Question

-x^2+6x-11=0; 3x^2-x-2=0; 0,64x^2+1,6x+1=0; -5x^2+2x-2,5=0

Answer 1

Вроде все правильно, но лучше проверить

Answer 2

To solve each of the quadratic equations you provided, we can use the quadratic formula, which states that for a quadratic equation in the form of ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Let's solve each equation one by one:

1. -x^2 + 6x - 11 = 0:

Comparing this equation with ax^2 + bx + c = 0, we have a = -1, b = 6, and c = -11. Substituting these values into the quadratic formula, we get:

x = (-6 ± √(6^2 - 4(-1)(-11))) / (2(-1)) x = (-6 ± √(36 - 44)) / (-2) x = (-6 ± √(-8)) / (-2)

Since we have a negative value under the square root, this equation does not have real solutions. The solutions would be complex numbers.

2. 3x^2 - x - 2 = 0:

Comparing this equation with ax^2 + bx + c = 0, we have a = 3, b = -1, and c = -2. Substituting these values into the quadratic formula, we get:

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

The solutions are: x = (1 + 5) / 6 = 6 / 6 = 1 x = (1 - 5) / 6 = -4 / 6 = -2/3

Therefore, the solutions for this equation are x = 1 and x = -2/3.

3. 0.64x^2 + 1.6x + 1 = 0:

To avoid working with decimal values, we can multiply the entire equation by 100 to eliminate the decimals:

64x^2 + 160x + 100 = 0

Comparing this equation with ax^2 + bx + c = 0, we have a = 64, b = 160, and c = 100. Substituting these values into the quadratic formula, we get:

x = (-160 ± √(160^2 - 4(64)(100))) / (2(64)) x = (-160 ± √(25600 - 25600)) / 128 x = (-160 ± √0) / 128 x = (-160 ± 0) / 128

The solutions are: x = (-160 + 0) / 128 = -160 / 128 = -5/4 x = (-160 - 0) / 128 = -160 / 128 = -5/4

Therefore, the solutions for this equation are x = -5/4 and x = -5/4.

4. -5x^2 + 2x - 2.5 = 0:

Comparing this equation with ax^2 + bx + c = 0, we have a = -5, b = 2, and c = -2.5. Substituting these values into the quadratic formula, we get:

x = (-2 ± √(2^2 - 4(-5)(-2.5))) / (2(-5)) x = (-