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 $(\quad)$.
(A) $a b \geqslant \frac{1}{8}$
(B) $a b \leqslant \frac{1}{8}$
(C) $a b \geqslant \frac{1}{4}$
(D) $a b \leqslant \frac{1}{4}$
5.B.

Since the equation has real solutions, we have $b^{2}-4 a c \geqslant 0$. According to the problem, we have
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}=b^{2}-4 a c
$$

or $\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}=b^{2}-4 a c$.
Let $u=\sqrt{b^{2}-4 a c}$, we get the equation
$2 a u^{2}-u+b=0$ or $2 a u^{2}+u+b=0$.
Since $u=\sqrt{b^{2}-4 a c}$ is a solution, the discriminant of the equation is non-negative, i.e., $1-8 a b \geqslant 0$. Therefore, $a b \leqslant \frac{1}{8}$.