# Difference between revisions of "2015 AMC 10A Problems/Problem 23"

## Problem

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

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}}\ 17\qquad\textbf{(E)}\ 18$ (Error compiling LaTeX. ! Extra }, or forgotten \$.)

## 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 $$(a - 4)^2 = k^2 + 16.$$ Hence $(a-4)^2 - k^2 = 16$ 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 $\textbf{(C)}$.

## Solution 2

Let $r_1$ and $r_2$ be the integer zeroes of the quadratic.

Since the coefficent of the $x^2$ term is $1$, the quadratic can be written as $(x - r_1)(x - r_2)$ or $x^2 - (r_1 + r_2)x + r_1r_2)$.

By comparing this with $x^2 - ax + a^2$, $r_1 + r_2 = a$ and $r_1r_2 = 2a$.

Plugging the first equation in the second, $r_1r_2 = 2 (r_1 + r_2)$. Rearranging gives $r_1r_2 - 2r_1 - 2r_2 = 0$.

This can be factored as $(r_1 - 2)(r_2 - 2) = 4$.

These factors can be: $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)$.

We want the number of distinct $a = r_1 + r_2$, and these factors gives $a = {-1, 8, 9}$.

So the answer is $-1 + 8 + 9 = \boxed{\textbf{(C) }16}$.