Why does squaring $\sqrt{x+2}\ge x$ misses an interval? 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?
 A: Note that $\sqrt{x+2} \geq 0$ for all $x\in \mathbb R$ such that the LHS is well defined. Now, notice that for $x < 0$ we automatically have $$\sqrt{x+2} \geq 0 > x$$ whenever the LHS is well defined. So, all negative $x$'s such that LHS is well defined are part of the solution of the inequality. Since we are dealing with square root, $x+2$ needs to be nonnegative, i.e. the LHS is well defined for $x\geq -2$. We conclude that $$x\in [-2,0) \implies \sqrt{x+2} > x.$$
We are done with the case when $x < 0$, so let us assume that $x\geq 0$. Since both $\sqrt{x+2}$ and $x$ are nonnegative and squaring is strictly increasing function on $[0,+\infty)$, we have $$\sqrt{x+2} \geq x \iff \sqrt{x+2}^2 \geq x^2,\quad x\geq 0.$$ Your calculation then shows that it is necessary that $x\in [-1,2]$. However, since we assumed at the start of the case that $x\geq 0$, we only take the nonnegative part of $[-1,2]$, i.e. the solution of the inequality in this case is $[0,2]$.
To finish, we take the union of the solutions for both cases, i.e. the solution is $$[-2,0)\cup [0,2] = [-2,2].$$

In general, if you want to solve inequality $\displaystyle\sqrt{f(x)} \geq g(x)$, these are the steps:

*

*Determine when $f\geq 0$ and denote that set with $\mathcal D_{f\geq 0}$. In your example $f(x) = x+2$, so $\mathcal D_{f\geq 0} = [-2,+\infty)$.

*Determine when $g\geq 0$ and $g < 0$ and denote those sets with $\mathcal D_{g\geq 0}$ and $\mathcal D_{g < 0}$, respectively. In your example $g(x) = x$, so $\mathcal D_{g\geq 0} = [0,+\infty)$ and $\mathcal D_{g < 0} = (-\infty, 0)$.

*Split into two cases, the first being $x \in \mathcal D_{f\geq 0}\cap \mathcal D_{g\geq 0}$ and the second being $x\in \mathcal D_{f\geq 0}\cap \mathcal D_{g < 0}.$ In your example $\mathcal D_{f\geq 0}\cap \mathcal D_{g\geq 0} =  [0,+\infty)$ and $\mathcal D_{f\geq 0}\cap \mathcal D_{g < 0} = [-2,0)$.

*Solve the first case by squaring, since both sides are nonnegative, i.e.
$$ \sqrt{f(x)} \geq g(x) \iff f(x) \geq g(x)^2,\quad x\in \mathcal D_{f\geq 0}\cap \mathcal D_{g\geq 0}.$$ When you get solution, don't forget to restrict it only to $x$'s belonging to this case. (Just like we restricted $[-1,2]$ to $[0,2]$.) Denote the set of solutions as $\mathcal S_1$. In your example $\mathcal S_1 =  [0,2]$.

*The second case is automatic since for $x\in \mathcal D_{f\geq 0}\cap \mathcal D_{g< 0}$ we have $$\sqrt{f(x)} \geq 0 > g(x).$$ So, all $x$'s in this case are part of the solution, so $\mathcal S_2 = \mathcal D_{f\geq 0}\cap \mathcal D_{g< 0}$. In your example $\mathcal S_2 = [-2,0)$.

*Take the union $\mathcal S_1 \cup \mathcal S_2$.

The reason why we have to split into cases is that squaring is not strictly increasing on whole $\mathbb R$, but only for nonnegative reals. This is important to remember in general, $a > b \iff a^2 > b^2$ works only for $a,b \geq 0$, and otherwise one needs to consider cases before squaring.
A: 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.
A: I think you miss something here. when $x<0$ is the part of the solution because $$\sqrt {x+2}\geq 0 , x\leq0 $$  for clarifying try to second power of both sides $$0.1 >-0.5$$ so
$$(-\infty,o] \cap [-2,\infty) =[-2,0]$$ is the part of the  solution.
hope it can helps you
