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