Question

二次方程的两个根都是质数。 $k$ 可能取值的个数是多少？ $x^2 - 63x + k = 0$

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

$0$

$1$

$2$

$4$

多于四个

more than four

如果两个根是质数 $p$ 和 $q$ 则由韦达定理 $p + q = 63$ ，且 $pq = k$ 。

因为 $63$ 是奇数，所以一个质数必须是偶数，即 $2$ ，另一个是 $61$ 。两者都是质数，所以 $k = 2\cdot 61 = 122$ 是唯一可能值。

因此，正确答案是 B。

If the roots are primes $p$ and $q,$ then by Vieta's formulas $p + q = 63$ and $pq = k.$

Since $63$ is odd, one prime must be even, namely $2,$ and the other is $61.$ Both are prime, so $k = 2\cdot 61 = 122$ is the only possible value.

Thus, the correct answer is B.