Question 22·25 Super-Hard SAT Advanced Math Questions·Advanced Math

No Desmos needed. A root-product fact will give the answer without finding four individual roots. Expand the given square:

Cancel the opposite constants:

In the composition $f(f(x))$, the whole output $x^2-4x$ goes into the rule again. Set that result equal to the given output:

Subtract $4$:

The highest power comes from $(x^2)^2=x^4$. The other parts have no $x^4$, so its leading coefficient, the number multiplying the highest power, is $1$. That makes this a monic fourth-degree equation.

The expression $x^2-4x$ has no term without $x$. Squaring it or multiplying it by $-4$ cannot create one. So the constant term of the equation is the remaining $-4$.

Call the four roots, the values that make the equation zero, $r_1,r_2,r_3,r_4$. Because the leading coefficient is $1$, the left side factors as $(x-r_1)(x-r_2)(x-r_3)(x-r_4)$. For a monic fourth-degree equation, the product of its four roots is its constant term. Taking the constant from each factor gives:

Four negative signs cancel:

So the product of the four solutions is $-4$. Choice C.

Let $u=f(x)=x^2-4x$. The outer function must send this input $u$ to $4$, so:

Subtract $4$:

Call the two solutions for $u$ $u_1$ and $u_2$. For a monic quadratic, the product of its roots equals its constant term, because $(u-u_1)(u-u_2)$ has constant term $u_1u_2$. Here, the two possible inner outputs have product:

For either possible output $u_i$, set the inner rule equal to it:

Subtract $u_i$:

The two $x$-roots of this quadratic multiply to its constant term, $-u_i$. Each branch supplies two roots: the problem gives four real solutions in all.

Multiply the pair products:

So the product of all four $x$-solutions is $-4$. Choice C.