Complex Numbers And Quadratic Equations question 1
Question: If both the roots of $ k(6x^{2}+3)+rx+2x^{2}-1=0 $ and $ 6k(2x^{2}+1)+px+4x^{2}-2=0 $ are common, then $ 2r-p $ is equal to [MNR 1983]
Options:
A) -1
B) 0
C) 1
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation can be written as $ (6k+2)x^{2}+rx+3k-1=0 $ …..(i) and $ 2(6k+2)x^{2}+px+2(3k-1)=0 $ …..(ii) Condition for common roots is $ \frac{12k+4}{6k+2} $ $ =\frac{p}{r}=\frac{6k-2}{3k-1}=2 $ or $ 2r-p=0 $
BETA
