Algebra Examples

$\frac{x^{2} - 4}{x^{2} + 36} \geq 0$

Step 1

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

$x^{2} - 4 = 0$

$x^{2} + 36 = 0$

Step 2

Add $4$ to both sides of the equation.

$x^{2} = 4$

Step 3

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

$x = \pm \sqrt{4}$

Step 4

Simplify $\pm \sqrt{4}$ .

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Step 4.1

Rewrite $4$ as $2^{2}$ .

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

Step 4.2

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

$x = \pm 2$

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.

$x = 2$

Step 5.2

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

$x = - 2$

Step 5.3

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

$x = 2, - 2$

Step 6

Subtract $36$ from both sides of the equation.

$x^{2} = - 36$

Step 7

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

$x = \pm \sqrt{- 36}$

Step 8

Simplify $\pm \sqrt{- 36}$ .

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Step 8.1

Rewrite $- 36$ as $- 1 (36)$ .

$x = \pm \sqrt{- 1 (36)}$

Step 8.2

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{36}$

Step 8.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{36}$

Step 8.4

Rewrite $36$ as $6^{2}$ .

$x = \pm i \cdot \sqrt{6^{2}}$

Step 8.5

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

$x = \pm i \cdot 6$

Step 8.6

Move $6$ to the left of $i$ .

$x = \pm 6 i$

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.

$x = 6 i$

Step 9.2

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

$x = - 6 i$

Step 9.3

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

$x = 6 i, - 6 i$

Step 10

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

$x = 2, - 2$

$x = 6 i, - 6 i$

Step 11

Consolidate the solutions.

$x = 2, - 2$

Step 12

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 2$

$x > 2$

Step 13

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 13.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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Step 13.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 13.1.2

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

$\frac{{(- 4)}^{2} - 4}{{(- 4)}^{2} + 36} \geq 0$

Step 13.1.3

The left side $0.\overline{230769}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 13.2

Test a value on the interval $- 2 < x < 2$ to see if it makes the inequality true.

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Step 13.2.1

Choose a value on the interval $- 2 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 13.2.2

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

$\frac{{(0)}^{2} - 4}{{(0)}^{2} + 36} \geq 0$

Step 13.2.3

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

False

Step 13.3

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 13.3.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 13.3.2

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

$\frac{{(4)}^{2} - 4}{{(4)}^{2} + 36} \geq 0$

Step 13.3.3

The left side $0.\overline{230769}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 13.4

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

$x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

$x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

Step 14

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq 2$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or x \geq 2$

Interval Notation:

$(- \infty, - 2] \cup [2, \infty)$

Step 16