Solve the Inequality for x x^2-4x-24>=0

$x^{2} - 4 x - 24 \geq 0$

Convert the inequality to an equation.

$x^{2} - 4 x - 24 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 4$ , and $c = - 24$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot - 24)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4$ by $- 24$ .

$x = \frac{4 \pm \sqrt{16 + 96}}{2 \cdot 1}$

Add $16$ and $96$ .

$x = \frac{4 \pm \sqrt{112}}{2 \cdot 1}$

Rewrite $112$ as $4^{2} \cdot 7$ .

Factor $16$ out of $112$ .

$x = \frac{4 \pm \sqrt{16 (7)}}{2 \cdot 1}$

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

$x = \frac{4 \pm \sqrt{4^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 4 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 4 \sqrt{7}}{2}$

Simplify $\frac{4 \pm 4 \sqrt{7}}{2}$ .

$x = 2 \pm 2 \sqrt{7}$

Consolidate the solutions.

$x = 2 + 2 \sqrt{7}, 2 - 2 \sqrt{7}$

Use each root to create test intervals.

$x < 2 - 2 \sqrt{7}$

$2 - 2 \sqrt{7} < x < 2 + 2 \sqrt{7}$

$x > 2 + 2 \sqrt{7}$

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 $x < 2 - 2 \sqrt{7}$ to see if it makes the inequality true.

Choose a value on the interval $x < 2 - 2 \sqrt{7}$ and see if this value makes the original inequality true.

$x = - 6$

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

${(- 6)}^{2} - 4 \cdot - 6 - 24 \geq 0$

The left side $36$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $2 - 2 \sqrt{7} < x < 2 + 2 \sqrt{7}$ to see if it makes the inequality true.

Choose a value on the interval $2 - 2 \sqrt{7} < x < 2 + 2 \sqrt{7}$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} - 4 \cdot 0 - 24 \geq 0$

The left side $- 24$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 2 + 2 \sqrt{7}$ to see if it makes the inequality true.

Choose a value on the interval $x > 2 + 2 \sqrt{7}$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

${(10)}^{2} - 4 \cdot 10 - 24 \geq 0$

The left side $36$ 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.

$x < 2 - 2 \sqrt{7}$ True

$2 - 2 \sqrt{7} < x < 2 + 2 \sqrt{7}$ False

$x > 2 + 2 \sqrt{7}$ True

$x < 2 - 2 \sqrt{7}$ True

$2 - 2 \sqrt{7} < x < 2 + 2 \sqrt{7}$ False

$x > 2 + 2 \sqrt{7}$ True

The solution consists of all of the true intervals.

$x \leq 2 - 2 \sqrt{7}$ or $x \geq 2 + 2 \sqrt{7}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq 2 - 2 \sqrt{7} or x \geq 2 + 2 \sqrt{7}$

Interval Notation:

$(- \infty, 2 - 2 \sqrt{7}] \cup [2 + 2 \sqrt{7}, \infty)$

Solve the Inequality for x x^2-4x-24>=0

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