1.5a Lösning 6b

\[\begin{align} {1+x \over 1} & = {1 \over x} & | \;\; \cdot x{\color{White} x}\\ x \cdot (1 + x) & = 1 \\ x + x^2 & = 1 & | \;\; -1 \\ x^2 + x - 1 & = 0 \\ x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1} \\ x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4} \\ x_1 & = -{1 \over 2} + {1 \over 2} \cdot \sqrt{5}&=-{1 \over 2}\,(1 - \sqrt{5})&=-{1-sqrt{5} \over 2} \\ x_2 & = -{1 \over 2} - {1 \over 2} \cdot \sqrt{5} \\ \end{align}\]