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Math Help - Square root madness!

  1. #1
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    Square root madness!

    Find \sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}
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    Hello,
    Quote Originally Posted by usagi_killer View Post
    Find \sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}
    Note that 17-4 \sqrt{15}=5+2^2 \cdot 3-2 \cdot \sqrt{5}\cdot (2 \sqrt{3})=(\sqrt{5})^2+(2\sqrt{3})^2-2 \cdot \sqrt{5}\cdot (2 \sqrt{3})=(\sqrt{5}-2\sqrt{3})^2
    (I don't have a particular method for this one... maybe solve for a and b in : aČ+bČ=17 and ab=2sqrt(15))

    Hence :
    \sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}=\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+|\sqrt{5}-2\sqrt{3}|}}

    But 2 \sqrt{3}=\sqrt{12}>\sqrt{5}
    So |\sqrt{5}-2\sqrt{3}|=2\sqrt{3}-\sqrt{5}

    This gives :
    \sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+2\sqrt{3}-\sqrt{5}}}=\sqrt{-\sqrt{3}+\sqrt{4+2\sqrt{3}}}

    Note that 4+2\sqrt{3}=(\sqrt{3})^2+1+2 \sqrt{3}=(\sqrt{3}+1)^2

    So your expression is :
    \sqrt{-\sqrt{3}+\sqrt{3}+1}=\boxed{1}


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