Question Number 163861 by mathlove last updated on 11/Jan/22
Answered by Ar Brandon last updated on 11/Jan/22
$${x}=\sqrt{\mathrm{5}}+\frac{\mathrm{1}}{{x}}\Rightarrow{x}=\frac{\sqrt{\mathrm{5}}\pm\sqrt{\mathrm{9}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\mathrm{2}}\:\left({x}>\mathrm{0},\:\sqrt{\mathrm{5}}−\mathrm{3}<\mathrm{0}\right) \\ $$