Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ Find all real numbers such that 
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier method to solve this?

Multiply with $\sqrt{x-1/x}-\sqrt{1-1/x}$ to get
$$x-1=(x-\frac1x)-(1-\frac1x) =x\left(\sqrt{x-1/x}-\sqrt{1-1/x}\right)$$
so $$1-\frac1x = \sqrt{x-1/x}-\sqrt{1-1/x}$$
and by adding
$$ x+1-\frac1x =2\sqrt{x-\frac1x}$$
Now with $z=x-\frac1x$, this is simply
$$z+1=2\sqrt z,$$
a quadratic in $\sqrt z$ with (double) solution $\sqrt z=1$, so $z=1$ and we need to solve
$$ x-\frac1x=1$$
i.e. $$ x^2-x-1=0.$$
This has solutions $x=\frac{1\pm\sqrt 5}2$, but only the positive value is possible.
