Find the value of $x=\sqrt{15+\sqrt{15+\sqrt{15+\cdots}}}$.
3
2
$1+\frac{\sqrt{61}}{2}$
4
Correct answer: $1+\frac{\sqrt{61}}{2}$
Solution
Let $x=\sqrt{15+x}$. Squaring gives $x^2=15+x$, or $x^2-x-15=0$. Solving this quadratic gives $x=\frac{1\pm\sqrt{61}}{2}$, and the positive value is $\frac{1+\sqrt{61}}{2}$.