Strange Incredible Identity

  • Thread starter mathslover
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  • #1
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How can we proceed to prove the following identity ?
 

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  • #2
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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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