algebra precalculus - Why is $sqrt{x} =- 1$ extraneous?

An equation is extraneous if (at least to my understanding) it has no valid solutions. The example my math teacher used was $\sqrt{x}=-1$, citing this proof

$$\sqrt{x}=-1 \\ x=1$$

They then stated that $\sqrt{1} \ne -1$, and therefor the equation is extraneous. While this wasn't initially a problem, I seemed to accept it for some reason, I now realize that $\sqrt{1} = \pm 1$, and therefor -1 is a square root of 1, so why is the initial equation extraneous?

Am I missing something major here or was my math teacher wrong?

Answer

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.