4. Given positive numbers $p, q, a, b, c$, where $p \neq q$. If $p, a, q$ form a geometric sequence, and $p, b, c, q$ form an arithmetic sequence, then the quadratic equation $b x^{2}-2 a x+c=0(\quad)$.
(A) has no real roots
(B) has two equal real roots
(C) has two distinct real roots with the same sign
(D) has two real roots with opposite signs
4. (A).

From the problem, we know that $p q=a^{2}, 2 b=p+c, 2 c=q+b$. From the last two equations, we get $b=\frac{2 p+q}{3}, c=\frac{p+2 q}{3}$. Therefore, we have
$$
\begin{aligned}
b c & =\frac{p+p+q}{3} \cdot \frac{p+q+q}{3} \\
& \geqslant \sqrt[3]{p^{2} q} \cdot \sqrt[3]{p q^{2}}=p q=a^{2} .
\end{aligned}
$$

Since $p \neq q$, it follows that $b c>a^{2}$. The discriminant of the equation $\Delta=4 a^{2}$ $-4 b c<0$. Therefore, the equation has no real roots.