x-4-x-3-x-2-x-1-0-

Question Number 173839 by Khalmohmmad last updated on 19/Jul/22

$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$

Commented by mr W last updated on 19/Jul/22

x_k =cos ((2kπ)/5)+i sin ((2kπ)/5) (k=1,2,3,4)

$${x}_{{k}} =\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{\mathrm{5}}+{i}\:\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{\mathrm{5}}\:\:\:\:\left({k}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right) \\ $$

Answered by mahdipoor last updated on 19/Jul/22

(x^4 +...+1)(x−1)=0×(x−1)⇒ x^5 −1=0⇒x=1 but x is root of (x−1) ⇒⇒wbitout answer

$$\left({x}^{\mathrm{4}} +…+\mathrm{1}\right)\left({x}−\mathrm{1}\right)=\mathrm{0}×\left({x}−\mathrm{1}\right)\Rightarrow \\ $$$${x}^{\mathrm{5}} −\mathrm{1}=\mathrm{0}\Rightarrow{x}=\mathrm{1} \\ $$$${but}\:{x}\:{is}\:{root}\:{of}\:\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow\Rightarrow{wbitout}\:{answer} \\ $$

Answered by Rasheed.Sindhi last updated on 19/Jul/22

x^4 +x^3 +x^2 +x+1=0 (x−1)(x^4 +x^3 +x^2 +x+1)=0 x^5 −1=0 x^5 =1 x is fifth root of unity Let δ is one 5th root of unity. All the 5th roots are δ ,δ^2 ,δ^3 ,δ^4 , δ^5 =1 x=1 is root of x−1 and is not root of the given equation. ∴ δ ,δ^2 ,δ^3 ,δ^4 are the roots of the given equation.

$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{5}} −\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{5}} =\mathrm{1} \\ $$$${x}\:{is}\:{fifth}\:{root}\:{of}\:{unity} \\ $$$${Let}\:\delta\:{is}\:{one}\:\mathrm{5}{th}\:{root}\:{of}\:{unity}. \\ $$$${All}\:{the}\:\mathrm{5}{th}\:{roots}\:{are} \\ $$$$\delta\:,\delta^{\mathrm{2}} ,\delta^{\mathrm{3}} ,\delta^{\mathrm{4}} ,\:\delta^{\mathrm{5}} =\mathrm{1} \\ $$$${x}=\mathrm{1}\:{is}\:{root}\:{of}\:{x}−\mathrm{1}\:{and}\:{is}\:{not}\:{root} \\ $$$${of}\:{the}\:{given}\:{equation}. \\ $$$$\therefore\:\delta\:,\delta^{\mathrm{2}} ,\delta^{\mathrm{3}} ,\delta^{\mathrm{4}} \:{are}\:{the}\:{roots}\:{of}\:{the}\:{given} \\ $$$${equation}. \\ $$

Answered by BaliramKumar last updated on 20/Jul/22

x^4 +x^3 +x^2 +x+1=(x^2 +ax+1)(x^2 +bx+1)=0 x^4 +x^3 +x^2 +x+1=x^4 +(a+b)x^3 +(ab+2)x^2 +(a+b)x+1=0 a+b=1 ab+2=1 a = (((√5)+1)/2) b = −(((√5)−1)/2) x^4 +x^3 +x^2 +x+1=[x^2 +((((√5)+1)/2))x+1][x^2 −((((√5)−1)/2))x+1]=0 x^2 +((((√5)+1)/2))x + 1 = 0 & x^2 −((((√5)−1)/2))x + 1 = 0 .....................

$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{bx}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}={x}^{\mathrm{4}} +\left({a}+{b}\right){x}^{\mathrm{3}} +\left({ab}+\mathrm{2}\right){x}^{\mathrm{2}} +\left({a}+{b}\right){x}+\mathrm{1}=\mathrm{0} \\ $$$${a}+{b}=\mathrm{1}\: \\ $$$${ab}+\mathrm{2}=\mathrm{1} \\ $$$${a}\:=\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:{b}\:=\:−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\left[{x}^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right){x}+\mathrm{1}\right]\left[{x}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right){x}+\mathrm{1}\right]=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right){x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\:\:\:\:\:\:\&\:\:\:\:\:\:{x}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right){x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$………………… \\ $$

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