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