Question

(x-2)^2=x^2 2х-4=х^2

Answer 1

Ответ:

(x-2)^2=x^2

2х-4=х^2

Объяснение:

………………

Answer 2

Объяснение:

1. x² - 4x + 4 - x² = 0

4x = 4

x = 1

2. x² - 2x + 4 = 0

дискриминант отрицательный => не имеет корней

Answer 3

To solve the equation $(x-2)^2 = x^2$, you can start by expanding the left side of the equation and then simplifying:

$(x-2)^2 = x^2$

Expanding the left side using the square of a binomial:

$x^2 - 4x + 4 = x^2$

Now, subtract $x^2$ from both sides of the equation to isolate the terms with "x":

$-4x + 4 = 0$

Next, subtract 4 from both sides:

$-4x = -4$

Now, divide both sides by -4 to solve for x:

$x = -4 / -4$

$x = 1$

So, the solution to the equation $(x-2)^2 = x^2$ is $x = 1$.

As for the equation $2x - 4 = x^2$, it's a quadratic equation, and you can solve it by rearranging it into the standard quadratic form:

$x^2 - 2x + 4 = 0$

Now, you can solve it using the quadratic formula:

The quadratic formula is $x = (-b ± √(b^2 - 4ac)) / (2a)$, where in this case, $a = 1$, $b = -2$, and $c = 4$.

$x = (-(-2) ± √((-2)^2 - 4(1)(4))) / (2(1))$

$x = (2 ± √(4 - 16)) / 2$

$x = (2 ± √(-12)) / 2$

The discriminant ($b^2 - 4ac$) is negative, which means that there are no real solutions to this equation. The solutions are complex numbers:

$x = (2 ± √(-12)) / 2$

$x = (2 ± 2i√3) / 2$

Now, you can simplify by dividing both the numerator and denominator by 2:

$x = 1 ± i√3$

So, the solutions to the equation $2x - 4 = x^2$ are $x = 1 + i√3$ and $x = 1 - i√3$, which are complex numbers.