Complex Numbers And Quadratic Equations question 528
Question: If $ \alpha ,\beta $ be the roots of the equation $ 2x^{2}-2(m^{2}+1)x+m^{4}+m^{2}+1=0 $ , then $ {{\alpha }^{2}}+{{\beta }^{2}} $ =
Options:
A) 0
B) 1
C) m
D) $ m^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \alpha +\beta =\frac{2(m^{2}+1)}{2}=m^{2}+1 $ …..(i) and $ \alpha \beta =\frac{m^{4}+m^{2}+1}{2} $ …..(ii) Therefore $ {{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta $ $ ={{(m^{2}+1)}^{2}}-2\frac{(m^{4}+m^{2}+1)}{2} $ $ =m^{4}+2m^{2}+1-m^{4}-m^{2}-1=m^{2} $
BETA
