Question

5х²-х=0 , 3x²-1,2x=0 , 1/4y²+y=0​

Answer 1

Объяснение:

решение на фото прилагаю

Answer 2

To solve each of these quadratic equations, we can use the quadratic formula:

For a quadratic equation of the form ax² + bx + c = 0, the quadratic formula is given by:

x = (-b ± √(b² - 4ac)) / (2a)

Let's solve each equation one by one:

1. 5x² - x = 0: a = 5, b = -1, c = 0

Using the quadratic formula: x = (-(-1) ± √((-1)² - 4 * 5 * 0)) / (2 * 5) x = (1 ± √(1 - 0)) / 10 x = (1 ± 1) / 10

So the solutions are: x₁ = (1 + 1) / 10 = 2 / 10 = 0.2 x₂ = (1 - 1) / 10 = 0 / 10 = 0

The solutions for the equation 5x² - x = 0 are x = 0 and x = 0.2.

2. 3x² - 1.2x = 0: a = 3, b = -1.2, c = 0

Using the quadratic formula: x = (-(-1.2) ± √((-1.2)² - 4 * 3 * 0)) / (2 * 3) x = (1.2 ± √(1.44 - 0)) / 6 x = (1.2 ± √(1.44)) / 6

So the solutions are: x₁ = (1.2 + √(1.44)) / 6 x₂ = (1.2 - √(1.44)) / 6

The solutions for the equation 3x² - 1.2x = 0 depend on the values inside the square root.

3. (1/4)y² + y = 0: a = 1/4, b = 1, c = 0

Using the quadratic formula: y = (-(1) ± √((1)² - 4 * (1/4) * 0)) / (2 * (1/4)) y = (-1 ± √(1 - 0)) / (1/2) y = (-1 ± √(1)) / (1/2) y = (-1 ± 1) / (1/2)

So the solutions are: y₁ = (-1 + 1) / (1/2) = 0 / (1/2) = 0 y₂ = (-1 - 1) / (1/2) = -2 / (1/2) = -4

The solutions for the equation (1/4)y² + y = 0 are y = 0 and y = -4.