Squaring left part of equation is an equivalent transformation if the right part is zero and I limit variable to its original domain?

Question

Is equation $\sqrt{f(x)}=0$ exactly eqivalent to a system of an equation $f(x)=0$ and inequality $f(x)\geqslant 0$ - meaning only those x solutions (of $f(x)=0$) qualify which being substituted into f(x) make it non-negative?

Answer 1

Yes, it is, though that inequality immediately becomes redundant the moment you equate $f(x)$ to a square. The important inequality is the one on the RHS, namely $$\sqrt{A}= B\iff\begin{cases} A=B^2\\ B\ge 0\end{cases}$$

$B\ge 0$, in your case, is just the tautology $0\ge 0$.

For inequations \begin{align}\sqrt A\ge B&\iff \begin{cases}A\ge 0\\ B\le 0\end{cases}\lor \begin{cases}A\ge B^2\\ B\ge 0\end{cases}\\\sqrt A\le B&\iff\begin{cases}B\ge 0\\ A\le B^2\\ A\ge 0\end{cases}\end{align}