For example, let $8 a$ be a real number greater than zero. It is known that there exists a unique real number $k$ such that the quadratic equation in $x$
$$
x^{2}+\left(k^{2}+a k\right) x+1999+k^{2}+a k=0
$$

has two roots that are both prime numbers. Find the value of $a$.
Solution: Let the two prime roots of the equation be \( p \) and \( q \). By the relationship between roots and coefficients, we have
\[
\begin{array}{l}
p+q=-(k^{2}+a k), \\
p q=1999+k^{2}+a k .
\end{array}
\]
Adding (1) and (2), we get \( p+q+p q=1999 \).

Thus, \((p+1)(q+1)=2^{4} \times 5^{3}\).
From (3), it is clear that \( p \) and \( q \) are not 2, so they must be odd.

Therefore, \(\frac{p+1}{2}\) and \(\frac{q+1}{2}\) are both integers, and \(\frac{p+1}{2} \cdot \frac{q+1}{2} = 2^{2} \times 5^{3}\).

If \(\frac{p+1}{2}\) is odd, then \(\frac{p+1}{2} = 5^{r} (r=1,2,3)\), thus, \( p = 2 \times 5^{r} - 1 \) is a composite number, which is a contradiction. Therefore, \(\frac{p+1}{2}\) must be even. Similarly, \(\frac{q+1}{2}\) is also even. So, \(\frac{p+1}{4}\) and \(\frac{q+1}{4}\) are both integers, and \(\frac{p+1}{4} \cdot \frac{q+1}{4} = 5^{3}\).
Assume \( p \leqslant q \), then \(\frac{p+1}{4} = 1\) or 5.
When \(\frac{p+1}{4} = 1\), \(\frac{q+1}{4} = 5^{3}\), we get \( p = 3 \), \( q = 499 \), both of which are prime numbers.

When \(\frac{p+1}{4} = 5\), \(\frac{q+1}{4} = 5\), we get \( p = 19 \), \( q = 99 \), which is a composite number, not meeting the requirement.
In summary, \( p = 3 \), \( q = 499 \).
From (1), we get \( k^{2} + a k + 502 = 0 \).
According to the problem, equation (4) has a unique real solution.
Thus, \(\Delta = a^{2} - 4 \times 502 = 0\). Therefore, \( a = 2 \sqrt{502} \).