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?

Clearly, $x\ne0$
$$\sqrt{x-1/x} + \sqrt{1 - 1/x} = x$$
$$\implies \sqrt{x^2-1}+\sqrt{x-1}=x\sqrt x$$
$$\implies \sqrt{x-1}(\sqrt{x+1}+1)=x\sqrt x$$
$$\implies \sqrt{x-1}\frac{x+1-x}{\sqrt{x+1}-1}=x\sqrt x\text{ (rationalizing the numerator) }$$
$$\implies \sqrt{x-1}=\sqrt x(\sqrt{x+1}-1)=\sqrt{x^2+x}-\sqrt x$$
$$\implies \sqrt x+ \sqrt{x-1}=\sqrt x(\sqrt{x+1}-1)=\sqrt{x^2+x}$$
Squaring we get, $$x+x-1+2\sqrt{x^2-x}=x^2+x\iff 2\sqrt{x^2-x}=x^2-x+1$$
Putting $x^2-x=a^2,a^2=2a-1\implies (a-1)^2=0\implies a=1$ 
