For example, $5 \alpha$ is a real number greater than zero. It is known that there exists a unique real number $k$, such that the equation about $x$
$$
x^{2}+\left(k^{2}+\alpha k\right) x+1999+k^{2}+\alpha k=0
$$
has two roots that are prime numbers. Find the value of $\alpha$.
Analysis: Since $\alpha, k$ are real numbers, the discriminant method cannot solve the problem.

Let the two roots of the equation be $x_{1}, x_{2}$, and $x_{1} \leqslant x_{2}, x_{1}, x_{2}$ are both prime numbers, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-k^{2}-\alpha k, \\
x_{1} x_{2}=1999+k^{2}+\alpha k .
\end{array}\right.
$$

Eliminating the parameters, we get
$$
x_{1}+x_{2}+x_{1} x_{2}=1999 \text {, }
$$

which means $\left(x_{1}+1\right)\left(x_{2}+1\right)=2000=2^{4} \times 5^{3}$.
Obviously, $x_{1} \neq 2$. Therefore, $x_{1}+1, x_{2}+1$ are both even and $x_{1}+1 \leqslant x_{2}+1$. Thus, the only possibilities are:
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x _ { 1 } + 1 = 2 ^ { 2 } , } \\
{ x _ { 2 } + 1 = 2 ^ { 2 } \times 5 ^ { 3 } ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}+1=2^{3}, \\
x_{2}+1=2 \times 5^{3} ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ x _ { 1 } + 1 = 2 \times 5 , } \\
{ x _ { 2 } + 1 = 2 ^ { 3 } \times 5 ^ { 2 } ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}+1=2^{2} \times 5^{2}, \\
x_{2}+1=2^{2} \times 5 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ x _ { 1 } + 1 = 2 ^ { 2 } \times 5 , } \\
{ x _ { 2 } + 1 = 2 ^ { 2 } \times 5 ^ { 2 } ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}+1=2^{3} \times 5, \\
x_{2}+1=2 \times 5^{2} .
\end{array}\right.\right.
\end{array}
$$

The only solution that fits the problem is $\left\{\begin{array}{l}x_{1}=3, \\ x_{2}=499 \text {. }\end{array}\right.$
Thus, $3+499=-k^{2}-\alpha k$.
Since there is a unique $k$, the equation $k^{2}+\alpha k+502$ $=0$ has two equal roots.
The discriminant $\Delta=\alpha^{2}-4 \times 502=0$, solving for $\alpha$ we get
$$
\alpha=2 \sqrt{502} \text {. }
$$

Note: The key to applying Vieta's formulas lies in eliminating the parameters, first finding the solutions of the equation, and after eliminating the parameters, paying attention to the use of factorization.