Find Parameter in Quadratic Equation Given Relations Between Roots

Given the quadratic equation: $x^2 - 3x + r = 0$

Since $\alpha$ and $\beta$ are the roots of this equation, by Vieta's formulas, we have: $\alpha + \beta = 3 \quad \text{--- (1)}$ $\alpha \beta = r \quad \text{--- (2)}$

We are also given that $\frac{\alpha}{2}$ and $2\beta$ are the roots of the equation: $x^2 + 3x + r = 0$

Using Vieta's formulas for this equation: $\frac{\alpha}{2} + 2\beta = -3 \implies \alpha + 4\beta = -6 \quad \text{--- (3)}$ $\left(\frac{\alpha}{2}\right)(2\beta) = \alpha\beta = r \quad \text{(which is consistent with (2))}$

Now, subtract equation (1) from equation (3): $(\alpha + 4\beta) - (\alpha + \beta) = -6 - 3$ $3\beta = -9 \implies \beta = -3$

Substituting $\beta = -3$ into equation (1): $\alpha + (-3) = 3 \implies \alpha = 6$

Using equation (2), we find $r$: $r = \alpha \beta = 6 \times (-3) = -18$

Now, let the roots of the equation $x^2 + 6x - m = 0$ be $p$ and $q$, where: $p = 2\alpha + \beta + 2r$ $q = \alpha - 2\beta - \frac{r}{2}$

Substituting the values of $\alpha$, $\beta$, and $r$: $p = 2(6) + (-3) + 2(-18) = 12 - 3 - 36 = -27$ $q = 6 - 2(-3) - \frac{-18}{2} = 6 + 6 + 9 = 21$

For the quadratic equation $x^2 + 6x - m = 0$, the product of the roots is given by $-m$: $p \cdot q = -m$ $(-27) \times 21 = -m$ $-567 = -m \implies m = 567$

Thus, the value of $m$ is $567$.