Question

Express as a continued fraction the positive root of the following equation $x^2+2x-1=0$.

Answer 1

Given equation is $x^2+2x-1=0$.

$\Rightarrow x=\frac{-2\pm \sqrt{8}}{2}=\frac{-2\pm 2\sqrt{2}}{2}$

$=-1-\sqrt{2},\sqrt{2}-1$

Therefore, positive root is $\sqrt{2}-1$.

$\Rightarrow \sqrt{2}-1=\frac{1}{\sqrt{2}+1};$

$\Rightarrow \sqrt{2}+1=2+\sqrt{2}-1=2+\frac{1}{\sqrt{2}+1}$

Thus $\sqrt{2}-1=\frac{1}{2+\frac{1}{2+\frac{1}{2+....}}}$