Confusion on finding $y$ in $y^2=x^2$ If $y^2=x^2$, then $y=\pm x$ or $y=\pm \mid x\mid$, I am really confused!!!
I have a feeling that both of them are true. But intuition is not enough I guess. Need justification.
 A: I often find it helpful to rewrite equations like $$y^2=x^2$$ as $$y^2-x^2=(y+x)(y-x)=0$$
Then one of the two factors must be zero i.e. $y+x=0$ or $y-x=0$ that is $y=-x$ or $y=x$, which we express as $y=\pm x$
If instead we take the square root directly with the convention that the principal value of the square root is positive we get $y=|x|$. Of course we know that there is also a negative solution $y=-|x|$, and if we want both we put $y=\pm |x|$.
These two ways of addressing the problem give two different expressions for the answer, but the answers are two different ways of expressing the same thing. If $x\gt 0$ we have $x=|x|, -x=-|x|$ and if $x\lt 0$ it is $x=-|x|, -x=|x|$. 
If $y=x=0$ we have just one value. But that special case clearly doesn't cause a problem.
A: hint
both are true since 
$$y^2=x^2=|x|^2$$
and
$$y^2=x^2\implies (y-x)(y+x)=0$$
and $$(y+|x|)(y-|x|)=0$$
thus
$y=\pm x$ or $y=\pm |x|$.
A: $0=y^2-x^2=(y-x)(y+x)$ implies that $y-x=0$ or $y+x=0$ or equivalently that $y=x$ or $y=-x$. 
You could write the solution as: $$y\in\{-x,x\}\text{ or abbreviated }y=\pm x$$ 
Also we have $\{-x,x\}=\{-|x|,|x|\}$ so another way of writing the solution is: $$y\in\{-|x|,|x|\}\text{ or abbreviated }y=\pm |x|$$
A: If $x^2=y^2$ thus $(x-y)(x+y)=0$ so $x=\pm y$
Jet the second equation has the same solution (draw a graph). 
