580. Найдите корни уравнения и выполните проверку по теореме, обратной теореме Виета: a) x² – 15x – 16 = 0; б) m² – 6m – 11 = 0; в) 12x² – 4x – 1 = 0; г) t² – 6 = 0; д) 5x² – 18x = 0; е) 2y² – 41 = 0.

580. Найдите корни уравнения и выполните проверку по теореме, обратной теореме Виета:

a) $$x^2 - 15x - 16 = 0;$$

$$D = b^2 - 4ac = (-15)^2 - 4 \cdot 1 \cdot (-16) = 225 + 64 = 289;$$

$$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{15 + \sqrt{289}}{2} = \frac{15 + 17}{2} = \frac{32}{2} = 16;$$

$$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{15 - \sqrt{289}}{2} = \frac{15 - 17}{2} = \frac{-2}{2} = -1.$$

Проверка:

$$x_1 + x_2 = 16 + (-1) = 15;$$

$$-\frac{b}{a} = -\frac{-15}{1} = 15.$$

$$x_1 \cdot x_2 = 16 \cdot (-1) = -16;$$

$$\frac{c}{a} = \frac{-16}{1} = -16.$$

б) $$m^2 - 6m - 11 = 0;$$

$$D = b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot (-11) = 36 + 44 = 80;$$

$$m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{80}}{2} = \frac{6 + 4\sqrt{5}}{2} = 3 + 2\sqrt{5};$$

$$m_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{80}}{2} = \frac{6 - 4\sqrt{5}}{2} = 3 - 2\sqrt{5}.$$

Проверка:

$$m_1 + m_2 = 3 + 2\sqrt{5} + 3 - 2\sqrt{5} = 6;$$

$$-\frac{b}{a} = -\frac{-6}{1} = 6.$$

$$m_1 \cdot m_2 = (3 + 2\sqrt{5}) \cdot (3 - 2\sqrt{5}) = 9 - 4 \cdot 5 = 9 - 20 = -11;$$

$$\frac{c}{a} = \frac{-11}{1} = -11.$$

в) $$12x^2 - 4x - 1 = 0;$$

$$D = b^2 - 4ac = (-4)^2 - 4 \cdot 12 \cdot (-1) = 16 + 48 = 64;$$

$$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{64}}{2 \cdot 12} = \frac{4 + 8}{24} = \frac{12}{24} = \frac{1}{2};$$

$$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{64}}{2 \cdot 12} = \frac{4 - 8}{24} = \frac{-4}{24} = -\frac{1}{6}.$$

Проверка:

$$x_1 + x_2 = \frac{1}{2} + (-\frac{1}{6}) = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3};$$

$$-\frac{b}{a} = -\frac{-4}{12} = \frac{4}{12} = \frac{1}{3}.$$

$$x_1 \cdot x_2 = \frac{1}{2} \cdot (-\frac{1}{6}) = -\frac{1}{12};$$

$$\frac{c}{a} = \frac{-1}{12} = -\frac{1}{12}.$$

г) $$t^2 - 6 = 0;$$

$$t^2 = 6;$$

$$t_1 = \sqrt{6};$$

$$t_2 = -\sqrt{6}.$$

Проверка:

$$t_1 + t_2 = \sqrt{6} + (-\sqrt{6}) = 0;$$

$$-\frac{b}{a} = -\frac{0}{1} = 0.$$

$$t_1 \cdot t_2 = \sqrt{6} \cdot (-\sqrt{6}) = -6;$$

$$\frac{c}{a} = \frac{-6}{1} = -6.$$

д) $$5x^2 - 18x = 0;$$

$$x(5x - 18) = 0;$$

$$x_1 = 0;$$

$$5x - 18 = 0;$$

$$5x = 18;$$

$$x_2 = \frac{18}{5} = 3,6.$$

Проверка:

$$x_1 + x_2 = 0 + 3,6 = 3,6;$$

$$-\frac{b}{a} = -\frac{-18}{5} = \frac{18}{5} = 3,6.$$

$$x_1 \cdot x_2 = 0 \cdot 3,6 = 0;$$

$$\frac{c}{a} = \frac{0}{5} = 0.$$

е) $$2y^2 - 41 = 0;$$

$$2y^2 = 41;$$

$$y^2 = \frac{41}{2};$$

$$y_1 = \sqrt{\frac{41}{2}};$$

$$y_2 = -\sqrt{\frac{41}{2}}.$$

Проверка:

$$y_1 + y_2 = \sqrt{\frac{41}{2}} - \sqrt{\frac{41}{2}} = 0;$$

$$-\frac{b}{a} = -\frac{0}{2} = 0.$$

$$y_1 \cdot y_2 = \sqrt{\frac{41}{2}} \cdot (-\sqrt{\frac{41}{2}}) = -\frac{41}{2};$$

$$\frac{c}{a} = \frac{-41}{2} = -\frac{41}{2}.$$

Ответ:

• a) 16 и -1;
• б) $$3 + 2\sqrt{5}$$ и $$3 - 2\sqrt{5};$$
• в) 1/2 и -1/6;
• г) $$\sqrt{6}$$ и $$\sqrt{6};$$
• д) 0 и 3,6;
• е) $$\sqrt{\frac{41}{2}}$$ и $$\sqrt{\frac{41}{2}}$$.