## Nested Square Roots 2

Find the value of $\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}$
Source: NCTM Mathematics Teacher, October 2006

Solution
Let $x=\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}$
By substitution
$x=\sqrt{16+x}$
Squaring both sides
$x^2=16+x$
$x^2-x-16=0$
Using the quadratic formula to solve for $x$
$x=\dfrac{1\pm\sqrt{65}}{2}$
$\dfrac{1-\sqrt{65}}{2}$ is an extraneous solution because the expression cannot be negative.
$\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}=\dfrac{1+\sqrt{65}}{2}$

Answer: $\dfrac{1+\sqrt{65}}{2}$