Question

1)2х^2+2х+11=5. 2)х^2+х+1=5

Answer 1

1. Метод дискриминанта.
2x^2+2x+11-5=0
2x^2+2x+6=0
D=4-48<0

2.х^2+х+1=5
х^2+х+1-5=0
х^2+х-4=0
D=1+16= \sqrt{17}
x1=1-sqrt{17}/2
x2=1+sqrt{17}/2

Answer 2

1) 2x^2 + 2x + 11 = 5

To solve the equation 2x^2 + 2x + 11 = 5, we can start by rearranging the equation to set it equal to zero:

2x^2 + 2x + 11 - 5 = 0 2x^2 + 2x + 6 = 0

Now we have a quadratic equation in the form ax^2 + bx + c = 0, where a = 2, b = 2, and c = 6.

We can solve this quadratic equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values of a, b, and c: x = (-2 ± √(2^2 - 4*2*6)) / (2*2) x = (-2 ± √(4 - 48)) / 4 x = (-2 ± √(-44)) / 4 x = (-2 ± 2i√11) / 4 x = (-1 ± i√11) / 2

So, the solutions to the equation 2x^2 + 2x + 11 = 5 are: x = (-1 + i√11) / 2 and x = (-1 - i√11) / 2

2) x^2 + x + 1 = 5

To solve the equation x^2 + x + 1 = 5, we can again rearrange the equation to set it equal to zero:

x^2 + x + 1 - 5 = 0 x^2 + x - 4 = 0

Now we have another quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = 1, and c = -4.

We can solve this quadratic equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values of a, b, and c: x = (-1 ± √(1^2 - 4*1*(-4))) / (2*1) x = (-1 ± √(1 + 16)) / 2 x = (-1 ± √17) / 2

So, the solutions to the equation x^2 + x + 1 = 5 are: x = (-1 + √17) / 2 and x = (-1 - √17) / 2