Question

(3х-1)в квадрате+(х-2)(х+2)=1

Answer 1

Решение на фото .......

Answer 2

To solve the equation (3x-1)^2 + (x-2)(x+2) = 1, we'll expand and simplify the equation:

Expanding the equation:

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

Simplifying the equation:

(9x^2 - 6x + 1) + (x^2 - 4) = 1

Combining like terms:

9x^2 - 6x + 1 + x^2 - 4 = 1

Combining the x^2 terms and the constant terms:

10x^2 - 6x - 3 = 1

Rearranging the equation:

10x^2 - 6x - 4 = 0

Now we have a quadratic equation in the form ax^2 + bx + c = 0, where a = 10, b = -6, and c = -4.

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

x = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values into the formula:

x = (-(-6) ± √((-6)^2 - 4 * 10 * (-4))) / (2 * 10)

Simplifying further:

x = (6 ± √(36 + 160)) / 20

x = (6 ± √196) / 20

x = (6 ± 14) / 20

Now we can calculate the two possible solutions for x:

x1 = (6 + 14) / 20 = 20 / 20 = 1

x2 = (6 - 14) / 20 = -8 / 20 = -0.4

Therefore, the solutions to the equation are x = 1 and x = -0.4.