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Math Help - algebra proof

  1. #1
    mms
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    algebra proof

    prove that <br />
2a\sqrt {\frac{1}<br />
{{6 + 4\sqrt 3 }}} = a\left( {3^{1/4} - 3^{ - 1/4} } \right)<br /> <br />
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  2. #2
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    Hello, mms!

    This requires Olympic-level gymnastics . . .


    Prove that: . 2a\sqrt {\frac{1}<br />
{6 + 4\sqrt 3 }} \:=\: a\left(3^{\frac{1}{4}} - 3^{ -\frac{1}{4}}\right)<br />
    The left side is: . \frac{2a}{\sqrt{6 + 4\sqrt{3}}} .[1]

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

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


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

    Multiply by \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: . \frac{a(3^{\frac{1}{2}} - 1)}{3^{\frac{1}{4}}} \;=\;a\left(3^{\frac{1}{4}} - 3^{-\frac{1}{4}}\right)


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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, mms!

    This requires Olympic-level gymnastics . . .



    Pretty impressive. How did you determine this:
     (2\sqrt{3}+4) \;=\;(\sqrt{3}+1)^2

    It's true, but I would not have been able to figure that out, or even realize I would have to do that for this proof!
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  4. #4
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    Hello QM deFuturo
    Quote Originally Posted by QM deFuturo View Post
    Pretty impressive. How did you determine this:
     (2\sqrt{3}+4) \;=\;(\sqrt{3}+1)^2

    It's true, but I would not have been able to figure that out, or even realize I would have to do that for this proof!
    Note that:

    • (a+b)^2 = a^2 + 2ab + b^2

    and

    • (\sqrt3)^2 = 3

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

    Put a = \sqrt3 and b = 1 in the expansion of (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
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