Does this equality hold: $\frac{\sqrt{2|x|-x^2}-0}x = \sqrt{\pm\frac2x-1}$?

Question

Is the following proposition true for real numbers?

$$\frac{\sqrt{2|x|-x^2}-0}{x}=\begin{cases}\sqrt{\frac{2}{x}-1}&,\;\;\;x>0\\{}\\\sqrt{-\frac{2}{x}-1}&,\;\;\;x<0\end{cases}$$

I claim not, since if $x=-1$ the left hand side is $-1$, but the right hand side is $1$.

Answer 1

When you do the step $$\frac{\sqrt{\cdots}}x=\sqrt{\frac{\cdots}{x^2}}$$

You are losing the sign on the $x$. This means that if $x$ was negative or positive the result is the same, which is wrong.

Your counterexample is fine, and julien's remark is exactly the meaning of the point I am trying to make.

Remember that $x=\sqrt{x^2}$ if and only if $x$ is positive, that is $\sqrt{x^2}=|x|$ rather than just $x$. And for negative numbers this could be a problem.

Answer 2

For every $x\neq 0$ $$ \sqrt{2|x|-x^2}=\sqrt{x^2}\sqrt{2\frac{|x|}{x^2}-1}=|x|\sqrt{2\frac{|x|}{x^2}-1}. $$ Now treat the separate cases: $|x|=x$ if $x>0$ and $|x|=-x$ if $x<0$.