Solve for x 4(x+2)^2<=0

$4 {(x + 2)}^{2} \leq 0$

Step 1

Divide each term in $4 {(x + 2)}^{2} \leq 0$ by $4$ and simplify.

Divide each term in $4 {(x + 2)}^{2} \leq 0$ by $4$ .

$\frac{4 {(x + 2)}^{2}}{4} \leq \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 {(x + 2)}^{2}}{4} \leq \frac{0}{4}$

Divide ${(x + 2)}^{2}$ by $1$ .

${(x + 2)}^{2} \leq \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

${(x + 2)}^{2} \leq 0$

Step 2

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

$\sqrt{{(x + 2)}^{2}} \leq \sqrt{0}$

Step 3

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

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

Simplify the right side.

Simplify $\sqrt{0}$ .

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

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

Pull terms out from under the radical.

$| x + 2 | \leq | 0 |$

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

$| x + 2 | \leq 0$

Step 4

Write $| x + 2 | \leq 0$ as a piecewise.

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

$x + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

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

$x + 2 \leq 0$

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 $- 1$ .

$- (x + 2) \leq 0$

Write as a piecewise.

${\begin{cases} x + 2 \leq 0 & x \geq - 2 \\ - (x + 2) \leq 0 & x < - 2 \end{cases}$

Simplify $- (x + 2) \leq 0$ .

Apply the distributive property.

${\begin{cases} x + 2 \leq 0 & x \geq - 2 \\ - x - 1 \cdot 2 \leq 0 & x < - 2 \end{cases}$

Multiply $- 1$ by $2$ .

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

Step 5

Solve $x + 2 \leq 0$ when $x \geq - 2$ .

Subtract $2$ from both sides of the inequality.

$x \leq - 2$

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

$x = - 2$

Step 6

Solve $- x - 2 \leq 0$ when $x < - 2$ .

Solve $- x - 2 \leq 0$ for $x$ .

Add $2$ to both sides of the inequality.

$- x \leq 2$

Divide each term in $- x \leq 2$ by $- 1$ and simplify.

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $- 1$ .

$x \geq - 2$

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

No solution

Step 7

Find the union of the solutions.

$x = - 2$

Step 8

Solve for x 4(x+2)^2<=0

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