Visa källkod för 1.5a Lösning 6b

<math>\begin{align} {1+x \over 1}  & = {1 \over x} \qquad\qquad | \;\, \cdot x{\color{White} xx}  \\
                   x \cdot (1 + x) & = 1                                                       \\
                          x + x^2  & = 1           \qquad\qquad | \, -1                        \\
                       x^2 + x - 1 & = 0                                                       \\
                           x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1}                 \\
                           x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4}                       \\
                            x_1    & = -{1 \over 2} + {1 \over 2} \cdot \sqrt{5}={1 \over 2}\,(-1 + \sqrt{5})={\sqrt{5}-1 \over 2} = 0,618033989\cdots \\
                            x_2    & = -{1 \over 2} - {1 \over 2} \cdot \sqrt{5}=-{1 \over 2}\,(1 + \sqrt{5})=-\,{1+\sqrt{5} \over 2}\; {\rm :negativ} \\
                             g     & = {\sqrt{5}-1 \over 2} \approx 0,618033989
       \end{align}</math>