Solve the Inequality for x (x^2-9)/(x+5)<0

$\frac{x^{2} - 9}{x + 5} < 0$

Step 1

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

$x^{2} - 9 = 0$

$x + 5 = 0$

Step 2

Add $9$ to both sides of the equation.

$x^{2} = 9$

Step 3

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

$x = \pm \sqrt{9}$

Step 4

Simplify $\pm \sqrt{9}$ .

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

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

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

$x = \pm 3$

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.

$x = 3$

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

$x = - 3$

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

$x = 3, - 3$

Step 6

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 7

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

$x = 3, - 3$

$x = - 5$

Step 8

Consolidate the solutions.

$x = 3, - 3, - 5$

Step 9

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

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

Step 10

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 3$

$- 3 < x < 3$

$x > 3$

Step 11

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.

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

The left side $- 18.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = - 4$

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

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

The left side $7$ is not less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

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

The left side $- 1.8$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 6$

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

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

The left side $2.\overline{45}$ 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.

$x < - 5$ True

$- 5 < x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

$x < - 5$ True

$- 5 < x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

Step 12

The solution consists of all of the true intervals.

$x < - 5$ or $- 3 < x < 3$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 5 or - 3 < x < 3$

Interval Notation:

$(- \infty, - 5) \cup (- 3, 3)$

Step 14

Solve the Inequality for x (x^2-9)/(x+5)<0

Related Questions

Support LCMGCF.com by disabling your adblocker.