Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 335

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\[ax^{2} + bx + c = 0\]

\[x^{2} + \frac{b}{a}x + \frac{c}{a} = 0\]

\[По\ теореме\ Виета:\]

\[x_{1} + x_{2} = - \frac{b}{a};\ \ \ x_{1} \cdot x_{2} = \frac{c}{a}.\]

\[x^{2} + px + q = 0\]

\[y_{1} = \frac{1}{x_{1}};\ \ \ y_{2} = \frac{1}{x_{2}};\]

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

\[= - \frac{b}{a}\ :\frac{c}{a} = - \frac{b}{a} \cdot \frac{a}{c} = - \frac{b}{c};\]

\[q = \frac{1}{x_{1}} \cdot \frac{1}{x_{2}} = 1\ :\frac{c}{a} = \frac{a}{c}.\]

\[Получаем\ уравнение:\]

\[x^{2} + \frac{b}{c}x + \frac{a}{c} = 0\ \ \ \ | \cdot c\]

\[cx^{2} + bx + a = 0.\]