Convert to Interval Notation x^2-x-6>0

$x^{2} - x - 6 > 0$

Step 1

Convert the inequality to an equation.

$x^{2} - x - 6 = 0$

Step 2

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$(x - 3) (x + 2) = 0$

Step 3

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

$x - 3 = 0$

$x + 2 = 0$

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 $x + 2$ 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 $(x - 3) (x + 2) = 0$ true.

$x = 3, - 2$

Step 7

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 3$

$x > 3$

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 $x < - 2$ to see if it makes the inequality true.

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

$x = - 4$

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

${(- 4)}^{2} - (- 4) - 6 > 0$

The left side $14$ is greater than the right side $0$ , which means that the given statement is always true.

True

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.

${(0)}^{2} - (0) - 6 > 0$

The left side $- 6$ is not greater 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.

${(6)}^{2} - (6) - 6 > 0$

The left side $24$ 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.

$x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

$x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

Step 9

The solution consists of all of the true intervals.

$x < - 2$ or $x > 3$

Step 10

Convert the inequality to interval notation.

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

Step 11

Convert to Interval Notation x^2-x-6>0

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