AMC12 2022 B

AMC12 2022 B · Q4

AMC12 2022 B · Q4. It mainly tests Vieta / quadratic relationships (basic), Divisibility & factors.

For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?

常数 $k$ 有多少个值，使得多项式 $x^{2}+kx+36$ 有两个不同的整数根？

(A) 6 6

(B) 8 8

(D) 14 14

(E) 16 16

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By Vieta's Formulas, we have $p+q=-k$ and $pq=36.$ It follows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are \[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\] Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{\textbf{(B) }8}$ values of $k,$ corresponding to $\mp37,\mp20,\mp15,\mp13,$ respectively.

设 $p$ 和 $q$ 是 $x^{2}+kx+36$ 的根。由维达公式，$p+q=-k$，$pq=36$。 \[\] 因此 $p$ 和 $q$ 必须是 $36$ 的不同因子。$\{p,q\}$ 的可能值为 \[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\] \[\] 每个无序对给出唯一的 $k$ 值。因此，有 $\boxed{\textbf{(B) }8}$ 个 $k$ 值，分别对应 $\mp37,\mp20,\mp15,\mp13$。

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