ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 769

\[\boxed{\text{769.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[3x^{2} + bx + 10 = 0,\ \ \]

\[x_{1} - x_{2} = 4\frac{1}{3} \Longrightarrow x_{1} = \frac{13}{3} + x_{2}\]

\[x^{2} + \frac{b}{3}x + \frac{10}{3} = 0\ \Longrightarrow\]

\[\left\{ \begin{matrix} x_{1} + x_{2} = - \frac{b}{3} \\ x_{1}x_{2} = \frac{10}{3}\text{\ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} \frac{13}{3} + x_{2} + x_{2} = - \frac{b}{3} \\ \left( \frac{13}{3} + x_{2} \right)x_{2} = \frac{10}{3}\text{\ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} \frac{13}{3} + 2x_{2} = - \frac{b}{3}\ \ \ | \cdot 3 \\ \frac{13}{3}x_{2} + x_{2}^{2} = \frac{10}{3}\ \ \ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} 13 + 6x_{2} = - b \\ 13x_{2} + 3x_{2}^{2} = 10 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b = - 6x_{2} - 13\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x_{2}^{2} + 13x_{2} - 10 = 0* \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[*D = 169 + 120 = 289 = 17^{2}\]

\[x_{1,2} = \frac{- 13 \pm 17}{6} = \frac{2}{3};\ - 5\]

\[при\ x_{1} = \frac{2}{3};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x_{2} = - 5\]

\[b_{1} = - 6 \cdot \frac{2}{3} - 13 = - 17;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\text{\ \ \ }b_{2} = - 6 \cdot ( - 5) - 13 = 17\]

\[Ответ:b = \pm 17.\ \]