If both the roots of the equations $k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0$ and $6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0$ are common, then $2 r - p$ is equal to-

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Solution

The correct option is C $0$
Given equations
$k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0$
$\Rightarrow (6 k + 2) x^{2} + r x + 3 k - 1 = 0$ .....(1)
Other equation $6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0$
$\Rightarrow 2 (6 k + 2) x^{2} + p x + 2 (3 k - 1) = 0$ ....(2)
Let the common root be $α$
$α^{2} (6 k + 2) + r α + 3 k - 1 = 0$ ... (3)
$2 α^{2} (6 k + 2) + p α + 6 k - 2 = 0$ ... (4)
Now multiplying $(3)$ by $2$ and subtracting $(3)$ and $(4)$ we get,
$2 r - p = 0$

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