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

Решим уравнение вида ax² + bx + c = 0, используя формулу корней:

$$x = \frac{-b \pm \sqrt{D}}{2a}$$, где D = b² - 4ac - дискриминант.

a) 5x² - 11x + 2 = 0

a = 5, b = -11, c = 2

D = (-11)² - 4 × 5 × 2 = 121 - 40 = 81

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

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

б) 2p² + 7p - 30 = 0

a = 2, b = 7, c = -30

D = (7)² - 4 × 2 × (-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) 9у² - 30у + 25 = 0

a = 9, b = -30, c = 25

D = (-30)² - 4 × 9 × 25 = 900 - 900 = 0

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

г) 35x² + 2x - 1 = 0

a = 35, b = 2, c = -1

D = (2)² - 4 × 35 × (-1) = 4 + 140 = 144

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

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

д) 2y² - y - 5 = 0

a = 2, b = -1, c = -5

D = (-1)² - 4 × 2 × (-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

a = 16, b = -8, c = 1

D = (-8)² - 4 × 16 × 1 = 64 - 64 = 0

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

Ответ: а) x₁ = 2, x₂ = 1/5; б) p₁ = 5/2, p₂ = -6; в) y = 5/3; г) x₁ = 1/7, x₂ = -1/5; д) y₁ = (1 + √41)/4, y₂ = (1 - √41)/4; e) x = 1/4.