Question

(x^2-2x)^2-7(x^2-2x)-8=0

Answer 1

$a-b+c=0\Rightarrow x=\left \{ -1;-\frac{c}{a} \right \}\\x^2-2x=\left \{ -1;8 \right \}\\\left[ \begin{gathered} x^2-2x+1=0; \\ x^2-2x-8=0 \\ \end{gathered} \right.\Rightarrow x=\left \{ -2;1;4 \right \}$

Answer 2

To solve the quadratic equation $(x^2 - 2x)^2 - 7(x^2 - 2x) - 8 = 0$, we can use a substitution to simplify the equation. Let's set $y = x^2 - 2x$, so the equation becomes:

$y^2 - 7y - 8 = 0$

Now we have a simple quadratic equation in terms of $y$. We can solve this equation using factoring, completing the square, or the quadratic formula. In this case, I'll use factoring:

$y^2 - 7y - 8 = 0$ $(y - 8)(y + 1) = 0$

Now we have two possible values for $y$:

1. $y - 8 = 0 \implies y = 8$
2. $y + 1 = 0 \implies y = -1$

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

1. $x^2 - 2x = 8$
2. $x^2 - 2x = -1$

Solving the first equation $x^2 - 2x = 8$:

$x^2 - 2x - 8 = 0$ $(x - 4)(x + 2) = 0$

This gives us two possible values for $x$:

1. $x - 4 = 0 \implies x = 4$
2. $x + 2 = 0 \implies x = -2$

Solving the second equation $x^2 - 2x = -1$:

$x^2 - 2x + 1 = 0$ $(x - 1)^2 = 0$

This equation has a double root at $x = 1$.

So, the solutions to the original equation $(x^2 - 2x)^2 - 7(x^2 - 2x) - 8 = 0$ are $x = 4$, $x = -2$, and $x = 1$ (with multiplicity 2).