x-1-x-lt-4-find-the-solution-

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

∣ x + \frac{1}{x} ∣ < 4

find the solution

Commented by jagoll last updated on 05/Mar/20

thank you both

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

i f x > 0 :

x + \frac{1}{x} < 4

x^{2} - 4 x + 1 < 0

{(x - 2)}^{2} - 3 < 0

- \sqrt{3} < x - 2 < \sqrt{3}

\Rightarrow 2 - \sqrt{3} < x < 2 + \sqrt{3}

i f x < 0 :

x + \frac{1}{x} > - 4

x^{2} + 4 x + 1 < 0

{(x + 2)}^{2} - 3 < 0

- \sqrt{3} < x + 2 < \sqrt{3}

- 2 - \sqrt{3} < x < - 2 + \sqrt{3}

s o l u t i o n :

- 2 - \sqrt{3} < x < - 2 + \sqrt{3} \lor 2 - \sqrt{3} < x < 2 + \sqrt{3}

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