = 0$

By the Zero Product Property, either

,

in which case

,

or

,

in which case

and 5 are the boundaries of the intervals, which are $(-\infty , -5)$ , , and $(5, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , -5)$ : Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(-6- 5) (-6+ 5 $)^{2}$ \ge 0$

$(-11) (-1 $)^{2}$ \ge 0$

$(-11) (1 ) \ge 0$

$-11 \ge 0$

False - do not include $(-\infty , -5)$

: Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(0- 5) (0+ 5 $)^{2}$ \ge 0$

$( - 5) ( 5 $)^{2}$ \ge 0$

$( - 5) (2 5 ) \ge 0$

$-125 \ge 0$

False - do not include

$(5, \infty)$ : Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(6- 5) (6+ 5 $)^{2}$ \ge 0$

$(1) (11 $)^{2}$ \ge 0$

$(1) (121 ) \ge 0$

$121 \ge 0$

True - include $(5, \infty)$

Also, the boundary values themselves should be included, because the inequality symbol $\ge$ allows for equality.

The solution set is therefore $\left \{ x | x = -5 \textrm{ or } x \ge 5 \right \}$ .

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Question

Give the solution set of the inequality

$(x- $4)^{2}$ (x+ 4 )>0$

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Answer

The boundaries of the intervals in the solution set of the polynomial inequality

$(x- $4)^{2}$ (x+ 4 )>0$

are the zeroes of the equation

$(x- $4)^{2}$ (x+ 4 )= 0$

By the Zero Product Property, either

,

in which case

,

or

,

in which case

and 4 are the boundaries of the intervals, which are $(-\infty , -4)$ , , and $(4, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , -4)$ : Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(-5- $4)^{2}$ (-5+ 4 )> 0$

False - do not include $(-\infty , -4)$

: Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(0- $4)^{2}$ (0+ 4 )>0$

$( - $4)^{2}$ (4) > 0$

True - include

$(4, \infty)$ : Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(5- $4)^{2}$ (5+ 4 )>0$

True - include $(4, \infty)$

Also, the boundary values themselves should not be included, because the inequality symbol does not allow for equality.

The solution set is therefore $\left \{ x | -4 < x < 4 \textrm{ or }x> 4 \ \right \}$

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Question

Give the solution set of the inequality

Tap to reveal answer

Answer

To solve the polynomial inequality

it is necessary to get it into standard form by writing all terms at the same side. Subtract from both sides:

The values of that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for in the equation:

Factor the polynomial by grouping and taking out the GCFs:

The polynomial cannot be factored further, as , as the sum of squares, is prime. By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

or

,

which has no real solution.

Therefore, the only boundary of the intervals is 7, which divides the real numbers into two intervals, $(-\infty , 7)$ and $(7, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , 7)$ : Choose

$0 ^{3} + 4 (0) >7 \cdot $0^{2}$ + 28$

False - do not include $(-\infty , 7)$

$(7, \infty)$ : Choose

$512+ 32 >7 \cdot 64 + 28$

True - include $(7, \infty)$

Also, since the inequality symbol does not allow for equality, 7 is not included in the solution. The solution set is $\left \{ x | x > 7\right \}$ .

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Question

Solve

Tap to reveal answer

Answer

First we substract on each side

Now we can factor the left hand side

From this, we can see that the roots are at , and .

To make this inequality true, we will have two answers.

$5<x<\infty$

$-\infty<x<-2$

← Didn't Know|Knew It →

Question

Which is correct graph of the following function?

Tap to reveal answer

Answer

First graph

Since the inequality is $\geq$ , this means that the line is included in the solution, as indicated by a solid line.

Shade the area outside of the parabola since it is a negative parabola, but greater than or equal to.

← Didn't Know|Knew It →

Question

When is the inequality true?

Tap to reveal answer

Answer

We can rewrite this as . In order to solve the inequality we need to factor the polynomial. Taking a couple educated guesses we find that this polynomial factors as . So now we have our inequality written as

.

When is this true? When both linear factors are positive, and also when they are both negative.

$x-9 > 0 \rightarrow x > 9$

$x-9 < 0 \rightarrow x < 9$

$x+7 > 0 \rightarrow x > -7$

$x+7 < 0 \rightarrow x < -7$

If we match them up then we do get that the inequality is satisfied when and when .

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

Tap to reveal answer

Answer

To solve the polynomial inequality

it is necessary to get it into standard form by writing all terms at the same side. Subtract from both sides:

The values of that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for in the equation:

Factor the polynomial by grouping and taking out the GCFs:

is the difference of squares and can be factored according to a pattern:

By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

,

in which case

,

or

in which case

.

, , and 2 are the boundaries of the intervals, which are $(-\infty, -7)$ , , , and $(2, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty, -7)$ : Choose

$-512 + 7 \cdot 64 < -32+ 28$

True - include $(-\infty, -7)$

: Choose

$-27 + 7 \cdot 9 < -12 + 28$

False - do not include

: Choose

True - include

$(2, \infty)$ : Choose

$27 + 7 \cdot 9 < 12+ 28$

False - do not include $(2, \infty)$

Also, the boundary values themselves should not be included, because the inequality symbol does not allow for equality.

The solution set is $\left \{ x| x < -7 \textrm{ or } -2 < x < 2 \right \}$

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

Tap to reveal answer

Answer

The boundaries of the intervals in the solution set of the polynomial inequality

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

are the zeroes of the equation

$(x- 5) (x+ 5 $)^{2}$ = 0$

By the Zero Product Property, either

,

in which case

,

or

,

in which case

and 5 are the boundaries of the intervals, which are $(-\infty , -5)$ , , and $(5, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , -5)$ : Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(-6- 5) (-6+ 5 $)^{2}$ \ge 0$

$(-11) (-1 $)^{2}$ \ge 0$

$(-11) (1 ) \ge 0$

$-11 \ge 0$

False - do not include $(-\infty , -5)$

: Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(0- 5) (0+ 5 $)^{2}$ \ge 0$

$( - 5) ( 5 $)^{2}$ \ge 0$

$( - 5) (2 5 ) \ge 0$

$-125 \ge 0$

False - do not include

$(5, \infty)$ : Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(6- 5) (6+ 5 $)^{2}$ \ge 0$

$(1) (11 $)^{2}$ \ge 0$

$(1) (121 ) \ge 0$

$121 \ge 0$

True - include $(5, \infty)$

Also, the boundary values themselves should be included, because the inequality symbol $\ge$ allows for equality.

The solution set is therefore $\left \{ x | x = -5 \textrm{ or } x \ge 5 \right \}$ .

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$(x- $4)^{2}$ (x+ 4 )>0$

Tap to reveal answer

Answer

The boundaries of the intervals in the solution set of the polynomial inequality

$(x- $4)^{2}$ (x+ 4 )>0$

are the zeroes of the equation

$(x- $4)^{2}$ (x+ 4 )= 0$

By the Zero Product Property, either

,

in which case

,

or

,

in which case

and 4 are the boundaries of the intervals, which are $(-\infty , -4)$ , , and $(4, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , -4)$ : Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(-5- $4)^{2}$ (-5+ 4 )> 0$

False - do not include $(-\infty , -4)$

: Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(0- $4)^{2}$ (0+ 4 )>0$

$( - $4)^{2}$ (4) > 0$

True - include

$(4, \infty)$ : Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(5- $4)^{2}$ (5+ 4 )>0$

True - include $(4, \infty)$

Also, the boundary values themselves should not be included, because the inequality symbol does not allow for equality.

The solution set is therefore $\left \{ x | -4 < x < 4 \textrm{ or }x> 4 \ \right \}$

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

Tap to reveal answer

Answer

To solve the polynomial inequality

it is necessary to get it into standard form by writing all terms at the same side. Subtract from both sides:

The values of that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for in the equation:

Factor the polynomial by grouping and taking out the GCFs:

The polynomial cannot be factored further, as , as the sum of squares, is prime. By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

or

,

which has no real solution.

Therefore, the only boundary of the intervals is 7, which divides the real numbers into two intervals, $(-\infty , 7)$ and $(7, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , 7)$ : Choose

$0 ^{3} + 4 (0) >7 \cdot $0^{2}$ + 28$

False - do not include $(-\infty , 7)$

$(7, \infty)$ : Choose

$512+ 32 >7 \cdot 64 + 28$

True - include $(7, \infty)$

Also, since the inequality symbol does not allow for equality, 7 is not included in the solution. The solution set is $\left \{ x | x > 7\right \}$ .

← Didn't Know|Knew It →

Question

Solve

Tap to reveal answer

Answer

First we substract on each side

Now we can factor the left hand side

From this, we can see that the roots are at , and .

To make this inequality true, we will have two answers.

$5<x<\infty$

$-\infty<x<-2$

← Didn't Know|Knew It →

Question

Which is correct graph of the following function?

Tap to reveal answer

Answer

First graph

Since the inequality is $\geq$ , this means that the line is included in the solution, as indicated by a solid line.

Shade the area outside of the parabola since it is a negative parabola, but greater than or equal to.

← Didn't Know|Knew It →

Question

When is the inequality true?

Tap to reveal answer

Answer

We can rewrite this as . In order to solve the inequality we need to factor the polynomial. Taking a couple educated guesses we find that this polynomial factors as . So now we have our inequality written as

.

When is this true? When both linear factors are positive, and also when they are both negative.

$x-9 > 0 \rightarrow x > 9$

$x-9 < 0 \rightarrow x < 9$

$x+7 > 0 \rightarrow x > -7$

$x+7 < 0 \rightarrow x < -7$

If we match them up then we do get that the inequality is satisfied when and when .

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

Tap to reveal answer

Answer

To solve the polynomial inequality

it is necessary to get it into standard form by writing all terms at the same side. Subtract from both sides:

The values of that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for in the equation:

Factor the polynomial by grouping and taking out the GCFs:

is the difference of squares and can be factored according to a pattern:

By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

,

in which case

,

or

in which case

.

, , and 2 are the boundaries of the intervals, which are $(-\infty, -7)$ , , , and $(2, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty, -7)$ : Choose

$-512 + 7 \cdot 64 < -32+ 28$

True - include $(-\infty, -7)$

: Choose

$-27 + 7 \cdot 9 < -12 + 28$

False - do not include

: Choose

True - include

$(2, \infty)$ : Choose

$27 + 7 \cdot 9 < 12+ 28$

False - do not include $(2, \infty)$

Also, the boundary values themselves should not be included, because the inequality symbol does not allow for equality.

The solution set is $\left \{ x| x < -7 \textrm{ or } -2 < x < 2 \right \}$

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

Tap to reveal answer

Answer

The boundaries of the intervals in the solution set of the polynomial inequality

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

are the zeroes of the equation

$(x- 5) (x+ 5 $)^{2}$ = 0$

By the Zero Product Property, either

,

in which case

,

or

,

in which case

and 5 are the boundaries of the intervals, which are $(-\infty , -5)$ , , and $(5, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , -5)$ : Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(-6- 5) (-6+ 5 $)^{2}$ \ge 0$

$(-11) (-1 $)^{2}$ \ge 0$

$(-11) (1 ) \ge 0$

$-11 \ge 0$

False - do not include $(-\infty , -5)$

: Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(0- 5) (0+ 5 $)^{2}$ \ge 0$

$( - 5) ( 5 $)^{2}$ \ge 0$

$( - 5) (2 5 ) \ge 0$

$-125 \ge 0$

False - do not include

$(5, \infty)$ : Choose

$(x- 5) (x+ 5 $)^{2}$ \ge 0$

$(6- 5) (6+ 5 $)^{2}$ \ge 0$

$(1) (11 $)^{2}$ \ge 0$

$(1) (121 ) \ge 0$

$121 \ge 0$

True - include $(5, \infty)$

Also, the boundary values themselves should be included, because the inequality symbol $\ge$ allows for equality.

The solution set is therefore $\left \{ x | x = -5 \textrm{ or } x \ge 5 \right \}$ .

← Didn't Know|Knew It →

Question

Give the solution set of the inequality

$(x- $4)^{2}$ (x+ 4 )>0$

Tap to reveal answer

Answer

The boundaries of the intervals in the solution set of the polynomial inequality

$(x- $4)^{2}$ (x+ 4 )>0$

are the zeroes of the equation

$(x- $4)^{2}$ (x+ 4 )= 0$

By the Zero Product Property, either

,

in which case

,

or

,

in which case

and 4 are the boundaries of the intervals, which are $(-\infty , -4)$ , , and $(4, \infty)$ . To determine which intervals belong to the solution set, choose one value from each, substitute for in the original inequality, and determine whether it is true.

$(-\infty , -4)$ : Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(-5- $4)^{2}$ (-5+ 4 )> 0$

False - do not include $(-\infty , -4)$

: Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(0- $4)^{2}$ (0+ 4 )>0$

$( - $4)^{2}$ (4) > 0$

True - include

$(4, \infty)$ : Choose

$(x- $4)^{2}$ (x+ 4 )> 0$

$(5- $4)^{2}$ (5+ 4 )>0$

True - include $(4, \infty)$

Also, the boundary values themselves should not be included, because the inequality symbol does not allow for equality.

The solution set is therefore $\left \{ x | -4 < x < 4 \textrm{ or }x> 4 \ \right \}$

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