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.

Let $t=\sqrt{2x-1}\geq 0$, then $x= {t^2+1\over 2}$ so $$y = \sqrt{t^2+2t+1\over 2} = {t+1\over \sqrt{2}}$$
Now is easy to finish.
If I didn't make any mistake, then inverse function is

$$y=x^2-x\sqrt{2}+1$$

