Решите уравнение: x^2+20x+91=0

\[\ \ x² + 20x + 91 = 0\]

\[x_{1} + x_{2} = - 20\]

\[x_{1} \cdot x_{2} = 91 \Longrightarrow x_{1} = - 7\ \ и\ \ \ x_{2} = - 13\]

\[Ответ:\ \ x_{1} = - 7\ \ и\ \ x_{2} = - 13.\]

\[Пусть\ b\ см - одна\ сторона\ \]

\[прямоугольника.\]

\[По\ условию\ задачи,\ периметр\ 26\ см\ и\ \ \]

\[площадь\ равна\ 36\ \ см^{2}\text{.\ \ }\]

\[Составим\ уравнение:\]

\[2 \cdot \left( b + \frac{36}{b} \right) = 26\]

\[b^{\backslash b} + \frac{36}{b} = 13^{\backslash b}\]

\[b^{2} - 13b + 36 = 0\]

\[b_{1} + b_{2} = 13\]

\[b_{1} \cdot b_{2} = 36 \Longrightarrow b_{1} = 9\ \ и\ \ b_{2} = 4.\]

\[3)\ 13 - 9 = 4\ (см).\]

\[4)\ 13 - 4 = 9\ (см).\]

\[Ответ:стороны\ равны\ 4\ см\ и\ 9\ см.\]

\[x^{2} + \text{px} + 56 = 0\ \ \ и\ \ \ x_{1} = - 4\]

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

\[- 4 - 14 = - p\]

\[- 18 = - p\]

\[p = 18.\]

\[x_{1} \cdot x_{2} = 56\]

\[x_{2} \cdot ( - 4) = 56\]

\[x_{2} = - 14.\]

\[Ответ:\ \ x_{2} = - 14\ \ и\ \ p = 18.\]

\[\ 9x^{2} - 7x - 2 = 0\]

\[D = b^{2} - 4ac = 49 - 4 \cdot 9 \cdot ( - 2) =\]

\[= 49 + 2 = 121\]

\[x_{1} = \frac{7 + 11}{18} = \frac{18}{18} = 1\]

\[x_{2} = \frac{7 - 11}{18} = \frac{- 4}{18} = - \frac{2}{9}\]

\[Ответ:\ \ \ x_{1} = 1\ \ \ и\ \ \ x_{2} = - \frac{2}{9}.\]