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

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.

• Of course I am stuck at this point. I am finding the domain for the $sqrt/2-sqrtx$ – user137452 Sep 27 '14 at 20:45
• Just to clarify, are you trying to solve for $x$ in $\sqrt{x}\leq 2$? – Sujaan Kunalan Sep 27 '14 at 20:46
• yes, that is what I am looking for. I already have the answer [0,4] however, I do not understand how the domain was determined. – user137452 Sep 27 '14 at 20:48
• "Domain" is ordinarily a property of a function. The "domain" of the (real) square root function is nonnegative real numbers. You seem to be trying to combine the domain of the square root function with the satisfaction of the inequality, and with that in mind it's pretty obvious how you'd get to interval $[0,4]$. – hardmath Sep 28 '14 at 2:42

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 $0\le x\le 4$.
• Yes, both sides were squared. We can do this since both sides are non-negative. Think of how $\sqrt{4}<\sqrt{9}\implies 4<9$, but if a side is negative, we can no longer square both sides: $-3<2$ does not mean that $(-3)^2<2^2$, since $9<4$ is not true. – Sujaan Kunalan Sep 27 '14 at 20:54