Question

1) 3x² – 5x + 2 = 0 2) 7x² + 12x + 5 = 0 3) x² + 3x – 10 = 0 4) x² + x - 2 = 0 5) 4x² + 20x + 25 = 0 6) x² – 3x + 5 = 0 7) x² + 2x = 0 8) x² – 5 = 0

Answer 1

Объяснение:

Тут как бы все через дискриминант, кроме последних 2х примеров, в 7 вынести х за скобки, а в 8 просто 5 перенести в правую сторону (я не мастер объяснений)

Answer 2

It looks like you have provided a list of quadratic equations, and you would like to find their solutions. Quadratic equations are typically solved using the quadratic formula:

For an equation in the form ax² + bx + c = 0, the solutions (roots) are given by:

x = (-b ± √(b² - 4ac)) / (2a)

Let's solve each of the equations you've listed:

1. 3x² – 5x + 2 = 0 a = 3, b = -5, c = 2

x = (-(-5) ± √((-5)² - 4 * 3 * 2)) / (2 * 3) x = (5 ± √(25 - 24)) / 6 x = (5 ± √1) / 6

So, the solutions are: x₁ = (5 + 1) / 6 = 6/6 = 1 x₂ = (5 - 1) / 6 = 4/6 = 2/3

2. 7x² + 12x + 5 = 0 a = 7, b = 12, c = 5

x = (-12 ± √(12² - 4 * 7 * 5)) / (2 * 7) x = (-12 ± √(144 - 140)) / 14 x = (-12 ± √4) / 14

So, the solutions are: x₁ = (-12 + 2) / 14 = -10/14 = -5/7 x₂ = (-12 - 2) / 14 = -14/14 = -1

3. x² + 3x – 10 = 0 a = 1, b = 3, c = -10

x = (-3 ± √(3² - 4 * 1 * (-10))) / (2 * 1) x = (-3 ± √(9 + 40)) / 2 x = (-3 ± √49) / 2

So, the solutions are: x₁ = (-3 + 7) / 2 = 4/2 = 2 x₂ = (-3 - 7) / 2 = -10/2 = -5

4. x² + x - 2 = 0 a = 1, b = 1, c = -2

x = (-1 ± √(1² - 4 * 1 * (-2))) / (2 * 1) x = (-1 ± √(1 + 8)) / 2 x = (-1 ± √9) / 2

So, the solutions are: x₁ = (-1 + 3) / 2 = 2/2 = 1 x₂ = (-1 - 3) / 2 = -4/2 = -2

5. 4x² + 20x + 25 = 0 a = 4, b = 20, c = 25

x = (-20 ± √(20² - 4 * 4 * 25)) / (2 * 4) x = (-20 ± √(400 - 400)) / 8 x = (-20 ± √0) / 8

Since the discriminant is zero, there is one real solution: x = -20/8 = -5/2

6. x² – 3x + 5 = 0 a = 1, b = -3, c = 5

x = (3 ± √((-3)² - 4 * 1 * 5)) / (2 * 1) x = (3 ± √(9 - 20)) / 2 x = (3 ± √(-11)) / 2

Since the discriminant is negative, there are no real solutions. The solutions are complex.

7. x² + 2x = 0 a = 1, b = 2, c = 0

x = (-2 ± √(2² - 4 * 1 * 0)) / (2 * 1) x = (-2 ± √(4)) / 2 x = (-2 ± 2) / 2

So, the solutions are: x₁ = (-2 + 2) / 2 = 0/2 = 0 x₂ = (-2 - 2) / 2 = -4/2 = -2

8. x² – 5 = 0 a = 1, b = 0, c = -5

x = (0 ± √(0² - 4 * 1 * (-5))) / (2 * 1) x = (0 ± √(0 + 20)) / 2 x = (0 ± √20) / 2

So, the solutions are: x₁ = (0 + √20) / 2 = √20/2 = √5 x₂ = (0 - √20) / 2 = -√20/2 = -√5

These are the solutions to the given quadratic equations.