Solve $x=\sqrt{x-x^{-1}}+\sqrt{1-x^{-1}}$ Find all real numbers of $x$ such that $$x=\left(x-\frac 1x\right)^{\frac 12}+\left(1-\frac 1x\right)^{\frac 12}$$
My Attempt: Square both sides to get$$x^2=x-\frac 1x+1-\frac 1x+2\sqrt{\left(x-\frac 1x\right)\left(1-\frac 1x\right)}$$
Then move all the terms to the left side except for the square root and then square the equation. But I predict the terms to get really ugly really fast.
So I'm wondering if there is an elegant way of obtaining the solution without going through much of a hassle.
 A: Multiply the original equation 
$$
(x-x^{-1})^{1/2}+(1-x^{-1})^{1/2}=x\tag{1}
$$
by $(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}$ to get
$$
x-1=x((x-x^{-1})^{1/2}-(1-x^{-1})^{1/2})
$$
That is
$$
(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}=1-x^{-1}\tag{2}
$$
Denote $a=(x-x^{-1})^{1/2}$ and $b=(1-x^{-1})^{1/2}$. Then $(1)$ and $(2)$ can be rewritten as 
$$
a+b=x\tag{3}
$$
$$ 
a-b=b^2\tag{4}
$$ 
From the very definition of $a$ and $b$ we have $a^2-b^2=x-1$. With $(3)$ we get $$
a^2-b^2=a+b-1\tag{5}
$$ 
Subtracting $(4)$ from $(5)$ we obtain $a^2-a=a-1$. Hence $a=1$. The rest is clear.
A: The RHS of the given equation is $\geq0$, hence we need only consider solutions $x\geq0$. In this range the square roots on the RHS are only defined when $x\geq1$. We therefore may restrict our search to $x\geq1$. 
Put $$x-{1\over x}=:u^2,\qquad 1-{1\over x}=:v^2$$
with $u\geq0$, $v\geq0$. Then $u+v=x$ and
$$u-v={u^2-v^2\over u+v}={x-1\over x}=1-{1\over x}\ .$$
This leads to $2v=x+{1\over x}-1$ and therefore
$$\left(x+{1\over x}-1\right)^2=4v^2=4-{4\over x}\ .$$
After multiplying by $x^2$ and collecting terms we then obtain
$$0=x^4-2x^3-x^2+2x+1=\bigl(x^2-x-1\bigr)^2\ .$$
The equation $x^2-x-1=0$ has the solutions $\xi:={\sqrt{5}+1\over2}\doteq1.618$ and $-{1\over\xi}<0$. It is easy to check that $\xi$ is indeed a solution (and therewith the only solution) of the original problem:
Note that $\xi-1={1\over\xi}$. Therefore $\xi-{1\over\xi}=1$, and $$1-{1\over\xi}={\xi-1\over\xi}={1\over\xi^2}\ .$$
It follows that
$$\sqrt{\xi-{1\over\xi}}+\sqrt{1-{1\over\xi}}=1+{1\over\xi}=\xi\ .\qquad\square$$
