Difference between revisions of "2000 AIME II Problems/Problem 13"
(minor wik) |
|||
Line 29: | Line 29: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 20:31, 4 July 2013
Problem
The equation has exactly two real roots, one of which is
, where
,
and
are integers,
and
are relatively prime, and
. Find
.
Solution
We may factor the equation as:[1]
Now for real
. Thus the real roots must be the roots of the equation
. By the quadratic formula the roots of this are:
Thus , and so the final answer is
.
^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of with half of the polynomial's degree (in this case, divide through by
), and then to use one of the substitutions
. In this case, the substitution
gives
and
, which reduces the polynomial to just
. Then one can backwards solve for
.
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.