Question

(x² +x)²+2(x²+x) - 8=0​

Answer 1

Ответ:x=-2;x=1

Объяснение:

Пусть х^2+х=t

t^2+2t-8=0

D=4+32=36

t1=-2-6/2=-4(не подходит)

t2=-2+6/2=2

х^2+х=2

х^2+х-2=0

D=1+8=9

x1=-1-3/2=-2

x2=-1+3/2=1

Answer 2

To solve the quadratic equation $(x^2 + x)^2 + 2(x^2 + x) - 8 = 0$, we can use a substitution to simplify the expression. Let's substitute $y = x^2 + x$. Then the equation becomes a quadratic equation in terms of $y$:

$y^2 + 2y - 8 = 0$

Now, we can solve this quadratic equation for $y$ using either factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

The quadratic formula is given by: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $y^2 + 2y - 8 = 0$, we have $a = 1$, $b = 2$, and $c = -8$.

Substitute the values into the quadratic formula:

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

$y = \frac{-2 \pm \sqrt{4 + 32}}{2}$

$y = \frac{-2 \pm \sqrt{36}}{2}$

$y = \frac{-2 \pm 6}{2}$

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

1. $y = \frac{-2 + 6}{2} = 2$
2. $y = \frac{-2 - 6}{2} = -4$

Remember, $y = x^2 + x$. So, we have two equations to solve for $x$:

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

Now, let's solve these equations one by one:

1. $x^2 + x = 2$

Subtracting 2 from both sides: $x^2 + x - 2 = 0$

Factor the quadratic equation: $(x+2)(x-1) = 0$

Now, set each factor to zero and solve for $x$:

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

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

Adding 4 to both sides: $x^2 + x + 4 = 0$

This equation does not have real solutions since the discriminant $b^2 - 4ac$ is negative.

Therefore, the solutions for the original equation $(x^2 + x)^2 + 2(x^2 + x) - 8 = 0$ are $x = -2$ and $x = 1$. The second equation $x^2 + x = -4$ has no real solutions.