11. Let $f(x)=x^{2}+6 x+c$ for all real numbers $x$, where $c$ is some real number. For what values of $c$ does $f(f(x))$ have exactly 3 distinct real roots?
be the case that $f\left(x_{1}\right)=r_{1}$. As a result, $f(f(x))$ would have at most two roots, thus not satisfying the problem condition. Hence $f$ has two distinct roots. Let them be $r_{1} \neq r_{2}$.
Since $f(f(x))$ has just three distinct roots, either $f(x)=r_{1}$ or $f(x)=r_{2}$ has one distinct root. Assume without loss of generality that $r_{1}$ has one distinct root. Then $f(x)=x^{2}+6 x+c=r_{1}$ has one root, so that $x^{2}+6 x+c-r_{1}$ is a square polynomial. Therefore, $c-r_{1}=9$, so that $r_{1}=c-9$. So $c-9$ is a root of $f$. So $(c-9)^{2}+6(c-9)+c=0$, yielding $c^{2}-11 c+27=0$, or $\left(c-\frac{11}{2}\right)^{2}=\frac{13}{2}$. This results to $c=\frac{11 \pm \sqrt{13}}{2}$.
If $c=\frac{11-\sqrt{13}}{2}, f(x)=x^{2}+6 x+\frac{11-\sqrt{13}}{2}=\left(x+\frac{7+\sqrt{13}}{2}\right)\left(x+\frac{5-\sqrt{13}}{2}\right)$. We know $f(x)=\frac{-7-\sqrt{13}}{2}$ has a double root, -3 . Now $\frac{-5+\sqrt{13}}{2}>\frac{-7-\sqrt{13}}{2}$ so the second root is above the vertex of the parabola, and is hit twice.
If $c=\frac{11+\sqrt{13}}{2}, f(x)=x^{2}+6 x+\frac{11+\sqrt{13}}{2}=\left(x+\frac{7-\sqrt{13}}{2}\right)\left(x+\frac{5+\sqrt{13}}{2}\right)$. We know $f(x)=\frac{-7+\sqrt{13}}{2}$ has a double root, -3 , and this is the value of $f$ at the vertex of the parabola, so it is its minimum value. Since $\frac{-5-\sqrt{13}}{2}<\frac{-7+\sqrt{13}}{2}, f(x)=\frac{-5-\sqrt{13}}{2}$ has no solutions. So in this case, $f$ has only one real root.

So the answer is $c=\frac{11-\sqrt{13}}{2}$.
Note: In the solutions packet we had both roots listed as the correct answer. We noticed this oversight on the day of the test and awarded points only for the correct answer.