Forcing an absolute value of x after a square root operation Given the following two equations:
$$
f(x) = x \\
g^2(x) = 2x
$$
I need to find the $(x,y)$ coordinates for when they meet.
So after performing the square root operation, we have:
$$
f(x) = x \\
g(x) = \pm\sqrt{2\lvert x\rvert}
$$
So when trying to find the coordinates we get the following:
$$
f(x) = g(x) \\
x = \pm\sqrt{2\lvert x\rvert} \\
x \mp\sqrt{2\lvert x\rvert} = 0 \\
\sqrt{\lvert x\rvert} \cdot (\sqrt{\lvert x\rvert} \mp \sqrt{2} ) = 0 \\
\Rightarrow \sqrt{\lvert x\rvert} = \pm\sqrt{2} \quad\quad /(\quad)^2 \\
\Rightarrow \lvert x\rvert = 2 \\
\Rightarrow x = \pm2 \\
\Rightarrow f(\pm2) = \pm2
$$
Which doesn't make sense because that $g(x)$ isn't defined at $(-2,-2)$.
So what is wrong with the equations?
 A: There's nothing wrong with your equations other than the fact that they aren't equations. An equation is an expression which says that the left side of it is equal to the right, while the expression 
$$x=\pm\sqrt{2|x|}$$
really means

$x$ is equal to $\sqrt{2|x|}$ or $x$ is equal to $-\sqrt{2|x|}$.

Your entire reasoning therefore starts by assuming that $f(x) = x$ and $g^2(x) = 2x$ and concludes by showing that if $g(x) = f(x)$, then $x$ is equal to $2$ or $x$ is equal to $-2$.
This is a completely true statement, just like the statement "the moon is round or the moon is made of cheese" is a true statement.
A: \begin{align}
f(x)&=g(x)\\
x&=\pm\sqrt{2x}\tag{1}\\
x^2&=2x\tag{2}\\
x^2-2x&=0\\
x(x-2)&=0\\
x&=0,\,2
\end{align}
Now check if these are indeed solutions to the original equation $(1)$. (They are.)

Back to your proposed solution set. Note that $x=-2$ is not a solution of $(1)$ since the left-hand side is $-2$, but the right-hand side is $\pm\sqrt{2(-2)}$ which not a real number.
More generally, for equations involving radicals, squaring both sides to obtain a different equation may introduce extraneous roots to the original equation. That is why you need to check to see if the roots of the new equation are indeed roots of the original equation. 
For example, with $\sqrt{x-1}=x-7$, squaring both sides and rearranging leads to $x^2-15x+50=0$ which has solutions $x=10,5$. Now, $x=10$ is indeed a solution of the original equation, but $x=5$ is not. 
