Solve the Inequality for x (x^2-x-12)/(x+5)>=0

$\frac{x^{2} - x - 12}{x + 5} \geq 0$

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

$x^{2} - x - 12 = 0$

$x + 5 = 0$

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$(x - 4) (x + 3) = 0$

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

$x - 4 = 0$

$x + 3 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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 $(x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

$x = 4, - 3$

$x = - 5$

Consolidate the solutions.

$x = 4, - 3, - 5$

Find the domain of $\frac{x^{2} - x - 12}{x + 5}$ .

Set the denominator in $\frac{x^{2} - x - 12}{x + 5}$ 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.

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

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 3$

$- 3 < x < 4$

$x > 4$

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.

$\frac{{(- 8)}^{2} - (- 8) - 12}{(- 8) + 5} \geq 0$

The left side $- 20$ is less than the right side $0$ , which means that the given statement is false.

False

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.

$\frac{{(- 4)}^{2} - (- 4) - 12}{(- 4) + 5} \geq 0$

The left side $8$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} - (0) - 12}{(0) + 5} \geq 0$

The left side $- 2.4$ is less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 6$

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

$\frac{{(6)}^{2} - (6) - 12}{(6) + 5} \geq 0$

The left side $1.\overline{63}$ 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.

$x < - 5$ False

$- 5 < x < - 3$ True

$- 3 < x < 4$ False

$x > 4$ True

$x < - 5$ False

$- 5 < x < - 3$ True

$- 3 < x < 4$ False

$x > 4$ True

The solution consists of all of the true intervals.

$- 5 < x \leq - 3$ or $x \geq 4$

The result can be shown in multiple forms.

Inequality Form:

$- 5 < x \leq - 3 or x \geq 4$

Interval Notation:

$(- 5, - 3] \cup [4, \infty)$

Solve the Inequality for x (x^2-x-12)/(x+5)>=0

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