Problem:
Find the positive solution to
x2−10x−291​+x2−10x−451​−x2−10x−692​=0
Solution:
Let x2−10x=y. The equation in the problem then becomes
y−291​+y−451​−y−692​=0
from which
y−291​−y−691​=y−691​−y−451​
and
(y−29)(y−69)−40​=(y−45)(y−69)24​
follows. This equation has y=39 as its only solution. We then note that x2−10x=39 is satisfied by the positive number 13​.
The problems on this page are the property of the MAA's American Mathematics Competitions