Solve for x x^2<=81

$x^{2} \leq 81$

Step 1

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

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

Step 2

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{81}$

Simplify the right side.

Simplify $\sqrt{81}$ .

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

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

Pull terms out from under the radical.

$| x | \leq | 9 |$

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

$| x | \leq 9$

Step 3

Write $| x | \leq 9$ 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.

$x \leq 9$

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 $- 1$ .

$- x \leq 9$

Write as a piecewise.

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

Step 4

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

$0 \leq x \leq 9$

Step 5

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

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $9$ by $- 1$ .

$x \geq - 9$

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

$- 9 \leq x < 0$

Step 6

Find the union of the solutions.

$- 9 \leq x \leq 9$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$- 9 \leq x \leq 9$

Interval Notation:

$[- 9, 9]$

Step 8

Solve for x x^2<=81

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