Solve the Inequality for x x^2>=16

$x^{2} \geq 16$

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

$\sqrt{x^{2}} \geq \sqrt{16}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{16}$

Simplify the right side.

Simplify $\sqrt{16}$ .

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

$| x | \geq \sqrt{4^{2}}$

Pull terms out from under the radical.

$| x | \geq | 4 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$| x | \geq 4$

Write $| x | \geq 4$ 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.

$x \geq 4$

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 $- 1$ .

$- x \geq 4$

Write as a piecewise.

${\begin{cases} x \geq 4 & x \geq 0 \\ - x \geq 4 & x < 0 \end{cases}$

Find the intersection of $x \geq 4$ and $x \geq 0$ .

$x \geq 4$

Divide each term in $- x \geq 4$ by $- 1$ and simplify.

Divide each term in $- x \geq 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{4}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{4}{- 1}$

Simplify the right side.

Divide $4$ by $- 1$ .

$x \leq - 4$

Find the union of the solutions.

$x \leq - 4$ or $x \geq 4$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 4 or x \geq 4$

Interval Notation:

$(- \infty, - 4] \cup [4, \infty)$

Solve the Inequality for x x^2>=16

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