Solve the Inequality for x (x+2)/(x^2+3x-18)<=0

$\frac{x + 2}{x^{2} + 3 x - 18} \leq 0$

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

$x + 2 = 0$

$x^{2} + 3 x - 18 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Factor $x^{2} + 3 x - 18$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 18$ and whose sum is $3$ .

$- 3, 6$

Write the factored form using these integers.

$(x - 3) (x + 6) = 0$

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

$x - 3 = 0$

$x + 6 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x - 3) (x + 6) = 0$ true.

$x = 3, - 6$

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

$x = - 2$

$x = 3, - 6$

Consolidate the solutions.

$x = - 2, 3, - 6$

Find the domain of $\frac{x + 2}{x^{2} + 3 x - 18}$ .

Set the denominator in $\frac{x + 2}{x^{2} + 3 x - 18}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 3 x - 18 = 0$

Solve for $x$ .

Factor $x^{2} + 3 x - 18$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 18$ and whose sum is $3$ .

$- 3, 6$

Write the factored form using these integers.

$(x - 3) (x + 6) = 0$

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

$x - 3 = 0$

$x + 6 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x - 3) (x + 6) = 0$ true.

$x = 3, - 6$

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

$(- \infty, - 6) \cup (- 6, 3) \cup (3, \infty)$

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 2$

$- 2 < x < 3$

$x > 3$

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.

$\frac{(- 8) + 2}{{(- 8)}^{2} + 3 (- 8) - 18} \leq 0$

The left side $- 0.\overline{27}$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

$\frac{(- 4) + 2}{{(- 4)}^{2} + 3 (- 4) - 18} \leq 0$

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

False

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

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

$x = 0$

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

$\frac{(0) + 2}{{(0)}^{2} + 3 (0) - 18} \leq 0$

The left side $- 0.\overline{1}$ 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.

$\frac{(6) + 2}{{(6)}^{2} + 3 (6) - 18} \leq 0$

The left side $0.\overline{2}$ 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.

$x < - 6$ True

$- 6 < x < - 2$ False

$- 2 < x < 3$ True

$x > 3$ False

$x < - 6$ True

$- 6 < x < - 2$ False

$- 2 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$x < - 6$ or $- 2 \leq x < 3$

The result can be shown in multiple forms.

Inequality Form:

$x < - 6 or - 2 \leq x < 3$

Interval Notation:

$(- \infty, - 6) \cup [- 2, 3)$

Solve the Inequality for x (x+2)/(x^2+3x-18)<=0

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