Find the positive value of $x$ satisfying the given equation $${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.

HINT
We can try with
$${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
$$\sqrt {x+1}{\sqrt {x- 1\over x}} + {\sqrt{x-1\over x}} = x$$
$$(\sqrt {x+1}+1){\sqrt {x- 1\over x}} = x$$
$${\sqrt {x- 1\over x}} = \frac{x}{\sqrt {x+1}+1}\frac{\sqrt {x+1}-1}{\sqrt {x+1}-1}=\sqrt {x+1}-1$$
$${\sqrt {x- 1}} =\sqrt {x^2+x}-\sqrt x$$
and from here we can square to eliminate the square roots.
Recall to check at the end the conditions for the existence related to the original equation


*

*${{x^2- 1\over x}}\ge 0$

*${{x- 1\over x}}\ge 0$

*$x\neq 0$
