Solve (x+4)^2leq16

$$( x + 4 ) ^ { 2 } \leq 16$$

Solve for x

$$x\in \begin{bmatrix}-8,0\end{bmatrix}$$

Solution Steps

Factor out x.

$$x\left(x+8\right)\leq 0$$

For the product to be ≤0, one of the values x+8 and x has to be ≥0 and the other has to be ≤0. Consider the case when x+8\geq 0 and x\leq 0.

$$x+8\geq 0 x\leq 0$$

The solution satisfying both inequalities is x\in \left[-8,0\right].

$$x\in \begin{bmatrix}-8,0\end{bmatrix}$$

Consider the case when x+8\leq 0 and x\geq 0.

$$x\geq 0 x+8\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-8,0\end{bmatrix}$$