ГДЗ по алгебре 8 класс Макарычев Задание 581

\[\textbf{а)}\ x^{2} - 2x - 9 = 0\]

\[D_{1} = 1 + 9 = 10\]

\[x_{1} = 1 + \sqrt{10};\]

\[x_{2} = 1 - \sqrt{10}.\]

\[1)\ x_{1} + x_{2} = 2:\]

\[1 + \sqrt{10} + 1 - \sqrt{10} = 2\]

\[2 = 2.\]

\[2)\ x_{1}x_{2} = - 9:\]

\[\left( 1 + \sqrt{10} \right)\left( 1 - \sqrt{10} \right) = - 9\]

\[1^{2} - \left( \sqrt{10} \right)^{2} = - 9\]

\[1 - 10 = - 9\]

\[- 9 = - 9.\]

\[\textbf{б)}\ 3x^{2} - 4x - 4 = 0\]

\[D_{1} = 2^{2} + 3 \cdot 4 = 16\]

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

\[1)\ x_{1} + x_{2} = \frac{4}{3}:\]

\[2 + \left( - \frac{2}{3} \right) = \frac{4}{3}\]

\[\frac{4}{3} = \frac{4}{3}.\]

\[2)\ x_{1}x_{2} = - \frac{4}{3}:\]

\[2 \cdot \left( - \frac{2}{3} \right) = - \frac{4}{3}\]

\[- \frac{4}{3} = - \frac{4}{3}\text{.\ \ }\]

\[\textbf{в)}\ 2x^{2} + 7x - 6 = 0\]

\[D = 49 + 48 = 97\]

\[x_{1} = \frac{- 7 + \sqrt{97}}{4};\]

\[x_{2} = \frac{- 7 - \sqrt{97}}{4}.\]

\[1)\ x_{1} + x_{2} = - 3,5:\]

\[\frac{- 7 + \sqrt{97}}{4} + \frac{- 7 - \sqrt{97}}{4} = - 3,5\ \ \ \ \ | \cdot 4\]

\[- 7 + \sqrt{97} - 7 - \sqrt{97} = - 14\]

\[- 14 = - 14.\]

\[2)\ x_{1} \cdot x_{2} = - 3:\]

\[\frac{- 7 + \sqrt{97}}{4} \cdot \frac{- 7 - \sqrt{97}}{4} = - 3\]

\[\frac{( - 7)^{2} - {\sqrt{97}}^{2}}{16} = - 3\]

\[\frac{49 - 97}{16} = - 3\]

\[- \frac{48}{16} = - 3\]

\[- 3 = - 3.\]

\[\textbf{г)}\ 2x^{2} + 9x + 8 = 0\]

\[D = 81 - 64 = 17\]

\[x_{1} = \frac{- 9 + \sqrt{17}}{4};\]

\[x_{2} = \frac{- 9 - \sqrt{17}}{4}.\]

\[1)\ x_{1} + x_{2} = - 4,5:\]

\[\frac{- 9 + \sqrt{17}}{4} + \frac{- 9 - \sqrt{17}}{4} = 4,5\ \ \ \ | \cdot 4\]

\[- 9 + \sqrt{17} - 9 - \sqrt{17} = - 18\]

\[- 18 = - 18.\]

\[2)\ x_{1}x_{2} = 4:\]

\[\frac{- 9 + \sqrt{17}}{4} \cdot \frac{- 9 - \sqrt{17}}{4} = 4\]

\[\frac{( - 9)^{2} - {\sqrt{17}}^{2}}{16} = 4\]

\[\frac{81 - 17}{16} = 4\]

\[\frac{64}{16} = 4\]

\[4 = 4.\ \ \]