If the equation $k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0$ and $6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0$ have symmetrical roots, then $2 r - p$ equals

1

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0

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Solution

The correct option is D $0$
$k ({6 x}^{2} + 3) + r x + ({2 x}^{2} - 1) = 0$
$\Rightarrow (6 k + 2) x^{2} + r x + (3 k - 1) = 0$ ... $(1)$
$6 k ({2 x}^{2} + 1) + p x + {4 x}^{2} - 2 = 0$
$\Rightarrow (12 k + 4) x^{2} + p x + (6 k - 2) = 0$ ... $(2)$
From $(1)$ and $(2)$
$\frac{6 k + 2}{12 k + 4} = \frac{r}{p} = \frac{(3 k - 1)}{(6 k - 2)}$
$\Rightarrow 2 r = p \Rightarrow 2 r - p = 0$

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