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

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

Step 1

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

$x^{2} + 2 x = 0$

$x^{2} - 9 = 0$

Step 2

Factor $x$ out of $x^{2} + 2 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x + 2 x = 0$

Factor $x$ out of $2 x$ .

$x \cdot x + x \cdot 2 = 0$

Factor $x$ out of $x \cdot x + x \cdot 2$ .

$x (x + 2) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 2 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

The final solution is all the values that make $x (x + 2) = 0$ true.

$x = 0, - 2$

Step 7

Add $9$ to both sides of the equation.

$x^{2} = 9$

Step 8

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

$x = \pm \sqrt{9}$

Step 9

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 10

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 11

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

$x = 0, - 2$

$x = 3, - 3$

Step 12

Consolidate the solutions.

$x = 0, - 2, 3, - 3$

Step 13

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

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

$x^{2} - 9 = 0$

Solve for $x$ .

Add $9$ to both sides of the equation.

$x^{2} = 9$

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

$x = \pm \sqrt{9}$

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.

$x = \pm 3$

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.

$x = 3, - 3$

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

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

Step 14

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 2$

$- 2 < x < 0$

$0 < x < 3$

$x > 3$

Step 15

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 $x < - 3$ 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.

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

The left side $0.\overline{8}$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < - 2$ and see if this value makes the original inequality true.

$x = - 2.5$

Replace $x$ with $- 2.5$ in the original inequality.

$\frac{{(- 2.5)}^{2} + 2 (- 2.5)}{{(- 2.5)}^{2} - 9} \leq 0$

The left side $- 0.\overline{45}$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

$\frac{{(- 1)}^{2} + 2 (- 1)}{{(- 1)}^{2} - 9} \leq 0$

The left side $0.125$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

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

The left side $- 1.6$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ 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.

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

The left side $1.\overline{7}$ 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.

$x < - 3$ False

$- 3 < x < - 2$ True

$- 2 < x < 0$ False

$0 < x < 3$ True

$x > 3$ False

$x < - 3$ False

$- 3 < x < - 2$ True

$- 2 < x < 0$ False

$0 < x < 3$ True

$x > 3$ False

Step 16

The solution consists of all of the true intervals.

$- 3 < x \leq - 2$ or $0 \leq x < 3$

Step 17

The result can be shown in multiple forms.

Inequality Form:

$- 3 < x \leq - 2 or 0 \leq x < 3$

Interval Notation:

$(- 3, - 2] \cup [0, 3)$

Step 18

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

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