Complex Numbers And Quadratic Equations question 562
Question: Let $ \alpha ,{{\alpha }^{2}} $ be the roots of $ x^{2}+x+1=0 $ , then the equation whose roots are $ {{\alpha }^{31}},{{\alpha }^{62}} $ is [AMU 1999]
Options:
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 $
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Answer:
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 $ .
BETA
