Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $
then find value of 2x-1
I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.
 A: I assume you mean
$$ x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4\pm\ldots}}}}}$$
so that
$$ \begin{align}x&=\sqrt{4+\sqrt{4-x}}\\
 x^2&=4+\sqrt{4-x}\\
 (x^2-4)^2&=4-x\\
 0 &= x^4-8x^2+x+12= (x^2-x-3)(x^2+x-4)\end{align}$$
Since clearly $x\ge \sqrt 4=2$, the second factor is $x^2+x-4\ge2>0$, which leaves us with the positive solution $x=\frac{1+\sqrt{13}}{2}\approx 2.3027756 $ from the first factor.
A: We have
$$-\left((x^2-4)^2-4\right)=x\iff  x^4-8x^2+x+12= (x^2+x-4)(x^2-x-3) =0$$
solve this equation and notice that $x\ge2$ we find that the acceptable answer is 
$$x=\frac12\sqrt{13}+\frac12$$
hence
$$2x-1=\sqrt{13}$$
A: Since $$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}}},$$ you have $$x^2=4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}},$$ so $$(x^2-4)^2=4-\sqrt{4+\sqrt{4-\cdots}}.$$ Hence, $$(x^2-4)^2=4-x,$$ which you can try to solve.
A: Edit: I misread the excercise but still the method I used applies. The first step has to be done twice and then a polynomial equation of degree 4 has to be solved.
$x=\sqrt{ \left( 4+...\right.} \underbrace{\Rightarrow}_{\text{square and }-4} x^{2}-4=\sqrt{ \left( 4+...\right.}=x \Rightarrow 0=x^{2}-x-4$
which is an easy to solve polynomial equation, the solutions are $\frac{1}{2}+\sqrt{\frac{1}{4}+4}$ and $\frac{1}{2}-\sqrt{\frac{1}{4}+4}$.
A: Solution without complex factorization:
Let,
$ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $
$ y= \sqrt{4-\sqrt{4+\sqrt{ 4-\dots}}} $
Clearly,
$x = \sqrt{4 + y} \implies x^2 - y = 4$    (i)
$y = \sqrt{4 - x} \implies y^2 + x = 4$    (ii)
from (i) and (ii) we get,
$x^2 - y = y^2 + x$
$\implies (x+y)(x-y) = x+y$
Since, $x,y \ge 0$, we have $x+y \ne 0$
So, $x = y + 1$   (iii)
Putting (iii) in (i) or (ii) we get:
$x = \frac{\sqrt{13}+1}{2}$.
