If the equations k(6x^2 + 3) + rx + 2x^2 - 1 | vedclass

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(D) Given equations are:$k(6x^2 + 3) + rx + 2x^2 - 1 = 0 \implies (6k + 2)x^2 + rx + (3k - 1) = 0$  $(i)$$6k(2x^2 + 1) + px + 4x^2 - 2 = 0 \implies (1

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(D) Given equations are:$k(6x^2 + 3) + rx + 2x^2 - 1 = 0 \implies (6k + 2)x^2 + rx + (3k - 1) = 0$  $(i)$$6k(2x^2 + 1) + px + 4x^2 - 2 = 0 \implies (12k + 4)x^2 + px + (6k - 2) = 0$  $(ii)$Notice that equation $(ii)$ can be written as $2[(6k + 2)x^2 + (p/2)x + (3k - 1)] = 0$,which simplifies to $(6k + 2)x^2 + (p/2)x + (3k - 1) = 0$  $(iii)$Since the equations have common roots,and the coefficients of $x^2$ and the constant terms are identical in $(i)$ and $(iii)$,the equations must be identical for the roots to be the same.Comparing the coefficients of $x$ in $(i)$ and $(iii)$,we get $r = p/2$.Therefore,$2r = p$,which implies $2r - p = 0$.

If the equations $k(6x^2 + 3) + rx + 2x^2 - 1 = 0$ and $6k(2x^2 + 1) + px + 4x^2 - 2 = 0$ have common roots,then $2r - p = \dots$

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