Complex Numbers And Quadratic Equations question 30
Question: The value of p for which both the roots of the equation $ 4x^{2}-20px+(25p^{2}+15p-66)=0 $ are less than 2, lies in
Options:
A) $ (4/5,\ 2) $
B) $ (2,\infty ) $
C) $ (-1,-4/5) $
D) $ (-\infty ,-1) $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ f(x)=4x^{2}-20px+(25p^{2}+15p-66)=0 $ …..(i) The roots of (i) are real if $ b^{2}-4ac=400p^{2}-16(25p^{2}+15p-66) $ $ =16(66-15p)\ge 0 $
Þ $ p\le 22/5 $ …..(ii) Both roots of (i) are less than 2. Therefore $ f(2)>0 $ and sum of roots < 4.
Þ $ {{4.2}^{2}}-20p.2+(25p^{2}+15p-66)>0 $ and $ \frac{20p}{4} $ <4
Þ $ p^{2}-p-2>0 $ and $ p<4/5 $
Þ $ (p+1)(p-2)>0 $ and $ p<4/5 $
Þ $ p<-1 $ or $ p>2 $ and $ p<4/5 $
Þ $ p<-1 $ …..(iii) From (ii) and (iii), we get $ p<-1 $ i.e. $ p\in (-\infty ,-1) $ .
BETA
