Question Number 100660 by bobhans last updated on 28/Jun/20

∣x^2 −x∣ < 2+x . find solution set.

Commented bybramlex last updated on 28/Jun/20

since ∣x^2 −x∣ ≥0 then 2+x must be >0  (1) 2+x > 0 ; x >−2  (2) (x^2 −x−x−2)(x^2 −x+x+2) <0  (x^2 −2x−2)(x^2 +2)<0  (x−1)^2 −((√3))^2  < 0  (x−1−(√3))(x−1+(√3)) <0  1−(√3) < x < 1+(√3)  solution set we get from (1)∩(2)  ∴ 1−(√3) < x < 1+(√3)