Решебник по алгебре 8 класс Мерзляк Задание 751

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\[1)\ x² - x - 12 = 0\]

\[x_{1} + x_{2} = 1,\ \ x_{1} = 4\]

\[x_{1} \cdot x_{2} = - 12,\ \ x_{2} = - 3\]

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[2)\ x² + 2x - 35 = 0\]

\[x_{1} + x_{2} = - 2,\ \ x_{1} = - 7\]

\[x_{1} \cdot x_{2} = - 35,\ \ x_{2} = 5\]

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[3)\ 3x² - 16x + 5 = 0\]

\[D = 256 - 4 \cdot 3 \cdot 5 = 196\]

\[x = \frac{16 \pm 14}{6}\]

\[x_{1} = 5;\ \ \ \ x_{2} = \frac{1}{3}\]

\[Ответ:\ x = 5;\ \ x = \frac{1}{3}.\]

\[4)\ 16x² - 24x + 3 = 0\]

\[D = 576 - 4 \cdot 3 \cdot 16 = 384\]

\[x = \frac{24 \pm 8\sqrt{6}}{32} = \frac{3 \pm \sqrt{6}}{4}\]

\[x_{1} = \frac{3 + \sqrt{6}}{4};\ \ \ \ \ \ \ \ x_{2} = \frac{3 - \sqrt{6}}{4}\]

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[5)\ 4x² + 28x + 49 = 0\]

\[D = 784 - 4 \cdot 4 \cdot 49 = 0\]

\[x = - \frac{28}{8} = - 3,5\]

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[6)\ 3x² + 21x - 90 = 0\]

\[D = 441 + 4 \cdot 3 \cdot 90 = 1521\]

\[x = \frac{- 21 \pm 39}{6}\]

\(x_{1} = - 10\); \(x_{2} = 3\)

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]