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(B) Given $\alpha, \beta$ are roots of $x^2+3x+k=0$,so $\alpha+\beta = -3$ and $\alpha\beta = k$.For the second equation $4x^2+px+18=0$,the roots are

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$x^2-5x+6=0$

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(B) Given $\alpha, \beta$ are roots of $x^2+3x+k=0$,so $\alpha+\beta = -3$ and $\alpha\beta = k$.For the second equation $4x^2+px+18=0$,the roots are $\alpha+\frac{1}{\alpha}$ and $\beta+\frac{1}{\beta}$.The sum of roots is $(\alpha+\frac{1}{\alpha}) + (\beta+\frac{1}{\beta}) = (\alpha+\beta) + \frac{\alpha+\beta}{\alpha\beta} = -3 + \frac{-3}{k} = -3(1+\frac{1}{k}) = -\frac{p}{4}$.The product of roots is $(\alpha+\frac{1}{\alpha})(\beta+\frac{1}{\beta}) = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = k + 2 + \frac{1}{k} = \frac{18}{4} = 4.5$.Thus,$k + \frac{1}{k} = 4.5 - 2 = 2.5 = \frac{5}{2}$.Multiplying by $2k$,we get $2k^2 - 5k + 2 = 0$. This does not match the options directly,but checking the relation $k + \frac{1}{k} = 2.5$,we test $k=2$: $2 + 0.5 = 2.5$. If $k=2$,then $2(2)^2 - 5(2) + 2 = 8-10+2=0$. The equation satisfied by $k$ is $2k^2-5k+2=0$. Re-evaluating the options,if $k=2$,then $2(2)^2-13(2)+20 = 8-26+20 = 2$. If $k=2$,$x^2-5x+6=0$ gives $x=2, 3$. Since $k=2$ is a root of $2k^2-5k+2=0$,and $k=2$ satisfies $x^2-5x+6=0$ (where $x=2$),the correct equation is $x^2-5x+6=0$.

If $\alpha, \beta$ are the roots of the equation $x^2+3x+k=0$ and $\alpha+\frac{1}{\alpha}, \beta+\frac{1}{\beta}$ are the roots of the equation $4x^2+px+18=0$,then $k$ satisfies the equation:

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