Algebra Examples

$\frac{4}{x^{2} - 25} > 0$

Step 1

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

$x^{2} - 25 = 0$

Step 2

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 3

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

$x = \pm \sqrt{25}$

Step 4

Simplify $\pm \sqrt{25}$ .

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Step 4.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 4.2

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

$x = \pm 5$

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.

$x = 5$

Step 5.2

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

$x = - 5$

Step 5.3

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

$x = 5, - 5$

Step 6

Consolidate the solutions.

$x = 5, - 5$

Step 7

Find the domain of $\frac{4}{x^{2} - 25}$ .

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Step 7.1

Set the denominator in $\frac{4}{x^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 25 = 0$

Step 7.2

Solve for $x$ .

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Step 7.2.1

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 7.2.2

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

$x = \pm \sqrt{25}$

Step 7.2.3

Simplify $\pm \sqrt{25}$ .

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Step 7.2.3.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 7.2.3.2

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

$x = \pm 5$

Step 7.2.4

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

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Step 7.2.4.1

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

$x = 5$

Step 7.2.4.2

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

$x = - 5$

Step 7.2.4.3

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

$x = 5, - 5$

Step 7.3

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

$(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)$

Step 8

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 5$

$x > 5$

Step 9

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 9.1

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

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Step 9.1.1

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

$x = - 8$

Step 9.1.2

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

$\frac{4}{{(- 8)}^{2} - 25} > 0$

Step 9.1.3

The left side $0.\overline{102564}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.2

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

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Step 9.2.1

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

$x = 0$

Step 9.2.2

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

$\frac{4}{{(0)}^{2} - 25} > 0$

Step 9.2.3

The left side $- 0.16$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 9.3

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

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Step 9.3.1

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

$x = 8$

Step 9.3.2

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

$\frac{4}{{(8)}^{2} - 25} > 0$

Step 9.3.3

The left side $0.\overline{102564}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.4

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

$x < - 5$ True

$- 5 < x < 5$ False

$x > 5$ True

$x < - 5$ True

$- 5 < x < 5$ False

$x > 5$ True

Step 10

The solution consists of all of the true intervals.

$x < - 5$ or $x > 5$

Step 11

Convert the inequality to interval notation.

$(- \infty, - 5) \cup (5, \infty)$

Step 12