Решение квадратных уравнений вида ax² + bx + c = 0, может быть найдено с использованием формулы корней:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
a) 3x² - 7x + 4 = 0
- a = 3, b = -7, c = 4
- D = (-7)² - 4 × 3 × 4 = 49 - 48 = 1
- x₁ = (7 + √1) / (2 × 3) = 8 / 6 = 4 / 3
- x₂ = (7 - √1) / (2 × 3) = 6 / 6 = 1
б) 5x² - 8x + 3 = 0
- a = 5, b = -8, c = 3
- D = (-8)² - 4 × 5 × 3 = 64 - 60 = 4
- x₁ = (8 + √4) / (2 × 5) = 10 / 10 = 1
- x₂ = (8 - √4) / (2 × 5) = 6 / 10 = 3 / 5
в) 3x² - 13x + 14 = 0
- a = 3, b = -13, c = 14
- D = (-13)² - 4 × 3 × 14 = 169 - 168 = 1
- x₁ = (13 + √1) / (2 × 3) = 14 / 6 = 7 / 3
- x₂ = (13 - √1) / (2 × 3) = 12 / 6 = 2
г) 2y² - 9y + 10 = 0
- a = 2, b = -9, c = 10
- D = (-9)² - 4 × 2 × 10 = 81 - 80 = 1
- y₁ = (9 + √1) / (2 × 2) = 10 / 4 = 5 / 2
- y₂ = (9 - √1) / (2 × 2) = 8 / 4 = 2
д) 5y² - 6y + 1 = 0
- a = 5, b = -6, c = 1
- D = (-6)² - 4 × 5 × 1 = 36 - 20 = 16
- y₁ = (6 + √16) / (2 × 5) = 10 / 10 = 1
- y₂ = (6 - √16) / (2 × 5) = 2 / 10 = 1 / 5
e) 4x² + x - 33 = 0
- a = 4, b = 1, c = -33
- D = (1)² - 4 × 4 × (-33) = 1 + 528 = 529
- x₁ = (-1 + √529) / (2 × 4) = 22 / 8 = 11 / 4
- x₂ = (-1 - √529) / (2 × 4) = -24 / 8 = -3
ж) y² - 10y - 24 = 0
- a = 1, b = -10, c = -24
- D = (-10)² - 4 × 1 × (-24) = 100 + 96 = 196
- y₁ = (10 + √196) / (2 × 1) = 24 / 2 = 12
- y₂ = (10 - √196) / (2 × 1) = -4 / 2 = -2
з) p² + p - 90 = 0
- a = 1, b = 1, c = -90
- D = (1)² - 4 × 1 × (-90) = 1 + 360 = 361
- p₁ = (-1 + √361) / (2 × 1) = 18 / 2 = 9
- p₂ = (-1 - √361) / (2 × 1) = -20 / 2 = -10
Ответ:
a) x₁ = 4/3, x₂ = 1;
б) x₁ = 1, x₂ = 3/5;
в) x₁ = 7/3, x₂ = 2;
г) y₁ = 5/2, y₂ = 2;
д) y₁ = 1, y₂ = 1/5;
e) x₁ = 11/4, x₂ = -3;
ж) y₁ = 12, y₂ = -2;
з) p₁ = 9, p₂ = -10.