Solve for x x^2-1>=1

$x^{2} - 1 \geq 1$

Move all terms not containing $x$ to the right side of the inequality.

Add $1$ to both sides of the inequality.

$x^{2} \geq 1 + 1$

Add $1$ and $1$ .

$x^{2} \geq 2$

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

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

Simplify the left side.

Pull terms out from under the radical.

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

Write $| x | \geq \sqrt{2}$ 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.

$x \geq \sqrt{2}$

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 $- 1$ .

$- x \geq \sqrt{2}$

Write as a piecewise.

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

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

$x \geq \sqrt{2}$

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

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

$\frac{- x}{- 1} \leq \frac{\sqrt{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{\sqrt{2}}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{\sqrt{2}}{- 1}$

Simplify the right side.

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

$x \leq - 1 \cdot \sqrt{2}$

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

$x \leq - \sqrt{2}$

Find the union of the solutions.

$x \leq - \sqrt{2}$ or $x \geq \sqrt{2}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \sqrt{2} or x \geq \sqrt{2}$

Interval Notation:

$(- \infty, - \sqrt{2}] \cup [\sqrt{2}, \infty)$

Solve for x x^2-1>=1

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