What is the algebra involved in solving the inequality $\sqrt x\le 2$

Question

I would like to know how one would solve $\sqrt x\le 2$ algebraically.

• How do you get rid of the radical sign?
• Do you square both sides? Why is this allowed to do in an inequality?

I already have the answer $[0,4]$, however I do not understand how this was determined.

Answer 1

First of all, $x\ge 0$ is needed. Then, since both sides are non-negative, we have$$\sqrt x\le 2\iff (\sqrt x)^2\le 2^2\iff x\le 4.$$ So, the answer is $0\le x\le 4$.