Complex Numbers And Quadratic Equations question 687
Question: Let $ \alpha ,\beta $ be the roots of $ x^{2}+x+1=0. $ Then the equation whose roots are $ {{\alpha }^{229}} $ and $ {{\alpha }^{1004}} $ is
Options:
A) $ x^{2}-x-1=0 $
B) $ x^{2}-x+1=0 $
C) $ x^{2}+x-1=0 $
D) $ x^{2}+x+1=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
The roots of $ {x^{2}} + x +1 = 0 $ are $ \omega $ and $ {{\omega }^{2}} $ [see the cube roots of unity in complex numbers] Let $ \alpha = \omega , \beta = {{\omega }^{2}} $ Now $ {{\alpha }^{229}} = {{\omega }^{229}} = {{\omega }^{228}}.\omega = {{( {{\omega }^{3}} )}^{76}} .{{\omega }^{2}} $ $ = \omega = \alpha ( \because {{\omega }^{3}} = 1 ) $ $ {{\alpha }^{1004}}={{\omega }^{1002}}.{{\omega }^{2}}={{({{\omega }^{3}})}^{334}}.{{\omega }^{2}}={{\omega }^{2}}=\beta $
$ \therefore $ equation with roots $ {{\alpha }^{229}} $ and $ {{\alpha }^{1004}} $ is same as the equation with roots $ \alpha $ and $ \beta $ i.e. the original equation.
BETA
