Let α, β be the roots of the equation x^2 - 3x + r = 0 , and...

From the first equation $x^2 - 3x + r = 0$, the sum and product of the roots are:

$\alpha + \beta = 3$

$\alpha \beta = r$

From the second equation $x^2 + 3x + r = 0$, the sum and product of the roots are:

$\dfrac{\alpha}{2} + 2\beta = -3$

\left(\dfrac{\alpha}{2}\right)(2\beta) = r \Rightarrow \alpha \beta = r$<br><br>We have a system of linear equations in terms of$\alpha and \beta$:<br><br>$\alpha + \beta = 3$<br><br>$\alpha + 4\beta = -6$<br><br>Subtracting the first equation from the second gives:<br><br>$3\beta = -9 \Rightarrow \beta = -3$<br><br>Substituting$\beta = -3$into$\alpha + \beta = 3$gives:<br><br>$\alpha = 6$<br><br>Now, we can find$r$:<br><br>$r = \alpha \beta = 6 \times (-3) = -18$<br><br>The roots of the third equation$x^2 + 6x - m = 0$are given as$2\alpha + \beta + 2r$and$\alpha - 2\beta - \dfrac{r}{2}$. Let us calculate their values:<br><br>$R_1 = 2(6) + (-3) + 2(-18) = 12 - 3 - 36 = -27$<br><br>$R_2 = 6 - 2(-3) - \dfrac{-18}{2} = 6 + 6 + 9 = 21$<br><br>The product of the roots of the equation$x^2 + 6x - m = 0$is$-m$. Therefore:<br><br>$-m = R_1 R_2 = (-27)(21)$<br><br>$-m = -567 \Rightarrow m = 567$<br><br>Answer:$567