Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 135

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\[1)\ y = x^{2} - 4x + 3\]

\[x^{2} - 4x + 3 = 0\]

\[D_{1} = 4 - 3 = 1\]

\[x_{1} = 2 + 1 = 3;\ \ x_{2} = 2 - 1 = 1.\]

\[Ответ:x = 3;x = 1.\]

\[2)\ y = x^{2} + x - 6\]

\[x^{2} + x - 6 = 0\]

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = - 3;\ \ x_{2} = 2.\]

\[Ответ:x = - 3;x = 2.\]

\[3)\ y = 3x^{2} + 5x - 2\]

\[3x^{2} + 5x - 2 = 0\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{- 5 + 7}{6} = \frac{2}{6} = \frac{1}{3};\ \ \]

\[x_{2} = \frac{- 5 - 7}{6} = - 2.\]

\[Ответ:x = - 2;\ \ x = \frac{1}{3}.\]

\[4)\ y = - 3x^{2} + 7x - 2\]

\[- 3x^{2} + 7x - 2 = 0\]

\[3x^{2} - 7x + 2 = 0\]

\[D = 49 - 24 = 25\]

\[x_{1} = \frac{7 + 5}{6} = 2;\ \ \ \]

\[x_{2} = \frac{7 - 5}{6} = \frac{1}{3}.\]

\[Ответ:x = \frac{1}{3};\ \ x = 2.\]