Background
I'm not completely sure if these rules can be applied outside $\mathbb{R}$, but let's assume real numbers only for the sake of simplicity.
We have Exponent Properties, which then can be extended to Fractional Exponents.
Then we have this and this statements, which declare that:
$$_{(1)}\sqrt{(x ^ 2)} = |x|$$ $$_{(2)}{({\sqrt{x}}) ^ 2} = x$$
Problem
Assuming all of the above is correct, then any of the following procedures would be valid to counter-argument aforementioned statements: By following exponent properties, and/or then finding a connection between above both (seemingly exclusive) sentences, we obtain:
$$_{(3)}\sqrt{(x ^ 2)} = {({x ^ 2}) ^ \frac{1}{2}} = {x ^ {2\frac{1}{2}}} = {x ^ \frac{2}{2}} = {x ^ 1} = x$$ $$_{(4)}\sqrt{(x ^ 2)} = \sqrt{(xx)} = \sqrt{x}\sqrt{x} = {(\sqrt{x}) ^ 2} = x |_{by (2)}$$
Personally, I find (4) to be more interesting since it clearly shows a connection between both cases, becoming valid to counter-argument (1) and (2) simultaneously, by following the chain of operations from either side.
On the other hand, I have the slight feeling that, since $\sqrt{x}$ can be both positive and negative (thus becoming $\pm\sqrt{x}$), maybe that has something to do with my point of view, yet one of my referenced sources explicitly state that $\sqrt{x}$, by convention, only picks the positive root; possibly I'm overlooking something else.
Given all of the above, the question is:
What's the true value of $\sqrt{(x ^ 2)}$ and ${({\sqrt{x}}) ^ 2}$
Edits
#0
Ironically enough, even before posting my question, reading through the info of fake-proofs while tagging this answer of mine, I found a piece of valuable information to the question itself, specifically:
An example of a fake proof is $$1=\sqrt{-1\cdot-1}=\sqrt{-1}\sqrt{-1}=i^2=-1$$ which fails because $\sqrt{xy}=\sqrt x\sqrt y$ does not hold if $x$ or $y$ is negative.
However I will proceed by posting my question anyways: That way, It'll allow for more insight to be provided by others for future reference.