<s>[INST] Square root of x plus square root of x plus ... I want to confirm if my answer in this problem is correct: 
$$\sqrt{(x + \sqrt{(x + ...))}} = (1 + \sqrt{53}) / 2 $$
Solution:
$$x + \sqrt{(x + \sqrt{(x + ...))}} = (1 + \sqrt{53})^2 / 4 $$
$$x + (1 + \sqrt{53}) / 2 = (1 + 2\sqrt{53} + 53) / 4 $$
$$ x + (1 + \sqrt{53}) / 2 = (54 + 2\sqrt{53}) / 4 $$
$$x + (1 + \sqrt{53}) / 2 = (27 + \sqrt{53}) / 2 $$
$$2x + 1 + \sqrt{53} = 27 + \sqrt{53}$$ 
$2x = 26$
$x = 13$ 
I also tried solving something like 
$$\sqrt{x + \sqrt{x + \sqrt{x + \sqrt(x)}}} = (1 + \sqrt{53}) / 2 $$
and the answer is $x = 13.0006$ 
Please confirm if my answer is correct or not. Thanks!
 [/INST] Yes your answer is correct. Let's write it more readable: 
Let 
$$y=\sqrt{x+\sqrt{x+\cdots}}$$
hence the equation is
$$y=\frac{1+\sqrt{53}}{2}$$
so squaring the terms of the last equality gives
$$y^2=x+y=\frac{(1+\sqrt{53})^2}{4} $$
hence we have
$$x=y^2-y=\frac{(1+\sqrt{53})^2}{4}-\frac{1+\sqrt{53}}{2}=\frac{1+2\sqrt{53}+53-2-2\sqrt{53}}{4}=\frac{52}{4}=13$$
</s>