Over the past one or two years, at least two different teachers have told my children that $\sqrt 4$ is $2$ or $-2.$ I don't think this is useful, but if you want to define $\sqrt x$ as the set of solutions of $y^2 = x$, I guess you could. Of course you get into all kinds of confusion with very common constructions and notation, for example the graph of $x\mapsto \sqrt x$ (should we write $\max\sqrt x$?), Pythagoras' theorem or the root formula for quadratic equations, but let's say we accept that, and the common notation is just sloppy. In fact, the teacher is consistent, and says that the root formula should be written
$$x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$$
rather than
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
Now the teacher went one step further, saying that
$$x + \sqrt{7 - 3x} = 1$$
has two solutions: $x = -3$ (obviously), but also $x = 2$. This is found by writing
$$\sqrt{7 - 3x} = 1 - x$$
and squaring. Since it is assumed $\sqrt{7 - 3\cdot 3} = \pm 1$, we select the $-1$, and in this way this also is a solution to the original equation.
This seems obviously wrong to me, but I'm not sure how obvious that really is. If we really define the square root as a set of values (of 0, 1 or 2 elements) rather than as a (single-valued) function on non-negative real numbers, we could say that the solutions of an equation involving square roots are those values of $x$ so that from each of these sets we can select a value satisfying the equation. That would have strange consequences like
$$(1 + \sqrt x)^2 = -3$$
having the solution $x = 4$. Also you cannot factorize in terms of square roots anymore: $a - b$ is a single value, but $(\sqrt a + \sqrt b)(\sqrt a - \sqrt b)$ can have 8 different values.
Of course these last two are less defensible than the original example, for the teacher's example at least for each of the solutions a branch of the square root can be defined for which it is a solution, but they are not taught that you have to fix whether you want $\sqrt 4$ to be $2$ or $-2,$ but rather that it is both.
After this long introduction, my question: could it be reasonably defended that $x = 2$ is a valid solution of $x + \sqrt{7 - 3x} = 1$ alongside $x = -3$? In other words, if at an exam a student only produces $x = -3$, could they protest if full points are not awarded, even though the teacher taught them that both are solutions? After all, even though mathematical truth is non-negotiable, definitions and conventions are, and to a certain extent the teacher is free to fix these.