How should I solve the equation $\sqrt{x-\frac 1x}+\sqrt{1-\frac 1x}=x$ I could square both sides of the equation, but that ends up giving me a cubic to solve. What I need is a beginning approach to solve such questions, not the whole answer. 
Thanks 
 A: Note that $x = 0$ is clearly not a solution. Observe that:
\begin{align*}
&\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x \\
&\Rightarrow \left(\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\right)\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\
&\Rightarrow \left(\left(x - \frac{1}{x}\right) - \left(1 - \frac{1}{x}\right)\right) = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\
&\Rightarrow x - 1 = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\
&\Rightarrow \sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} = 1 - \frac{1}{x}
\end{align*} 
Adding both together yields:
\begin{align*}
&\left(\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\right) + \left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) = x + \left(1 - \frac{1}{x}\right) \\
&\Rightarrow 2\sqrt{x - \frac{1}{x}} = \left(x - \frac{1}{x}\right) + 1
\end{align*}
We substitute $y = \sqrt{x - \frac{1}{x}}$, and we see that:
\begin{align*}
2y = y^2 + 1 \Rightarrow (y - 1)^2 = 0 \Rightarrow y = 1
\end{align*}
It remains to solve $\sqrt{x - \frac{1}{x}} = 1$.
A: Hint: After squaring two times we get
$$- \left( {x}^{2}-x-1 \right) ^{2}=0$$
A: Square $\sqrt{x-1/x}=x-\sqrt{1-1/x}$ to get $x^2-x+1=2x\sqrt{1-1/x}$. Square again,
$$x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$$
which yields $x=\frac{1\pm\sqrt5}{2}$, of which only the positive root is the true  solution as required by the original equation
$$x=\frac{1+\sqrt5}{2}$$
