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Question Number 106655 by Algoritm last updated on 06/Aug/20

Commented by prakash jain last updated on 06/Aug/20

x_1 =kx_2 k^9 =−1⇒k=−1(e^(i2πm/9) ) (m=0..8) (I) k^(15) =−1 only applicable solution from (I) for k are k=−1,−e^(i2π/3) ,−e^(i4π/3) x_1 +x_2 =1 x_1 (1+k)=1⇒x_1 =(1/(1+k)) k=−1 is invalid ⇒k=−e^(2πi/3) ,−e^(4πi/3) c=x_1 x_2 =kx_1 ^2 =(k/((1+k)^2 )) where k=−e^(2πi/3) ,−e^(4πi/3)

x1=kx2k9=−1⇒k=−1(ei2πm/9)(m=0..8)(I)k15=−1onlyapplicablesolutionfrom(I)forkarek=−1,−ei2π/3,−ei4π/3x1+x2=1x1(1+k)=1⇒x1=11+kk=−1isinvalid⇒k=−e2πi/3,−e4πi/3c=x1x2=kx12=k(1+k)2wherek=−e2πi/3,−e4πi/3

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