$\begin{equation*} \begin{cases} \frac{x_1}{x_2} = -3 &\\ x_1 + x_2 = -\frac{b}{4} &\\ x_1x_2 = -\frac{27}{4} & \end{cases} \end{equation*}$
$\frac{x_1}{x_2} * x_1x_2 = x^2_{1} = -3 * (-\frac{27}{4}) = \frac{81}{4} = (\frac{9}{2})^2$
$x_1 = ±\frac{9}{2}$
$x_2 = -\frac{27}{4} : (±\frac{9}{2}) = ±\frac{27}{4} * \frac{9}{2} = ±\frac{3}{2}$
Имеем две пары корней:
$x_1 = \frac{9}{2}$
$x_2 = -\frac{3}{2}$
и
$x_1 = -\frac{9}{2}$
$x_2 = \frac{3}{2}$
$b = -4(x_1 + x_2)$
$b_1 = -4(\frac{9}{2} - \frac{3}{2}) = -12$
$b_2 = -4(-\frac{9}{2} + \frac{3}{2}) = 12$
Ответ: ±12