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2015 AMC 10A Problems/Problem 23

Revision as of 23:33, 13 February 2017 by Mathwizard12345 (talk | contribs) (→‎Solution 1)

Problem

The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$

Solution 1

By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.

Because the zeros are integral, the discriminant of the function, $a^2 - 8a$, is a perfect square, say $k^2$. Then adding 16 to both sides and completing the square yields \[(a - 4)^2 = k^2 + 16.\] Therfore $(a-4)^2 - k^2 = 16$ and \[((a-4) - k)((a-4) + k) = 16.\] Let $(a-4) - k = u$ and $(a-4) + k = v$; then, $a-4 = \dfrac{u+v}{2}$ and so $a = \dfrac{u+v}{2} + 4$. Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect ($u + v$), $(2, 8), (4, 4), (-2, -8), (-4, -4)$, yields $a = 9, 8, -1, 0$. These $a$ sum to $16$, so our answer is $\boxed{\textbf{(C) }16}$.

Solution 2

Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. Since the coefficient of the $x^2$ term is $1$, the quadratic can be written as \[(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2\]

By comparing this with $x^2 - ax + 2a$, \[r_1 + r_2 = a\text{ and }r_1r_2 = 2a.\]

Plugging the first equation in the second, \[r_1r_2 = 2 (r_1 + r_2).\] Rearranging gives \[r_1r_2 - 2r_1 - 2r_2 = 0\implies (r_1 - 2)(r_2 - 2) = 4.\] These factors can be$(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),$ or $(-2, -2).$

We want the number of distinct $a = r_1 + r_2$, and these factors gives $a = -1, 0, 8, 9$. So the answer is $-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}$.

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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