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.
 A: Let's see how insurmountable this sextic equation is.
Start by deriving the equation.  First multiply by $\sqrt{x}$, square both sides, and isolate the remaining radical:
$(x^2-1)+2\sqrt{(x^2-1)(x-1)}+(x-1)=x^3$
$2\sqrt{(x^2-1)(x-1)}=x^2-x^2-x+2$
Square again, expand and collect to get our monster:
$x^6-2x^5-x^4+2x^3+x^2=0$
Clearly since $x$ must be positive (certainly not zero, which the fractional radicands in the original equations would forbid), the factor $x^2$ can be divided out.  We are down to degree $4$:
$x^4-2x^3-x^2+2x+1=0$
Next observe that this quartic equation has the following property:
(Linear coefficient)/(Cubic coefficient) $=c$
(Constant)/(Quartic coefficient)$=c^2$
When this occurs, the roots of the quartic equation occur in pairs giving the product $c$ and thus we must have a factorization:
$(x^2+ax+c)(x^2+bx+c)$
Here, $c=-1$ and so:
$x^4-2x^3-x^4+2x+1=0=(x^2+ax-1)(x^2+bx-1)$
Expanding the right side and matching like terms leads to two independent equations, thus:
$a+b=-2; b=-2-a$
$ab=1$
Then $a(-2-a)=1, a^2+2a+1=0$ and we have the root $a=-1$. Thus $b=-1$ and our polynomial is now reduced to a single, squared factor:
$(x^2-x-1)^2=0$
And so, from the quadratic formula,
$x=\frac{1+\sqrt{5}}{2}$
Which the reader can verify, checks out! Note that in the checking process we should find $(x^2-1)/x=1$.
A: 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


*

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

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

*$x\neq 0$
A: Checking other answers:
Note that $\frac{x^2-1}{x}\ge 0;\frac{x-1}{x}\ge 0$ and $x>0$ imply $x>1$. 
If $x^2=x+1 \ \ (1)$, then:
$$ x^2-1=x \Rightarrow \frac{x-1}{x}=\frac{1}{x+1},\\
{\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x \ \ \ \ \ \ \ \ \ \ (2) \ \ \ \ \ \Rightarrow \\
{\sqrt {(x+1)- 1\over x}} + {\sqrt{1\over x+1}} = \sqrt{x+1} \Rightarrow \\
1+\frac1{\sqrt{x+1}}=\sqrt{x+1} \Rightarrow \\
\sqrt{x+1}+1=x+1 \Rightarrow \\
x+1=x^2.$$
So, $(1)$ and $(2)$ are equivalent.
