Solve the Inequality for x 2x^2-1>6x

$2 x^{2} - 1 > 6 x$

Subtract $6 x$ from both sides of the inequality.

$2 x^{2} - 1 - 6 x > 0$

Convert the inequality to an equation.

$2 x^{2} - 1 - 6 x = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 6$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (2 \cdot - 1)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot - 1}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{6 \pm \sqrt{36 - 8 \cdot - 1}}{2 \cdot 2}$

Multiply $- 8$ by $- 1$ .

$x = \frac{6 \pm \sqrt{36 + 8}}{2 \cdot 2}$

Add $36$ and $8$ .

$x = \frac{6 \pm \sqrt{44}}{2 \cdot 2}$

Rewrite $44$ as $2^{2} \cdot 11$ .

Factor $4$ out of $44$ .

$x = \frac{6 \pm \sqrt{4 (11)}}{2 \cdot 2}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 11}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{11}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{6 \pm 2 \sqrt{11}}{4}$

Simplify $\frac{6 \pm 2 \sqrt{11}}{4}$ .

$x = \frac{3 \pm \sqrt{11}}{2}$

Consolidate the solutions.

$x = \frac{3 + \sqrt{11}}{2}, \frac{3 - \sqrt{11}}{2}$

Use each root to create test intervals.

$x < \frac{3 - \sqrt{11}}{2}$

$\frac{3 - \sqrt{11}}{2} < x < \frac{3 + \sqrt{11}}{2}$

$x > \frac{3 + \sqrt{11}}{2}$

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 $x < \frac{3 - \sqrt{11}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{3 - \sqrt{11}}{2}$ and see if this value makes the original inequality true.

$x = - 3$

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

$2 {(- 3)}^{2} - 1 > 6 (- 3)$

The left side $17$ is greater than the right side $- 18$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{3 - \sqrt{11}}{2} < x < \frac{3 + \sqrt{11}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{3 - \sqrt{11}}{2} < x < \frac{3 + \sqrt{11}}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

$2 {(0)}^{2} - 1 > 6 (0)$

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{3 + \sqrt{11}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{3 + \sqrt{11}}{2}$ and see if this value makes the original inequality true.

$x = 6$

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

$2 {(6)}^{2} - 1 > 6 (6)$

The left side $71$ is greater than the right side $36$ , which means that the given statement is always true.

True

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

$x < \frac{3 - \sqrt{11}}{2}$ True

$\frac{3 - \sqrt{11}}{2} < x < \frac{3 + \sqrt{11}}{2}$ False

$x > \frac{3 + \sqrt{11}}{2}$ True

$x < \frac{3 - \sqrt{11}}{2}$ True

$\frac{3 - \sqrt{11}}{2} < x < \frac{3 + \sqrt{11}}{2}$ False

$x > \frac{3 + \sqrt{11}}{2}$ True

The solution consists of all of the true intervals.

$x < \frac{3 - \sqrt{11}}{2}$ or $x > \frac{3 + \sqrt{11}}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{3 - \sqrt{11}}{2} or x > \frac{3 + \sqrt{11}}{2}$

Interval Notation:

$(- \infty, \frac{3 - \sqrt{11}}{2}) \cup (\frac{3 + \sqrt{11}}{2}, \infty)$

Solve the Inequality for x 2x^2-1>6x

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