TITLE: Why does squaring $\sqrt{x+2}\ge x$ misses an interval?
QUESTION [6 upvotes]: If I square both sides of $\sqrt{x+2}\ge x$ I get,
$$x^2-x-2\le0\quad\Rightarrow\quad (x-2)(x+1)\le0\quad\Rightarrow\quad x\in[-1,2]$$
But the interval $[-2,-1)$ should be included in the solution which is missed here. I'm wondering what's going wrong in the above approach that misses an interval?

REPLY [3 votes]: Perhaps all has been said already.
0)$\sqrt{2+x} \ge 0$ for $x\ge - 2;$
1)$\sqrt{2+x} \ge x$ for $x \in [-2,0];$
3)Consider $x>0$
We have for $a \ge b >0:$
$a^2\ge b^2$ $\iff$ $a \ge b$ (reasoning?).
Square the inequality where both sides are positive.
$2+x \ge x^2;$
$0 \ge (x^2-x-2);$
$0 \ge (x-2)(x+1);$
$x \in [-1,2];$
Recall we consider x>0, hence
$x \in (0,2]$
(formally $[-1,2]\cap (0,\infty)$).
4)Putting together 2) and 3):
$x \in [-2,0]\cup (0,2]=[-2,2],$ and we are done.