Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com