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.
Best Answer
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 : $\sqrt{x+2} \geq x \implies x \in [-2,2]$.