prove that $\displaystyle

2a\sqrt {\frac{1}

{{6 + 4\sqrt 3 }}} = a\left( {3^{1/4} - 3^{ - 1/4} } \right)

$

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- Aug 22nd 2009, 03:11 PMmmsalgebra proof
prove that $\displaystyle

2a\sqrt {\frac{1}

{{6 + 4\sqrt 3 }}} = a\left( {3^{1/4} - 3^{ - 1/4} } \right)

$ - Aug 22nd 2009, 05:48 PMSoroban
Hello, mms!

This requires Olympic-level gymnastics . . .

Quote:

Prove that: .$\displaystyle 2a\sqrt {\frac{1}

{6 + 4\sqrt 3 }} \:=\: a\left(3^{\frac{1}{4}} - 3^{ -\frac{1}{4}}\right)

$

Note that: .$\displaystyle 6 + 4\sqrt{3} \:=\:\sqrt{3}(2\sqrt{3}+4) \;=\;\sqrt{3}(\sqrt{3}+1)^2$

. . . . . . Hence: .$\displaystyle \sqrt{6+4\sqrt{3}} \;=\;\sqrt[4]{3}(\sqrt{3}+1) $

Then [1] becomes: .$\displaystyle \frac{2a}{\sqrt[4]{3}(\sqrt{3}+1)}$

Multiply by $\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}-1}\!:\quad \frac{2a}{\sqrt[4]{3}(\sqrt{3}+1)}\cdot\frac{\sqrt{3}-1}{\sqrt{3}-1} \;=\;\frac{2a(\sqrt{3}-1)}{\sqrt[4]{3}\,(2)} \;=\;\frac{a(\sqrt{3}-1)}{\sqrt[4]{3}} $

Therefore: .$\displaystyle \frac{a(3^{\frac{1}{2}} - 1)}{3^{\frac{1}{4}}} \;=\;a\left(3^{\frac{1}{4}} - 3^{-\frac{1}{4}}\right)$

- Aug 22nd 2009, 09:51 PMQM deFuturo
- Aug 23rd 2009, 12:44 AMGrandad
Hello QM deFuturoNote that:

- $\displaystyle (a+b)^2 = a^2 + 2ab + b^2$

and

- $\displaystyle (\sqrt3)^2 = 3$

Then notice that $\displaystyle 2\sqrt3+4 = 3 + 2\sqrt3 + 1 = (\sqrt3)^2 + 2\sqrt3 + 1$.

Put $\displaystyle a = \sqrt3$ and $\displaystyle b = 1$ in the expansion of $\displaystyle (a+b)^2$, and you're there. I'm afraid this sort of cleverness only comes with years of practice. I (and I suspect Soroban also) have been doing these things since Noah was a boy.

Grandad