2015 AMC 10A Problems/Problem 23
Contents
Problem
The zeroes of the function 
are integers .What is the sum of the possible values of a?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}}\ 17\qquad\textbf{(E)}\ 18$ (Error compiling LaTeX. Unknown error_msg)
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
![\[(a - 4)^2 = k^2 + 16.\]](http://latex.artofproblemsolving.com/0/f/0/0f0fc54cac5d6efbe535f2c0592f14c11a50bde5.png)
Hence 
and
![\[((a-4) - k)((a-4) + k) = 16.\]](http://latex.artofproblemsolving.com/8/0/3/8035bbf221fd1d0b33f061a4dea96f3e9d32f20d.png)
Let 
and 
; then, 
and so 
. Listing all possible 
pairs (not counting transpositions because this does not affect 
), 
, yields 
. However, 
and 
do not work because the problem states that there are "zeros" of the function that are "integers", which clearly signifies more than one root. Thus the only that work are 
and 
. These sum to 
, so our answer is 
.
Solution 2
Let 
and 
be the integer zeroes of the quadratic.
Since the coefficent of the 
term is 
, the quadratic can be written as 
or 
.
By comparing this with 
, 
and 
.
Plugging the first equation in the second, 
. Rearranging gives 
.
This can be factored as 
.
These factors can be: 
.
We want the number of distinct 
, and these factors gives 
.
So the answer is 
.
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 | ||
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