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3x-3-x-2-m-2-x-4-0-equation-root-x-1-x-2-x-3-and-x-1-1-x-2-1-x-3-find-m-

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Question Number 160777 by HongKing last updated on 06/Dec/21
3x^3 + x^2 + (m + 2)∙x + 4 = 0 equation root x_1 ; x_2 ; x_3 and x_1 = (1/x^2 ) + (1/x^3 ) find m=?
3x3+x2+(m+2)⋅x+4=0equationrootx1;x2;x3andx1=1x2+1x3findm=?
Commented by MJS_new last updated on 06/Dec/21
do you mean x_1 =(1/x_2 )+(1/x_3 ) ?
doyoumeanx1=1x2+1x3?
Commented by HongKing last updated on 06/Dec/21
Sorry dear Sir, yes
SorrydearSir,yes
Commented by HongKing last updated on 06/Dec/21
answer: -10
answer:−10
Answered by TheSupreme last updated on 06/Dec/21
3(x−(1/a)−(1/b))(x−a)(x−b)= 3(x−((b+a)/(ab)))(x^2 −ax−bx+ab)= =3(x^3 −ax^2 −bx^2 +abx−((b+a)/(ab)) x^2 +(((b+a)^2 )/(ab))x−b−a) { ((−a−b−((b+a)/(ab))=1)),((ab+(((b+a)^2 )/(ab))=m+2)),((−b−a=(4/3))) :} { ((−b−a=(4/3))),((4(1+(1/(ab)))=(1/3)→ab)),((−(((12)/(11)))−(((16)/9)/((12)/(11)))=m+2)) :}=−((12)/(11))
3(x−1a−1b)(x−a)(x−b)=3(x−b+aab)(x2−ax−bx+ab)==3(x3−ax2−bx2+abx−b+aabx2+(b+a)2abx−b−a){−a−b−b+aab=1ab+(b+a)2ab=m+2−b−a=43{−b−a=434(1+1ab)=13→ab−(1211)−1691211=m+2=−1211
Answered by mr W last updated on 06/Dec/21
x_1 +x_2 +x_3 =−(1/3) x_1 x_2 x_3 =−(4/3) x_1 =(1/x_2 )+(1/x_3 )=((x_2 +x_3 )/(x_2 x_3 )) ⇒x_2 +x_3 =x_1 x_2 x_3 =−(4/3) ⇒x_2 +x_3 +x_1 =−(4/3)+x_1 ⇒−(1/3)=−(4/3)+x_1 ⇒x_1 =1 3×1^3 +1^2 +(m+2)×1+4=0 m+10=0 ⇒m=−10
x1+x2+x3=−13x1x2x3=−43x1=1x2+1x3=x2+x3x2x3⇒x2+x3=x1x2x3=−43⇒x2+x3+x1=−43+x1⇒−13=−43+x1⇒x1=13×13+12+(m+2)×1+4=0m+10=0⇒m=−10
Commented by HongKing last updated on 06/Dec/21
thank you my dear Sir cool
thankyoumydearSircool
Answered by mnjuly1970 last updated on 06/Dec/21
−(4/3) + x_( 1) = ((−1)/3) ⇒ x_( 1) = 1 3+1 +m+2+4 =0 m= −10
−43+x1=−13⇒x1=13+1+m+2+4=0m=−10
Commented by HongKing last updated on 06/Dec/21
cool thank you my dear Sir
coolthankyoumydearSir

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