Problem:
Monic quadratic polynomials P(x) and Q(x) have the property that P(Q(x)) has zeros at x=−23,−21,−17, and -15 , and Q(P(x)) has zeros at x= −59,−57,−51, and -49 . What is the sum of the minimum values of P(x) and Q(x) ?
Answer Choices:
A. −100
B. −82
C. −73
D. −64
E. 0
Solution:
Because both P(Q(x)) and Q(P(x)) have four distinct real zeros, both P(x) and Q(x) must have two distinct real zeros, so there are real numbers h1,k1,h2, and k2 such that P(x)=(x−h1)2−k12 and Q(x)=(x−h2)2−k22. The zeros of P(Q(x)) occur when Q(x)=h1±k1. The solutions of each equation are equidistant from h2, so h2=−19. It follows that Q(−15)−Q(−17)= (16−k22)−(4−k22)=12, and also Q(−15)−Q(−17)=2k1, so k1=6. Similarly h1=−54, so 2k2=P(−49)−P(−51)=(25−k12)−(9−k12)=16, and k2=8. Thus the sum of the minimum values of P(x) and Q(x) is −k12−k22=−100.
The problems on this page are the property of the MAA's American Mathematics Competitions