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\[\alpha ,\,\,\beta \]are the roots of\[{{x}^{2}}+x+1=0\] \[\Rightarrow \] \[{{x}^{2}}-(\omega +{{\omega }^{2}})x+1=0\] \[\Rightarrow \] \[(x-\omega )(x-{{\omega }^{2}})=0\] \[\therefore \] \[\alpha =\omega ,\,\,\beta ={{\omega }^{2}}\] Now,\[{{\alpha }^{19}}={{\omega }^{19}}={{({{\omega }^{3}})}^{6}}\omega =\omega \] and \[{{\beta }^{7}}={{({{\omega }^{2}})}^{7}}={{({{\omega }^{3}})}^{4}}{{\omega }^{2}}={{\omega }^{2}}\] \[\therefore \]Equation whose roots are \[{{\alpha }^{19}}=\omega \]and\[{{\beta }^{7}}={{\omega }^{2}}\]is \[{{x}^{2}}-(\omega +{{\omega }^{2}})x+\omega \cdot {{\omega }^{2}}=1\] \[\therefore \] \[{{x}^{2}}+x+1=0\]
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