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$BC^2$ 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!

TizzleRizzle
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  • You can take out an additional factor of $\sqrt2$, and then use Joseph's idea to simplify the square root of $7-4\sqrt3$. – Dave Radcliffe May 31 '22 at 16:56
  • I don't see how his works, since you cannot just split a square root into addition and subtraction. I am trying to get BC, i already have $BC^2$ – TizzleRizzle May 31 '22 at 17:00
  • I have edited your header to agree with what you appear to be asking for in the body of the post. Please let me know if you intended something else. –  Jun 03 '22 at 01:56

2 Answers2

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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 $a \cdot b = 3 \cdot 16^2$.

Joseph Camacho
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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 $[2(\pm\sqrt6,\mp\frac{2\sqrt6}{3})]^2=56-32\sqrt3$

Piquito
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  • Why does the $4(14-8\sqrt3)$ change into $(a+b\sqrt)^2$? I understand the 4 = 2 squared but I don't understand how you can just square that. – TizzleRizzle Jun 01 '22 at 13:46
  • If you want to have the square root of $a^2b$, the problem you have to solve is just the calculation of $\sqrt b$ and your solution will be $a\sqrt b$ – Piquito Jun 02 '22 at 15:59