Precalculus Examples

$\frac{49 x^{2} - 81}{x^{2} - 64} \leq 0$

Step 1

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

$49 x^{2} - 81 = 0$

$x^{2} - 64 = 0$

Step 2

Add $81$ to both sides of the equation.

$49 x^{2} = 81$

Step 3

Divide each term in $49 x^{2} = 81$ by $49$ and simplify.

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Step 3.1

Divide each term in $49 x^{2} = 81$ by $49$ .

$\frac{49 x^{2}}{49} = \frac{81}{49}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $49$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{49 x^{2}}{49} = \frac{81}{49}$

Step 3.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{81}{49}$

Step 4

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

$x = \pm \sqrt{\frac{81}{49}}$

Step 5

Simplify $\pm \sqrt{\frac{81}{49}}$ .

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Step 5.1

Rewrite $\sqrt{\frac{81}{49}}$ as $\frac{\sqrt{81}}{\sqrt{49}}$ .

$x = \pm \frac{\sqrt{81}}{\sqrt{49}}$

Step 5.2

Simplify the numerator.

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Step 5.2.1

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

$x = \pm \frac{\sqrt{9^{2}}}{\sqrt{49}}$

Step 5.2.2

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

$x = \pm \frac{9}{\sqrt{49}}$

Step 5.3

Simplify the denominator.

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Step 5.3.1

Rewrite $49$ as $7^{2}$ .

$x = \pm \frac{9}{\sqrt{7^{2}}}$

Step 5.3.2

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

$x = \pm \frac{9}{7}$

Step 6

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

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Step 6.1

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

$x = \frac{9}{7}$

Step 6.2

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

$x = - \frac{9}{7}$

Step 6.3

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

$x = \frac{9}{7}, - \frac{9}{7}$

Step 7

Add $64$ to both sides of the equation.

$x^{2} = 64$

Step 8

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

$x = \pm \sqrt{64}$

Step 9

Simplify $\pm \sqrt{64}$ .

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Step 9.1

Rewrite $64$ as $8^{2}$ .

$x = \pm \sqrt{8^{2}}$

Step 9.2

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

$x = \pm 8$

Step 10

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

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Step 10.1

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

$x = 8$

Step 10.2

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

$x = - 8$

Step 10.3

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

$x = 8, - 8$

Step 11

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

$x = \frac{9}{7}, - \frac{9}{7}$

$x = 8, - 8$

Step 12

Consolidate the solutions.

$x = \frac{9}{7}, - \frac{9}{7}, 8, - 8$

Step 13

Find the domain of $\frac{49 x^{2} - 81}{x^{2} - 64}$ .

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Step 13.1

Set the denominator in $\frac{49 x^{2} - 81}{x^{2} - 64}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 64 = 0$

Step 13.2

Solve for $x$ .

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Step 13.2.1

Add $64$ to both sides of the equation.

$x^{2} = 64$

Step 13.2.2

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

$x = \pm \sqrt{64}$

Step 13.2.3

Simplify $\pm \sqrt{64}$ .

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Step 13.2.3.1

Rewrite $64$ as $8^{2}$ .

$x = \pm \sqrt{8^{2}}$

Step 13.2.3.2

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

$x = \pm 8$

Step 13.2.4

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

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Step 13.2.4.1

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

$x = 8$

Step 13.2.4.2

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

$x = - 8$

Step 13.2.4.3

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

$x = 8, - 8$

Step 13.3

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

$(- \infty, - 8) \cup (- 8, 8) \cup (8, \infty)$

Step 14

Use each root to create test intervals.

$x < - 8$

$- 8 < x < - \frac{9}{7}$

$- \frac{9}{7} < x < \frac{9}{7}$

$\frac{9}{7} < x < 8$

$x > 8$

Step 15

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 15.1

Test a value on the interval $x < - 8$ to see if it makes the inequality true.

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Step 15.1.1

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Step 15.1.2

Replace $x$ with $- 10$ in the original inequality.

$\frac{49 {(- 10)}^{2} - 81}{{(- 10)}^{2} - 64} \leq 0$

Step 15.1.3

The left side $133.86\overline{1}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 15.2

Test a value on the interval $- 8 < x < - \frac{9}{7}$ to see if it makes the inequality true.

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Step 15.2.1

Choose a value on the interval $- 8 < x < - \frac{9}{7}$ and see if this value makes the original inequality true.

$x = - 5$

Step 15.2.2

Replace $x$ with $- 5$ in the original inequality.

$\frac{49 {(- 5)}^{2} - 81}{{(- 5)}^{2} - 64} \leq 0$

Step 15.2.3

The left side $- 29.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 15.3

Test a value on the interval $- \frac{9}{7} < x < \frac{9}{7}$ to see if it makes the inequality true.

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Step 15.3.1

Choose a value on the interval $- \frac{9}{7} < x < \frac{9}{7}$ and see if this value makes the original inequality true.

$x = 0$

Step 15.3.2

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

$\frac{49 {(0)}^{2} - 81}{{(0)}^{2} - 64} \leq 0$

Step 15.3.3

The left side $1.265625$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 15.4

Test a value on the interval $\frac{9}{7} < x < 8$ to see if it makes the inequality true.

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Step 15.4.1

Choose a value on the interval $\frac{9}{7} < x < 8$ and see if this value makes the original inequality true.

$x = 5$

Step 15.4.2

Replace $x$ with $5$ in the original inequality.

$\frac{49 {(5)}^{2} - 81}{{(5)}^{2} - 64} \leq 0$

Step 15.4.3

The left side $- 29.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 15.5

Test a value on the interval $x > 8$ to see if it makes the inequality true.

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Step 15.5.1

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Step 15.5.2

Replace $x$ with $10$ in the original inequality.

$\frac{49 {(10)}^{2} - 81}{{(10)}^{2} - 64} \leq 0$

Step 15.5.3

The left side $133.86\overline{1}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 15.6

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

$x < - 8$ False

$- 8 < x < - \frac{9}{7}$ True

$- \frac{9}{7} < x < \frac{9}{7}$ False

$\frac{9}{7} < x < 8$ True

$x > 8$ False

$x < - 8$ False

$- 8 < x < - \frac{9}{7}$ True

$- \frac{9}{7} < x < \frac{9}{7}$ False

$\frac{9}{7} < x < 8$ True

$x > 8$ False

Step 16

The solution consists of all of the true intervals.

$- 8 < x \leq - \frac{9}{7}$ or $\frac{9}{7} \leq x < 8$

Step 17

Convert the inequality to interval notation.

$(- 8, - \frac{9}{7}] \cup [\frac{9}{7}, 8)$

Step 18