We talk about an extraneous or spurious solution when we solve an algebraic equation by raising both sides to some power. The map $x\mapsto x^n$ is not injective, hence once the original problem has been reduced to finding the roots of some polynomial, it is not granted that every root of such polynomial is indeed a solution of the original equation. For instance
$$ \sqrt{x-1} = 7-x \tag{1}$$
implies
$$ x-1 = (x-7)^2 \tag{2}$$
and
$$ p(x) = (x-7)^2-(x-1) = x^2-15x+50 = 0 \tag{3} $$
but while $x=5$ is an actual solution of $(1)$, the other root of $p(x)$, i.e. $x=10$, is a spurious solution, because it fulfills $(2)$ but not $(1)$.
It is enough to recall that the very definition of the square root function over the real numbers:
Def. $\sqrt{x}$ is the only non-negative real number whose square equals $x$.
In particular the maximal domain of the square root function is the set of non-negative real numbers, and over such set the square root function is non-negative. So $\sqrt{1}=1$, not $\pm 1$.
Over the set of complex numbers, for any $z\neq 0$ there are two opposite numbers whose squares equal $z$: in such context we write $\sqrt{z}=\pm w$ by meaning that both $w$ and $-w$ are roots of the polynomial $q(t)=t^2-z$, i.e. we regard $\sqrt{\cdot}$ as a multi-valued function: not a function, strictly speaking.