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Strange Incredible Identity

  1. Dec 9, 2007 #1
    How can we proceed to prove the following identity ?
     

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  2. jcsd
  3. Dec 9, 2007 #2
    The original expression :
    [tex]\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \
    \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}\ }\ }[/tex]

    First, note that 16=11+5 .

    So, we can rewrite the last part :

    [tex]\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \
    \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{ ( \ \sqrt{11-2\sqrt{29}\ }\ +\ \sqrt{5}\ ) ^2\ }[/tex]

    Or:

    [tex]\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \
    \sqrt{11+2\sqrt{29}\ }\ +\ (\ \sqrt{11-2\sqrt{29}\ }\ +\ \sqrt{5}\ )[/tex]

    Thats the same as:

    [tex]\sqrt{22+2\sqrt{5}\ }\ = \ \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{11-2\sqrt{29}\ }[/tex]

    Squaring both sides, we get:

    [tex]22+2\sqrt{5}\ = 22\ +\ 2\sqrt{11^2\ -\ (2\sqrt{29})^2\ }[/tex]

    Or:

    [tex]22+2\sqrt{5}\ = 22\ +\ 2\sqrt{121\ -\ 4*29\ }[/tex]

    Which is true.

    :smile:
     
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