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

∣x^2 −x∣ < 2+x . find solution set.

$$\mid{x}^{\mathrm{2}} −{x}\mid\:<\:\mathrm{2}+{x}\:.\:{find}\:{solution}\:{set}. \\ $$

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)

$${since}\:\mid{x}^{\mathrm{2}} −{x}\mid\:\geqslant\mathrm{0}\:{then}\:\mathrm{2}+{x}\:{must}\:{be}\:>\mathrm{0} \\ $$ $$\left(\mathrm{1}\right)\:\mathrm{2}+{x}\:>\:\mathrm{0}\:;\:{x}\:>−\mathrm{2} \\ $$ $$\left(\mathrm{2}\right)\:\left({x}^{\mathrm{2}} −{x}−{x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} −{x}+{x}+\mathrm{2}\right)\:<\mathrm{0} \\ $$ $$\left({x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)<\mathrm{0} \\ $$ $$\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:<\:\mathrm{0} \\ $$ $$\left({x}−\mathrm{1}−\sqrt{\mathrm{3}}\right)\left({x}−\mathrm{1}+\sqrt{\mathrm{3}}\right)\:<\mathrm{0} \\ $$ $$\mathrm{1}−\sqrt{\mathrm{3}}\:<\:{x}\:<\:\mathrm{1}+\sqrt{\mathrm{3}} \\ $$ $${solution}\:{set}\:{we}\:{get}\:{from}\:\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right) \\ $$ $$\therefore\:\mathrm{1}−\sqrt{\mathrm{3}}\:<\:{x}\:<\:\mathrm{1}+\sqrt{\mathrm{3}}\: \\ $$

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