How can we proceed to prove the following identity ?
Dec 9, 2007 #1 mathslover 17 0 How can we proceed to prove the following identity ? Attachments sum.jpg 9.5 KB Views: 363
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Dec 9, 2007 #2 Rogerio 403 1 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.
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.