Why can you not cancel-down the square within a square-root?

Question

$y = \sqrt{x²+5x+6.25} \iff y = \sqrt{(x+2.5)²} $

but $\iff y = x+2.5$ together with $y = -x-2.5$ is wrong.

Why? What are the underlying rules?

Answer 1

The convention is that the square root function, written either as $\sqrt{x}$ or $x^{0.5}$ takes only the positive value. So if you're trying to state that x is the solution to $x^2=4$, then you need to specify that $x=\pm \sqrt{4} = \pm 2$.

It's fair to say that the convention doesn't necessarily hold - for example, when talking about functions on the complex numbers (where it's hard to talk about "positive" and "negative" in the first place) then the square root function is indeed multiple valued, and if you only want to get one value out of it you need to specify how you're going to pick one over the other.

Answer 2

Who says that's wrong? If $$y=(x^2+5x+6.25)^{0.5}$$ then $$y=((x+2.5)^2)^{0.5}=\left|x+2.5\right|$$ So $y=x+2.5$ (if $x\geq{-2.5}$) or $y={-x}-2.5$ (if $x\leq{-2.5}$) . I suppose it depends on what you mean by "together with".