Problem:
Suppose that the roots of x3+3x2+4x−11=0 are a,b, and c, and that the roots of x3+rx2+sx+t=0 are a+b,b+c, and c+a. Find t.
Solution:
The first equation implies that a+b+c=−3. The second equation implies that t=−(a+b)(b+c)(c+a). It follows that t=−(−3−c)(−3−a)(−3−b), which expands to t=27+9(a+b+c)+3(ab+bc+ca)+abc. The first equation implies that ab+bc+ca=4 and abc=11, hence that t=27−27+12+11=23.
OR
The first equation implies that a+b+c=−3. It follows that the roots of the second equation are −3−c,−3−a, and −3−b. These are also the roots of the equation (−x−3)3+3(−x−3)2+4(−x−3)−11=0, obtained by replacing x by −x−3 in the first equation. The leading coefficient of this equation is −1 and the constant term is (−3)3+3(−3)2+4(−3)−11=−23; thus t=23.
The problems on this page are the property of the MAA's American Mathematics Competitions