<s>[INST] Oops, $y^2=2x^2+C$ is not the same as $y=\sqrt{2x^2}+C$ Oops, $y^2=2x^2+C$ is not the same as $y=\sqrt{2x^2}+C$ 
I almost slipped and just assumed $\sqrt{C}=C$ but when you take the square root of both sides, you are really ending up with $y=\sqrt{2x^2+C}$  And you can't just take the C out of the radical.  Right?
 [/INST] First, note that it's not true that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}, \; \forall a,b \in \mathbb{R}$, as, for example: $$\underbrace{\sqrt{1+1}}_{=\sqrt{2}} \neq \underbrace{\sqrt{1} + \sqrt{1}}_{=2}$$ 
hence, in general, $\sqrt{2x^2+C} \neq \sqrt{2x^2} + \sqrt{C}$, except for some specific values of $x$ and $C$.
There is also this common misconception that $\sqrt{x^2}=x,\; \forall x \in \mathbb{R}$. This equality is only true if $x \geq 0$ !
What we know is that
$$\sqrt{x^2}=| x |,\; \forall x \in \mathbb{R}.$$
So, when you take the square root of both sides of $y^2=2x^2+C$, you end up with:
$$|y|=\sqrt{2x^2+C}.$$
Thus, the answer to your question is: no, $y^2=2x^2+C$ is not the same as $y=\sqrt{2x^2}+C$ and it's not even the same as $y=\sqrt{2x^2+C}$.
When in doubt, you can always have a look at some plots: (I used Mathematica, but you could get similar pictures using WolframAlpha)


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*Plot of $y^2=2x^2+C$, for $C=2$





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*Plot of $|y|=\sqrt{2x^2+C}$, for $C=2$





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*Plot of $y=\sqrt{2x^2+C}$, for $C=2$





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*Plot of $y=\sqrt{2x^2}+C$, for $C=2$



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