534. Найдите корни уравнения: a) 5x² - 11x + 2 = 0; б) 2p² + 7p - 30 = 0; в) 9y² - 30y + 25 = 0; г) 35x² + 2x - 1 = 0; д) 2у2 - у - 5 = 0; e) 16x² - 8x + 1 = 0.

Решение:

a) 5x² - 11x + 2 = 0

$$D = (-11)^2 - 4 \cdot 5 \cdot 2 = 121 - 40 = 81$$

$$x_1 = \frac{-(-11) + \sqrt{81}}{2 \cdot 5} = \frac{11 + 9}{10} = \frac{20}{10} = 2$$

$$x_2 = \frac{-(-11) - \sqrt{81}}{2 \cdot 5} = \frac{11 - 9}{10} = \frac{2}{10} = \frac{1}{5}$$

б) 2p² + 7p - 30 = 0

$$D = (7)^2 - 4 \cdot 2 \cdot (-30) = 49 + 240 = 289$$

$$p_1 = \frac{-7 + \sqrt{289}}{2 \cdot 2} = \frac{-7 + 17}{4} = \frac{10}{4} = \frac{5}{2}$$

$$p_2 = \frac{-7 - \sqrt{289}}{2 \cdot 2} = \frac{-7 - 17}{4} = \frac{-24}{4} = -6$$

в) 9y² - 30y + 25 = 0

$$D = (-30)^2 - 4 \cdot 9 \cdot 25 = 900 - 900 = 0$$

$$y = \frac{-(-30)}{2 \cdot 9} = \frac{30}{18} = \frac{5}{3}$$

г) 35x² + 2x - 1 = 0

$$D = (2)^2 - 4 \cdot 35 \cdot (-1) = 4 + 140 = 144$$

$$x_1 = \frac{-2 + \sqrt{144}}{2 \cdot 35} = \frac{-2 + 12}{70} = \frac{10}{70} = \frac{1}{7}$$

$$x_2 = \frac{-2 - \sqrt{144}}{2 \cdot 35} = \frac{-2 - 12}{70} = \frac{-14}{70} = -\frac{1}{5}$$

д) 2у2 - у - 5 = 0

$$D = (-1)^2 - 4 \cdot 2 \cdot (-5) = 1 + 40 = 41$$

$$y_1 = \frac{-(-1) + \sqrt{41}}{2 \cdot 2} = \frac{1 + \sqrt{41}}{4}$$

$$y_2 = \frac{-(-1) - \sqrt{41}}{2 \cdot 2} = \frac{1 - \sqrt{41}}{4}$$

e) 16x² - 8x + 1 = 0

$$D = (-8)^2 - 4 \cdot 16 \cdot 1 = 64 - 64 = 0$$

$$x = \frac{-(-8)}{2 \cdot 16} = \frac{8}{32} = \frac{1}{4}$$

Ответ:

a) $$x_1 = 2$$, $$x_2 = \frac{1}{5}$$

б) $$p_1 = \frac{5}{2}$$, $$p_2 = -6$$

в) $$y = \frac{5}{3}$$

г) $$x_1 = \frac{1}{7}$$, $$x_2 = -\frac{1}{5}$$

д) $$y_1 = \frac{1 + \sqrt{41}}{4}$$, $$y_2 = \frac{1 - \sqrt{41}}{4}$$

e) $$x = \frac{1}{4}$$