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Math Help - [SOLVED] Simplifying an expression with radicals

  1. #1
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    [SOLVED] Simplifying an expression with radicals

    The problem is the following:

    \sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}

    I have to simplify it, but I have no idea how to start. Any help is appreciated. Thanks!

    Pharod
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  2. #2
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    Hello Pharod

    Welcome to Math Help Forum!
    Quote Originally Posted by Pharod View Post
    The problem is the following:

    \sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}

    I have to simplify it, but I have no idea how to start. Any help is appreciated. Thanks!

    Pharod
    Let x = \sqrt{a+b}+\sqrt{a-b}, where a = 8 and b = 2\sqrt{10+2\sqrt5}

    Then x^2 = a+b+2\sqrt{a^2-b^2}+a-b
    =2a+2\sqrt{a^2-b^2}

    =16+2\sqrt{24-8\sqrt5}

    =4(4+\sqrt{6-2\sqrt5})
    Also 6-2\sqrt5 = (\sqrt5 - 1)^2

    \Rightarrow \sqrt{6-2\sqrt5} = \sqrt5 - 1

    \Rightarrow x^2 = 4(4+\sqrt5-1)

    \Rightarrow x = 2\sqrt{3+\sqrt5}

    I don't think it will simplify any more.

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post

    \vdots
    \Rightarrow x = 2\sqrt{3+\sqrt5}

    I don't think it will simplify any more.
    It will go a bit further: (1+\sqrt5)^2 = 6+2\sqrt5, and so 2\sqrt{3+\sqrt5} = \sqrt2\sqrt{6+2\sqrt5} = \sqrt2(1+\sqrt5) = \sqrt2+\sqrt{10}
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  4. #4
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    Hey, thank you both for leading me. I knew it had to be something with squaring, I just couldn't get to it for some reason.

    Pharod
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  5. #5
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    Hello, Grandad!

    x \:=\: 2\sqrt{3+\sqrt5}

    I don't think it will simplify any more.

    3 + \sqrt{5} \:=\:\frac{(1 + \sqrt{5})^2}{2} \;=\;2\cdot\frac{(1+\sqrt{5})^2}{4}

    . . . . . .  \;=\;2\left(\underbrace{\frac{1+\sqrt{5}}{2}}_{\te  xt{Golden Mean}}\right)^2 \;=\;2\,\phi^2


    Therefore: . x \;=\;2\sqrt{2}\,\phi

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