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Question Number 67844 by aliesam last updated on 01/Sep/19

x^2 +∣x∣−6=0

x2+∣x∣−6=0

Commented by gunawan last updated on 01/Sep/19

case 1 x<0 x^2 −x−6=0 (x−3)(x+2)=0 x=3 ∨x=−2 solution (2)^2 +∣−2∣−6=0 (3)^2 +∣3∣−6≠0 solution x=−2 case 2 x >0 x^2 +x−6=0 (x+3)(x−2)=0 x=−3 ∨ x=2 2^2 +∣2∣−6=0 (−3)^2 +∣−3∣−6≠0 solution is x=2 or x=−2

case1x<0x2−x−6=0(x−3)(x+2)=0x=3∨x=−2solution(2)2+∣−2∣−6=0(3)2+∣3∣−6≠0solutionx=−2case2x>0x2+x−6=0(x+3)(x−2)=0x=−3∨x=222+∣2∣−6=0(−3)2+∣−3∣−6≠0solutionisx=2orx=−2

Commented by mathmax by abdo last updated on 01/Sep/19

(e)⇒∣x∣^2 +∣x∣−6 let ∣x∣ =t with t≥0 ⇒t^2 +t−6 =0 Δ =1−4(−6) =25 ⇒t_1 =((−1+5)/2) =2 and t_2 =((−1−5)/2) =−3<0(to eliminate) and ∣x∣ =2 ⇒x =2 or x =−2 .

(e)⇒∣x∣2+∣x∣−6let∣x∣=twitht⩾0⇒t2+t−6=0Δ=1−4(−6)=25⇒t1=−1+52=2andt2=−1−52=−3<0(toeliminate)and∣x∣=2⇒x=2orx=−2.

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