ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 712

\[\boxed{\text{712.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} y = x^{2} - 8x + 16 \\ 2x - 3y = 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \Longrightarrow \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} y = x^{2} - 8x + 16\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x - 3 \cdot \left( x^{2} - 8x + 16 \right) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = x^{2} - 8x + 16\ \ \ \ \ \ \ \\ - 3x^{2} + 26x - 48 = 0 \\ \end{matrix} \right.\ \]

\[3x^{2} - 26x + 48 = 0\]

\[D_{1} = 13^{2} - 3 \cdot 48 = 25\]

\[x_{1} = \frac{13 + 5}{3} = 6;\ \ \ \]

\[x_{2} = \frac{13 - 5}{3} = \frac{8}{3} = 2\frac{2}{3}.\]

\[1)\ x_{1} = 6;\ \ y_{1} = 4;\]

\[2)\ x_{2} = 2\frac{2}{3};\ \ \ y_{2} = 1\frac{7}{9}.\]

\[Ответ:пересекает\ в\ \]

\[точках\ (6;4):\ \ \left( 2\frac{2}{3};1\frac{7}{9} \right).\]

\[\textbf{б)}\ y = 2x^{2} + 9x - 5\]

\[Пересекает\ ось\ x\ (при\ y = 0):\]

\[2x^{2} + 9x - 5 = 0\]

\[D = 81 + 40 = 121\]

\[x_{1} = \frac{- 9 + 11}{4} = 0,5;\ \ \]

\[x_{2} = \frac{- 9 - 11}{4} = - 5.\]

\[Пересекает\ ось\ y\ (при\ x = 0):\]

\[y = - 5.\]

\[Ответ:(0,5;0);( - 5;0);(0; - 5).\]