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Question-163861




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

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