Graph yy>2

$y \cdot y > 2$

Multiply $y$ by $y$ .

$y^{2} > 2$

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

$\sqrt{y^{2}} > \sqrt{2}$

Simplify the left side.

Pull terms out from under the radical.

$| y | > \sqrt{2}$

Write $| y | > \sqrt{2}$ as a piecewise.

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

$y \geq 0$

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

$y > \sqrt{2}$

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

$y < 0$

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

$- y > \sqrt{2}$

Write as a piecewise.

${\begin{cases} y > \sqrt{2} & y \geq 0 \\ - y > \sqrt{2} & y < 0 \end{cases}$

Find the intersection of $y > \sqrt{2}$ and $y \geq 0$ .

$y > \sqrt{2}$

Divide each term in $- y > \sqrt{2}$ by $- 1$ and simplify.

Divide each term in $- y > \sqrt{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} < \frac{\sqrt{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} < \frac{\sqrt{2}}{- 1}$

Divide $y$ by $1$ .

$y < \frac{\sqrt{2}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{2}}{- 1}$ .

$y < - 1 \cdot \sqrt{2}$

Rewrite $- 1 \cdot \sqrt{2}$ as $- \sqrt{2}$ .

$y < - \sqrt{2}$

Find the union of the solutions.

$y < - \sqrt{2}$ or $y > \sqrt{2}$

Graph yy>2

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