# 3 Ways of Evaluating Nested Square Roots

Nested square roots or nested radical problems are quite interesting to solve. The key skill for this question is to understand how the students can handle “…”. This enables us to set up a quadratic equation to evaluate its exact value using the quadratic formula,
$$x= \frac{-b \ \pm \sqrt{b^2-4ac}}{2a}$$.
Let’s look at the following examples for finding the nested square roots.

## Example 1

Evaluate $\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}}$

\begin{aligned} \require{AMSsymbols} \require{color} \text{Let } x &= \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}} &\color{green} {(1)} \\ x^2 &= 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} &\color{green}{\text{square both sides}} \\ x^2-2 &= \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} &\color{green}{\text{move 2 to the left}} \\ x^2-2 &= x &\color{green}{\text{replace the nested square root by } (1)} \\ x^2-x-2 &= 0 &\color{green}{\text{form a quadratic equation}} \\ x &= \frac{1 \pm \sqrt{(-1)^2-4 \times 1 \times (-2)}}{2} &\color{green}{\text{apply into quadratic formula}} \\ x &= \frac{1 \pm \sqrt{9}}{2} \\ x &= \frac{1 \pm 3}{2} \\ x &= \frac{1 + 3}{2} \text{ or } \frac{1-3}{2} \\ x &= 2 \text{ or } -1 \\ x &= 2 &\color{green} {\text{ as } \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}} > 0} \\ \therefore \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}} &= 2 \end{aligned}

## Example 2

Evaluate $\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}$

\begin{aligned} \require{AMSsymbols} \require{color} \text{Let } x &= \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\ldots}}}} &\color{green} {(1)} \\ x^2 &= 2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}} &\color{green}{\text{square both sides}} \\ x^2-2 &= -\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}} &\color{green}{\text{move 2 to the left}} \\ x^2-2 &= -x &\color{green}{\text{replace the nested square root by } (1)} \\ x^2 + x-2 &= 0 &\color{green}{\text{form a quadratic equation}} \\ x &= \frac{-1 \pm \sqrt{1^2-4 \times 1 \times (-2)}}{2} &\color{green}{\text{apply into quadratic formula}} \\ x &= \frac{-1 \pm \sqrt{9}}{2} \\ x &= \frac{-1 \pm 3}{2} \\ x &= \frac{-1 + 3}{2} \text{ or } \frac{-1-3}{2} \\ x &= 1 \text{ or } -2 \\ x &= 1 &\color{green} {\text{ as } \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}} > 0} \\ \therefore \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}} &= 1 \end{aligned}

## Example 3

Evaluate $\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}}$ using the double angle trigonometric property.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \cos\frac{\pi}{4} &= \frac{\sqrt{2}}{2} &\color{green}{(1)} \\ 2 \cos^2\frac{\pi}{8}-1 &= \cos\frac{\pi}{4} &\color{green}{\text{apply the double-angle formula}} \\ 2 \cos^2\frac{\pi}{8}-1 &= \frac{\sqrt{2}}{2} &\color{green}{\text{from (1)}} \\ 2 \cos^2\frac{\pi}{8} &= 1 + \frac{\sqrt{2}}{2} &\color{green}{\text{move -1 to the right hand side}} \\ 2 \cos^2\frac{\pi}{8} &= \frac{2 + \sqrt{2}}{2} &\color{green}{\text{single fraction}} \\ \cos^2\frac{\pi}{8} &= \frac{2 + \sqrt{2}}{4} &\color{green}{\text{divide both sides by 2}} \\ \cos\frac{\pi}{8} &= \frac{\sqrt{2 + \sqrt{2}}}{2} &\color{green}{(2)} \\ \cos\frac{\pi}{16} &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} &\color{green}{(3)} \\ \ldots \\ \cos\frac{\pi}{\infty} &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}}{2} &\color{green}{(4)} \\ \cos 0 &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}}{2} \\ 1 &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}}{2} \\ \therefore \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} &= 2 \end{aligned}