Simplifying repeating square roots. How can I solve the equation:
$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}} = \frac{1+\sqrt{53}}{2}$$
 A: Set $y = \sqrt{x + \sqrt{x + \dots}}$.
$y^2 = x+\sqrt{x+\dots} = x + y$
So $y^2 - y = x$.
But $\displaystyle y = \frac{1 + \sqrt{53}}{2}$
Now just substitute.

Addendum for people who want to see something a bit more rigorous.
We will take the infinite square root expression to be defined as the limit of the sequence defined recursively:
$$s(1) = \sqrt{x}$$
$$s(n+1) = \sqrt{x + s(n)}$$
We want to take $\lim_{n \to \infty} s(n)$.
For $x > 0, s(n) \le s(n+1)$ by induction on $n$:
$s(1) \le s(2)$ because $s(2) = \sqrt{x + s(1)} > \sqrt{x}$ since $s(1) > 0$.
If $s(n) \le s(n+1)$, then $s(n+2) = \sqrt{x + s(n+1)} \ge \sqrt{x+s(n)} = s(n+1)$.
So $s$ is monotonic.
Moreover, $s$ is bounded above by $\max(x,2)$.  This will also be proven by induction.
$s(1) = \sqrt{x} \le \max(x,1) \le \max(x,2)$
If $s(n) \le 2$, then $s(n+1) = \sqrt{x+2} \le \sqrt{4} \le 2$.
Similarly, if $2 < s(n) \le x$, then $s(n+1) = \sqrt{x+s(n)} \le \sqrt{2x} \le \sqrt{x^2} = x$
Since $s$ is bounded and monotonic, it has a limit as $n \to \infty$.
Since $f(y) = \sqrt{ x + y }$ is continuous, $$\lim_{n \to \infty} s(n) = \lim_{n \to \infty} s(n+1) = \lim_{n \to \infty} \sqrt{ x + s(n) } = \sqrt{ x + \lim_{n \to \infty} s(n)}$$
Justifying the reasoning above.
