2015 AMC 10A Problems/Problem 23: Difference between revisions
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These factors can be: <math>(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)</math>. | These factors can be: <math>(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)</math>. | ||
We want the number of distinct <math>a = r_1 + r_2</math>, and these factors gives <math>a = | We want the number of distinct <math>a = r_1 + r_2</math>, and these factors gives <math>a = -1, 0, 8, 9</math>. | ||
So the answer is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>. | So the answer is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>. | ||
Revision as of 18:50, 4 February 2015
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 
. These sum to 16, 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 First Problem |
Followed by Problem 2 | |
| 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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