Решебник по алгебре 9 класс Мерзляк Задание 969

\[\boxed{\mathbf{969\ (969).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[y = x^{2} + bx + 2\]

\[1)\ x² + bx + 2 = 0\ \]

\[D = b^{2} - 8\ \ \]

\[b^{2} - 8 = 0\]

\[b^{2} = 8 \Longrightarrow \ \ b = \pm 2\sqrt{2};\]

\[2)\ b² - 8 < 0\ \]

\[\ b^{2} < 8 \Longrightarrow \ \ b \in \left( - 2\sqrt{2};2\sqrt{2} \right);\]

\[3)\ \left\{ \begin{matrix} x_{1} + x_{2} = - b \\ x_{1}x_{2} = 2\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

\[x_{1} - x_{2} = 4;\ \ x_{1} = 4 + x_{2}\]

\[\left\{ \begin{matrix} 4 + x_{2} + x_{2} = - b \\ \left( 4 + x_{2} \right)x_{2} = 2\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} 4 + 2x_{2} = - b \\ 4x_{2} + x_{2}^{2} = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x_{2} = \frac{- b - 4}{2} \\ x_{1} = \frac{4 - b}{2}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[\frac{- b - 4}{2} \cdot \frac{4 - b}{2} = 2\ \ \ | \cdot 4\]

\[( - b - 4)(4 - b) = 8\]

\[- 4b + b^{2} - 16 + 4b = 8\]

\[b² = 24 \Longrightarrow \ \ b = \pm 2\sqrt{6}.\]