When are equations considered equivalent? I know of this definition: "equivalent equations - equations that have the same set of solutions". Is that all, or do we need to add something about the domain?  If two equations have the same solution but different domain, can we say that these equations are equivalent?
For example:

*

*$(x-2)(x^2+1)=0$
Domain: $\mathbb R$
$x_0=2$


*$\sqrt{x-2}=0$
Domain: $[2,\infty)$
$x_0=2$
Both equations have the same solution $x_0=2;$ however, for $x = -10$, the second equation doesn't exist while the first equation is logical false.
 A: *

*The statements \begin{align}∀x{\in}\mathbb
    R\:\:\Big((x-2)(x^2+1)=0\iff
    x=2\Big)\\∀x{\in}[2,\infty)\:\:\Big((x-2)(x^2+1)=0\iff\sqrt{x-2}=0\Big)\tag0\end{align}
are both true, and the conditional
equations
$$(x-2)(x^2+1)=0\tag1$$ and $$\sqrt{x-2}=0\tag2$$ are equivalent to
each other on the interval $\mathbf{[2,\infty)}.$


*Technically, equations $(1)$ and $(2)$ and the biconditional
\begin{align}(x-2)(x^2+1)=0\longleftrightarrow\sqrt{x-2}=0\end{align}
do not have definite truth values, so are
predicates rather
than statements.


*In the context of $\mathbb R,$ since the above biconditional is
well-formed only on $[2,\infty),$ the equivalence of equations $(1)$
and $(2)$ is tacitly understood as being quantified on that
interval.
In other words, if the domain of discourse hasn't been specified, then $$(x-2)(x^2+1)=0\iff\sqrt{x-2}=0$$ should be considered an abridgement of statement $(0).$
A: "The same set of solutions" says it all.
$$(x-2)(x^2+1)=0\iff x=2$$
and
$$\sqrt{x-2}=0\iff x=2,$$
that's it.
The extent of the domain is implicit in the definition, as a solution is valid only if it lies in the domain.

The situation would be different if you had additional restrictions on the domain, like if you had to solve $f(x)=0$, where 
$$f(x):[10,\infty)\to\mathbb R:x\to(x-2)(x^2+1)$$ (no solution).
