2015 AMC 12A Problems/Problem 18

Problem

The zeros 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

Solution 1

The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.

The quadratic formula gives the roots of the quadratic equation: $x = \frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg)

As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$ , for some nonnegative integer $k$ .

$a^2 - 8a = k^2$

$a(a - 8) = k^2$

$((a - 4) + 4)((a - 4) - 4) = k^2$

$(a - 4)^2 - 4^2 = k^2$

$(a - 4)^2 = k^2 + 4^2$

From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k, 4, |a - 4|)$ must be a Pythagorean triple unless $k = 0$ .

In the case $k = 0$ , the equation simplifies to $|a - 4| = 4$ . From this equation, we have $a = 0, 8$ . For both $a = 0$ and $a = 8$ , $\frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg) yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")

If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k, 4, |a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3, 4, 5)$ triple. Here, $k = 3$ and $|a - 4| = 5$ . Hence, $a = -1, 9$ . Again, $\frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg) yields two integers for both $a = -1$ and $a = 9$ , so these two values also satisfy the original constraints.

There are a total of four possible values for $a$ : $-1, 0, 8,$ and $9$ . Hence, the sum of all of the possible values of $a$ is 16 (C).

Solution 2

Let $m$ and $n$ be the roots of $x^2-ax+2a$

By Vieta's Formulas, $n + m = a$ and $mn = 2a$

Substituting gets us $n + m = \frac{mn}{2}$

$2n - mn + 2m = 0$

Using Simon's Favorite Factoring Trick:

$n(2-m) + 2m = 0$

$-n(2-m) - 2m = 0$

$-n(2-m) - 2m + 4 = 4$

$(2-n)(2-m) = 4$

This means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $a$ values of $-1, 0, 8,$ and $9$ . Adding these up gets 16 (C)

Solution 3

The quadratic formula gives $x = \frac{a \pm \sqrt{a(a-8)}}{2}$ . For $x$ to be an integer, it is necessary (and sufficient!) that $a(a-8)$ to be a perfect square. So we have $a(a-8) = b^2$ ; this is a quadratic in itself and the quadratic formula gives $a = 4 \pm \sqrt{16 + b^2}$

We want $16 + b^2$ to be a perfect square. From smartly trying small values of $b$ , we find $b = 0, b = 3$ as solutions, which correspond to $a = -1, 0, 8, 9$ . These are the only ones; if we want to make sure then we must hand check up to $b=8$ . Indeed, for $b \geq 9$ we have that the differences between consecutive squares are greater than $16$ so we can't have $b^2 + 16$ be a perfect square. So summing our values for $a$ we find 16 (C) as the answer.

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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