A) \[{{x}^{2}}-x+1=0\]
B) \[{{x}^{2}}+x-1=0\]
C) \[{{x}^{2}}+x+1=0\]
D) \[{{x}^{60}}+{{x}^{30}}+1=0\]
Correct Answer: C
Solution :
Give equation \[{{x}^{2}}+x+1=0\] Þ \[\alpha +{{\alpha }^{2}}=-1\] .....(i) and \[{{\alpha }^{3}}=1\] .....(ii) Now the equation whose roots are \[{{\alpha }^{31}}\]and \[{{\alpha }^{62}}\] \[\therefore \]\[{{\alpha }^{31}}+{{\alpha }^{62}}={{\alpha }^{31}}(1+{{\alpha }^{31}})\] Þ \[{{\alpha }^{31}}+{{\alpha }^{62}}={{\alpha }^{30}}.\alpha (1+{{\alpha }^{30}}.\alpha )\] \[{{\alpha }^{31}}+{{\alpha }^{62}}={{({{\alpha }^{3}})}^{10}}.\alpha \{1+{{({{\alpha }^{3}})}^{10}}.\alpha \}\] Þ \[{{\alpha }^{31}}+{{\alpha }^{62}}=\alpha (1+\alpha )\] Þ \[{{\alpha }^{31}}+{{\alpha }^{62}}=-1\] [from (i)] Again \[{{\alpha }^{31}}.{{\alpha }^{62}}={{\alpha }^{93}}\]Þ \[{{\alpha }^{31}}.{{\alpha }^{62}}={{[{{\alpha }^{3}}]}^{31}}=1\] Required equation is \[{{x}^{2}}-({{\alpha }^{31}}+{{\alpha }^{62}})x+{{\alpha }^{31}}.{{\alpha }^{62}}=0\] Þ \[{{x}^{2}}+x+1=0\]. Trick: \[\alpha =\frac{-1+i\sqrt{3}}{2}=\omega \] \[{{\alpha }^{2}}=\frac{-1-i\sqrt{3}}{2}={{\omega }^{2}}\] \[\therefore \,\,\,\,{{\alpha }^{31}}={{\omega }^{31}}=\omega \]and \[{{\alpha }^{62}}={{\omega }^{62}}={{\omega }^{2}}\] \[\therefore \] Equation is \[{{x}^{2}}+x+1=0\].
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