Why is $\sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} = 3$ This is a problem from SASMO Grade 8 (Secondary 2) Sample Questions.
Solve for $x$
$\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = 3$
Answer: $x=6$
I have tried this on a calculator: the more $x$ we add, the closer we are to $3$. But how can we prove it, and is there a way to figure this out by hand?

Let $x= \sqrt{6+ \sqrt{6+\sqrt{6+ \sqrt{\cdot\cdot\cdot}}}}$
then $x^2= 6+ \sqrt{6+ \sqrt{6+ \sqrt{\cdot\cdot\cdot}}}= 6+ x$.  Solve $x^2-x- 6= 0$.
Clearly this has to be positive.  What is the positive root of $x^2- x- 6= 0$?
(There is only one positive root and this clearly has to be positive.)
