solve-x-1-x-gt-2-

Question Number 95634 by i jagooll last updated on 26/May/20

solve ∣ x + \frac{1}{x} ∣ > 2

Answered by john santu last updated on 26/May/20

solution :

this is equivalent to ∣ \frac{x^{2} + 1}{x} ∣ > 2

\frac{x^{2} + 1}{∣ x ∣} > 2 [since x^{2} + 1 > 0]

x^{2} + 1 > 2 ∣ x ∣; x^{2} - 2 ∣ x ∣ + 1 > 0

∣ x ∣^{2} - 2 ∣ x ∣ + 1 > 0, {(∣ x ∣ - 1)}^{2} > 0

∣ x ∣ \neq 1 . answer all x e x c e p t x = 1 a n d x = - 1

Commented by peter frank last updated on 26/May/20

thank you

Commented by i jagooll last updated on 26/May/20

thank you