How to solve $\sqrt{x+2}\geq x$?

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.

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

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.