Precalculus Examples

$\frac{x^{2} - 81}{x^{4} - 81} < 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x^{2} - 81 = 0$

$x^{4} - 81 = 0$

Step 2

Add $81$ to both sides of the equation.

$x^{2} = 81$

Step 3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{81}$

Step 4

Simplify $\pm \sqrt{81}$ .

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Step 4.1

Rewrite $81$ as $9^{2}$ .

$x = \pm \sqrt{9^{2}}$

Step 4.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 9$

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 9$

Step 5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 9$

Step 5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 9, - 9$

Step 6

Add $81$ to both sides of the equation.

$x^{4} = 81$

Step 7

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{81}$

Step 8

Simplify $\pm \sqrt[4]{81}$ .

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Step 8.1

Rewrite $81$ as $3^{4}$ .

$x = \pm \sqrt[4]{3^{4}}$

Step 8.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

Step 9

The complete solution is the result of both the positive and negative portions of the solution.

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Step 9.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Step 9.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

Step 9.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

Step 10

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 9, - 9$

$x = 3, - 3$

Step 11

Consolidate the solutions.

$x = 9, - 9, 3, - 3$

Step 12

Find the domain of $\frac{x^{2} - 81}{x^{4} - 81}$ .

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Step 12.1

Set the denominator in $\frac{x^{2} - 81}{x^{4} - 81}$ equal to $0$ to find where the expression is undefined.

$x^{4} - 81 = 0$

Step 12.2

Solve for $x$ .

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Step 12.2.1

Add $81$ to both sides of the equation.

$x^{4} = 81$

Step 12.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{81}$

Step 12.2.3

Simplify $\pm \sqrt[4]{81}$ .

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Step 12.2.3.1

Rewrite $81$ as $3^{4}$ .

$x = \pm \sqrt[4]{3^{4}}$

Step 12.2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

Step 12.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 12.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Step 12.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

Step 12.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

Step 12.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Step 13

Use each root to create test intervals.

$x < - 9$

$- 9 < x < - 3$

$- 3 < x < 3$

$3 < x < 9$

$x > 9$

Step 14

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 14.1

Test a value on the interval $x < - 9$ to see if it makes the inequality true.

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Step 14.1.1

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Step 14.1.2

Replace $x$ with $- 12$ in the original inequality.

$\frac{{(- 12)}^{2} - 81}{{(- 12)}^{4} - 81} < 0$

Step 14.1.3

The left side $0.0030501$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 14.2

Test a value on the interval $- 9 < x < - 3$ to see if it makes the inequality true.

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Step 14.2.1

Choose a value on the interval $- 9 < x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Step 14.2.2

Replace $x$ with $- 6$ in the original inequality.

$\frac{{(- 6)}^{2} - 81}{{(- 6)}^{4} - 81} < 0$

Step 14.2.3

The left side $- 0.\overline{037}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 14.3

Test a value on the interval $- 3 < x < 3$ to see if it makes the inequality true.

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Step 14.3.1

Choose a value on the interval $- 3 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 14.3.2

Replace $x$ with $0$ in the original inequality.

$\frac{{(0)}^{2} - 81}{{(0)}^{4} - 81} < 0$

Step 14.3.3

The left side $1$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 14.4

Test a value on the interval $3 < x < 9$ to see if it makes the inequality true.

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Step 14.4.1

Choose a value on the interval $3 < x < 9$ and see if this value makes the original inequality true.

$x = 6$

Step 14.4.2

Replace $x$ with $6$ in the original inequality.

$\frac{{(6)}^{2} - 81}{{(6)}^{4} - 81} < 0$

Step 14.4.3

The left side $- 0.\overline{037}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 14.5

Test a value on the interval $x > 9$ to see if it makes the inequality true.

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Step 14.5.1

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Step 14.5.2

Replace $x$ with $12$ in the original inequality.

$\frac{{(12)}^{2} - 81}{{(12)}^{4} - 81} < 0$

Step 14.5.3

The left side $0.0030501$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 14.6

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 9$ False

$- 9 < x < - 3$ True

$- 3 < x < 3$ False

$3 < x < 9$ True

$x > 9$ False

$x < - 9$ False

$- 9 < x < - 3$ True

$- 3 < x < 3$ False

$3 < x < 9$ True

$x > 9$ False

Step 15

The solution consists of all of the true intervals.

$- 9 < x < - 3$ or $3 < x < 9$

Step 16

Convert the inequality to interval notation.

$(- 9, - 3) \cup (3, 9)$

Step 17