2015 AMC 10A Problems/Problem 23
Contents
Problem
The zeroes of the function
are integers. What is the sum of the possible values of
Solution 1
By Vieta's Formula,
is the sum of the integral zeros of the function, and so
is integral.
Because the zeros are integral, the discriminant of the function,
, is a perfect square, say
. Then adding 16 to both sides and completing the square yields
Therefore
and
Let
and
; then,
and so
. Listing all possible
pairs (not counting transpositions because this does not affect (
),
, yields
. These
sum to
, so our answer is
.
Solution 2
Let
and
be the integer zeroes of the quadratic. ,
By Vieta's Formulas,
Plugging the first equation in the second,
Rearranging gives
These factors
(ignoring order, because we want the sum of factors), can be
or
.
The sum of distinct
= (f_1+2) + (f_2+2)
\sum_a a = (5+4) + (-5+4) + (4+4) + (0+4) = \boxed{\textbf{(C) }16}$.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc10a/397
Video Solution
~savannahsolver
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.