A) \[{{x}^{2}}-x-1=0\]
B) \[{{x}^{2}}-x+1=0\]
C) \[{{x}^{2}}+x-1=0\]
D) \[{{x}^{2}}+x+1=0\]
Correct Answer: D
Solution :
Given \[{{x}^{2}}+x+1=0\] \ \[x=\frac{1}{2}[-1\pm i\sqrt{3}]=\frac{1}{2}(-1+i\sqrt{3}),\frac{1}{2}(-1-i\sqrt{3})\]\[=\omega ,{{\omega }^{2}}\] But\[{{\alpha }^{19}}={{\omega }^{19}}=\omega \]and \[{{\beta }^{7}}={{\omega }^{14}}={{\omega }^{2}}.\] Hence the equation will be same.
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