Complex Numbers And Quadratic Equations question 587
Question: If the roots of $ x^{2}-bx+c=0 $ are two consecutive integers, then $ b^{2}-4c $ is [RPET 1991; Kurukshetra CEE 1998; AIEEE 2005]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: A
Solution:
Let the roots are $ \alpha $ and $ \alpha +1 $ , then, sum of roots $ =2\alpha +1=b $ product of roots $ =\alpha (\alpha +1)=c $ Now, $ b^{2}-4c={{(2\alpha +1)}^{2}}-4\alpha (\alpha +1) $ $ =4{{\alpha }^{2}}+1+4\alpha -4{{\alpha }^{2}}-4\alpha $
$ \therefore \ b^{2}-4c=1 $ .
BETA
