Convert to Interval Notation (x^2-16)/(x^2-25)<0

$\frac{x^{2} - 16}{x^{2} - 25} < 0$

Step 1

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

$x^{2} - 16 = 0$

$x^{2} - 25 = 0$

Step 2

Add $16$ to both sides of the equation.

$x^{2} = 16$

Step 3

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

$x = \pm \sqrt{16}$

Step 4

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

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

$x = \pm 4$

Step 5

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.

$x = 4$

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

$x = - 4$

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

$x = 4, - 4$

Step 6

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 7

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

$x = \pm \sqrt{25}$

Step 8

Simplify $\pm \sqrt{25}$ .

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

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

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

$x = \pm 5$

Step 9

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.

$x = 5$

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

$x = - 5$

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

$x = 5, - 5$

Step 10

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

$x = 4, - 4$

$x = 5, - 5$

Step 11

Consolidate the solutions.

$x = 4, - 4, 5, - 5$

Step 12

Find the domain of $\frac{x^{2} - 16}{x^{2} - 25}$ .

Set the denominator in $\frac{x^{2} - 16}{x^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 25 = 0$

Solve for $x$ .

Add $25$ to both sides of the equation.

$x^{2} = 25$

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

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

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

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

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

$x = \pm 5$

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.

$x = 5$

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

$x = - 5$

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

$x = 5, - 5$

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

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

Step 13

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 4$

$- 4 < x < 4$

$4 < x < 5$

$x > 5$

Step 14

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 $x < - 5$ to see if it makes the inequality true.

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

$x = - 8$

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

$\frac{{(- 8)}^{2} - 16}{{(- 8)}^{2} - 25} < 0$

The left side $1.\overline{230769}$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - 4$ and see if this value makes the original inequality true.

$x = - 4.5$

Replace $x$ with $- 4.5$ in the original inequality.

$\frac{{(- 4.5)}^{2} - 16}{{(- 4.5)}^{2} - 25} < 0$

The left side $- 0.89473684$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 4 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} - 16}{{(0)}^{2} - 25} < 0$

The left side $0.64$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $4 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $4 < x < 5$ and see if this value makes the original inequality true.

$x = 4.5$

Replace $x$ with $4.5$ in the original inequality.

$\frac{{(4.5)}^{2} - 16}{{(4.5)}^{2} - 25} < 0$

The left side $- 0.89473684$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 8$

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

$\frac{{(8)}^{2} - 16}{{(8)}^{2} - 25} < 0$

The left side $1.\overline{230769}$ is not less 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.

$x < - 5$ False

$- 5 < x < - 4$ True

$- 4 < x < 4$ False

$4 < x < 5$ True

$x > 5$ False

$x < - 5$ False

$- 5 < x < - 4$ True

$- 4 < x < 4$ False

$4 < x < 5$ True

$x > 5$ False

Step 15

The solution consists of all of the true intervals.

$- 5 < x < - 4$ or $4 < x < 5$

Step 16

Convert the inequality to interval notation.

$(- 5, - 4) \cup (4, 5)$

Step 17

Convert to Interval Notation (x^2-16)/(x^2-25)<0

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