AMC12 2002 A

AMC12 2002 A · Q12

AMC12 2002 A · Q12. It mainly tests Quadratic equations, Primes & prime factorization.

Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is

二次方程 $x^2 - 63x + k = 0$ 的两个根都是质数。$k$ 的可能取值个数是

(A) 0 0

(B) 1 1

(D) 4 4

(E) more than four 超过四个

Answer

Correct choice: (B)

正确答案：（B）

Solution

Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$. It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$. Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's Formulas). We now have that the sum of the two roots is $63$ while the product is $k$. Since both roots are primes, one must be $2$, otherwise, the sum would be even. That means the other root is $61$ and the product must be $122$. Hence, our answer is $\boxed{\text{(B)}\ 1 }$.

考虑一个一般的二次方程，$x^2$ 的系数为 $1$，两根为 $r$ 和 $s$。它可分解为 $(x-r)(x-s)$，即 $x^2-(r+s)x+rs$。因此，两根之和等于 $x$ 项系数的相反数，两根之积等于常数项。（一般而言，这就是韦达定理。）现在两根之和为 $63$，两根之积为 $k$。由于两根都是质数，其中一个必须是 $2$，否则和将为偶数。于是另一个根为 $61$，乘积为 $122$。因此答案是 $\boxed{\text{(B)}\ 1 }$。

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