How to solve $\sqrt{x+2}\geq x$? 
How do you solve the inequality $$\sqrt{x+2}\geq{x}?$$

Now since ${x+2}$ is under the radical sign, it must be greater than or equal to ${0}$ to be defined.
So,
${x+2}\geq{0}$
Thus ${x}\geq{-2}$
Now keeping this in mind, we can solve the inequality by squaring both the sides:
${x+2}\geq{x^2}$
So ${x^2-x-2}\leq{0}$
Solving, ${(x-2)(x+1)}\leq{0}$
Therefore ${x}$ belongs to the interval ${[-1,2]}$.
As ${x}\geq{-2}$, the function is also defined.
Why does the answer say that ${x}$ belongs to ${[-2,2]}$, then?
Please feel free to point out the mistakes.
 A: Be careful : When you square, the inequality preserves its sign direction if both sides are positive.
Note that $\sqrt{x+2}$ is defined for $x \geq - 2$, so first you need to consider $x \geq 0$ and work as such :
$$\sqrt{x+2} \geq x \Rightarrow x+2 \geq x^2 \Leftrightarrow x^2-x-2 \leq 0 \Leftrightarrow (x-2)(x+1) \leq 0$$
This indeed yields $x \in [-1,2]$ if you also consider the negative values for which the derived inequality is satisfied . 
But if $x$ is negative $(-2 \leq x < 0)$, then the (positive) square root will always be bigger than the negative left-hand side. Thus, $[-2,0)$ will do the trick in that case.
Concluding : $\sqrt{x+2} \geq x \implies x \in [-2,2]$.
A: Clearly $x\geq -2$. 


*

*If $x<0$ then each $x\in [-2,0) $ is a solution (since negative number is always smaller than square root).

*Now if $x\geq 0$ then you can square it, so you get $$x^2-x-2 = (x-2)(x+1)\leq 0$$
So in this case every $x\in[0,2]$ is a solution. 


So finally, every $x\in [-2,2]$ is a solution.
A: Once you know $x \geqslant -2$, consider first $x \in [-2, 0)$.  The LHS is defined and non-negative, while the RHS is __________.
Next, consider the case $x \geqslant 0$, where you can freely square as you have done.  Here you should get $x \in [0, 2]$ as the solution.
Now the solution set is the union of these cases.
A: Let $a=\sqrt{x+2}\ge0$ for real $x$
We need  $$a\ge a^2-2\iff0\ge a^2-a-2=(a-2)(a+1)$$
$$\iff -1\le a\le2\ \ \ \ (1)$$
But we need to honor $a\ge0\ \ \ \ (2)$
Find the intersection of $(1),(2)$
A: Hint:
The inequation $\;\sqrt A\ge B$, on its domain (defined by the condition $A\ge 0$) is equivalent to
 $$A\ge B^2\quad\textbf{ or }\quad B\le 0.$$
A: Firstly, we will assume1 that $x$ is a real number since the field of complex numbers does not have an order with the usual addition and multiplication operations. For instance, if $i<2i$, then $0<i$ but squaring gives the contradiction2 $0<-1$. On the other hand, if $i>2i$, then $0>i$ or equivalently, $-i>0$ but squaring gives the same contradiction. This means that $\sqrt{x+2}\ge x$ does not naturally make sense whenever $\Im x\ne0$ or $\Im\sqrt{x+2}\ne0$. Note that there are other orderings such as $a\preceq b$ iff $\Re a\le\Re b$ and $\Im a\le\Im b$ but since they are incompatible with operations that govern a field, we will not pursue this here.3
Now that we have confidence that $x$ should be a real number, we can give our full attention to solving the inequality; that is, finding $x$.4 One approach at first glance is to square both sides of the inequality to get $x+2\ge x^2$, but in doing so, we have forgotten about what squaring means to the direction of an inequality. Since the direction is preserved only when both sides are non-negative, we have inadvertently omitted solutions when $x<0$.5 So we must split into two cases:

*

*$x+2\ge x^2$ and $x\ge0$;


*$\sqrt{x+2}\ge x$ and $x\le0$.
However, the second case is always true, since the square root must be non-negative as discussed in the first paragraph. Thus it only depends on the interval where $\sqrt{x+2}$ is defined, which is $x+2\ge0\iff x\ge-2$, and so we have the solution $x\in[-2,0]$ in this case.
The first case can be solved in multiple ways.6 For example, we can square both sides and rewrite the inequality as $x^2-x-2\le0$ from which we can complete the square to obtain $$(x-1/2)^2-1/4-2\le0\implies(x-1/2)^2\le9/4.$$ Taking square roots, we have $|x-1/2|\le3/2$ or $x\in[-1,2]$. And we recover the intended solution $x\in[-2,0]\cup[-1,2]=[-2,2]$.
Another approach is to use calculus. Let $f(x)=\sqrt{x+2}-x$ where $f$ is defined for all $x\ge-2$ due to the square root. Taking derivatives, we obtain $$f'(x)=(x+2)^{-1/2}/2-1$$ and stationary points occur when $f'(x^*)=0$. Solving this equation gives us $(x^*+2)^{-1/2}=1/2\implies x^*=-7/4$, which is the argmax since $$f''(-7/4)=-(-7/4+2)^{-3/2}/4<0.$$ Similarly, by solving $f'(x)>0$ and $f'(x)<0$, we find that $f$ is strictly increasing on $(-2,-7/4)$ and strictly decreasing on $(-7/4,+\infty)$. Since $f(-2)=2$ and $f(2)=0$ (by observation7), we can conclude that $f(x)\ge0$, and hence $\sqrt{x+2}\ge x$ only when $x\in[-2,2]$.

1 We love assuming things. Riemann hypothesis, anyone?
2 Citation needed. This is actually very far from easy to prove: indeed, for a very, very long time, humans did not understand what negative numbers were. So one can argue that the question of whether $0<-1$ is true was an open problem for millions of years.
3 Extension for interest: Quaternion square roots.
4 Here is a visual solution.
5 One way to eliminate this inconvenience is to assume that $x\ge0$.
6 Such as plugging it into a CAS or providing someone else with a financial incentive to do it for you. Just don't sue me after being busted for cheating... please.
7 This can be achieved by zooming in infinitely close to the intersection on the $x$-axis on Desmos. Or just assume that it is $2$. Or you can solve $\sqrt{x+2}=x$, but isn't that rather boring?
happy april fools y'all
A: You made a mistake squaring both sides without checking the sign first.  $1>-1$ but $1^2\not>(-1)^2$.
A: To square both sides of an equality/inequality and obtain an equivalent statement you must be certain that both sides have the same sign (in case of inequality and both sides negative you should swich the direction of the inequality though). In your case left side is always nonnegative, therefore for nonpositive $x$ the inequality is aleays fulfiled as the $LHS \ge 0 \ge RHS$. But since the inequality doesn't make sense for $x<-2$ only nonpositive numbers greater than it satysfiy this inequality.
A: $$\sqrt{x+2}\ge x$$
$$\implies \sqrt{x+2}\ge (x+2)-2$$
Let $\sqrt{x+2}=a$
$$\implies a\ge a^2-2$$
$$a^2-a-2\le0$$
Which gives $a\in [-1,2]$
thanks to @dxiv in the comments for pointing out, I was considering the lower limit to be -1 while squaring it which is wrong as explained in previous answers
Now $a$ is a square rooted term so its value cannot be negative, so
$$ 0 \le \sqrt{x+2} \le 2$$
$$ -2 \le x \le 2$$
Now, since $a$ is withing square root,
$$x+2 \ge 0 \implies x \ge -2$$
Combining the ranges for $x$ which we have got, we get
$$x \in [-2,2]$$
