Question Number 83621 by jagoll last updated on 04/Mar/20
$$\mid\:{x}+\frac{\mathrm{1}}{{x}}\mid\:<\:\mathrm{4}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$
Commented by jagoll last updated on 05/Mar/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$
Answered by mr W last updated on 04/Mar/20
$${if}\:{x}>\mathrm{0}: \\ $$$${x}+\frac{\mathrm{1}}{{x}}<\mathrm{4} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}<\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}<\mathrm{0} \\ $$$$−\sqrt{\mathrm{3}}<{x}−\mathrm{2}<\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}−\sqrt{\mathrm{3}}<{x}<\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${if}\:{x}<\mathrm{0}: \\ $$$${x}+\frac{\mathrm{1}}{{x}}>−\mathrm{4} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}<\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}<\mathrm{0} \\ $$$$−\sqrt{\mathrm{3}}<{x}+\mathrm{2}<\sqrt{\mathrm{3}} \\ $$$$−\mathrm{2}−\sqrt{\mathrm{3}}<{x}<−\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${solution}: \\ $$$$−\mathrm{2}−\sqrt{\mathrm{3}}<{x}<−\mathrm{2}+\sqrt{\mathrm{3}}\:\vee\:\mathrm{2}−\sqrt{\mathrm{3}}<{x}<\mathrm{2}+\sqrt{\mathrm{3}} \\ $$
Answered by john santu last updated on 04/Mar/20
