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2015 AMC 12A Problems/Problem 18

Revision as of 00:53, 5 February 2015 by Gina (talk | contribs)

Problem

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

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}}\ 17 \qquad\textbf{(E)}\ 18$ (Error compiling LaTeX. Unknown error_msg)

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)

See Also

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