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Question Number 122322 by EngLewis last updated on 15/Nov/20
∣x+1∣<∣x^2 +2x+2∣ solve for x
∣x+1∣<∣x2+2x+2∣solveforx
Answered by mathmax by abdo last updated on 15/Nov/20
we have x^2 +2x+2 =(x+1)^2 +1>0 ⇒∣x^2 +2x+2∣=x^2 +2x+2 e⇒∣x+1∣−x^2 −2x−2<0 let f(x)=∣x+1∣−x^2 −2x−2 x −∞ −1 +∞ ∣x+1∣ −x−1 0 x+1 f(x) −x^2 −3x−3 −x^2 −x−1 ⇒ f(x)= { ((−x^2 −3x−3 ,x≤−1)),((−x^2 −x−1 ,x≥−1)) :} case1 x≤−1 f(x)<0 ⇒−x^2 −3x−3<0 ⇒x^2 +3x+3>0 Δ=9−12 =−3 ⇒this polynom is slways >0 ⇒S_1 =]−∞ ,−1] case2 x≥1 f(x)<0 ⇒−x^2 −x−1<0 ⇒x^2 +x+1>0 this polynom is also >0 ⇒S_2 =[−1,+∞[ so S =∪ S_i =R
wehavex2+2x+2=(x+1)2+1>0⇒∣x2+2x+2∣=x2+2x+2e⇒∣x+1∣−x2−2x−2<0letf(x)=∣x+1∣−x2−2x−2x−∞−1+∞∣x+1∣−x−10x+1f(x)−x2−3x−3−x2−x−1⇒f(x)={−x2−3x−3,x⩽−1−x2−x−1,x⩾−1case1x⩽−1f(x)<0⇒−x2−3x−3<0⇒x2+3x+3>0Δ=9−12=−3⇒thispolynomisslways>0⇒S1=]−∞,−1]case2x⩾1f(x)<0⇒−x2−x−1<0⇒x2+x+1>0thispolynomisalso>0⇒S2=[−1,+∞[soS=∪Si=R
Answered by MJS_new last updated on 15/Nov/20
∣x+1∣<∣(x+1)^2 +1∣ ∣x+1∣<(x+1)^2 +1 t=x+1 ∣t∣<t^2 +1 (1) t<0 ⇒ −t<t^2 +1 ⇔ t^2 +t+1>0 always true (2) t>0 ⇒ t<t^2 +1 ⇔ t^2 −t+1>0 always true ⇒ solution is x∈R
∣x+1∣<∣(x+1)2+1∣∣x+1∣<(x+1)2+1t=x+1∣t∣<t2+1(1)t<0⇒−t<t2+1⇔t2+t+1>0alwaystrue(2)t>0⇒t<t2+1⇔t2−t+1>0alwaystrue⇒solutionisx∈R
Answered by mathmax by abdo last updated on 15/Nov/20
another way but eazy (x^2 +2x+2)^2 −(x+1)^2 =(x^2 +2x+2−x−1)(x^2 +2x+2+x+1) =(x^2 +x+1)( x^2 +3x+3) but x^2 +x+1>0 ∀x due yo Δ<0 x^2 +3x+3 >0 ∀x due to Δ<0 ⇒ ∀x ∈R (x+1)^2 <(x^2 +2x+2)^2 ⇒ ∣x+1∣<x^2 +2x+2
anotherwaybuteazy(x2+2x+2)2−(x+1)2=(x2+2x+2−x−1)(x2+2x+2+x+1)=(x2+x+1)(x2+3x+3)butx2+x+1>0∀xdueyoΔ<0x2+3x+3>0∀xduetoΔ<0⇒∀x∈R(x+1)2<(x2+2x+2)2⇒∣x+1∣<x2+2x+2
Answered by TANMAY PANACEA last updated on 16/Nov/20
x^2 +2x+2 (x+1)^2 +1 (x+1)^2 always positive so (x+1)^2 +1 always greater than (x+1)
x2+2x+2(x+1)2+1(x+1)2alwayspositiveso(x+1)2+1alwaysgreaterthan(x+1)

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