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Question Number 47581 by EagleEye1 last updated on 11/Nov/18

The root of the equation (2/(x−1)) + (1/(x+2)) + ((3x(x+1))/((x−1)(x+2))) = 0 among the following is

Therootoftheequation2x−1+1x+2+3x(x+1)(x−1)(x+2)=0amongthefollowingis

Commented by maxmathsup by imad last updated on 12/Nov/18

equation exist if x≠1 and x≠−2 . (e) ⇔ ((2x+4+x−1)/((x−1)(x+2))) +((3x^2 +3x)/((x−1)(x+2))) =0 ⇔((3x+3 +3x^2 +3x)/((x−1)(x+2))) =0 ⇔ 3x^2 +6x +3 =0 ⇔3(x^2 +2x+1)=0 ⇔(x+1)^2 =0 ⇔x=−1 .

equationexistifx≠1andx≠−2.(e)⇔2x+4+x−1(x−1)(x+2)+3x2+3x(x−1)(x+2)=0⇔3x+3+3x2+3x(x−1)(x+2)=0⇔3x2+6x+3=0⇔3(x2+2x+1)=0⇔(x+1)2=0⇔x=−1.

Answered by Rio Michael last updated on 11/Nov/18

((2(x+2)+1(x−1))/(x^2 +x−2))+((3x^2 +3)/(x^2 +x−2))=0 ((3x−3+3x^2 +3)/(x^2 +x−2))=0 ((3x^2 +3x)/(x^2 +x−2))=0 x=0 or x=−1 or x=1 or x=−2

2(x+2)+1(x−1)x2+x−2+3x2+3x2+x−2=03x−3+3x2+3x2+x−2=03x2+3xx2+x−2=0x=0orx=−1orx=1orx=−2

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