Solve the Inequality for x (x^2-4)/(x^2-36)<=0

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

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

$x^{2} - 4 = 0$

$x^{2} - 36 = 0$

Add $4$ to both sides of the equation.

$x^{2} = 4$

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

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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

$x = \pm 2$

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

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

$x = 2$

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

$x = - 2$

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

$x = 2, - 2$

Add $36$ to both sides of the equation.

$x^{2} = 36$

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

$x = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

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

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

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

$x = \pm 6$

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

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

$x = 6$

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

$x = - 6$

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

$x = 6, - 6$

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

$x = 2, - 2$

$x = 6, - 6$

Consolidate the solutions.

$x = 2, - 2, 6, - 6$

Find the domain of $\frac{x^{2} - 4}{x^{2} - 36}$ .

Set the denominator in $\frac{x^{2} - 4}{x^{2} - 36}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 36 = 0$

Solve for $x$ .

Add $36$ to both sides of the equation.

$x^{2} = 36$

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

$x = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

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

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

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

$x = \pm 6$

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

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

$x = 6$

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

$x = - 6$

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

$x = 6, - 6$

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

$(- \infty, - 6) \cup (- 6, 6) \cup (6, \infty)$

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 2$

$- 2 < x < 2$

$2 < x < 6$

$x > 6$

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

Test a value on the interval $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

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

$\frac{{(- 8)}^{2} - 4}{{(- 8)}^{2} - 36} \leq 0$

The left side $2.\overline{142857}$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 6 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

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

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

The left side $- 0.6$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

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

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

False

Test a value on the interval $2 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

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

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

The left side $- 0.6$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

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

$\frac{{(8)}^{2} - 4}{{(8)}^{2} - 36} \leq 0$

The left side $2.\overline{142857}$ is greater than the right side $0$ , which means that the given statement is false.

False

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

$x < - 6$ False

$- 6 < x < - 2$ True

$- 2 < x < 2$ False

$2 < x < 6$ True

$x > 6$ False

$x < - 6$ False

$- 6 < x < - 2$ True

$- 2 < x < 2$ False

$2 < x < 6$ True

$x > 6$ False

The solution consists of all of the true intervals.

$- 6 < x \leq - 2$ or $2 \leq x < 6$

The result can be shown in multiple forms.

Inequality Form:

$- 6 < x \leq - 2 or 2 \leq x < 6$

Interval Notation:

$(- 6, - 2] \cup [2, 6)$

Solve the Inequality for x (x^2-4)/(x^2-36)<=0

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