Complex Numbers And Quadratic Equations question 653
Question: If $ a,b,c\in \mathbf{R} $ and the equations $ a\ne 0, $ has real roots $ \alpha $ and $ \beta $ satisfying $ \alpha <-1 $ and $ \beta >1, $ then $ 1+\frac{c}{a}+| \frac{b}{a} | $ is
Options:
A) positive
B) negative
C) zero
D) none
Show Answer
Answer:
Correct Answer: B
Solution:
$ \alpha <-1.Let \alpha =-1-p $ $ \beta >1.Let \beta =1+q, p>0,q>0 $ Now $ 1+\frac{c}{a}+| \frac{b}{a} |=1+\alpha \beta +| -\alpha -\beta | $ $ =1+( 1+q )( -1-p )+| -1-p+1+q | $ $ =1-( 1+p+q+pq )+| q-p | $ $ = \begin{matrix} -p-q-pq+q-p=-2p-pq<0ifq>p \\ -p-q-pq+p-q=-2q-pq<0ifq<p \\ \end{matrix} . $
$ \therefore 1+\frac{c}{a}+| \frac{b}{a} |<0 $
BETA
