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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 by jagoll 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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