1.5a Lösning 6b

\[\begin{align} {1+x \over 1} & = {1 \over x} & | \;\; \cdot x \\ \sqrt{6 x + 10} & = x - 1 & | \; (\;\;\;)^2 \\ 6 x + 10 & = (x - 1)^2 \\ 6 x + 10 & = x^2 - 2 x + 1 \qquad\qquad & | - 10 \\ 6 x & = x^2 - 2 x - 9 \qquad\qquad & | - 6 x \\ 0 & = x^2 - 8 x - 9 \\ \end{align}\]

\[\begin{align} {1+x \over 1} & = {1 \over x} \\ x^2 - 8 x - 9 & = 0 \\ x_{1,2} & = 4 \pm \sqrt{16 + 9} \\ x_{1,2} & = 4 \pm 5 \\ x_1 & = 9 \\ x_2 & = - 1 \\ \end{align}\]