A) \[-1\]
B) 0
C) 1
D) 2
Correct Answer: B
Solution :
Given equation can be written as \[(6k+2){{x}^{2}}+rx+3k\,-1=0\] ... (i) and \[2\left( 6k + 2 \right){{x}^{2}} +px+ 2\left( 3k -1 \right) = 0\] ... (ii) Condition for common root is \[\frac{12\,k+4}{6\,k+2}=\frac{p}{r}=\frac{6k-2}{3k-1}=2\,\,\,or\,\,2r-p=0\]You need to login to perform this action.
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