Example 5 Given that $b^{2}-4 a c$ is a real root of the quadratic equation $a x^{2}+b x+c=0(a \neq 0)$. Then the range of values for $a b$ is ( ).
(A) $a b \geqslant \frac{1}{8}$
(B) $a b \leqslant \frac{1}{8}$
(C) $a b \geqslant \frac{1}{4}$
(D) $a b \leqslant \frac{1}{4}$
Given $x_{0}=b^{2}-4 a c \geqslant 0$ is a real root of the equation, we have
$$
a x_{0}^{2}+b x_{0}+c=0 \text {. }
$$

Rearranging into the discriminant form (i.e., equation (2)) and substituting $x_{0}$, we get
$$
\left(2 a x_{0}+b\right)^{2}-x_{0}=0,
$$

which can be factored as $\left[\left(2 a x_{0}+b\right)-\sqrt{x_{0}}\right]\left[\left(2 a x_{0}+b\right)+\sqrt{x_{0}}\right]=0$. 
Rearranging into the discriminant form (multiplying each bracket by $8a$), we get
$$
\begin{array}{l}
{\left[\left(4 a \sqrt{x_{0}}-1\right)^{2}-(1-8 a b)\right] .} \\
{\left[\left(4 a \sqrt{x_{0}}+1\right)^{2}-(1-8 a b)\right]=0 .}
\end{array}
$$

Thus, we have
$$
1-8 a b=\left(4 a \sqrt{x_{0}}-1\right)^{2} \geqslant 0 \text {, }
$$

or $1-8 a b=\left(4 a \sqrt{x_{0}}+1\right)^{2} \geqslant 0$.
In both cases, $a b \leqslant \frac{1}{8}$, and the condition for equality can be derived from equations (3) and (2):
$$
\left\{\begin{array}{l}
a b=\frac{1}{8}, \\
4 a \sqrt{x_{0}} \pm 1=0,
\end{array}\right.
$$

which simplifies to
$$
\left\{\begin{array}{l}a b=\frac{1}{8}, \\ b^{2}-4 a c=\frac{1}{16 a^{2}}\end{array}\right.
$$
Such $a, b, c$ do exist, for example, when $a=\frac{1}{2}, b=\frac{1}{4}, c=-\frac{3}{32}$, the equation $\frac{1}{2} x^{2}+\frac{1}{4} x-\frac{3}{32}=0$ indeed has a root $x_{0}=b^{2}-4 a c=\frac{1}{4}$.
Therefore, the range of $a b$ is $a b \leqslant \frac{1}{8}$.