Complex Numbers And Quadratic Equations question 708
Question: If the roots of the equation $ x^{2}-ax+b=0 $ are real and differ by a quantity which is less than $ c(c>0), $ then b lies between
Options:
A) $ \frac{a^{2}-c^{2}}{4} $ and $ \frac{a^{2}}{4} $
B) $ \frac{a^{2}+c^{2}}{4} $ and $ \frac{a^{2}}{4} $
C) $ \frac{a^{2}-c^{2}}{2} $ and $ \frac{a^{2}}{4} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given roots are real and distinct, then $ {a^{2}}-4b>0 $
$ \Rightarrow b<a^{2}/4 $ Again $ \alpha $ and $ \beta $ differ by a quantity less than $ c( c >0 ) $
$ \Rightarrow | \alpha -\beta |<cor{{(\alpha -\beta )}^{2}}<c^{2} $
$ \Rightarrow {{( \alpha +\beta )}^{2}}-4\alpha \beta <c^{2} or a^{2}-4b<c^{2} $ or $ \frac{a^{2}-c^{2}}{4}<b $
$ \Rightarrow \frac{a^{2}-c^{2}}{4}<b<\frac{a^{2}}{4} $ by (1) and (2)
BETA
