\[\boxed{\text{590\ (590).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[x^{2} + x + c = 0;\ \ \ x_{1} - x_{2} = 6\]
\[\left\{ \begin{matrix} x_{1} + x_{2} = - 1 \\ x_{1}x_{2} = c\ \ \ \ \ \ \ \ \ \\ x_{1} - x_{2} = 6\ \ \ \\ \end{matrix} \right.\ \ \ \ \Longrightarrow \ \ \ x_{1} = 6 + x_{2}\]
\[6 + x_{2} + x_{2} = - 1\]
\[2x_{2} = - 7\]
\[x_{2} = - 3,5.\]
\[x_{1} = 6 + x_{2} = 6 + ( - 3,5) = 2,5.\]
\[c = x_{1}x_{2} = ( - 3,5) \cdot 2,5 = - 8,75.\]
\[Ответ:c = - 8,75.\ \]
\[\boxed{\text{590.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
Пояснение.
Решение.
\[x^{2} - 3x + a = 0;\ \ x_{1}^{2} + x_{2}^{2} = 65\]
\[\left\{ \begin{matrix} \left( x_{1} + x_{2} \right)^{2} = 3^{2} \\ x_{1} \cdot x_{2} = a\ \ \ \ \ \ \ \ \ \\ x_{1}^{2} + x_{2}^{2} = 65\ \ \ \ \ \\ \end{matrix} \right.\ \ \]
\[x_{1}^{2} + 2x_{1}x_{2} + x_{2}^{2} = 9\]
\[\left( x_{1}^{2} + x_{2}^{2} \right) + 2x_{1}x_{2} = 9\]
\[65 + 2a = 9\]
\[2a = - 56\]
\[a = - 28\]
\[Ответ:a = - 28.\text{\ \ }\]