\[x^{2} + bx + 150 = 0;\ \ \ x_{1} = 6x_{2}\]
\[\left\{ \begin{matrix} x_{1} + x_{2} = - b \\ x_{1} \cdot x_{2} = 150 \\ \end{matrix} \right.\ \]
\[1)\ 6x_{2} \cdot x_{2} = 150\]
\[\ x_{2}^{2} = 25\]
\[x_{2} = \pm 5.\]
\[2)\ x_{1} = 6 \cdot ( \pm 5) = \pm 30.\]
\[3)\ \pm 30 \pm 5 = - b\]
\[- b = \pm 35\ \]
\[b = \pm 35.\]
\[Ответ:\ \ x_{1} = 30,\ x_{2} = 5,\ \]
\[b = - 35;\ \ \ x_{1} = - 30,\ x_{2} = - 5,\ \]
\[b = 35.\]