2015 AMC 10A Problem 23

Problem 23 of 25HarderAlgebra

The zeros of the function

f(x)=x^2-ax+2a

are integers. What is the sum of the possible values of

a?

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Solution

Let the zeros be

r

and

s.

Using Vieta’s formulas, we have that

a = r + s

and

2a = rs.

Then we get that

rs = 2(r + s),

which rearranges to

rs - 2r - 2s = 0

rs - 2r - 2s + 4 = 4

(r - 2)(s - 2) = 4.

The only possible pairs

(r - 2, s - 2)

that work are

(1, 4), (-1, -4), (4, 1), (-4, -1),

(2, 2), (-2, -2).

For any of these pairs, we have that

a = r - 2 + s - 2 + 4.

We want all the such unique values of

a.

We get that they are

-1, 0, 8, 9.

The sum of these values is

-1 + 8 + 9 = 16.

Thus, C is the correct answer.

Tagged: Vieta’s Formulas · Simon’s Favorite Factoring Trick · quadratic