Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 667

\[1)\ z^{2} - 2z + 5 = 0\]

\[D = 4 - 20 = - 16\]

\[z = \frac{2 \pm \sqrt{- 16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i.\]

\[Ответ:\ \ 1 - 2i;\ 1 + 2i.\]

\[2)\ z^{2} + 10z + 26 = 0\]

\[D = 100 - 104 = - 4\]

\[z = \frac{- 10 \pm \sqrt{- 4}}{2} = \frac{- 10 \pm 2i}{2} =\]

\[= - 5 \pm i.\]

\[Ответ:\ - 5 - i;\ - 5 + i.\]

\[3)\ 5z^{2} + 6z + 5 = 0\]

\[D = 36 - 100 = - 64\]

\[z = \frac{- 6 \pm \sqrt{- 64}}{2 \bullet 5} = \frac{- 6 \pm 8i}{10} =\]

\[= - \frac{3}{5} \pm \frac{4}{5}i.\]

\[Ответ:\ - \frac{3}{5} - \frac{4}{5}i;\ - \frac{3}{5} + \frac{4}{5}\text{i.}\]

\[4)\ 2z^{2} + 3z + 3 = 0\]

\[D = 9 - 24 = - 15\]

\[z = \frac{- 3 \pm \sqrt{- 15}}{2 \bullet 2} = \frac{- 3 \pm i\sqrt{15}}{4} =\]

\[= - \frac{3}{4} \pm \frac{\sqrt{15}}{4}i.\]

\[Ответ:\ - \frac{3}{4} - \frac{\sqrt{15}}{4}i;\ - \frac{3}{4} + \frac{\sqrt{15}}{4}\text{i.}\]