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!
Best Answer
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 $