Решите уравнение: x^2-16x+63=0

\[\ x^{2} - 16x + 63 = 0\]

\[x_{1} + x_{2} = 16\]

\[x_{1} \cdot x_{2} = 63 \Longrightarrow x_{1} = 7\ \ \ и\ \ x_{2} = 9\]

\[Ответ:\ \ x_{1} = 7\ \ \ и\ \ \ x_{2} = 9.\]

\[Пусть\ b\ см - одна\ сторона\ \]

\[прямоугольника.\]

\[По\ условию\ задачи,\ периметр\ равен\ \]

\[20\ см\ и\ площадь\ равна\ 24\ см^{2}.\]

\[Составим\ уравнение:\]

\[2 \cdot \left( b + \frac{24}{b} \right) = 20\]

\[b^{\backslash b} + \frac{24}{b} = 10^{\backslash b}\]

\[b^{2} - 10b + 24 = 0\]

\[b_{1} + b_{2} = 10\]

\[b_{1} \cdot b_{2} = 24\]

\[\Longrightarrow b_{1} = 4\ \ \ и\ b_{2} = 6.\]

\[3)\ 10 - 6 = 4\ (см).\]

\[4)\ 10 - 4 = 6\ (см).\]

\[Ответ:стороны\ равны\ 4\ см\ и\ 6\ см.\]

\[x^{2} + \text{px} - 18 = 0\ \ \ \ \ \ и\ \ \ \ x_{1} = - 9\]

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

\[- 9 + 2 = - p\]

\[- 7 = - p\]

\[p = 7.\]

\[x_{1} \cdot x_{2} = - 18\]

\[- 9 \cdot x_{2} = - 18\]

\[x_{2} = 2.\]

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

\[\ 3x² + 13x - 10 = 0\]

\[D = b^{2} - 4ac = 169 - 4 \cdot 3 \cdot ( - 10) =\]

\[= 169 + 120 = 289\]

\[x_{1} = \frac{- 13 - 17}{6} = - \frac{30}{6} = - 5\]

\[x_{2} = \frac{- 13 + 17}{6} = \frac{4}{6} = \frac{2}{3}\]

\[Ответ:x_{1} = - 5\ \ \ и\ \ x_{2} = \frac{2}{3}.\]