Question

Express as a continued fraction the positive root of the following equation $7x^2-8x-3=0$.

Answer 1

${7x}^2-8x-3=0$

$\Rightarrow x=\frac{8\pm \sqrt{64+84}}{14}$

$x=\frac{8\pm \sqrt{148}}{14}$

$x=\frac{8\pm 2\sqrt{37}}{14}$

$x=\frac{4\pm \sqrt{37}}{7}$

So, the positive root is $\frac{4+\sqrt{37}}{7}$

$\sqrt{37}=1+\sqrt{37}-1$

$=1+\frac{36}{1+\sqrt{37}}$

$=1+\frac{36}{1+1+\frac{36}{1+\sqrt{37}}}$

$=1+\frac{36}{2+\frac{36}{1+1+\frac{1}{1+\sqrt{37}}}}$

$=1+\frac{36}{2+\frac{36}{2+\frac{1}{2+...}}}$

$\therefore \frac{4}{7}+\frac{1}{7}[1+\frac{36}{2+\frac{36}{2+...}}]$