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Question Number 161500 by HongKing last updated on 18/Dec/21

x^6 - 6x^5 + ax^4 + bx^3 + cx^2 + dx + 1 = 0 all the roots of the equation are positive find a+b+c+d=?

$$\mathrm{x}^{\mathrm{6}} \:-\:\mathrm{6x}^{\mathrm{5}} \:+\:\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{bx}^{\mathrm{3}} \:+\:\mathrm{cx}^{\mathrm{2}} \:+\:\mathrm{dx}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{are}\:\mathrm{positive} \\ $$$$\mathrm{find}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=? \\ $$

Answered by mr W last updated on 19/Dec/21

f(x)=x^6 -6x^5 +ax^4 +bx^3 +cx^2 +dx+1=0 it has six roots, say x_1 ,x_2 ,x_3 ,...,x_6 . given: x_1 ,x_2 ,x_3 ,...,x_6 > 0 we have x_1 +x_2 +x_3 +...+x_6 =−(−6)=6 x_1 x_2 x_3 ...x_6 =1 ⇒the only possibility is x_1 =x_2 =x_3 =...=x_6 =1 ⇒f(x)=k(x−1)^6 with k=1, i.e. f(x)=x^6 -6x^5 +ax^4 +bx^3 +cx^2 +dx+1=(x−1)^6 with x=1: f(1)=1-6+a+b+c+d+1=(1−1)^6 =0 ⇒a+b+c+d=4 or (x−1)^6 =x^6 −6x^5 +15x^4 −20x^3 +15x^2 −6x+1 ⇒a=15, b=−20, c=15, d=−6 ⇒a+b+c+d=15−20+15−6=4

$${f}\left({x}\right)=\mathrm{x}^{\mathrm{6}} -\mathrm{6x}^{\mathrm{5}} +\mathrm{ax}^{\mathrm{4}} +\mathrm{bx}^{\mathrm{3}} +\mathrm{cx}^{\mathrm{2}} +\mathrm{dx}+\mathrm{1}=\mathrm{0} \\ $$$${it}\:{has}\:{six}\:{roots},\:{say}\:{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{x}_{\mathrm{3}} ,...,{x}_{\mathrm{6}} . \\ $$$${given}:\:{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{x}_{\mathrm{3}} ,...,{x}_{\mathrm{6}} \:>\:\mathrm{0} \\ $$$${we}\:{have}\: \\ $$$$\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +...+{x}_{\mathrm{6}} =−\left(−\mathrm{6}\right)=\mathrm{6} \\ $$$$\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} ...{x}_{\mathrm{6}} =\mathrm{1} \\ $$$$\Rightarrow{the}\:{only}\:{possibility}\:{is}\:\: \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{2}} ={x}_{\mathrm{3}} =...={x}_{\mathrm{6}} =\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)={k}\left({x}−\mathrm{1}\right)^{\mathrm{6}} \:{with}\:{k}=\mathrm{1},\:{i}.{e}. \\ $$$${f}\left({x}\right)=\mathrm{x}^{\mathrm{6}} -\mathrm{6x}^{\mathrm{5}} +\mathrm{ax}^{\mathrm{4}} +\mathrm{bx}^{\mathrm{3}} +\mathrm{cx}^{\mathrm{2}} +\mathrm{dx}+\mathrm{1}=\left({x}−\mathrm{1}\right)^{\mathrm{6}} \\ $$$${with}\:{x}=\mathrm{1}: \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}-\mathrm{6}+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{1}=\left(\mathrm{1}−\mathrm{1}\right)^{\mathrm{6}} =\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}+\boldsymbol{{d}}=\mathrm{4} \\ $$$$ \\ $$$${or} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{6}} ={x}^{\mathrm{6}} −\mathrm{6}{x}^{\mathrm{5}} +\mathrm{15}{x}^{\mathrm{4}} −\mathrm{20}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{1} \\ $$$$\Rightarrow{a}=\mathrm{15},\:{b}=−\mathrm{20},\:{c}=\mathrm{15},\:{d}=−\mathrm{6} \\ $$$$\Rightarrow{a}+{b}+{c}+{d}=\mathrm{15}−\mathrm{20}+\mathrm{15}−\mathrm{6}=\mathrm{4} \\ $$

Commented by Tawa11 last updated on 19/Dec/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by HongKing last updated on 19/Dec/21

cool my dear Sir thank you so much

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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