<s>[INST] Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?
 [/INST] The root function $f(x) = \sqrt{x}$ is defined to be nonnegative. 
So $\sqrt{2} = 1.4...$, however the equation $x^2 - 4=0$ has to solutions:
If $x$ is such a solution, then so is $-x$ since $(-x)^2-4 = (-1)^2x^2-4 = x^4-4 = 0$
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