Solve the Inequality for x -x^2+9>0

$- x^{2} + 9 > 0$

Step 1

Subtract $9$ from both sides of the inequality.

$- x^{2} > - 9$

Step 2

Divide each term in $- x^{2} > - 9$ by $- 1$ and simplify.

Divide each term in $- x^{2} > - 9$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} < \frac{- 9}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} < \frac{- 9}{- 1}$

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

$x^{2} < \frac{- 9}{- 1}$

Simplify the right side.

Divide $- 9$ by $- 1$ .

$x^{2} < 9$

Step 3

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

$\sqrt{x^{2}} < \sqrt{9}$

Step 4

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | < \sqrt{9}$

Simplify the right side.

Simplify $\sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$| x | < \sqrt{3^{2}}$

Pull terms out from under the radical.

$| x | < | 3 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | < 3$

Step 5

Write $| x | < 3$ 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.

$x < 3$

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 $- 1$ .

$- x < 3$

Write as a piecewise.

${\begin{cases} x < 3 & x \geq 0 \\ - x < 3 & x < 0 \end{cases}$

Step 6

Find the intersection of $x < 3$ and $x \geq 0$ .

$0 \leq x < 3$

Step 7

Solve $- x < 3$ when $x < 0$ .

Divide each term in $- x < 3$ by $- 1$ and simplify.

Divide each term in $- x < 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{3}{- 1}$

Divide $x$ by $1$ .

$x > \frac{3}{- 1}$

Simplify the right side.

Divide $3$ by $- 1$ .

$x > - 3$

Find the intersection of $x > - 3$ and $x < 0$ .

$- 3 < x < 0$

Step 8

Find the union of the solutions.

$- 3 < x < 3$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$- 3 < x < 3$

Interval Notation:

$(- 3, 3)$

Step 10

Solve the Inequality for x -x^2+9>0

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