Question

Решите 1) x^2+3x-4=0 2) x^2-4x+3=0 3) 3x^2+2x-5=0

Answer 1

Ответ:

Пошаговое объяснение:

Первые 2 уравнения решаются по теореме Виета

1) х1=-4 х2=1

2) х1=3 х2=1

3)D=4 - 4*3*(-5 )=64

х1=(-2+8)/6=1

х2=(-2-8)/6=-10/6

Answer 2

1. x^2 + 3x - 4 = 0:

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

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Here, a = 1, b = 3, and c = -4. Substituting these values into the formula, we get:

x = (-3 ± sqrt(3^2 - 4(1)(-4))) / 2(1)

x = (-3 ± sqrt(9 + 16)) / 2

x = (-3 ± sqrt(25)) / 2

x = (-3 ± 5) / 2

So the solutions are:

x = (-3 + 5) / 2 = 1

x = (-3 - 5) / 2 = -4

Therefore, the solutions to the equation x^2 + 3x - 4 = 0 are x = 1 and x = -4.

2. x^2 - 4x + 3 = 0:

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

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Here, a = 1, b = -4, and c = 3. Substituting these values into the formula, we get:

x = (4 ± sqrt((-4)^2 - 4(1)(3))) / 2(1)

x = (4 ± sqrt(16 - 12)) / 2

x = (4 ± sqrt(4)) / 2

So the solutions are:

x = (4 + 2) / 2 = 3

x = (4 - 2) / 2 = 1

Therefore, the solutions to the equation x^2 - 4x + 3 = 0 are x = 1 and x = 3.

3. 3x^2 + 2x - 5 = 0:

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

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Here, a = 3, b = 2, and c = -5. Substituting these values into the formula, we get:

x = (-2 ± sqrt(2^2 - 4(3)(-5))) / 2(3)

x = (-2 ± sqrt(4 + 60)) / 6

x = (-2 ± sqrt(64)) / 6

So the solutions are:

x = (-2 + 8) / 6 = 1

x = (-2 - 8) / 6 = -5/3

Therefore, the solutions to the equation 3x^2 + 2x - 5 = 0 are x = 1 and x = -5/3.