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

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

Commented byjagoll last updated on 05/Mar/20

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

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

$$ifx>0:$$ $$x+\frac{1}{x}<4$$ $$x^2-4x+1<0$$ $${(x-2)}^2-3<0$$ $$-\sqrt{3}<x-2<\sqrt{3}$$ $$\Rightarrow 2-\sqrt{3}<x<2+\sqrt{3}$$ $$$$ $$ifx<0:$$ $$x+\frac{1}{x}>-4$$ $$x^2+4x+1<0$$ $${(x+2)}^2-3<0$$ $$-\sqrt{3}<x+2<\sqrt{3}$$ $$-2-\sqrt{3}<x<-2+\sqrt{3}$$ $$$$ $$solution:$$ $$-2-\sqrt{3}<x<-2+\sqrt{3}\vee 2-\sqrt{3}<x<2+\sqrt{3}$$

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

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