lf 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 the value $\frac{p}{r}$ is

\frac{1}{2}

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Solution

The correct option is B $2$
Since the roots of the equations are common, the coefficients are in proportion.
The first equation can be written as $(6 k + 2) x^{2} + r x + 3 k - 1 = 0$
The second one as $(12 k + 4) x^{2} + p x + 6 k - 2 = 0 ∴ \frac{6 k + 2}{12 k + 4} = \frac{r}{p} = \frac{3 k - 1}{6 k - 2}$
$\Rightarrow \frac{3 k + 1}{6 k + 2} = \frac{3 k - 1}{6 k - 2} = \frac{3 k + 1 + 3 k - 1}{6 k + 2 + 6 k - 2} = \frac{1}{2}$
$\Rightarrow$ Each ratio $= \frac{1}{2}$
$∴ \frac{r}{p} = \frac{1}{2} \Rightarrow \frac{p}{r} = 2$

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