Complex Numbers And Quadratic Equations question 462
Question: If the roots of the equation $ x^{2}+2ax+b=0 $ are real and distinct and they differ by at most 2m then b lies in the interval
Options:
A) $ (a^{2},a^{2}+m^{2}) $
B) $ (a^{2}-m^{2},a^{2}) $
C) [ $ a^{2}-m^{2},a^{2} $ )
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let the roots be $ \alpha $ , $ \beta $
$ \therefore \alpha +\beta =-2a $ and $ \alpha \beta $ =b Given, $ | \alpha -\beta |\le 2m $ or $ {{| \alpha -\beta |}^{2}}\le {{(2m)}^{2}} $ or $ {{(\alpha +\beta )}^{2}}-4ab\le 4m^{2} $ or $ 4a^{2}-4b\le 4m^{2} $
$ \Rightarrow a^{2}-m^{2}\le b $ and discriminant $ D>0 $ or $ 4a^{2}-4b>0 $
$ \Rightarrow a^{2}-m^{2}\le b $ and $ b<a^{2} $ . Hence, $ b\in [a^{2}-m^{2},a^{2}). $
BETA
