Solve the Inequality for x (x+2)^2<25

${(x + 2)}^{2} < 25$

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

$\sqrt{{(x + 2)}^{2}} < \sqrt{25}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x + 2 | < \sqrt{25}$

Simplify the right side.

Simplify $\sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$| x + 2 | < \sqrt{5^{2}}$

Pull terms out from under the radical.

$| x + 2 | < | 5 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$| x + 2 | < 5$

Write $| x + 2 | < 5$ as a piecewise.

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

$x + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

In the piece where $x + 2$ is non-negative, remove the absolute value.

$x + 2 < 5$

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

$x + 2 < 0$

Subtract $2$ from both sides of the inequality.

$x < - 2$

In the piece where $x + 2$ is negative, remove the absolute value and multiply by $- 1$ .

$- (x + 2) < 5$

Write as a piecewise.

${\begin{cases} x + 2 < 5 & x \geq - 2 \\ - (x + 2) < 5 & x < - 2 \end{cases}$

Simplify $- (x + 2) < 5$ .

Apply the distributive property.

${\begin{cases} x + 2 < 5 & x \geq - 2 \\ - x - 1 \cdot 2 < 5 & x < - 2 \end{cases}$

Multiply $- 1$ by $2$ .

${\begin{cases} x + 2 < 5 & x \geq - 2 \\ - x - 2 < 5 & x < - 2 \end{cases}$

Solve $x + 2 < 5$ when $x \geq - 2$ .

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

Subtract $2$ from both sides of the inequality.

$x < 5 - 2$

Subtract $2$ from $5$ .

$x < 3$

Find the intersection of $x < 3$ and $x \geq - 2$ .

$- 2 \leq x < 3$

Solve $- x - 2 < 5$ when $x < - 2$ .

Solve $- x - 2 < 5$ for $x$ .

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

Add $2$ to both sides of the inequality.

$- x < 5 + 2$

Add $5$ and $2$ .

$- x < 7$

Divide each term in $- x < 7$ by $- 1$ and simplify.

Divide each term in $- x < 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{7}{- 1}$

Divide $x$ by $1$ .

$x > \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$x > - 7$

Find the intersection of $x > - 7$ and $x < - 2$ .

$- 7 < x < - 2$

Find the union of the solutions.

$- 7 < x < 3$

The result can be shown in multiple forms.

Inequality Form:

$- 7 < x < 3$

Interval Notation:

$(- 7, 3)$

Solve the Inequality for x (x+2)^2<25

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