### Instruction: Let $f(x) = x^2-3x$. For what values of $x$ is $f(f(x)) = f(x)$?  Enter all the solutions, separated by commas.### Response: I want to find the values of $x$ that make $f(f(x))$ and $f(x)$ equal, so I can start by setting them equal to each other and see what I get.
That means I need to plug in $f(x)$ into the expression for $f(x)$, which gives me $f(f(x)) = (f(x))^2-3f(x)$.
So I have $(f(x))^2-3f(x) = f(x)$, or $(x^2-3x)^2-3(x^2-3x) = x^2-3x$.
Now I have a quadratic equation in terms of $x^2-3x$, so I can try to solve for it by moving everything to one side and factoring.
I get $(x^2-3x)^2-4(x^2-3x) = 0$, or $(x^2-3x)(x^2-3x-4) = 0$.
This means that either $x^2-3x = 0$ or $x^2-3x-4 = 0$.
If $x^2-3x = 0$, then $x(x-3) = 0$, so $x = 0$ or $x = 3$ are solutions.
If $x^2-3x-4 = 0$, then I can use the quadratic formula to find the solutions.
I get $x = \frac{3 \pm \sqrt{9+16}}{2}$, or $x = \frac{3 \pm 5}{2}$, so $x = -1$ or $x = 4$ are solutions.