Equation solving involving terms inside roots How do I solve 

$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x$$ 

My Try I tried to take the first term as $t$ but then I had to square both sides twice and that led to a complex bi quadratic. I'm not sure even that'll solve the problem. 

guys please let me know what are different possible ways to tackle problems like these, and how would you go about on solving this particular one.
$$\left(x-\frac{1}{x}\right)^{1/2}-\left(1-\frac{1}{x}\right)^{1/2} = \frac{\left(x-\frac{1}{x}\right)-\left(1-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}}=\frac{x-1}{x}=1-\frac1x,$$ 
adding both equations gives $$2\left(x-\frac{1}{x}\right)^{1/2}=x-\frac{1}{x}+1.$$ That means $$x-\frac{1}{x}=1,$$ so we have $x=\phi$ (the Golden Ratio). The other solution $x=-\phi^{-1}$ does not satisfy the original equation, we must have $x&gt;0$.
Hint:take $$a=x+\frac1x ,b=x-\frac1x$$
so $$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x\\\sqrt{a}+\sqrt{b}=\frac{a+b}{2}$$