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 byjagoll 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 | ||