Complex Numbers And Quadratic Equations question 649
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
Options:
A) $ -1 $
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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 root is $ \frac{12k+4}{6k+2}=\frac{p}{r}=\frac{6k-2}{3k-1}=2 $ or $ 2r-p=0 $
BETA
