Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 646

\[1)\ 4z^{2} - 4z + 5 = 0\]

\[D = 16 - 80 = - 64:\]

\[z = \frac{4 \pm \sqrt{- 64}}{2 \bullet 4} = \frac{4 \pm 8i}{8} = \frac{1}{2} \pm i.\]

\[Ответ:\ \ \frac{1}{2} - i;\ \frac{1}{2} + i.\]

\[2)\ 9z^{2} + 18z + 10 = 0\]

\[D = 324 - 360 = - 36:\]

\[z = \frac{- 18 \pm \sqrt{- 36}}{2 \bullet 9} = \frac{- 18 \pm 6i}{18} =\]

\[= - 1 \pm \frac{1}{3}i.\]

\[Ответ:\ - 1 - \frac{1}{3}i;\ - 1 + \frac{1}{3}\text{i.}\]

\[3)\ z^{2} - 4z + 7 = 0\]

\[D = 16 - 28 = - 12\]

\[z = \frac{4 \pm \sqrt{- 12}}{2} = \frac{4 \pm 2\sqrt{3}i}{2} =\]

\[= 2 \pm \sqrt{3}i.\]

\[Ответ:\ \ 2 - \sqrt{3}i;\ 2 + \sqrt{3}\text{i.}\]

\[4)\ z^{2} + 2z + 6 = 0;\]

\[D = 4 - 24 = - 20\]

\[z = \frac{- 2 \pm \sqrt{- 20}}{2} = \frac{- 2 \pm 2\sqrt{5}i}{2} =\]

\[= - 1 \pm \sqrt{5}i.\]

\[Ответ:\ - 1 - \sqrt{5}i;\ - 1 + \sqrt{5}\text{i.}\]

\[5)\ z^{3} + 27 = 0\]

\[(z + 3)\left( z^{2} - 3z + 9 \right) = 0\]

\[D = 9 - 36 = - 27\]

\[z = \frac{3 \pm \sqrt{- 27}}{2} = \frac{3 \pm 3\sqrt{3}i}{2}.\]

\[Ответ:\ - 3;\ \frac{3 - 3\sqrt{3}i}{2};\ \frac{3 + 3\sqrt{3}i}{2}.\]

\[6)\ z^{3} = 8\]

\[z^{3} - 8 = 0\]

\[(z - 2)\left( z^{2} + 2z + 4 \right) = 0\]

\[D = 4 - 16 = - 12\]

\[z = \frac{- 2 \pm \sqrt{- 12}}{2} = \frac{- 2 \pm 2\sqrt{3}i}{2} =\]

\[= - 1 \pm \sqrt{3}i.\]

\[Ответ:\ \ 2;\ - 1 - \sqrt{3}i;\ - 1 + \sqrt{3}\text{i.}\]