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Question Number 63481 by naka3546 last updated on 04/Jul/19
Find the solution of inequality : x^2 + ∣x∣ > 6
Findthesolutionofinequality:x2+∣x∣>6
Commented by mathmax by abdo last updated on 04/Jul/19
(ine) ⇒∣x∣^2 +∣x∣−6>0 ⇒t^2 +t−6>0 Δ=1−4(−6)=25 ⇒ t_1 =((−1+5)/2) =2 and t_2 =((−1−5)/2) =−3 ⇒ ∣x∣^2 +∣x∣−6 =(∣x∣−2)(∣x∣+3) so (ine) ⇒{∣x∣−2}{∣x∣+3}>0 ⇒ ∣x∣−2 >0 (because ∣x∣ +3>0 ⇒ x>2 or x<−2 ⇒ x ∈]−∞,−2[∪]2,+∞[
(ine)⇒∣x∣2+∣x∣−6>0⇒t2+t−6>0Δ=1−4(−6)=25⇒t1=−1+52=2andt2=−1−52=−3⇒∣x∣2+∣x∣−6=(∣x∣−2)(∣x∣+3)so(ine)⇒{∣x∣−2}{∣x∣+3}>0⇒∣x∣−2>0(because∣x∣+3>0⇒x>2orx<−2⇒x∈]−∞,−2[∪]2,+∞[
Answered by MJS last updated on 04/Jul/19
x>0 x^2 +x>6 x^2 +x−6=0 ⇒ x=−3∨x=2 ⇒ x<−3∨x>2 but x>0 ⇒ x>2 x<0 x^2 −x>6 x^2 −x−6=0 ⇒ x=−2∨x=3 ⇒ x<−2∨x>3 but x<0 ⇒ x<−2 x^2 +∣x∣>6 ⇒ x<−2∨2<x ⇔ x∈]−∞; −2[∪]2; +∞[
x>0x2+x>6x2+x−6=0⇒x=−3∨x=2⇒x<−3∨x>2butx>0⇒x>2x<0x2−x>6x2−x−6=0⇒x=−2∨x=3⇒x<−2∨x>3butx<0⇒x<−2x2+∣x∣>6⇒x<−2∨2<x⇔x∈]−∞;−2[∪]2;+∞[

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