continued fraction of $\sqrt{10k+3}$

Question

Could you please help me to find the continued fraction of

$$\sqrt{10k+3}.$$

Where $k$ is a positive integer.

All the best,

Answer 1

\begin{equation*} x+1=\sqrt{10k+3} \end{equation*} \begin{equation*} x^{2}+2x+1=10k+3 \end{equation*} \begin{equation*} x^{2}+2x=10k+2 \end{equation*} \begin{equation*} x\left(x+2\right)=10k+2 \end{equation*} \begin{equation*} x=\dfrac{10k+2}{2+x} \end{equation*} \begin{equation*} x=\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dots}}} \end{equation*} Then, \begin{equation*} \sqrt{10k+3}=1+x \end{equation*} \begin{equation*} \sqrt{10k+3}=1+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dots}}} \end{equation*}