2015 AMC 12A Problems/Problem 18
Problem
The zeros of the function are integers. What is the sum of the possible values of
?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}}\ 17 \qquad\textbf{(E)}\ 18$ (Error compiling LaTeX. Unknown error_msg)
Solution
Solution 1
The problem asks us to find the sum of every integer value of such that the roots of
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 equals
, for some nonnegative integer
.
From this last equation, we are given a hint of the Pythagorean theorem. Thus, must be a Pythagorean triple unless
.
In the case , the equation simplifies to
. From this equation, we have
. For both
and
, $\frac{a \± \sqrt{a^2 - 8a}}{2}$ (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 is a positive integer, then only one Pythagorean triple could match the triple
because the only Pythagorean triple with a
as one of the values is the classic
triple. Here,
and
. Hence,
. Again, $\frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg) yields two integers for both
and
, so these two values also satisfy the original constraints.
There are a total of four possible values for :
and
. Hence, the sum of all of the possible values of
is 16 (C).
Solution 2
Let and
be the roots of
By Vieta's Formulas, and
Substituting gets us
Using Simon's Favorite Factoring Trick:
This means that the values for are
giving us
values of
and
. 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 |