Complex Numbers And Quadratic Equations question 37
Question: If $ \alpha $ and $ \beta $ , $ \alpha $ and $ \gamma $ , $ \alpha $ and $ \delta $ are the roots of the equations $ ax^{2}+2bx+c=0 $ , $ 2bx^{2}+cx+a=0 $ and $ cx^{2}+ax+2b=0 $ respectively, where $ a,b $ and $ c $ are positive real numbers, then $ \alpha +{{\alpha }^{2}} $ = [Kerala (Engg.) 2005]
Options:
A) - 1
B) 0
C) abc
D) $ a+2b+c $
E) abc
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ \alpha +\beta =-\frac{2b}{a} $ $ \gamma +\alpha =-\frac{c}{2b} $ , $ \alpha +\delta =-\frac{a}{c} $ and $ \alpha \beta =\frac{c}{a},\alpha \gamma =\frac{a}{2b},\alpha \delta =\frac{2b}{c} $
Þ $ \alpha +\delta =-\frac{1}{\alpha \beta },{{\alpha }^{2}}\beta +\alpha \beta \delta =-1 $ …..(i)
Þ $ \alpha +\beta =-\frac{1}{\alpha \gamma },{{\alpha }^{2}}\gamma +\alpha \beta \gamma =-1 $ …..(ii)
Þ $ \alpha +\gamma =-\frac{1}{\alpha \delta },{{\alpha }^{2}}\delta +\alpha \beta \gamma =-1 $ …..(iii) Solve equations (i), (ii) and (iii), we get $ \alpha =-1 $ $ \alpha +{{\alpha }^{2}}=(-1)+{{(-1)}^{2}} $ = $ -1+1=0 $ .
BETA
