How to take the square root of $BC$ where $BC^2$ equals $56-32\sqrt3$

Question

$BC^2$ equals $56-32\sqrt3$

what is the square root of BC?

The dimensions I used to get this far (and I confirmed that they are correct) are A,B,C, in a right triangle, where B is $\sqrt{(42-24\sqrt3)}$ and A is $2\sqrt2-\sqrt6$

I can simplify it down to $2\sqrt{(14-8\sqrt3)}$ but the answer is

something - $2\sqrt6$

All help is appreciated! thank you!

Answer 1

Write $BC = \sqrt{a} - \sqrt{b}$. Then $$BC^2 = (\sqrt{a} - \sqrt{b})^2 = a + b - 2\sqrt{ab} = 56 - 32\sqrt{3} = 56 - 2\sqrt{3 \cdot 16^2}.$$ Now, try to find $a$ and $b$ so that $a + b = 56$ and $a \cdot b = 3 \cdot 16^2$.

Answer 2

Working in $\mathbb Z[\sqrt3]$ we have $56-32\sqrt3=4(14-8\sqrt3)=2^2(a+b\sqrt3)^2$ which leads to the system $$a^2+3b^2=14\\ab=-4$$ whose solutions are $$(a,b)=(\pm2\sqrt2,\mp\sqrt2), \left(\pm\sqrt6,\mp\frac{2\sqrt6}{3}\right)$$ Then do you have $$\boxed{[2(2\sqrt2-\sqrt6)]^2=56-32\sqrt3}$$ as you can verify.

I leave as a question if with the other solution $\left(\pm\sqrt6,\mp\dfrac{2\sqrt6}{3}\right)$ we also have $[2(\pm\sqrt6,\mp\frac{2\sqrt6}{3})]^2=56-32\sqrt3$