radicals - Assuming convergence of the following series, find the value of $sqrt{6+sqrt{6+sqrt{6+...}}}$

Assuming convergence of the following series, find the value of $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$

I was advised to proceed with this problem through substitution but that does not seem to help unless I am substituting the wrong parts. If i substitute the $6$, well then i am just stuck with above.

Any ideas on how to proceed. Also, what is the purpose of stating that it is convergent.

Answer

Let $x=\sqrt{6+\sqrt{6+\sqrt{6+...}}}$, then observe that $x=\sqrt{6+x}$. Squaring both sides yields
$$x^2=x+6$$
, which is a quadratic formula. Solve it normally and choose the wise answer out of the 2 roots.