Complex Numbers And Quadratic Equations question 161
Question: If $ a=\cos (2\pi /7)+i\sin (2\pi /7), $ then the quadratic equation whose roots are $ \alpha =a+a^{2}+a^{4} $ and $ \beta =a^{3}+a^{5}+a^{6} $ is [RPET 2000]
Options:
A) $ x^{2}-x+2=0 $
B) $ x^{2}+x-2=0 $
C) $ x^{2}-x-2=0 $
D) $ x^{2}+x+2=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ a=\cos (2\pi /7)+i\sin (2\pi /7) $ $ a^{7}={{[\cos (2\pi /7)+i\sin (2\pi /7)]}^{7}} $ $ =\cos 2\pi +i\sin 2\pi =1 $ …..(i) $ S=\alpha +\beta =(a+a^{2}+a^{4})+(a^{3}+a^{5}+a^{6}) $ $ S=a+a^{2}+a^{3}+a^{4}+a^{5}+a^{6} $ $ =\frac{a(1-a^{6})}{1-a} $ $ S=\frac{a-a^{7}}{1-a}=\frac{a-1}{1-a}=-1 $ …..(ii) $ P=\alpha \beta =(a+a^{2}+a^{4})(a^{3}+a^{5}+a^{6}) $ $ =a^{4}+a^{6}+a^{7}+a^{5}+a^{7}+a^{8}+a^{7}+a^{9}+a^{10} $ $ =a^{4}+a^{6}+1+a^{5}+1+a+1+a^{2}+a^{3} $ (From eqn (i)] $ =3+(a+a^{2}+a^{3}+a^{4}+a^{5}+a^{6}) $ $ =3+S $ $ = $ $ 3-1=2 $ [From (ii)] Required equation is, $ x^{2}-Sx+P=0 $
Þ $ x^{2}+x+2=0 $ .
BETA
