A) \[{{x}^{2}}-2x+4=0\]
B) \[{{x}^{2}}-x+1=0\]
C) \[{{x}^{2}}+x+4=0\]
D) \[{{x}^{2}}+2x-4=0\]
E) \[{{x}^{2}}+2x+4=0\]
Correct Answer: E
Solution :
\[a={{e}^{i2\pi /3}}=\omega \] \[\therefore \] \[a+\frac{1}{{{a}^{2}}}={{\omega }^{2}}+\frac{1}{{{\omega }^{2}}}=\omega +\omega =2\omega \] Similarly, \[{{a}^{2}}+\frac{1}{{{a}^{4}}}={{\omega }^{2}}+\frac{1}{{{\omega }^{4}}}=2{{\omega }^{2}}\] \[\therefore \] \[a+\frac{1}{{{a}^{2}}}+{{a}^{2}}+\frac{1}{{{a}^{4}}}=2\omega +2{{\omega }^{2}}\] \[=2(\omega +{{\omega }^{2}})=-2\] and\[\left( a+\frac{1}{{{a}^{2}}} \right)\left( {{a}^{2}}+\frac{1}{{{a}^{4}}} \right)=2\omega .2{{\omega }^{2}}=4{{\omega }^{3}}=4\] \[\therefore \]Required equation is\[{{x}^{2}}+2x+4=0\]
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