Algebra Examples

$\frac{{(x - 5)}^{2}}{x^{2} - 9} \geq 0$

Step 1

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

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

$x^{2} - 9 = 0$

Step 2

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 3

Add $5$ to both sides of the equation.

$x = 5$

Step 4

Add $9$ to both sides of the equation.

$x^{2} = 9$

Step 5

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

$x = \pm \sqrt{9}$

Step 6

Simplify $\pm \sqrt{9}$ .

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Step 6.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Step 6.2

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

$x = \pm 3$

Step 7

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

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Step 7.1

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

$x = 3$

Step 7.2

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

$x = - 3$

Step 7.3

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

$x = 3, - 3$

Step 8

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

$x = 5$

$x = 3, - 3$

Step 9

Consolidate the solutions.

$x = 5, 3, - 3$

Step 10

Find the domain of $\frac{{(x - 5)}^{2}}{x^{2} - 9}$ .

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Step 10.1

Set the denominator in $\frac{{(x - 5)}^{2}}{x^{2} - 9}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 9 = 0$

Step 10.2

Solve for $x$ .

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Step 10.2.1

Add $9$ to both sides of the equation.

$x^{2} = 9$

Step 10.2.2

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

$x = \pm \sqrt{9}$

Step 10.2.3

Simplify $\pm \sqrt{9}$ .

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Step 10.2.3.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Step 10.2.3.2

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

$x = \pm 3$

Step 10.2.4

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

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Step 10.2.4.1

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

$x = 3$

Step 10.2.4.2

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

$x = - 3$

Step 10.2.4.3

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

$x = 3, - 3$

Step 10.3

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

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

Step 11

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 3$

$3 < x < 5$

$x > 5$

Step 12

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 12.1

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

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Step 12.1.1

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

$x = - 6$

Step 12.1.2

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

$\frac{{((- 6) - 5)}^{2}}{{(- 6)}^{2} - 9} \geq 0$

Step 12.1.3

The left side $4.\overline{481}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.2

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

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Step 12.2.1

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

$x = 0$

Step 12.2.2

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

$\frac{{((0) - 5)}^{2}}{{(0)}^{2} - 9} \geq 0$

Step 12.2.3

The left side $- 2.\overline{7}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 12.3

Test a value on the interval $3 < x < 5$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $3 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

Step 12.3.2

Replace $x$ with $4$ in the original inequality.

$\frac{{((4) - 5)}^{2}}{{(4)}^{2} - 9} \geq 0$

Step 12.3.3

The left side $0.\overline{142857}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.4

Test a value on the interval $x > 5$ to see if it makes the inequality true.

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Step 12.4.1

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Step 12.4.2

Replace $x$ with $8$ in the original inequality.

$\frac{{((8) - 5)}^{2}}{{(8)}^{2} - 9} \geq 0$

Step 12.4.3

The left side $0.1\overline{63}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.5

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

$x < - 3$ True

$- 3 < x < 3$ False

$3 < x < 5$ True

$x > 5$ True

$x < - 3$ True

$- 3 < x < 3$ False

$3 < x < 5$ True

$x > 5$ True

Step 13

The solution consists of all of the true intervals.

$x < - 3$ or $3 < x \leq 5$ or $x \geq 5$

Step 14

Combine the intervals.

$x < - 3 or x > 3$

Step 15

Convert the inequality to interval notation.

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

Step 16