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Question Number 62424 by aliesam last updated on 21/Jun/19

Commented by mathmax by abdo last updated on 21/Jun/19

1 is not root for this equatio (e) ⇔((1−x^5 )/(1−x)) =0 ⇔ x^5 =1 let x =re^(iθ) ( we take x from C) x^5 =1 ⇔ r^5 e^(5iθ) =e^(i2kπ) ⇔ r=1 and 5θ =2kπ ⇒θ =((2kπ)/5) k∈[0,4] the roots are x_k =e^(i((2kπ)/5)) x_0 is not roots ⇒k∈[1,4] x_1 =e^((i2π)/5) ,x_2 = e^(i(((4π)/5))) =e^(i(π−(π/5))) =−e^(−((iπ)/5)) x_3 = e^(i(((6π)/5))) =e^(i(π+(π/5))) =−e^((iπ)/5) x_4 =e^(i(((8π)/5))) =e^(i(2π−((2π)/5))) = e^(−((i2π)/5)) and we see that x_4 =x_1 ^− and x_3 =x_2 ^−

1isnotrootforthisequatio(e)⇔1−x51−x=0⇔x5=1letx=reiθ(wetakexfromC)x5=1⇔r5e5iθ=ei2kπ⇔r=1and5θ=2kπ⇒θ=2kπ5k∈[0,4]therootsarexk=ei2kπ5x0isnotroots⇒k∈[1,4]x1=ei2π5,x2=ei(4π5)=ei(π−π5)=−e−iπ5x3=ei(6π5)=ei(π+π5)=−eiπ5x4=ei(8π5)=ei(2π−2π5)=e−i2π5andweseethatx4=x−1andx3=x−2

Commented by aliesam last updated on 21/Jun/19

well done sir thank you

welldonesirthankyou

Answered by tanmay last updated on 21/Jun/19

another way devide both side by x^2 x^2 +(1/x^2 )+x+(1/x)+1=0 x+(1/x)=a→x^2 +(1/x^2 )=a^2 −2 a^2 −2+a+1=0 a^2 +a−1=0 a=((−1±(√(1+4)))/2)=((−1±(√5))/2) x+(1/x)=a x^2 −ax+1=0 x=((a±(√(a^2 −4)))/2) x=(((((−1+(√5))/2))±(√(((5+1−2(√5))/4)−4)))/2) x=((((−1+(√5))/2)±(√((−10−2(√5))/4)))/2) x=((−1+(√5) ±i(√(10+2(√5))))/4)→(x_1 ,x_2 ) x=((((−1−(√5))/2)±(√(((5+1+2(√5))/4)−4)))/2) x=((−1−(√5) ±(√(2(√5) −10)))/4) x=((−1−(√5) ±i(√(10−2(√5))))/4)(x_3 ,x_4 )

anotherwaydevidebothsidebyx2x2+1x2+x+1x+1=0x+1x=a→x2+1x2=a2−2a2−2+a+1=0a2+a−1=0a=−1±1+42=−1±52x+1x=ax2−ax+1=0x=a±a2−42x=(−1+52)±5+1−254−42x=−1+52±−10−2542x=−1+5±i10+254→(x1,x2)x=−1−52±5+1+254−42x=−1−5±25−104x=−1−5±i10−254(x3,x4)

Commented by Tawa1 last updated on 21/Jun/19

Commented by Tawa1 last updated on 21/Jun/19

Help me with this sir.

Helpmewiththissir.

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