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Math Help - Adding infinite square roots

  1. #1
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    Adding infinite square roots

    Hi, I have a 'different' (not necessarily difficult) algebra problem here. Please help!



    Any help is appreciated.

    Thanks in advance.

    BG
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Let the expression be s.

    s^2=6+s
    s^2-s-6=0
    (s-3)(s+2)=0
    s=3 or -2

    Rejecting the negative root,

    s=3
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by BG5965 View Post
    Hi, I have a 'different' (not necessarily difficult) algebra problem here. Please help!



    Any help is appreciated.

    Thanks in advance.

    BG
    Well to simplify things lets suppose that the sum does exist, then

    x=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}+ ...}}}

    Then notice that red part is the "same" as x

    x=\sqrt{6+{\color{red} \sqrt{6+\sqrt{6+\sqrt{6}+ ...}}}}

    so you get

    x=\sqrt{6+x} \iff x^2=6+x \iff x^2-x-6=0 \iff (x-3)(x+2)=0

    We reject the negative solution because we are using only the posative square root. so x=3
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  4. #4
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    Hi, the only thing I don't understand is:

    Quote Originally Posted by TheEmptySet View Post
    Then notice that red part is the "same" as x

    x=\sqrt{6+{\color{red} \sqrt{6+\sqrt{6+\sqrt{6}+ ...}}}}
    Can someone please explain that further? Thanks BG
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  5. #5
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    Quote Originally Posted by BG5965 View Post
    Hi, I have a 'different' (not necessarily difficult) algebra problem here. Please help!



    Any help is appreciated.

    Thanks in advance.

    BG
    This looks like part of an IB portfolio question. You can discuss the matter with me via pm.

    Thread closed.
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