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Question Number 161066 by blackmamba last updated on 11/Dec/21

x_1 ,x_2 be the roots of the equation x^2 +x+m=0 & x_1 ^5 +x_2 ^5 = 2021. Find the sum of the possible values of m.

$$\:{x}_{\mathrm{1}} \:,{x}_{\mathrm{2}} \:{be}\:{the}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$ $$\:\:\:\:\:\:{x}^{\mathrm{2}} +{x}+{m}=\mathrm{0}\:\&\:{x}_{\mathrm{1}} ^{\mathrm{5}} +{x}_{\mathrm{2}} ^{\mathrm{5}} \:=\:\mathrm{2021}. \\ $$ $$\:{Find}\:{the}\:{sum}\:{of}\:{the}\:{possible}\:{values} \\ $$ $$\:\:{of}\:{m}. \\ $$

Answered by cortano last updated on 11/Dec/21

⇒x^2 +x+m = 0 → { ((x_1 +x_2 =−1)),((x_1 x_2 = m)) :} ⇒x_1 ^2 +x_2 ^2 = 1−2m ⇒x_1 ^3 +x_2 ^3 = 3m−1 ⇒(x_1 ^3 +x_2 ^3 )(x_1 ^2 +x_2 ^2 )=(3m−1)(1−2m) ⇒x_1 ^5 +x_2 ^5 +(x_1 x_2 )^2 (x_1 +x_2 )=−6m^2 +5m−1 ⇒2021−m^2 +6m^2 −5m+1=0 ⇒5m^2 −5m+2022=0 ⇒Σ_(i=1) ^2 m_i = 1

$$\:\Rightarrow{x}^{\mathrm{2}} +{x}+{m}\:=\:\mathrm{0}\:\rightarrow\begin{cases}{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\mathrm{1}}\\{{x}_{\mathrm{1}} {x}_{\mathrm{2}} =\:{m}}\end{cases} \\ $$ $$\:\Rightarrow{x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} =\:\mathrm{1}−\mathrm{2}{m} \\ $$ $$\Rightarrow{x}_{\mathrm{1}} ^{\mathrm{3}} +{x}_{\mathrm{2}} ^{\mathrm{3}} \:=\:\mathrm{3}{m}−\mathrm{1} \\ $$ $$\Rightarrow\left({x}_{\mathrm{1}} ^{\mathrm{3}} +{x}_{\mathrm{2}} ^{\mathrm{3}} \right)\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} \right)=\left(\mathrm{3}{m}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{2}{m}\right) \\ $$ $$\Rightarrow{x}_{\mathrm{1}} ^{\mathrm{5}} +{x}_{\mathrm{2}} ^{\mathrm{5}} +\left({x}_{\mathrm{1}} {x}_{\mathrm{2}} \right)^{\mathrm{2}} \left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)=−\mathrm{6}{m}^{\mathrm{2}} +\mathrm{5}{m}−\mathrm{1} \\ $$ $$\Rightarrow\mathrm{2021}−{m}^{\mathrm{2}} +\mathrm{6}{m}^{\mathrm{2}} −\mathrm{5}{m}+\mathrm{1}=\mathrm{0} \\ $$ $$\Rightarrow\mathrm{5}{m}^{\mathrm{2}} −\mathrm{5}{m}+\mathrm{2022}=\mathrm{0} \\ $$ $$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{\mathrm{2}} {\sum}}{m}_{{i}} \:=\:\mathrm{1}\: \\ $$

Answered by TheSupreme last updated on 11/Dec/21

let,s x_1 =a x_2 =b ab=m a+b=1 (a+b)^5 =−1=a^5 +b^5 +5a^4 b+10a^3 b^2 +10a^2 b^3 +5ab^4 +b^5 −1=2021+5a^3 m−10m^2 +5mb^3 −2022=5m(a^3 +b^3 )−10m^2 (a^3 +b^3 )=(a+b)(a^2 −ab+b^2 )=−((a+b)^2 −3ab)=−(1−3m)=3m−1 −2022=5m(3m−1)−10m^2 −2022=5m^2 −5m no solutions Δ<0. sum all possible valye =0

$${let},{s}\:{x}_{\mathrm{1}} ={a}\:{x}_{\mathrm{2}} ={b} \\ $$ $${ab}={m} \\ $$ $${a}+{b}=\mathrm{1} \\ $$ $$\left({a}+{b}\right)^{\mathrm{5}} =−\mathrm{1}={a}^{\mathrm{5}} +{b}^{\mathrm{5}} +\mathrm{5}{a}^{\mathrm{4}} {b}+\mathrm{10}{a}^{\mathrm{3}} {b}^{\mathrm{2}} +\mathrm{10}{a}^{\mathrm{2}} {b}^{\mathrm{3}} +\mathrm{5}{ab}^{\mathrm{4}} +{b}^{\mathrm{5}} \\ $$ $$−\mathrm{1}=\mathrm{2021}+\mathrm{5}{a}^{\mathrm{3}} {m}−\mathrm{10}{m}^{\mathrm{2}} +\mathrm{5}{mb}^{\mathrm{3}} \\ $$ $$−\mathrm{2022}=\mathrm{5}{m}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)−\mathrm{10}{m}^{\mathrm{2}} \\ $$ $$\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)=\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)=−\left(\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{3}{ab}\right)=−\left(\mathrm{1}−\mathrm{3}{m}\right)=\mathrm{3}{m}−\mathrm{1} \\ $$ $$−\mathrm{2022}=\mathrm{5}{m}\left(\mathrm{3}{m}−\mathrm{1}\right)−\mathrm{10}{m}^{\mathrm{2}} \\ $$ $$−\mathrm{2022}=\mathrm{5}{m}^{\mathrm{2}} −\mathrm{5}{m} \\ $$ $${no}\:{solutions}\:\Delta<\mathrm{0}. \\ $$ $${sum}\:{all}\:{possible}\:{valye}\:=\mathrm{0} \\ $$

Commented bycortano last updated on 11/Dec/21

false

$${false} \\ $$

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