579. Решите уравнение и выполните проверку по теореме, обратной теореме Виета: a) x² – 2x – 9 = 0; б) 3t² – 4t – 4 = 0; в) 2z² + 7z – 6 = 0; г) 2t² + 9t + 8 = 0.

Решение:

a) $$x^2 - 2x - 9 = 0$$

$$D = (-2)^2 - 4(1)(-9) = 4 + 36 = 40$$

$$x_1 = \frac{2 + \sqrt{40}}{2} = \frac{2 + 2\sqrt{10}}{2} = 1 + \sqrt{10}$$

$$x_2 = \frac{2 - \sqrt{40}}{2} = \frac{2 - 2\sqrt{10}}{2} = 1 - \sqrt{10}$$

Проверка по теореме Виета:

$$x_1 + x_2 = (1 + \sqrt{10}) + (1 - \sqrt{10}) = 2$$

$$x_1 \cdot x_2 = (1 + \sqrt{10})(1 - \sqrt{10}) = 1 - 10 = -9$$

б) $$3t^2 - 4t - 4 = 0$$

$$D = (-4)^2 - 4(3)(-4) = 16 + 48 = 64$$

$$t_1 = \frac{4 + \sqrt{64}}{2(3)} = \frac{4 + 8}{6} = \frac{12}{6} = 2$$

$$t_2 = \frac{4 - \sqrt{64}}{2(3)} = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}$$

Проверка по теореме Виета:

$$t_1 + t_2 = 2 - \frac{2}{3} = \frac{4}{3}$$

$$t_1 \cdot t_2 = 2 \cdot (-\frac{2}{3}) = -\frac{4}{3}$$

в) $$2z^2 + 7z - 6 = 0$$

$$D = 7^2 - 4(2)(-6) = 49 + 48 = 97$$

$$z_1 = \frac{-7 + \sqrt{97}}{2(2)} = \frac{-7 + \sqrt{97}}{4}$$

$$z_2 = \frac{-7 - \sqrt{97}}{2(2)} = \frac{-7 - \sqrt{97}}{4}$$

Проверка по теореме Виета:

$$z_1 + z_2 = \frac{-7 + \sqrt{97}}{4} + \frac{-7 - \sqrt{97}}{4} = -\frac{14}{4} = -\frac{7}{2}$$

$$z_1 \cdot z_2 = \frac{-7 + \sqrt{97}}{4} \cdot \frac{-7 - \sqrt{97}}{4} = \frac{49 - 97}{16} = \frac{-48}{16} = -3$$

г) $$2t^2 + 9t + 8 = 0$$

$$D = 9^2 - 4(2)(8) = 81 - 64 = 17$$

$$t_1 = \frac{-9 + \sqrt{17}}{2(2)} = \frac{-9 + \sqrt{17}}{4}$$

$$t_2 = \frac{-9 - \sqrt{17}}{2(2)} = \frac{-9 - \sqrt{17}}{4}$$

Проверка по теореме Виета:

$$t_1 + t_2 = \frac{-9 + \sqrt{17}}{4} + \frac{-9 - \sqrt{17}}{4} = -\frac{18}{4} = -\frac{9}{2}$$

$$t_1 \cdot t_2 = \frac{-9 + \sqrt{17}}{4} \cdot \frac{-9 - \sqrt{17}}{4} = \frac{81 - 17}{16} = \frac{64}{16} = 4$$

Ответ: a) $$x_1 = 1 + \sqrt{10}$$, $$x_2 = 1 - \sqrt{10}$$; б) $$t_1 = 2$$, $$t_2 = -\frac{2}{3}$$; в) $$z_1 = \frac{-7 + \sqrt{97}}{4}$$, $$z_2 = \frac{-7 - \sqrt{97}}{4}$$; г) $$t_1 = \frac{-9 + \sqrt{17}}{4}$$, $$t_2 = \frac{-9 - \sqrt{17}}{4}$$.