Question

(х^2-2х)^2+3(х^2-2х)-4=0

Answer 1

Объяснение:

(х^2-2х)((х^2-2х)+3-4/х^2-2х=0

х=0

х-2=0

х=2

х^4-4х^3-4х^2+3х^2-6х-4=0

х^4-4х^3-х^2-6х-4=0

х(х^3-4х^2-х-6)-4=0

Answer 2

To solve the quadratic equation $(x^2 - 2x)^2 + 3(x^2 - 2x) - 4 = 0$, we can use the following steps:

Let's substitute $y = x^2 - 2x$, so the equation becomes:

$y^2 + 3y - 4 = 0$

Now we have a quadratic equation in terms of $y$. We can factor this quadratic equation:

$(y + 4)(y - 1) = 0$

Now, we solve for $y$ using each factor separately:

1. $y + 4 = 0$ $y = -4$

2. $y - 1 = 0$ $y = 1$

Remember, $y = x^2 - 2x$, so we can substitute back to find the solutions for $x$:

1. $x^2 - 2x = -4$ Rearranging and factoring: $x^2 - 2x + 4 = 0$ This is a quadratic equation with no real solutions because the discriminant ($b^2 - 4ac$) is negative.

2. $x^2 - 2x = 1$ Rearranging: $x^2 - 2x - 1 = 0$

Now we can solve this equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -2$, and $c = -1$.

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$ $x = \frac{2 \pm \sqrt{4 + 4}}{2}$ $x = \frac{2 \pm \sqrt{8}}{2}$ $x = 1 \pm \sqrt{2}$

So, the solutions for the original equation $(x^2 - 2x)^2 + 3(x^2 - 2x) - 4 = 0$ are $x = 1 + \sqrt{2}$ and $x = 1 - \sqrt{2}$.