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Question Number 7600 by Rohit last updated on 05/Sep/16

solve ∣((x^2 −3x−1)/(x^2 +x+1))∣<3 give solution

solve∣x2−3x−1x2+x+1∣<3givesolution

Answered by Yozzia last updated on 05/Sep/16

∣u(x)∣<a⇒−a<u(x)<a. ∴∣((x^2 −3x−1)/(x^2 +x+1))∣<3⇒−3<((x^2 −3x−1)/(x^2 +x+1))<3. Now, x^2 +x+1>0 ∀x∈R. To see this observe that x^2 +x+1=(x+(1/2))^2 +(3/4) and for ∀x∈R, min((x+(1/2))^2 +(3/4))=(3/4)>0. ⇒x^2 +x+1>0 ∀x∈R. ∴ −3(x^2 +x+1)<x^2 −3x−1<3(x^2 +x+1) −−−−−−−−−−−−−−−−−−−−−−−−−−−− For x^2 −3x−1>−3x^2 −3x−3 4x^2 +2>0⇒2x^2 +1>0 which is true ∀x∈R. −−−−−−−−−−−−−−−−−−−−−−−−− For x^2 −3x−1<3x^2 +3x+3 2x^2 +6x+4>0 x^2 +3x+2>0 (x+1)(x+2)>0⇒ x>−1 or x<−2 −−−−−−−−−−−−−−−−−−−−−−−−−− In all x>−1 or x<−2.

∣u(x)∣<a⇒−a<u(x)<a.∴∣x2−3x−1x2+x+1∣<3⇒−3<x2−3x−1x2+x+1<3.Now,x2+x+1>0∀x∈R.Toseethisobservethatx2+x+1=(x+12)2+34andfor∀x∈R,min((x+12)2+34)=34>0.⇒x2+x+1>0∀x∈R.∴−3(x2+x+1)<x2−3x−1<3(x2+x+1)−−−−−−−−−−−−−−−−−−−−−−−−−−−−Forx2−3x−1>−3x2−3x−34x2+2>0⇒2x2+1>0whichistrue∀x∈R.−−−−−−−−−−−−−−−−−−−−−−−−−Forx2−3x−1<3x2+3x+32x2+6x+4>0x2+3x+2>0(x+1)(x+2)>0⇒x>−1orx<−2−−−−−−−−−−−−−−−−−−−−−−−−−−Inallx>−1orx<−2.

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