2022 AMC 10B Problem 7

If the integer roots are $r$ and $s$, Vieta's formulas give $rs = 36$ and $r + s = -k$. So $k$ is determined by the unordered pair $\{r, s\}$, and each unordered pair with a different sum gives a different $k$.

Factor pairs of $36$ with both factors positive: $\{1,36\}, \{2,18\}, \{3,12\}, \{4,9\}, \{6,6\}$. The pair $\{6,6\}$ is excluded because the roots must be distinct. The four remaining pairs have sums $37, 20, 15, 13$, giving $k = -37, -20, -15, -13$.

Both roots may also be negative: $\{-1,-36\}, \{-2,-18\}, \{-3,-12\}, \{-4,-9\}$ give $k = 37, 20, 15, 13$. (Roots of opposite sign would have a negative product, impossible.)

All eight sums are different, so there are $8$ values of $k$.

The answer is $\boxed{\textbf{(B)}\ 8}$.

"Integer roots" plus a known constant term is a Vieta problem: the constant fixes the product, so the roots are a factor pair and the linear coefficient is just their (negated) sum. Sign symmetry doubles the count, and the equal-root pair is the only one to throw out.

• Forgetting that both roots can be negative (answer 4) or counting ordered pairs and doubling to 16.
• Including $k = \pm 12$ from the pair $\{6, 6\}$, which gives a repeated root, and answering 10 or 9.