<s>[INST] Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$ 
If inverse of the function $y=\sqrt{x+\sqrt{2x-1}}$ be equal to
$y=ax^2+bx+c$, then what is the value of $a^2+b^2+c^2$?
$A) 3$
$B) 4$
$C) 5$
$D) 6$

I tried to isolate $x$ from the equation $y=\sqrt{x+\sqrt{2x-1}}$:
$$y^2=x+\sqrt{2x-1}$$
$$(y^2-x)^2=2x-1$$
$$y^4-2xy^2+x^2=2x-1$$
$$y^2(y^2-2x)=-x^2+2x-1$$
But I still have $-2x$ in the parenthesis of the LHS.
 [/INST] You have\begin{align}\require{cancel}y=\sqrt{x+\sqrt{2x-1}}&\implies y^2=x+\sqrt{2x-1}\\&\implies y^2-x=\sqrt{2x-1}\\&\implies(y^2-x+1)^2=(y^2-x)^2+1+2(y^2-x)\end{align}But then\begin{align}(y^2-x+1)^2=2x-\cancel1+\cancel1+2\sqrt{2x-1}=2y^2,\end{align}and therefore $y^2-x+1=\pm\sqrt2y$. So, $x=y^2\pm\sqrt2y+1$.
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