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Question-189429

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Question Number 189429 by 073 last updated on 16/Mar/23
Commented by 073 last updated on 16/Mar/23
solution please
solutionplease
Answered by cortano12 last updated on 16/Mar/23
3∣x_1 ∣=∣x_2 ∣ ; x_1 ,x_2 <0 ⇒−3x_1 =−x_2 , x_2 =3x_1 ⇒x_1 +x_2 =−8 ⇒ { ((x_1 =−2)),((x_2 =−6)) :} ⇒f(x)=a(x+2)(x+6) ;(−4,8) ⇒8=a(−2)(2) ⇒a=−2 ∴ f(x)=−2(x^2 +8x+12) ⇒f(x)=−2x^2 −16x−24
3∣x1∣=∣x2∣;x1,x2<0⇒−3x1=−x2,x2=3x1⇒x1+x2=−8⇒{x1=−2x2=−6⇒f(x)=a(x+2)(x+6);(−4,8)⇒8=a(−2)(2)⇒a=−2∴f(x)=−2(x2+8x+12)⇒f(x)=−2x2−16x−24
Commented by 073 last updated on 16/Mar/23
nice solution thanks alot please another one
nicesolutionthanksalotpleaseanotherone

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