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I have to find the continued fraction form of sqrt(6). I have tried it, and have the answers but I can't get to the correct answer. If someone could help me that would be much appreciated. Thank you!

Patrick Feltes
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    Can you post your work so we can look it over and identify the issue? – Amzoti Nov 25 '14 at 01:41
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    Also http://math.stackexchange.com/questions/213683/calculate-the-continued-fraction-of-square-root – MJD Nov 25 '14 at 01:45

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Hint: $2 < \sqrt{6} < 3$ so it starts $2 + 1/\ldots$. If $\sqrt{6} = 2 + 1/x$ then $x = \dfrac{1}{-2+\sqrt{6}} = \dfrac{-2-\sqrt{6}}{4 - 6} = \ldots$

Robert Israel
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