Convert to Set Notation x^2-5<=0

$x^{2} - 5 \leq 0$

Solve $- \sqrt{5} \leq x \leq \sqrt{5}$ .

Add $5$ to both sides of the inequality.

$x^{2} \leq 5$

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

$\sqrt{x^{2}} \leq \sqrt{5}$

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{5}$

Write $| x | \leq \sqrt{5}$ 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.

$x \leq \sqrt{5}$

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 $- 1$ .

$- x \leq \sqrt{5}$

Write as a piecewise.

${\begin{cases} x \leq \sqrt{5} & x \geq 0 \\ - x \leq \sqrt{5} & x < 0 \end{cases}$

Find the intersection of $x \leq \sqrt{5}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{5}$

Solve $- x \leq \sqrt{5}$ when $x < 0$ .

Divide each term in $- x \leq \sqrt{5}$ by $- 1$ and simplify.

Divide each term in $- x \leq \sqrt{5}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{5}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{5}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{5}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{5}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{5}$

Rewrite $- 1 \cdot \sqrt{5}$ as $- \sqrt{5}$ .

$x \geq - \sqrt{5}$

Find the intersection of $x \geq - \sqrt{5}$ and $x < 0$ .

$- \sqrt{5} \leq x < 0$

Find the union of the solutions.

$- \sqrt{5} \leq x \leq \sqrt{5}$

Use the inequality $- \sqrt{5} \leq x \leq \sqrt{5}$ to build the set notation.

${x | - \sqrt{5} \leq x \leq \sqrt{5}}$

Convert to Set Notation x^2-5<=0

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