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Question Number 49461 by Pk1167156@gmail.com last updated on 07/Dec/18
The number of solutions for x from the equation x^2 −∣x∣−2=0 is
Thenumberofsolutionsforxfromtheequationx2−∣x∣−2=0is
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
∣x∣=x x>0 =−x x<0 x^2 −x−2=0 [considered x>0] x^2 −2x+x−2=0 x(x−2)+1(x−2)=0 (x−2)(x+1)=0 x=2 but we can not consider x=−1 because already cosidered x>0 now consider x<0 x^2 +x−2=0 x^2 +2x−x−2=0 x(x+2)−1(x+2)=0 (x+2)(x−1)=0 x=−2 but we can not consider x=1 because we already considerd x<0 solution is x=±2
∣x∣=xx>0=−xx<0x2−x−2=0[consideredx>0]x2−2x+x−2=0x(x−2)+1(x−2)=0(x−2)(x+1)=0x=2butwecannotconsiderx=−1becausealreadycosideredx>0nowconsiderx<0x2+x−2=0x2+2x−x−2=0x(x+2)−1(x+2)=0(x+2)(x−1)=0x=−2butwecannotconsiderx=1becausewealreadyconsiderdx<0solutionisx=±2
Answered by mr W last updated on 07/Dec/18
let t=∣x∣≥0 x^2 =∣x∣^2 ⇒t^2 −t−2=0 ⇒(t+1)(t−2)=0 ⇒t=−1 (<0⇒not ok) or t=2 (>0 ⇒ok) ⇒x=±2 ⇒2 solutions!
lett=∣x∣⩾0x2=∣x∣2⇒t2−t−2=0⇒(t+1)(t−2)=0⇒t=−1(<0⇒notok)ort=2(>0⇒ok)⇒x=±2⇒2solutions!

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