Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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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 19:50, 4 February 2015
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
Hence
and
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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