Solve the Inequality for x (x+6)/(x^2-5x-24)>=0

$\frac{x + 6}{x^{2} - 5 x - 24} \geq 0$

Step 1

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

$x + 6 = 0$

$x^{2} - 5 x - 24 = 0$

Step 2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 3

Factor $x^{2} - 5 x - 24$ 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 $- 24$ and whose sum is $- 5$ .

$- 8, 3$

Write the factored form using these integers.

$(x - 8) (x + 3) = 0$

Step 4

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

$x - 8 = 0$

$x + 3 = 0$

Step 5

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Step 6

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

The final solution is all the values that make $(x - 8) (x + 3) = 0$ true.

$x = 8, - 3$

Step 8

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

$x = - 6$

$x = 8, - 3$

Step 9

Consolidate the solutions.

$x = - 6, 8, - 3$

Step 10

Find the domain of $\frac{x + 6}{x^{2} - 5 x - 24}$ .

Set the denominator in $\frac{x + 6}{x^{2} - 5 x - 24}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 x - 24 = 0$

Solve for $x$ .

Factor $x^{2} - 5 x - 24$ 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 $- 24$ and whose sum is $- 5$ .

$- 8, 3$

Write the factored form using these integers.

$(x - 8) (x + 3) = 0$

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

$x - 8 = 0$

$x + 3 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 8) (x + 3) = 0$ true.

$x = 8, - 3$

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

$(- \infty, - 3) \cup (- 3, 8) \cup (8, \infty)$

Step 11

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 3$

$- 3 < x < 8$

$x > 8$

Step 12

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.

$\frac{(- 8) + 6}{{(- 8)}^{2} - 5 \cdot - 8 - 24} \geq 0$

The left side $- 0.025$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 6 < x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < - 3$ and see if this value makes the original inequality true.

$x = - 4$

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

$\frac{(- 4) + 6}{{(- 4)}^{2} - 5 \cdot - 4 - 24} \geq 0$

The left side $0.1\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 8$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 8$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) + 6}{{(0)}^{2} - 5 \cdot 0 - 24} \geq 0$

The left side $- 0.25$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$\frac{(10) + 6}{{(10)}^{2} - 5 \cdot 10 - 24} \geq 0$

The left side $0.\overline{615384}$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 6$ False

$- 6 < x < - 3$ True

$- 3 < x < 8$ False

$x > 8$ True

$x < - 6$ False

$- 6 < x < - 3$ True

$- 3 < x < 8$ False

$x > 8$ True

Step 13

The solution consists of all of the true intervals.

$- 6 \leq x < - 3$ or $x > 8$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$- 6 \leq x < - 3 or x > 8$

Interval Notation:

$[- 6, - 3) \cup (8, \infty)$

Step 15

Solve the Inequality for x (x+6)/(x^2-5x-24)>=0

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