Решебник по алгебре 9 класс Мерзляк Задание 217

\[\boxed{\text{217\ (217).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{2} - (4a - 2)x + 3a^{2} - 4a +\]

\[+ 1 = 0;\ \ x_{1}\ и\ x_{2} \in \lbrack - 2;8\rbrack.\]

\[D = (4a - 2)^{2} -\]

\[- 4 \cdot \left( 3a^{2} - 4a + 1 \right) =\]

\[= 16a^{2} - 16a + 4 - 12a^{2} +\]

\[+ 16a - 4 =\]

\[= 4a^{2}.\]

\[x_{1} = \frac{4a - 2 + 2a}{2} = \frac{6a - 2}{2} =\]

\[= 3a - 1\]

\[x_{2} = \frac{4a - 2 - 2a}{2} =\]

\[= \frac{2a - 2}{2} = a - 1\]

\[\left\{ \begin{matrix} - 2 \leq 3a - 1 \leq 8 \\ - 2 \leq a - 1 \leq 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} - 1 \leq 3a \leq 9\ \ \ \ \ \ \ |\ :3 \\ - 1 \leq a \leq 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - \frac{1}{3} \leq a \leq 3 \\ - 1 \leq a \leq 9 \\ \end{matrix} \right.\ \]

\(\ \)

\[Ответ:a \in \left\lbrack - \frac{1}{3};3 \right\rbrack\text{.\ }\]