A) \[{{x}^{2}}-x+1=0\]
B) \[{{x}^{2}}+x+1=0\]
C) \[{{x}^{2}}+x-1=0\]
D) \[{{x}^{2}}-x-1=0\]
E) \[2{{x}^{2}}+x+1=0\]
Correct Answer: B
Solution :
Given equation is\[{{x}^{2}}+x+1=0\]. \[\therefore \] \[x=\frac{-1\pm \sqrt{{{1}^{2}}-4}}{2.1}\] \[=\frac{-1\pm \sqrt{3}i}{2}\] \[\Rightarrow \] \[x=\omega ,{{\omega }^{2}}\] where \[\omega =\frac{-1+\sqrt{3}i}{2}\]and \[{{\omega }^{2}}=\frac{-1-\sqrt{3}i}{2}\] Let \[\alpha =\omega \]and\[\beta ={{\omega }^{2}}\] \[\therefore \] \[{{\alpha }^{22}}={{\omega }^{22}}=\omega \]and \[{{\beta }^{19}}={{({{\omega }^{2}})}^{19}}={{\omega }^{2}}\] Since, the roots of the required equation is \[\omega \]and\[{{\omega }^{2}}\]. Then the required equation is also \[{{x}^{2}}+x+1=0.\]
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