<s>[INST] 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...
 [/INST] 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 $x= {\frac{1 +\sqrt{5}}{2}}$
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