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

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Solution

The correct option is C
0

Given equations can be written as
$k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0 . . . (i)$ and
$2 (6 k + 2) x^{2} + p x + 2 (3 k - 1) = 0 . . . (i i)$

Condition for common roots is that the corresponding coefficients must be proportional.
$\Rightarrow \frac{12 k + 4}{6 k + 2} = \frac{p}{r}$
$\Rightarrow \frac{6 k - 2}{3 k - 1} = 2$
$\Rightarrow 2 r - p = 0$

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Similar questions

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

k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0

6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0

2 r - p

k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0

6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0

2 r - p

k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0

6 k (2 x^{2} + 1) + p x + (4 x^{2} - 2) = 0

\frac{p}{r}

k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0

6 k (2 x^{2} + 1) + p x + (4 x^{2} - 2) = 0, k \neq 0

\frac{p}{r}

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