Решебник по алгебре 8 класс Макарычев ФГОС Задание 1293

\[\boxed{\text{1293.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\left\{ \begin{matrix} x^{2} + (a - 1)x - 2a = 0 \\ x_{1}^{2}{+ x}_{2}^{2} = 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\ \right.\ \text{\ \ }\]

\[\ \left\{ \begin{matrix} x_{1}{+ x}_{2} = 1 - a \\ x_{1}{\cdot x}_{2} = - 2a \\ \end{matrix} \right.\ \]

\[x_{1}^{2}{+ x}_{2}^{2} = x_{1}^{2}{+ x}_{2}^{2} + 2x_{1}{\cdot x}_{2} -\]

\[- {2x}_{1}{\cdot x}_{2} = \left( x_{1}{+ x}_{2} \right)^{2} -\]

\[- 2x_{1}{\cdot x}_{2} = 9\]

\[(1 - a)^{2} - 2 \cdot ( - 2a) = 9\]

\[1 - 2a + a^{2} + 4a - 9 = 0\]

\[a^{2} + 2a - 8 = 0\]

\[D = 4 + 32 = 36\]

\[a_{1,2} = \frac{- 2 \pm 6}{2} = - 4;2.\]

\[- 4 \Longrightarrow не\ может\ \ быть.\]

\[Ответ:a = 2.\]