Solve the Inequality for x x^2-2x-11<=0

$x^{2} - 2 x - 11 \leq 0$

Convert the inequality to an equation.

$x^{2} - 2 x - 11 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 11$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 11)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 11}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 11}}{2 \cdot 1}$

Multiply $- 4$ by $- 11$ .

$x = \frac{2 \pm \sqrt{4 + 44}}{2 \cdot 1}$

Add $4$ and $44$ .

$x = \frac{2 \pm \sqrt{48}}{2 \cdot 1}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$x = \frac{2 \pm \sqrt{16 (3)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{4^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 4 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 4 \sqrt{3}}{2}$

Simplify $\frac{2 \pm 4 \sqrt{3}}{2}$ .

$x = 1 \pm 2 \sqrt{3}$

Consolidate the solutions.

$x = 1 + 2 \sqrt{3}, 1 - 2 \sqrt{3}$

Use each root to create test intervals.

$x < 1 - 2 \sqrt{3}$

$1 - 2 \sqrt{3} < x < 1 + 2 \sqrt{3}$

$x > 1 + 2 \sqrt{3}$

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 $x < 1 - 2 \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < 1 - 2 \sqrt{3}$ and see if this value makes the original inequality true.

$x = - 5$

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

${(- 5)}^{2} - 2 \cdot - 5 - 11 \leq 0$

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

False

Test a value on the interval $1 - 2 \sqrt{3} < x < 1 + 2 \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $1 - 2 \sqrt{3} < x < 1 + 2 \sqrt{3}$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} - 2 \cdot 0 - 11 \leq 0$

The left side $- 11$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 1 + 2 \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $x > 1 + 2 \sqrt{3}$ and see if this value makes the original inequality true.

$x = 7$

Replace $x$ with $7$ in the original inequality.

${(7)}^{2} - 2 \cdot 7 - 11 \leq 0$

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

False

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

$x < 1 - 2 \sqrt{3}$ False

$1 - 2 \sqrt{3} < x < 1 + 2 \sqrt{3}$ True

$x > 1 + 2 \sqrt{3}$ False

$x < 1 - 2 \sqrt{3}$ False

$1 - 2 \sqrt{3} < x < 1 + 2 \sqrt{3}$ True

$x > 1 + 2 \sqrt{3}$ False

The solution consists of all of the true intervals.

$1 - 2 \sqrt{3} \leq x \leq 1 + 2 \sqrt{3}$

The result can be shown in multiple forms.

Inequality Form:

$1 - 2 \sqrt{3} \leq x \leq 1 + 2 \sqrt{3}$

Interval Notation:

$[1 - 2 \sqrt{3}, 1 + 2 \sqrt{3}]$

Solve the Inequality for x x^2-2x-11<=0

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