Why is it that $\sqrt{4}$ is only $2$, not $\pm 2$?

Question

I understand that the numbers that, when squared, result in 4 are ±2 because both -2 and 2 squared result in four. However, when in a radical, why is it that $\sqrt 4$ is only 2? Why isn't the sign undetermined until it the number is taken out of the square root? Why is the ± needed in front of $\sqrt 4$ to denote that the solution set includes -2?

Answer 1

Because in many situations we want a function: given a single positive $x$, we want a single $\sqrt x$.

That's the reason why we have the convention that $\sqrt x$, for $x\geq0$, means the positive square root of $x$. It's not a big deal, since you can get the other one as $-\sqrt x$.

The convention does not extend beyond square roots of positive numbers. That's why it is (kind of) poor form to write $i=\sqrt{-1}$. There are two complex numbers whose square is $-1$: namely $i$ and $-i$. When one writes $\sqrt{-1}$, it is not clear which one to choose. So, when working with complex numbers, we use that $i^2=-1$, with no need for writing square roots at all.