Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 817

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

\[(y - 21)y = 100\]

\[y^{2} - 21y - 100 = 0\]

\[D = 441 + 400 = 841\]

\[y_{1} = \frac{21 - 29}{2} = - 4;\]

\[y_{2} = \frac{21 + 29}{2} = 25.\]

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

\[x^{2} + 2x + 1 = 0\]

\[(x + 1)^{2} = 0\]

\[x = - 1.\]

\[2)\ x^{2} + 2x - 3 = 25\]

\[x^{2} + 2x - 28 = 0\]

\[D = 4 + 112 = 116\]

\[x = \frac{- 2 \pm \sqrt{116}}{2} = \frac{- 2 \pm 2\sqrt{29}}{2} =\]

\[= - 1 \pm \sqrt{29}.\]

\[Ответ:\ - 1;\ - 1 \pm \sqrt{29}.\]

\[y = x^{2} - 5x + 4:\]

\[y(y + 2) = 4\]

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

\[D = 4 + 16 = 20\]

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

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

\[Подставим:\]

\[x^{2} - 5x + 4 = - 1 \pm \sqrt{5}\]

\[x^{2} - 5x + \left( 5 \pm \sqrt{5} \right) = 0\]

\[D = 5^{2} - 4\left( 5 \pm \sqrt{5} \right) =\]

\[= 25 - 20 \mp 4\sqrt{5} =\]

\[= 5 \mp 4\sqrt{5} = 5 + 4\sqrt{5}\]

\[x = \frac{5 \pm \sqrt{5 + 4\sqrt{5}}}{2}.\]

\[Ответ:\ \ \frac{5 \pm \sqrt{5 + 4\sqrt{5}}}{2}.\]