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?
 A: Multiply with $\sqrt{x-1/x}-\sqrt{1-1/x}$ to get
$$x-1=(x-\frac1x)-(1-\frac1x) =x\left(\sqrt{x-1/x}-\sqrt{1-1/x}\right)$$
so $$1-\frac1x = \sqrt{x-1/x}-\sqrt{1-1/x}$$
and by adding
$$ x+1-\frac1x =2\sqrt{x-\frac1x}$$
Now with $z=x-\frac1x$, this is simply
$$z+1=2\sqrt z,$$
a quadratic in $\sqrt z$ with (double) solution $\sqrt z=1$, so $z=1$ and we need to solve
$$ x-\frac1x=1$$
i.e. $$ x^2-x-1=0.$$
This has solutions $x=\frac{1\pm\sqrt 5}2$, but only the positive value is possible.
A: Hints:
$$x=\sqrt{x-\frac1x}+\sqrt{1-\frac1x}\implies x\sqrt x=\sqrt{x^2-1}+\sqrt{x-1}\implies$$
$$ x^3=x^2+x-2+2\sqrt{x^3-x^2-x+1}\implies(x^3-x^2-x+2)^2=4(x^3-x^2-x+1)$$
Well, now you can put $\,w:=x^3-x^2-x+1\;$ and solve
$$(w+1)^2=4w\iff (w-1)^2=0\iff w=1\;\ldots\;\text{and etc.}\ldots$$
A: Let's multiply
$$
\sqrt{x-1/x} + \sqrt{1 - 1/x} = x\tag{1}
$$
by $(\sqrt{x-1/x} - \sqrt{1 - 1/x})$. Then we get
$$
x-1=x(\sqrt{x-1/x} - \sqrt{1 - 1/x})
$$
$$
\sqrt{x-1/x} - \sqrt{1 - 1/x}=1-1/x\tag{2}
$$
Sum up $(1)$ and $(2)$ to get
$$
2\sqrt{x-1/x}=x-1/x+1
$$
Now make the substitution $y=\sqrt{x-1/x}$. The rest is clear.
A: 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 $x^2-x=a^2,a^2=2a-1\implies (a-1)^2=0\implies a=1$ 
