Question

(X^2+2x)^2-2(x^2+2x)=0

Answer 1

(х²+2х)(х²+2х-2)=0

[х²+2х=0
[х²+2х-2=0

1)х(х+2)=0
х1=0;х2=-2

2)х²+2х-2=0
Д=4+8=12
х=(-2±√12)/2=-1±√3
х3=-1+√3
х4=-1-√3

Answer 2

Let's solve the equation:

$(x^2 + 2x)^2 - 2(x^2 + 2x) = 0$

To simplify the equation, let's factor out the common term $(x^2 + 2x)$:

$x^2 + 2x \left( (x^2 + 2x) - 2 \right) = 0$

Now, combine like terms inside the parentheses:

$x^2 + 2x(x^2 + 2x - 2) = 0$

Next, let's solve the quadratic equation inside the parentheses:

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

The quadratic equation can be solved using the quadratic formula:

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

where $ax^2 + bx + c$ is the standard form of the quadratic equation.

In this case, $a = 1$, $b = 2$, and $c = -2$, so the solutions for $x$ are:

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

$x = \frac{-2 \pm \sqrt{4 + 8}}{2}$

$x = \frac{-2 \pm \sqrt{12}}{2}$

$x = \frac{-2 \pm 2\sqrt{3}}{2}$

Now, let's simplify further:

$x = -1 \pm \sqrt{3}$

So, the two solutions to the original equation are:

$x = -1 + \sqrt{3}$

$x = -1 - \sqrt{3}$