Problem:
Real numbers r and s are roots of p(x)=x3+ax+b, and r+4 and s−3 are roots of q(x)=x3+ax+b+240. Find the sum of all possible values of ∣b∣.
Solution:
Because the coefficient of x2 in both p(x) and q(x) is 0, the remaining root of p(x) is t=−r−s, and the remaining root of q(x) is t−1. The coefficients of x in p(x) and q(x) are both equal to a, and equating the two coefficients gives
rs+st+tr=(r+4)(s−3)+(s−3)(t−1)+(t−1)(r+4)
from which t=4r−3s+13. Furthermore, b=−rst, so
b+240=−rst+240=−(r+4)(s−3)(t−1)
from which rs−4st+3tr−3r+4s+12t−252=0. Substituting t=4r−3s+13 gives
12r2−24rs+12s2+84r−84s−96=0
which is equivalent to (r−s)2+7(r−s)−8=0, and the solutions for r−s are 1 and −8. If r−s=1, then the roots of p(x) are r,s=r−1, and t=4r−3s+13=r+16. Because the sum of the roots is 0,r=−5. In this case the roots are −5,−6, and 11, and b=−rst=−330. If r−s=−8, then the roots of p(x) are r,s=r+8, and t=4r−3s+13=r−11. In this case the roots are 1,9, and −10, and b=−rst=90. Therefore the requested sum is ∣−330∣+∣90∣=420.
The problems on this page are the property of the MAA's American Mathematics Competitions