Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | The zeroes of the function <math>f(x)=x^2-ax+2a</math> are integers .What is the sum of the possible values of a? | + | The zeroes of the function <math>f(x)=x^2-ax+2a</math> are integers. What is the sum of the possible values of <math>a?</math> |
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18</math> | <math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18</math> | ||
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==Solution 2== | ==Solution 2== | ||
− | Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. | + | Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. Since the coefficient of the <math>x^2</math> term is <math>1</math>, the quadratic can be written as <cmath>(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2</cmath> |
− | + | By comparing this with <math>x^2 - ax + 2a</math>, <cmath>r_1 + r_2 = a\text{ and }r_1r_2 = 2a.</cmath> | |
− | + | Plugging the first equation in the second, <cmath>r_1r_2 = 2 (r_1 + r_2).</cmath> Rearranging gives <cmath>r_1r_2 - 2r_1 - 2r_2 = 0\implies (r_1 - 2)(r_2 - 2) = 4.</cmath> These factors can be<math>(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),</math> or <math>(-2, -2).</math> | |
− | + | 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>. | |
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− | We want the number of distinct <math>a = r_1 + r_2</math>, and these factors gives <math>a = -1, 0, 8, 9</math>. | ||
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− | So the answer is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>. | ||
==See Also== | ==See Also== |
Revision as of 11:10, 20 February 2016
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
Hence
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. Since the coefficient of the
term is
, the quadratic can be written as
By comparing this with ,
Plugging the first equation in the second, Rearranging gives
These factors can be
or
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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