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...
 A: HINT: after squaring one times we get
$$2\sqrt{x-\frac{1}{x}}\sqrt{1-\frac{1}{x}}=x^2-x+\frac{2}{x}-1$$
can you finish this?
squaring this one more times we get
$$4\left(x-\frac{1}{x}\right)\left(1-\frac{1}{x}\right)=\left(x^2-x+\frac{2}{x}-1\right)^2$$
expanding the left Hand side we obtain $$4\,x-4-4\,{x}^{-1}+4\,{x}^{-2}$$ and the right Hand side is given by $${x}^{4}-2\,{x}^{3}+6\,x-{x}^{2}-3+4\,{x}^{-2}-4\,{x}^{-1}$$ have you got this?
Bringung all together we obtain this equation
$$0=x^4-2x^3-x^2+2x+1$$
A: Divide by $x$:
$$\sqrt{\frac1x-\frac1{x^3}}+\sqrt{\frac1{x^2}-\frac1{x^3}}=1$$
Change $t=1/x$:
$$\sqrt{t-t^3}+\sqrt{t^2-t^3}=1$$
Square and rearrange:
$$2\sqrt{t^3-t^4-t^5+t^6}=t+t^2-2t^3-1$$
Square and rearrange again:
$$4t^6-4t^5-4t^4+4t^3=4t^6-4t^5-3t^4+6t^3-t^2-2t+1$$
Finally we get
$$t^4+2t^3-t^2-2t+1=0$$
That is,
$$(t^2+t-1)^2=0$$
A: $x\geq1$ 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 $x=\frac{1+\sqrt5}{2}$.
A: Squaring both sides:
$x^{2} - x +\frac{2}{x} - 1 = 2\sqrt{(x-\frac{1}{x})(1-\frac{1}{x})}$
Squaring again we get:
$x^{4} - 2x^{3} - x^{2} + \frac{4}{x^{2}} + 6x -\frac{4}{x} - 3 = 4(\frac{1}{x^{2}} + x - \frac{1}{x} - 1)$
We can reduce this to:
$x^{4} -2x^{3} -x^{2} +2x + 1 = 0$
Notice that this factorises into $(x^{2} - x -1)^{2} = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{5}}{2}$
Note this is the 'golden ratio' $\phi = \frac{1 + \sqrt{5}}{2}$ with the property that: $\frac{1}{\phi} = \phi - 1$
Since we have squared our initial equation thus introducing extra solutions, we must test each of these in the original equation.
This leads us to reject $x = \frac{1 - \sqrt{5}}{2}$
Therefore $x= {\frac{1 +\sqrt{5}}{2}}$
