How do I solve quadratic double inequalities? I have two questions involving quadratic double inequalities. 
Firstly, what are the steps to get the solution for the following?
$0\le(x+2)^2\le4$ 
My my thought was to separate the inequality into 
$0\le(x+2)^2$ and $(x+2)^2\le4$
Which would then allow me to take the square root of each side of the and. 
$0\le x+2$ and $0\ge x+2$ and $x+2 \le 2$ and $x+2 \ge -2$
Which could be simplified to $x+2 \le 2$ and $x+2 \ge -2$
And combined into
$-2 \le x+2 \le 2$
Is this the proper way to think about/solve this problem? Is there a better way to approach it?
Secondly, what strategy could I use to square both sides of 
$-2\le x+2\le2$ to get back $0\le(x+2)^2\le4$ 
Thanks!
 A: For real $x,$
$$(x+2)^2\ge0$$
So, the problem reduces to $$(x+2)^2\le4\iff x(x+4)\le0$$
$\implies$
either
$x\ge0$ and $x+4\le0\iff 0\le x\le-4$ which is impossible
or $x\le0$ and $x+4\ge0\implies -4\le x\le0 $
A: $$0 \le (x+2)^2 \le 4$$ implies $$-2 \le x+2 \le 2.$$  Therefore, $$-4 \le x \le 0.$$  I don't see why it needs to be any more complicated than that.
A: Option:
1) Set $y=x+2$;
$0 \le y^2 \le 4.$
$f(y)=√y$ is an increasing function:
$0\le \sqrt{y^2} \le 2$;
With $ \sqrt{y^2}=|y| $ we get:
$0\le |y| \le 2$;
$-2 \le y\le 2$, or $-2 \le x+2 \le 2$.
2) Rewrite:
$-2 \le y \le 2$ as  $|y| \le 2$.
Note $|y| \ge 0$.
Hence:
$|y| \le 2 $ implies
$ |y||y|\le 2 |y| \le 2\cdot 2$, .
$ y^2 \le 4.$
A: As $0 \leq (x+2)^2$, we know that any real number squared is always positive so this can be ignored.
As $4 \geq (x+2)^2$, then over here $x+2$ is between $-2$ and $2$, so $x$ is between $-4$ and $0$, respectively. 
$-4 \leq x \leq 0$.
A: You can say $\sqrt{(x+2)^2} = |x+2|$
$|x+2|\le2$
It is also trivial that $0\le (x+2)^2$
