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

∣x^2 −x∣ < 2+x . find solution set.

∣x2−x∣<2+x.findsolutionset.

Commented by bramlex 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)

since∣x2−x∣⩾0then2+xmustbe>0(1)2+x>0;x>−2(2)(x2−x−x−2)(x2−x+x+2)<0(x2−2x−2)(x2+2)<0(x−1)2−(3)2<0(x−1−3)(x−1+3)<01−3<x<1+3solutionsetwegetfrom(1)∩(2)∴1−3<x<1+3

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