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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=?
$$\mathrm{3x}^{\mathrm{3}} \:+\:\mathrm{x}^{\mathrm{2}} \:+\:\left(\mathrm{m}\:+\:\mathrm{2}\right)\centerdot\mathrm{x}\:+\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{equation}\:\mathrm{root}\:\:\mathrm{x}_{\mathrm{1}} \:;\:\mathrm{x}_{\mathrm{2}} \:;\:\mathrm{x}_{\mathrm{3}} \\ $$$$\mathrm{and}\:\:\mathrm{x}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} } \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{m}}=? \\ $$
Commented by MJS_new last updated on 06/Dec/21
do you mean x_1 =(1/x_2 )+(1/x_3 ) ?
$$\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:{x}_{\mathrm{1}} =\frac{\mathrm{1}}{{x}_{\mathrm{2}} }+\frac{\mathrm{1}}{{x}_{\mathrm{3}} }\:? \\ $$
Commented by HongKing last updated on 06/Dec/21
Sorry dear Sir, yes
$$\mathrm{Sorry}\:\mathrm{dear}\:\mathrm{Sir},\:\mathrm{yes} \\ $$
Commented by HongKing last updated on 06/Dec/21
answer: -10
$$\mathrm{answer}:\:-\mathrm{10} \\ $$
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))
$$\mathrm{3}\left({x}−\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)\left({x}−{a}\right)\left({x}−{b}\right)= \\ $$$$\mathrm{3}\left({x}−\frac{{b}+{a}}{{ab}}\right)\left({x}^{\mathrm{2}} −{ax}−{bx}+{ab}\right)= \\ $$$$=\mathrm{3}\left({x}^{\mathrm{3}} −{ax}^{\mathrm{2}} −{bx}^{\mathrm{2}} +{abx}−\frac{{b}+{a}}{{ab}}\:{x}^{\mathrm{2}} +\frac{\left({b}+{a}\right)^{\mathrm{2}} }{{ab}}{x}−{b}−{a}\right) \\ $$$$\begin{cases}{−{a}−{b}−\frac{{b}+{a}}{{ab}}=\mathrm{1}}\\{{ab}+\frac{\left({b}+{a}\right)^{\mathrm{2}} }{{ab}}={m}+\mathrm{2}}\\{−{b}−{a}=\frac{\mathrm{4}}{\mathrm{3}}}\end{cases} \\ $$$$\begin{cases}{−{b}−{a}=\frac{\mathrm{4}}{\mathrm{3}}}\\{\mathrm{4}\left(\mathrm{1}+\frac{\mathrm{1}}{{ab}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\rightarrow{ab}}\\{−\left(\frac{\mathrm{12}}{\mathrm{11}}\right)−\frac{\frac{\mathrm{16}}{\mathrm{9}}}{\frac{\mathrm{12}}{\mathrm{11}}}={m}+\mathrm{2}}\end{cases}=−\frac{\mathrm{12}}{\mathrm{11}} \\ $$
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
$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} =−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}}{{x}_{\mathrm{2}} }+\frac{\mathrm{1}}{{x}_{\mathrm{3}} }=\frac{{x}_{\mathrm{2}} +{x}_{\mathrm{3}} }{{x}_{\mathrm{2}} {x}_{\mathrm{3}} } \\ $$$$\Rightarrow{x}_{\mathrm{2}} +{x}_{\mathrm{3}} ={x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} =−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{1}} =−\frac{\mathrm{4}}{\mathrm{3}}+{x}_{\mathrm{1}} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{\mathrm{3}}=−\frac{\mathrm{4}}{\mathrm{3}}+{x}_{\mathrm{1}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{3}×\mathrm{1}^{\mathrm{3}} +\mathrm{1}^{\mathrm{2}} +\left({m}+\mathrm{2}\right)×\mathrm{1}+\mathrm{4}=\mathrm{0} \\ $$$${m}+\mathrm{10}=\mathrm{0} \\ $$$$\Rightarrow{m}=−\mathrm{10} \\ $$
Commented by HongKing last updated on 06/Dec/21
thank you my dear Sir cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$
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
$$\:\:\:−\frac{\mathrm{4}}{\mathrm{3}}\:+\:\:{x}_{\:\mathrm{1}} \:=\:\frac{−\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{x}_{\:\mathrm{1}} \:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\mathrm{3}+\mathrm{1}\:+{m}+\mathrm{2}+\mathrm{4}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{m}=\:−\mathrm{10} \\ $$
Commented by HongKing last updated on 06/Dec/21
cool thank you my dear Sir
$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

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