Q:

When is $\sqrt{a^2}=\pm a$ and when is $\sqrt{a^2}=a$?

When we derive some formula and have to do huge algebraic expansions that deal with raising powers we use exponent rules mindlessly and we never write down the $\pm$ symbol. Why is this right?
My reasoning is that we write $\pm$ when were looking for a set of solutions (e.g., $x^2=4\implies x=\pm2$) but not when we know the exact identity of $x$. If we know the exact identity of a symbol we can use exponent rules. For example, $2=\sqrt{2^2}$ seems more valid than $2=\sqrt{4}$ (since in the former case we're clearly undoing an operation on the number 2, which could've been a symbol like $a$), if that makes sense at all. I know its a stupid thing to say, but well, I'd like to see what you think.
One problem with the $\pm$ notation is that we won't be able to do things like $a\sqrt{4}=\sqrt{a^24}$ where tricks like these are done all the time.

A:

Actually, never. $\sqrt{b}$, by definition, is the nonnegative solution to the equation $x^2 = b$.
This means that, by definition, $\sqrt{a^2} = |a|$ for all real values of $a$.

A:

When you write $x^2=4\implies x=\pm2$, what exactly justifies that implication?
You start with two equal expressions: $x^2=4$.
You are allowed to apply the same function to both sides of the equation, and still have equality. So $\sqrt{x^2}=\sqrt{4}$.
Now here's where many students have been taught poorly. They've been taught that the $\sqrt{}$ "cancels" the ${}^2$, leaving $x$ on the left. And they've been taught that $\sqrt{4}=\pm2$. Both of these are incorrect. What really  should be the next line is $$|x|=2$$ because $\sqrt{4}$ is unambiguously $2$, and $\sqrt{x^2}$ is, for all real $x$, $|x|$.
So now that you know $|x|=2$, what must $x$ be? How many real numbers are there with absolute value $2$? Just $2$ and $-2$. So $x$ could be either one of these. And we are left with the sloppy summary $x=\pm2$.

