СРОЧНО СРОЧНО НАДО СЕГОДНЯ РЕШИТЬ КОНТРОЛЬНУЮ ДАЮ 5 ЗВЁЗДЫ

Ответ:

4x² - 20 = 0 => 4x² = 20 => x² = 5 => x = ±√5

3x² + 5x = 0 => x(3x + 5) = 0 => x = 0 or x = -5/3

(x-2)(x-5) > 0 => x ∈ (-∞, 2) ∪ (5, ∞)

7x² - 22x + 3 = 0 => x = (22 ± √(22² - 473)) / (2*7) => x₁ ≈ 0.42, x₂ ≈ 3.15

7x² - 6x + 2 = 0 => x = (6 ± √(6² - 472)) / (2*7) => x₁ ≈ 0.49 + 0.37i, x₂ ≈ 0.49 - 0.37i

4x² + 12x + 9 = 0 => x = (-12 ± √(12² - 449)) / (2*4) => x = -1.5

Let's assume the quadratic equation is ax² + bx + c = 0.

According to the problem, we have:

sum of the roots = -b/a = 6

product of the roots = c/a = 4

Using these two equations, we can express b and c in terms of a:

b = -6a and c = 4a

Substituting these values in the quadratic equation, we get:

ax² - 6ax + 4a = 0

Dividing both sides by a (assuming a is not equal to 0), we get:

x² - 6x + 4 = 0

This is the required quadratic equation