How to simplify the nested radical $\sqrt{1 - \frac{\sqrt{3}}{2}}$ by hand? I was solving a Mock Mathcounts Contest Mock contest (.pdf) written by a user on the Art of Problem Solving Forums. In problem #24 the only thing I couldn't do by hand was simplify the radical mentioned above. Note that the contest should involve math topics accessible to a high performing Mathcounts middle school student.

How do you simplify $$\sqrt{1 - \frac{\sqrt{3}}{2}}$$

 A: Proposed another way to collection
You can also use formulas
$$\sqrt{a\pm\sqrt b}=\sqrt{\frac{a + \sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a - \sqrt{a^2-b}}{2}}$$
Then
$$\sqrt{1-\frac{\sqrt3}{2}}=\frac12\sqrt{4-\sqrt{12}}=\frac12 \cdot\left(\sqrt{\frac{4 + \sqrt{4^2-12}}{2}}-\sqrt{\frac{4 - \sqrt{4^2-12}}{2}} \right)=$$
$$=\frac12 \left(\sqrt{\frac{4 + \sqrt{4}}{2}}-\sqrt{\frac{4 - \sqrt{4}}{2}} \right)=\frac12 \left(\sqrt{\frac{6}{2}}-\sqrt{\frac{2}{2}} \right)=\frac 12(\sqrt3-1)$$
A: One should note the perfect square: $(\sqrt 3-1)^2=4-2\sqrt3$, hence
$$\sqrt{1-\frac{\sqrt3}{2}}=\sqrt{\frac{2-\sqrt3}{2}}=\sqrt{\frac{4-2\sqrt3}{4}}=\frac{\sqrt{(\sqrt3)^2-2\cdot \sqrt 3\cdot 1+1^2}}{2}=\frac{\sqrt{(\sqrt3-1)^2}}{2}=\color{red}{\frac{\sqrt 3-1}{2}}$$
A: One may write
$$
\begin{align}
\sqrt{1 - \dfrac{\sqrt{3}}{2}}&=\dfrac1{\sqrt{2}}\sqrt{2 - \sqrt{3}}\\\\
&=\dfrac{\sqrt{2}}2\sqrt{2 - \sqrt{3}}\\\\
&=\dfrac12\sqrt{4 - 2\sqrt{3}}\\\\
&=\dfrac12\sqrt{(\sqrt{3}-1)^2}\\\\
&=\dfrac{\sqrt{3}-1}2.
\end{align}
$$
A: $$\sqrt{1-\frac{\sqrt 3}{2}}=\sqrt{\frac{2-\sqrt 3}{2}}=\sqrt{\frac{4-2\sqrt 3}{4}}=\sqrt{\frac{3+1-2\sqrt{3\times 1}}{4}}=\sqrt{\frac{(\sqrt 3-\sqrt 1)^2}{4}}$$
