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Question Number 83621 by jagoll last updated on 04/Mar/20

∣ x+(1/x)∣ < 4   find the solution

$$\mid\:{x}+\frac{\mathrm{1}}{{x}}\mid\:<\:\mathrm{4}\: \\ $$ $$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$

Commented byjagoll last updated on 05/Mar/20

thank you both

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$

Answered by mr W last updated on 04/Mar/20

if x>0:  x+(1/x)<4  x^2 −4x+1<0  (x−2)^2 −3<0  −(√3)<x−2<(√3)  ⇒2−(√3)<x<2+(√3)    if x<0:  x+(1/x)>−4  x^2 +4x+1<0  (x+2)^2 −3<0  −(√3)<x+2<(√3)  −2−(√3)<x<−2+(√3)    solution:  −2−(√3)<x<−2+(√3) ∨ 2−(√3)<x<2+(√3)

$${if}\:{x}>\mathrm{0}: \\ $$ $${x}+\frac{\mathrm{1}}{{x}}<\mathrm{4} \\ $$ $${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}<\mathrm{0} \\ $$ $$\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}<\mathrm{0} \\ $$ $$−\sqrt{\mathrm{3}}<{x}−\mathrm{2}<\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\mathrm{2}−\sqrt{\mathrm{3}}<{x}<\mathrm{2}+\sqrt{\mathrm{3}} \\ $$ $$ \\ $$ $${if}\:{x}<\mathrm{0}: \\ $$ $${x}+\frac{\mathrm{1}}{{x}}>−\mathrm{4} \\ $$ $${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}<\mathrm{0} \\ $$ $$\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}<\mathrm{0} \\ $$ $$−\sqrt{\mathrm{3}}<{x}+\mathrm{2}<\sqrt{\mathrm{3}} \\ $$ $$−\mathrm{2}−\sqrt{\mathrm{3}}<{x}<−\mathrm{2}+\sqrt{\mathrm{3}} \\ $$ $$ \\ $$ $${solution}: \\ $$ $$−\mathrm{2}−\sqrt{\mathrm{3}}<{x}<−\mathrm{2}+\sqrt{\mathrm{3}}\:\vee\:\mathrm{2}−\sqrt{\mathrm{3}}<{x}<\mathrm{2}+\sqrt{\mathrm{3}} \\ $$

Answered by john santu last updated on 04/Mar/20

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