<s>[INST] How to solve the following radical equation?$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$ 
How to solve the following radical equation?
  $$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$$

I'm looking for an easy way to solve it. 
 [/INST] If $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$
Then:
$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}}}}}=x$
In general:
$x=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\ldots}}}}$
Which means that:
$2-(x^2-2)^2=x$
$x^4-4x^2+4+x-2=0$
$x^4-x^3+x^3-x^2-3x^2+3x-2x+2=0$
$(x-1)(x^3+x^2-3x-2)=0$
$(x-1)(x^3+2x^2-x^2-x-2x-2)=0$
$(x-1)(x+2)(x^2-x-2)=0$
$(x-1)(x+2)\left(x - \dfrac{1 \pm \sqrt{5}}{2}\right)$
So the 'possible' answers for x would be :
$\left(1,-2 ,\phi,\dfrac{1}{\phi} \right)$
All are too small to be the answer of the original question except for $\phi$, the golden ratio. So that's the answer.
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