Question

Решите квадратные уравнения: 1) y^2+8y+15=0 2) 8x-2x^2+3=0 3) 10x^2+5x-0.6=0

Answer 1

Ответ:

1) ]y^{2}+8y+15=0

D=64-60={4}=2

x1=(-8+2)/2=-3

x2=(-8-2)/2=-5

Answer 2

1. y^2 + 8y + 15 = 0

To solve this quadratic equation, we can use the factoring method:

y^2 + 8y + 15 = 0

(y + 3)(y + 5) = 0

Therefore, y = -3 or y = -5.

2. 8x - 2x^2 + 3 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = -2, b = 8, and c = 3.

x = (-8 ± sqrt(8^2 - 4(-2)(3))) / 2(-2)

x = (-8 ± sqrt(64 + 24)) / (-4)

x = (-8 ± sqrt(88)) / (-4)

x = (-8 ± 2sqrt(22)) / (-4)

Simplifying the expression, we get:

x = 2 ± (1/2)sqrt(22)

Therefore, x = 2 + (1/2)sqrt(22) or x = 2 - (1/2)sqrt(22).

3. 10x^2 + 5x - 0.6 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 10, b = 5, and c = -0.6.

x = (-5 ± sqrt(5^2 - 4(10)(-0.6))) / 2(10)

x = (-5 ± sqrt(25 + 24)) / 20

x = (-5 ± sqrt(49)) / 20

x = (-5 ± 7) / 20

Simplifying the expression, we get:

x = 2/5 or x = -3/2.

Therefore, the solutions to the given quadratic equations are:

1. y = -3 or y = -5
2. x = 2 + (1/2)sqrt(22) or x = 2 - (1/2)sqrt(22)
3. x = 2/5 or x = -3/2.