Solve the Inequality for x x^2+6x>-8

$x^{2} + 6 x > - 8$

Convert the inequality to an equation.

$x^{2} + 6 x = - 8$

Add $8$ to both sides of the equation.

$x^{2} + 6 x + 8 = 0$

Factor $x^{2} + 6 x + 8$ 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 $8$ and whose sum is $6$ .

$2, 4$

Write the factored form using these integers.

$(x + 2) (x + 4) = 0$

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

$x + 2 = 0$

$x + 4 = 0$

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.

$x = - 2$

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.

$x = - 4$

The final solution is all the values that make $(x + 2) (x + 4) = 0$ true.

$x = - 2, - 4$

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 2$

$x > - 2$

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.

${(- 6)}^{2} + 6 (- 6) > - 8$

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

True

Test a value on the interval $- 4 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < - 2$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

${(- 3)}^{2} + 6 (- 3) > - 8$

The left side $- 9$ is not greater than the right side $- 8$ , which means that the given statement is false.

False

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.

$x = 0$

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

${(0)}^{2} + 6 (0) > - 8$

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

True

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

$x < - 4$ True

$- 4 < x < - 2$ False

$x > - 2$ True

$x < - 4$ True

$- 4 < x < - 2$ False

$x > - 2$ True

The solution consists of all of the true intervals.

$x < - 4$ or $x > - 2$

The result can be shown in multiple forms.

Inequality Form:

$x < - 4 or x > - 2$

Interval Notation:

$(- \infty, - 4) \cup (- 2, \infty)$

Solve the Inequality for x x^2+6x>-8

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