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)(log2​y1​+log2​y2​+log2​y3​)= (1/111)log2​(y1​y2​y3​)=(1/111)log2​4=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