Question

Решить уравнение (x^2 + x)^2 +2(x^2 + x) - 8 = 0

Answer 1

Замена:
х² + х = t
t² + 2t - 8 = 0
Согласно теореме Виета:
t1 + t2 = -2
t1t2 = -8
=>
t1 = -4
t2 = 2
Возвращаемся к замене:
х² + х = -4
x² + x + 4 = 0
D = b² - 4ac = 1 - 16 = -15 =>
Действительных корней уравнения не существует
х² + х = 2
х² + х - 2 = 0
По теореме Виета:
х1 + х2 = -1
х1х2 = -2
=>
х1 = -2; х2 = 1
Ответ: -2; 1

Answer 2

To solve the quadratic equation $(x^2 + x)^2 + 2(x^2 + x) - 8 = 0$, let's perform the following steps:

Step 1: Substitute $y = x^2 + x$ to simplify the equation. Step 2: Solve the resulting quadratic equation for $y$. Step 3: After finding $y$, find the corresponding values of $x$ using the relationship $y = x^2 + x$.

Let's start:

Step 1: Let $y = x^2 + x$. The equation becomes: $y^2 + 2y - 8 = 0$

Step 2: Now, we have a simple quadratic equation in terms of $y$. To solve it, we can factor or use the quadratic formula.

Factoring: The equation $y^2 + 2y - 8 = 0$ can be factored as follows: $(y + 4)(y - 2) = 0$

Step 3: Now, we have two possible values for $y$:

1. $y + 4 = 0$ => $y = -4$
2. $y - 2 = 0$ => $y = 2$

Now, we need to find the corresponding values of $x$ for each value of $y$:

For $y = -4$: $x^2 + x = -4$

Since $x^2 + x$ is a sum of two squares, it cannot be negative for any real value of $x$. Therefore, there are no real solutions for $y = -4$ and, consequently, no real solutions for the original equation.

For $y = 2$: $x^2 + x = 2$

Now, let's solve for $x$:

$x^2 + x - 2 = 0$

This is another quadratic equation. We can factor it as follows: $(x + 2)(x - 1) = 0$

Now, we have two possible values for $x$:

1. $x + 2 = 0$ => $x = -2$
2. $x - 1 = 0$ => $x = 1$

So, the real solutions to the original equation are $x = -2$ and $x = 1$.