Solve the Inequality for x x^2<81

$x^{2} < 81$

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

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

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 $9$ .

$| x | < 9$

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.

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

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

$0 \leq x < 9$

Solve $- x < 9$ when $x < 0$ .

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 $- 1$ .

$x > - 9$

Find the intersection of $x > - 9$ and $x < 0$ .

$- 9 < x < 0$

Find the union of the solutions.

$- 9 < x < 9$

The result can be shown in multiple forms.

Inequality Form:

$- 9 < x < 9$

Interval Notation:

$(- 9, 9)$

Solve the Inequality for x x^2<81

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