ГДЗ по алгебре 9 класс Макарычев Задание 96

\[\boxed{\text{96\ (96).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = 2x^{2}\]

\[\textbf{а)}\ y = 50:\ \]

\[2x^{2} = 50,\ \]

\[\ x^{2} = 25,\ \]

\[\ x = \pm 5.\]

\[(5;50);( - 5;50) \Longrightarrow точки\ \]

\[пересечения.\]

\[\textbf{б)}\ y = 100:\]

\[2x^{2} = 100,\ \]

\[\ x^{2} = 50,\ \]

\[\ x = \pm \sqrt{50} = \pm 5\sqrt{2}.\]

\[\left( 5\sqrt{2};100 \right);\left( - 5\sqrt{2};100 \right) \Longrightarrow\]

\[\Longrightarrow точки\ пересечения.\]

\[\textbf{в)}\ y = - 8:\]

\[2x^{2} = - 8,\ \ \]

\[x^{2} = - 4 \Longrightarrow графики\ не\ \]

\[пересекаются;\]

\[\textbf{г)}\ y = 14x - 20:\]

\[2x^{2} = 14x - 20\]

\[2x^{2} - 14x + 20 = 0\]

\[x^{2} - 7x + 10 = 0\]

\[D = 7^{2} - 4 \cdot 10 = 49 - 40 = 9\]

\[x_{1} = \frac{7 + 3}{2} = 5;\ \ \ x_{2} =\]

\[= \frac{7 - 3}{2} = 2.\]

\[y(5) = 14 \cdot 5 - 20 =\]

\[= 70 - 20 = 50;\]

\[y(2) = 14 \cdot 2 - 20 =\]

\[= 28 - 20 = 8;\]

\[(5;50);\ \ (2;8) - точки\ \]

\[пересечения.\]