2015 AMC 10A Problem 23

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23.

The zeros of the function f(x)=x2ax+2af(x)=x^2-ax+2a are integers. What is the sum of the possible values of a?a?

77

88

1616

1717

1818

Answer: C
Concepts:Vieta’s FormulasSimon’s Favorite Factoring Trickquadratic
Difficulty rating: 1660
Solution:

Let the zeros be rr and s.s. Using Vieta's formulas, we have that a=r+sa = r + s and 2a=rs.2a = rs.

Then we get that rs=2(r+s), rs = 2(r + s), which rearranges to rs2r2s=0 rs - 2r - 2s = 0 rs2r2s+4=4 rs - 2r - 2s + 4 = 4 (r2)(s2)=4. (r - 2)(s - 2) = 4.

The only possible pairs (r2,s2)(r - 2, s - 2) that work are (1,4),(1,4),(4,1),(4,1), (1, 4), (-1, -4), (4, 1), (-4, -1), (2,2),(2,2). (2, 2), (-2, -2). For any of these pairs, we have that a=r2+s2+4. a = r - 2 + s - 2 + 4. We want all the such unique values of a.a. We get that they are 1,0,8,9. -1, 0, 8, 9. The sum of these values is 1+8+9=16.-1 + 8 + 9 = 16.

Thus, C is the correct answer.

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