Tricky Entrance Exam Questions

I came across this problem on a YouTube channel. The problem was purported to be a Harvard University entrance exam question.$$ \text{Simplify } \sqrt{\sqrt{121}-\sqrt{120}}$$Once you see the method, it's easy enough so let's start to simplify:$$

\begin{align}

\sqrt{\sqrt{121}-\sqrt{120}} &= \sqrt{11 - 2 \cdot \sqrt{30}} \\

&= \sqrt{11 - 2 \cdot \sqrt{6} \cdot \sqrt{5}} \\

&= \sqrt{6 - 2 \cdot \sqrt{6} \cdot \sqrt{5} + 5} \\

&= \sqrt{(\sqrt{6})^2 - 2 \cdot \sqrt{6} \cdot \sqrt{5} + (\sqrt{5})^2} \\

&= \sqrt{(\sqrt{6} - \sqrt{5})^2} \\

&= \sqrt{6} - \sqrt{5}

\end{align}

$$Here's another one:$$ \text{Simplify } \sqrt{\sqrt{36}-\sqrt{20}}$$The approach is exactly the same:

$$
\begin{align}

\sqrt{\sqrt{36}-\sqrt{20}} &= \sqrt{6 - 2 \cdot \sqrt{5}} \\

&= \sqrt{6- 2 \cdot \sqrt{5} \cdot \sqrt{1}} \\

&= \sqrt{5 - 2 \cdot \sqrt{5} \cdot \sqrt{1} + 1} \\

&= \sqrt{(\sqrt{5})^2 - 2 \cdot \sqrt{5} \cdot \sqrt{1} + (\sqrt{1})^2} \\

&= \sqrt{(\sqrt{5} - \sqrt{1})^2} \\

&= \sqrt{5} - \sqrt{1}\\
&=\sqrt{5}-1

\end{align}

$$