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If the equations $k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0$ and $6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0$ have both roots common, then 2r-p =________ $(k \neq \pm \frac{1}{3})$

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answer is C.

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Detailed Solution

$(16 x + 2) x^{2} + r x + 3 k - 1 = 0 (12 k + 4) x^{2} + p x + 6 k - 2 = 0 \frac{6 k + 2}{12 k + 4} = \frac{r}{p} = \frac{3 k - 1}{6 k - 2} \Rightarrow \frac{r}{p} = \frac{1}{2} 2 r - p = 0$

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