Решите уравнение: x^2-2x-35=0

\[\ x² - 2x - 35 = 0\]

\[x_{1} + x_{2} = 2\]

\[x_{1} \cdot x_{2} = - 35 \Longrightarrow x_{1} = 7\ \ \ и\ \ \ \ x_{2} = - 5.\]

\[Ответ:x_{1} = 7\ \ \ и\ \ x_{2} = - 5.\]

\[Пусть\ b\ см - одна\ сторона\ \]

\[прямоугольника.\]

\[По\ условию\ задачи,\ периметр\ 30\ см\ и\ \]

\[площадь\ равна\ 56\ см^{2}\text{.\ }\]

\[Составим\ уравнение:\]

\[2 \cdot \left( b + \frac{56}{b} \right) = 30\]

\[b^{\backslash b} + \frac{56}{b} = 15^{\backslash b}\]

\[b^{2} - 15b + 56 = 0\]

\[b_{1} + b_{2} = 15\]

\[b_{1} \cdot b_{2} = 56 \Longrightarrow b_{1} = 8\ \ \ и\ \ b_{2} = 7.\]

\[3)\ 15 - 8 = 7\ (см).\]

\[4)\ 15 - 7 = 8\ (см).\]

\[Ответ:стороны\ равны\ 7\ см\ и\ 8\ см.\]

\[x^{2} + 11x + q = 0\ \ \ \ и\ \ \ x_{1} = - 7\]

\[x_{1} + x_{2} = - 11\]

\[- 7 + x_{2} = - 11\]

\[x_{2} = - 4.\]

\[x_{1} \cdot x_{2} = q\]

\[- 7 \cdot ( - 4) = q\]

\[q = 28\]

\[Ответ:x_{2} = - 4\ \ \ и\ \ q = 28.\]

\[\ 7x² - 9x + 2 = 0\]

\[D = b^{2} - 4ac = 81 - 4 \cdot 7 \cdot 2 = 81 - 56 =\]

\[= 25\]

\[x_{1} = \frac{9 + 5}{14} = \frac{14}{14} = 1\]

\[x_{2} = \frac{9 - 5}{14} = \frac{4}{14} = \frac{2}{7}\]

\[Ответ:\ \ x_{1} = 1\ \ и\ \ x_{2} = \frac{2}{7}.\]