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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)

$${x}_{\mathrm{1}} ={kx}_{\mathrm{2}} \\ $$$${k}^{\mathrm{9}} =−\mathrm{1}\Rightarrow{k}=−\mathrm{1}\left({e}^{{i}\mathrm{2}\pi{m}/\mathrm{9}} \right)\:\:\left({m}=\mathrm{0}..\mathrm{8}\right)\:\left({I}\right) \\ $$$${k}^{\mathrm{15}} =−\mathrm{1} \\ $$$${only}\:{applicable}\:{solution}\:{from}\:\left({I}\right) \\ $$$${for}\:{k}\:{are} \\ $$$${k}=−\mathrm{1},−{e}^{{i}\mathrm{2}\pi/\mathrm{3}} ,−{e}^{{i}\mathrm{4}\pi/\mathrm{3}} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\mathrm{1} \\ $$$${x}_{\mathrm{1}} \left(\mathrm{1}+{k}\right)=\mathrm{1}\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}+{k}} \\ $$$${k}=−\mathrm{1}\:\mathrm{is}\:\mathrm{invalid} \\ $$$$\Rightarrow{k}=−{e}^{\mathrm{2}\pi{i}/\mathrm{3}} ,−{e}^{\mathrm{4}\pi{i}/\mathrm{3}} \\ $$$${c}={x}_{\mathrm{1}} {x}_{\mathrm{2}} ={kx}_{\mathrm{1}} ^{\mathrm{2}} =\frac{{k}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} } \\ $$$${where}\:{k}=−{e}^{\mathrm{2}\pi{i}/\mathrm{3}} ,−{e}^{\mathrm{4}\pi{i}/\mathrm{3}} \\ $$

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