Solve the Inequality for x x^2+x-12>=0

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

Step 1

Convert the inequality to an equation.

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

Step 2

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 $1$ .

$- 3, 4$

Write the factored form using these integers.

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

Step 3

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

$x - 3 = 0$

$x + 4 = 0$

Step 4

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.

$x = 3$

Step 5

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6

The final solution is all the values that make $(x - 3) (x + 4) = 0$ true.

$x = 3, - 4$

Step 7

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 3$

$x > 3$

Step 8

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 $x < - 4$ 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.

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

The left side $18$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

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

The left side $- 12$ is less than the right side $0$ , which means that the given statement is false.

False

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.

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

The left side $30$ 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.

$x < - 4$ True

$- 4 < x < 3$ False

$x > 3$ True

$x < - 4$ True

$- 4 < x < 3$ False

$x > 3$ True

Step 9

The solution consists of all of the true intervals.

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

Step 10

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 11

Solve the Inequality for x x^2+x-12>=0

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