Solve for d |d+2|>=6

$| d + 2 | \geq 6$

Step 1

Write $| d + 2 | \geq 6$ as a piecewise.

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

$d + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$d \geq - 2$

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

$d + 2 \geq 6$

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

$d + 2 < 0$

Subtract $2$ from both sides of the inequality.

$d < - 2$

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

$- (d + 2) \geq 6$

Write as a piecewise.

${\begin{cases} d + 2 \geq 6 & d \geq - 2 \\ - (d + 2) \geq 6 & d < - 2 \end{cases}$

Simplify $- (d + 2) \geq 6$ .

Apply the distributive property.

${\begin{cases} d + 2 \geq 6 & d \geq - 2 \\ - d - 1 \cdot 2 \geq 6 & d < - 2 \end{cases}$

Multiply $- 1$ by $2$ .

${\begin{cases} d + 2 \geq 6 & d \geq - 2 \\ - d - 2 \geq 6 & d < - 2 \end{cases}$

Step 2

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

Subtract $2$ from both sides of the inequality.

$d \geq 6 - 2$

Subtract $2$ from $6$ .

$d \geq 4$

Step 3

Solve $- d - 2 \geq 6$ for $d$ .

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

Add $2$ to both sides of the inequality.

$- d \geq 6 + 2$

Add $6$ and $2$ .

$- d \geq 8$

Divide each term in $- d \geq 8$ by $- 1$ and simplify.

Divide each term in $- d \geq 8$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- d}{- 1} \leq \frac{8}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{d}{1} \leq \frac{8}{- 1}$

Divide $d$ by $1$ .

$d \leq \frac{8}{- 1}$

Simplify the right side.

Divide $8$ by $- 1$ .

$d \leq - 8$

Step 4

Find the union of the solutions.

$d \leq - 8$ or $d \geq 4$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$d \leq - 8 or d \geq 4$

Interval Notation:

$(- \infty, - 8] \cup [4, \infty)$

Step 6

Solve for d |d+2|>=6

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