Solve the Inequality for x x^2-81>0

$x^{2} - 81 > 0$

Step 1

Add $81$ to both sides of the inequality.

$x^{2} > 81$

Step 2

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} > \sqrt{81}$

Step 3

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | > \sqrt{81}$

Simplify the right side.

Simplify $\sqrt{81}$ .

Rewrite $81$ as $9^{2}$ .

$| x | > \sqrt{9^{2}}$

Pull terms out from under the radical.

$| x | > | 9 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $9$ is $9$ .

$| x | > 9$

Step 4

Write $| x | > 9$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x > 9$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x > 9$

Write as a piecewise.

${\begin{cases} x > 9 & x \geq 0 \\ - x > 9 & x < 0 \end{cases}$

Step 5

Find the intersection of $x > 9$ and $x \geq 0$ .

$x > 9$

Step 6

Divide each term in $- x > 9$ by $- 1$ and simplify.

Divide each term in $- x > 9$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} < \frac{9}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} < \frac{9}{- 1}$

Divide $x$ by $1$ .

$x < \frac{9}{- 1}$

Simplify the right side.

Divide $9$ by $- 1$ .

$x < - 9$

Step 7

Find the union of the solutions.

$x < - 9$ or $x > 9$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x < - 9 or x > 9$

Interval Notation:

$(- \infty, - 9) \cup (9, \infty)$

Step 9

Solve the Inequality for x x^2-81>0

Related Questions

Support LCMGCF.com by disabling your adblocker.