Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$

Question

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.

Answer 1

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

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

Answer 2

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, $x=y^2\pm\sqrt2y+1$.

Answer 3

Since we are told that the inverse function is quadratic, we know that the curve function $ \ y \ = \ \sqrt{x+\sqrt{2x-1}} \ \ $ is the "upper half" of a "horizontal" parabola (which is not difficult to verify). We can find one point on the curve by considering that its domain is $ \ x \ \ge \ \frac12 \ \ ; $ the coordinates of the vertex for the horizontal parabola are then $ \ x \ = \ \frac12 \ , \ y \ = \ \sqrt{\frac12 \ + \ 0} \ = \ \frac{\sqrt2}{2} \ \ . $ The vertex of the parabola corresponding to the inverse function is therefore $ \ x \ = \ \frac{\sqrt2}{2} \ , \ y \ = \ \frac12 \ \ , $ so we can write the "vertex form" of its curve equation as $ \ y \ = \ a·\left(x - \frac{\sqrt2}{2} \right)^2 + \frac12 \ \ . $

We can next choose any second point on the "horizontal" parabola that is convenient to calculate, say, $ \ x \ = \ 1 \ \Rightarrow \ y \ = \ \sqrt{1 + \sqrt{2-1}} \ = \ \sqrt2 \ \ , $ which tells us that $ \ (\sqrt2 \ , \ 1) \ $ is a point on the inverse-function parabola. Hence, $$ 1 \ \ = \ \ a·\left(\sqrt2 - \frac{\sqrt2}{2} \right)^2 \ + \ \frac12 \ \ = \ \ a·\left( \frac{\sqrt2}{2} \right)^2 \ + \ \frac12 \ \ = \ \ \frac12·a \ + \ \frac12 \ \ \Rightarrow \ \ a \ = \ 1 \ \ . $$ The "standard form" of the inverse function is given by $ \ y \ = \ \left(x - \frac{\sqrt2}{2} \right)^2 + \frac12 \ = \ x^2 \ - \ \sqrt2·x \ + \ 1 \ \ , $ for which the correct choice is $ \ \mathbf{(B)} \ \ [ \ 1^2 + (-\sqrt2)^2 + 1^2 \ = \ 4 \ ] \ \ . $