4. A student solved a quadratic equation $a x^{2}+b x+c=0$ that has no real roots, but due to misreading the sign of a term, obtained the roots as $\frac{1 \pm \sqrt{31977}}{4}$. Then the value of $\frac{b+c}{a}$ is ( ).
(A) 1998
(B) 1999
(C) 1998 or 1999
(D) 1999 or 2000
4. (C).

The symbol that was misread can only be the quadratic term or the constant term, not the linear term. The equation solved by the student is
$-a x^{2}+b x+c=0$ or $a x^{2}+b x-c=0$.
By the relationship between roots and coefficients, we know
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ - \frac { b } { - a } = \frac { 1 + \sqrt { 31977 } } { 4 } + \frac { 1 - \sqrt { 31977 } } { 4 } , } \\
{ \frac { c } { - a } = \frac { 1 + \sqrt { 31977 } } { 4 } \times \frac { 1 - \sqrt { 31977 } } { 4 } }
\end{array} \text { or } \left\{\begin{array}{l}
-\frac{b}{a}=\frac{1+\sqrt{31977}}{4}+\frac{1-\sqrt{31977}}{4}, \\
\frac{-c}{a}=\frac{1+\sqrt{31977}}{4} \times \frac{1-\sqrt{31977}}{4} .
\end{array}\right.\right. \\
\text { Then } \frac{b+c}{a}=1999 \text { or } 1998 .
\end{array}
$$