Solve the Inequality for x -x^2+9>=9

$- x^{2} + 9 \geq 9$

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

Subtract $9$ from both sides of the inequality.

$- x^{2} \geq 9 - 9$

Subtract $9$ from $9$ .

$- x^{2} \geq 0$

Divide each term in $- x^{2} \geq 0$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{0}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$x^{2} \leq 0$

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

$\sqrt{x^{2}} \leq \sqrt{0}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

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

Simplify the right side.

Simplify $\sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$| x | \leq \sqrt{0^{2}}$

Pull terms out from under the radical.

$| x | \leq | 0 |$

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

$| x | \leq 0$

Write $| x | \leq 0$ 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.

$x \leq 0$

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 $- 1$ .

$- x \leq 0$

Write as a piecewise.

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

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

$x = 0$

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

Divide each term in $- x \leq 0$ by $- 1$ and simplify.

Divide each term in $- x \leq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{0}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$x \geq 0$

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

No solution

Find the union of the solutions.

$x = 0$

Solve the Inequality for x -x^2+9>=9

Related Questions

Support LCMGCF.com by disabling your adblocker.