1.5a Lösning 6b

\[\begin{align} {1+x \over 1} & = {1 \over x} & | \;\; \cdot x{\color{White} x}\\ x \cdot (1 + x) & = 1 \\ x + x^2 & = 1 & | \;\; -1 \\ x^2 + x - 1 & = 0 \\ x_{1,2} & = -{1 \over 2} \pm \sqrt{{1 \over 4} + 1} \\ x_{1,2} & = -{1 \over 2} \pm \sqrt{5 \over 4} \\ x_{1,2} & = -{1 \over 2} \pm {1 \over 2} \cdot \sqrt{5} \\ x_1 & = 9 \\ x_2 & = - 1 \\ \end{align}\]