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I asked a similar question yesterday, but did not get a satisfactory response. So let me restate $$\frac {\sqrt x}{2}=-1$$ $$\implies\sqrt x=-2$$ We can square with the restriction $\sqrt x<0$ $$x=4,\sqrt x<0$$And indeed we see that $\sqrt 4 =\pm 2$ and given the condition $\sqrt x<0$ we get $-2/2=-1$. One possible explanation might be we only take the positive square root, to maintain square root as a function. In this case my question is why cannot square root be a function in this case(when $-ve$ values are included)? Another explanation might be that for $x$ to be a solution, then both positive and negative values must equal $-1$ when divided by $+2$. Is this a issue with terminology?

P.S. The question arose from @eisenstein's comments on @lonestudent's solution to my previous post.

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    From https://math.stackexchange.com/a/1618736/42969 (of which your previous question was closed as a duplicate): “Remember $\sqrt x$, when it is defined, denotes the non-negative square root of $x$” – So $x=4,\sqrt x<0$ has no solutions in the real number, and $\sqrt 4 = \pm 2$ is wrong. – Martin R Mar 14 '21 at 06:14
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    @MartinR I know that, and hence I added the terminology tag. But my question is not a duplicate of the aforementioned posts, since I ask more than what those questions ask. –  Mar 14 '21 at 06:19

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