Solve the Inequality for x x^2+5x-12<x+8

$x^{2} + 5 x - 12 < x + 8$

Step 1

Move all terms containing $x$ to the left side of the inequality.

Subtract $x$ from both sides of the inequality.

$x^{2} + 5 x - 12 - x < 8$

Subtract $x$ from $5 x$ .

$x^{2} + 4 x - 12 < 8$

Step 2

Move all terms to the left side of the equation and simplify.

Subtract $8$ from both sides of the inequality.

$x^{2} + 4 x - 12 - 8 < 0$

Subtract $8$ from $- 12$ .

$x^{2} + 4 x - 20 < 0$

Step 3

Convert the inequality to an equation.

$x^{2} + 4 x - 20 = 0$

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

Substitute the values $a = 1$ , $b = 4$ , and $c = - 20$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 20)}}{2 \cdot 1}$

Step 6

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 20}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 20$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 20}}{2 \cdot 1}$

Multiply $- 4$ by $- 20$ .

$x = \frac{- 4 \pm \sqrt{16 + 80}}{2 \cdot 1}$

Add $16$ and $80$ .

$x = \frac{- 4 \pm \sqrt{96}}{2 \cdot 1}$

Rewrite $96$ as $4^{2} \cdot 6$ .

Factor $16$ out of $96$ .

$x = \frac{- 4 \pm \sqrt{16 (6)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 4 \pm \sqrt{4^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 4 \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 4 \sqrt{6}}{2}$

Simplify $\frac{- 4 \pm 4 \sqrt{6}}{2}$ .

$x = - 2 \pm 2 \sqrt{6}$

Step 7

Consolidate the solutions.

$x = - 2 + 2 \sqrt{6}, - 2 - 2 \sqrt{6}$

Step 8

Use each root to create test intervals.

$x < - 2 - 2 \sqrt{6}$

$- 2 - 2 \sqrt{6} < x < - 2 + 2 \sqrt{6}$

$x > - 2 + 2 \sqrt{6}$

Step 9

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 $x < - 2 - 2 \sqrt{6}$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2 - 2 \sqrt{6}$ and see if this value makes the original inequality true.

$x = - 9$

Replace $x$ with $- 9$ in the original inequality.

${(- 9)}^{2} + 5 (- 9) - 12 < (- 9) + 8$

The left side $24$ is not less than the right side $- 1$ , which means that the given statement is false.

False

Test a value on the interval $- 2 - 2 \sqrt{6} < x < - 2 + 2 \sqrt{6}$ to see if it makes the inequality true.

Choose a value on the interval $- 2 - 2 \sqrt{6} < x < - 2 + 2 \sqrt{6}$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} + 5 (0) - 12 < (0) + 8$

The left side $- 12$ is less than the right side $8$ , which means that the given statement is always true.

True

Test a value on the interval $x > - 2 + 2 \sqrt{6}$ to see if it makes the inequality true.

Choose a value on the interval $x > - 2 + 2 \sqrt{6}$ and see if this value makes the original inequality true.

$x = 5$

Replace $x$ with $5$ in the original inequality.

${(5)}^{2} + 5 (5) - 12 < (5) + 8$

The left side $38$ is not less than the right side $13$ , which means that the given statement is false.

False

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

$x < - 2 - 2 \sqrt{6}$ False

$- 2 - 2 \sqrt{6} < x < - 2 + 2 \sqrt{6}$ True

$x > - 2 + 2 \sqrt{6}$ False

$x < - 2 - 2 \sqrt{6}$ False

$- 2 - 2 \sqrt{6} < x < - 2 + 2 \sqrt{6}$ True

$x > - 2 + 2 \sqrt{6}$ False

Step 10

The solution consists of all of the true intervals.

$- 2 - 2 \sqrt{6} < x < - 2 + 2 \sqrt{6}$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$- 2 - 2 \sqrt{6} < x < - 2 + 2 \sqrt{6}$

Interval Notation:

$(- 2 - 2 \sqrt{6}, - 2 + 2 \sqrt{6})$

Step 12

Solve the Inequality for x x^2+5x-12<x+8

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