Problem:
The equation
2333x−2+2111x+2=2222x+1+1
has three real roots. Given that their sum is m/n, where m and n are relatively prime positive integers, find m+n.
Solution:
Let y=2111x. The given equation is equivalent to (1/4)y3+4y=2y2+1, which can be simplified to y3−8y2+16y−4=0. Since the roots of the given equation are real, the roots of the last equation must be positive. Let the roots of the given equation be x1,x2, and x3, and let the roots of the equation in y be y1,y2, and y3. Then x1+x2+x3=(1/111)(log2y1+log2y2+log2y3)= (1/111)log2(y1y2y3)=(1/111)log24=2/111, and m+n=113.
Note: It can be verified that y3−8y2+16y−4=0 has three positive roots by sketching a graph.
The problems on this page are the property of the MAA's American Mathematics Competitions