Answer
$\left\{-2,-1,3,4\right\}$
Work Step by Step
Let $z=
x^2-2x
$. Then the given equation, $
(x^2-2x)^2=11(x^2-2x)-24
,$ is equivalent to
\begin{align*}
z^2&=11z-24
.\end{align*}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{align*}
z^2-11z+24&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-8)(z-3)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving the variable, then
\begin{array}{l|r}
z-8=0 & z-3=0
\\
z=8 & z=3
.\end{array}
Since $z=
x^2-2x
$, then by back-substitution,
\begin{array}{l|r}
x^2-2x=8 & x^2-2x=3
\\
x^2-2x-8=0 & x^2-2x-3=0
.\end{array}
Using factoring of trinomials, the equations above are equivalent to
\begin{array}{l|r}
(x-4)(x+2)=0 & (x-3)(x+1)=0
.\end{array}
Equating each factor to zero (Zero Product Property) and solving the variable, then
\begin{array}{c|c|c|c}
x-4=0 & x+2=0 & x-3=0 & x+1=0
\\
x=4 & x=-2 & x=3 & x=-1
.\end{array}
Hence, the solution set of the equation $
(x^2-2x)^2=11(x^2-2x)-24
$ is $\left\{-2,-1,3,4\right\}$.
