2022 AMC 10B Problem 7

Attempt Problem 7 of the 2022 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

For how many values of the constant kk will the polynomial x2+kx+36x^{2}+kx+36 have two distinct integer roots?
 6 \ 6
 8 \ 8
 9 \ 9
 14 \ 14
 16 \ 16
Answer: B
Concepts:Vieta’s Formulasfactorsystematic listing
Small hint:
Different root pairs give different values of k
Big hint:
Integer roots must be factor pairs of 3636
Solution:
Let the roots be r,s.r,s. Expanding (x−r)(x−s)(x-r)(x-s) gives x2−(r+s)x+rs.x^2-(r+s)x+rs. Comparing coefficients with the given polynomial yields rs=36rs=36 and r+s=−k.r+s=-k.
Therefore, we need rr and ss distinct such that rs=36.rs = 36. All the possible factor pairs are ±{1,36},±{2,18},±{3,12} \pm\{1,36\},\pm\{2,18\},\pm\{3,12\} and ±{4,9}.\pm\{4,9\}.
Each of these unordered pairs produces a unique value for k,k, so there are 88 possible values for k.k.
Thus, B is the correct answer.
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