Quadratic equation find all the real values of $x$ Find all real values of $x$ such that 
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...

$x\geq1$ and $1$ is not root. 
Hence we can rewrite our equation in the following form:
$$x-1=x\left(\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}\right)$$ or
$$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}=1-\frac{1}{x}$$ and  with the given we obtain
$$2\sqrt{x-\frac{1}{x}}=x+1-\frac{1}{x}$$ or
$$x-\frac{1}{x}-2\sqrt{x-\frac{1}{x}}+1=0$$ or
$$\left(\sqrt{x-\frac{1}{x}}-1\right)^2=0$$ or
$$x^2-x-1=0,$$
which gives $x=\frac{1+\sqrt5}{2}$.
