2010 AMC 12B Problem 23

If $P$ had only one real root, then $P(Q(x))=0$ would have at most two real solutions, not four. Thus $P$ has two distinct real roots, and the same argument applies to $Q.$ Write $P(x)=(x-h_1)^2-k_1^2$ and $Q(x)=(x-h_2)^2-k_2^2,$ with $k_1,k_2\gt0$ and minimum values $-k_1^2$ and $-k_2^2.$ The zeros of $P(Q(x))$ occur where $Q(x)=h_1\pm k_1;$ their four solutions are symmetric about $h_2,$ so $h_2$ is the average $\tfrac{-23-21-17-15}{4}=-19.$ Then $Q(-15)-Q(-17)$ $=(16-k_2^2)-(4-k_2^2)$ $=12,$ and this difference equals $2k_1,$ so $k_1=6.$ Symmetrically, $h_1=\tfrac{-59-57-51-49}{4}=-54,$ and $P(-49)-P(-51)$ $=(25-k_1^2)-(9-k_1^2)$ $=16=2k_2,$ so $k_2=8.$ The sum of the minimum values is $-k_1^2-k_2^2=-36-64=-100.$ Thus, the correct answer is A.