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(D) $1$) From the equation $x^2 - 3x + r = 0$, we have $\alpha + \beta = 3$ and $\alpha\beta = r$.$2$) From the equation $x^2 + 3x + r = 0$, the roots

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$567$

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(D) $1$) From the equation $x^2 - 3x + r = 0$, we have $\alpha + \beta = 3$ and $\alpha\beta = r$.$2$) From the equation $x^2 + 3x + r = 0$, the roots are $\frac{\alpha}{2}$ and $2\beta$. Thus, $\frac{\alpha}{2} + 2\beta = -3$ and $(\frac{\alpha}{2})(2\beta) = r$, which simplifies to $\alpha\beta = r$. This is consistent.$3$) Multiplying the first equation by $2$, we get $\alpha + 4\beta = -6$. Subtracting $\alpha + \beta = 3$ from this gives $3\beta = -9$, so $\beta = -3$. Substituting back, $\alpha = 6$.$4$) Then $r = \alpha\beta = 6(-3) = -18$.$5$) The roots of $x^2 + 6x - m = 0$ are $x_1 = 2\alpha + \beta + 2r = 2(6) - 3 + 2(-18) = 12 - 3 - 36 = -27$ and $x_2 = \alpha - 2\beta - \frac{r}{2} = 6 - 2(-3) - \frac{-18}{2} = 6 + 6 + 9 = 21$.$6$) The product of the roots is $x_1 x_2 = (-27)(21) = -567$. Since the equation is $x^2 + 6x - m = 0$, the product of the roots is $-m$. Therefore, $-m = -567$, which implies $m = 567$.

Let $\alpha, \beta$ be the roots of the equation $x^2 - 3x + r = 0$, and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2 + 3x + r = 0$. If the roots of the equation $x^2 + 6x = m$ are $2\alpha + \beta + 2r$ and $\alpha - 2\beta - \frac{r}{2}$, then $m$ is equal to:

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