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Math Help - Simplify [√(3+√5) + √(4-√15)] / [√(3+√5) - √(4-√15)]

  1. #1
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    Simplify [√(3+√5) + √(4-√15)] / [√(3+√5) - √(4-√15)]

    Simplify

    <br />
\frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}

    EDIT: Perfect!
    Last edited by mathwizard; June 17th 2008 at 01:24 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mathwizard View Post
    Simplify

    <br />
\frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}
    Before I type it all up is the answer

    (\sqrt{3}-1)\cdot\sqrt{5}+\sqrt{3}-2?
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  3. #3
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    Hello, mathwizard!

    Simplify: . <br />
\frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}
    Those expressions made me suspicious . . .


    Note that: . (1 + \sqrt{5})^2 \:=\:1 + 2\sqrt{5} + 5 \:=\:6 + 2\sqrt{5} \:=\:2(3 + \sqrt{5})
    . . Hence: . 3 + \sqrt{5} \:=\:\frac{(1 + \sqrt{5})^2}{2}

    Note that: . (\sqrt{5}-\sqrt{3})^2 \:=\:5 - 2\sqrt{15} + 3 \:=\:8 - 2\sqrt{15} \:=\:2(4-\sqrt{15})
    . . Hence: . 4 - \sqrt{15} \:=\:\frac{(\sqrt{5}-\sqrt{3})^2}{2}


    The numerator is: . \sqrt{\frac{(1 + \sqrt{5})^2}{2}} + \sqrt{\frac{(\sqrt{5}-\sqrt{3})^2}{2}} \; = \;\frac{1+\sqrt{5}}{\sqrt{2}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} \;=\; \frac{1 + 2\sqrt{5} - \sqrt{3}}{\sqrt{2}}


    The denominator is: . \sqrt{\frac{(1 + \sqrt{5})^2}{2}} - \sqrt{\frac{(\sqrt{5}-\sqrt{3})^2}{2}} \;=\; \frac{1+\sqrt{5}}{\sqrt{2}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} \;=\;\frac{1 + \sqrt{3}}{\sqrt{2}}


    The problem becomes: . \frac{\dfrac{1 +2\sqrt{5} - \sqrt{3}}{\sqrt{2}}} {\dfrac{1 + \sqrt{3}}{\sqrt{2}}} \;=\;\frac{1 + 2\sqrt{5}-\sqrt{3}}{1 + \sqrt{3}}

    Rationalize: . \frac{1 + 2\sqrt{5}-\sqrt{3}}{1 + \sqrt{3}}\cdot{\color{blue}\frac{1-\sqrt{3}}{1-\sqrt{3}}}

    . . which simplifies to: . \frac{-2\sqrt{3} + 2\sqrt{5} - 2\sqrt{15} + 4}{-2} \;\;=\;\;\boxed{\sqrt{3} - \sqrt{5} + \sqrt{15} -2}


    We agree, Mathstud28 !!

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