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For the form $\sqrt{b}$, I could just apply the recursive quadratic formula: $$P_{k+1} = a_kQ_k - P_k$$ $$Q_{k+1} = \dfrac{d - P^2_{k+1}}{Q_k}$$ $$\alpha_k = \dfrac{P_k + \sqrt{d}}{Q_k}$$ $$a_k = \lfloor \alpha_k \rfloor$$

In this case, we have a coefficient namely $a$, so what's $d$? Is it still $b$?

Thanks,

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I think I figured it out, $a\sqrt{b} = \sqrt{a^2b}$. –  Chan May 2 '11 at 0:24

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up vote 2 down vote accepted

(So you can have something to "accept"...)

$a \sqrt{b} = \sqrt{a^2b}$ - Chan

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@Chaz: Thank you. –  Chan Jun 19 '11 at 21:28
    
@Chan: Hah! I answer even when no one is asking! –  The Chaz 2.0 Jun 19 '11 at 23:25

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