3. Given a quadratic equation $a x^{2}+b x+c=0$ $(a \neq 0, c \neq 0)$, the sum, difference, product, and quotient of its roots are $p$, $q$, $r$, and $s$, respectively. They have the following dependency relations:
(1) $p^{2}-q^{2}=4 r$,
(2) $\left(\frac{s+1}{p}\right)^{2}=\frac{s}{r}$,
(3) $\left(\frac{s-1}{q}\right)^{2}=\frac{s}{r}$,
(4) $\left(\frac{s+1}{s-1}\right)^{2}=\left(\frac{p}{q}\right)^{2}$.
Among them, the correct relations are ( ) in number.
(A) 1
(B) 2
(C) 3
(D) 4
3. D.

Let's assume the two roots of $a x^{2}+b x+c=0(a \neq 0, c \neq 0)$ are $x_{1}$ and $x_{2}$, then we have
$$
x_{1}+x_{2}=-\frac{b}{a}=p, x_{1} x_{2}=\frac{c}{a}=r .
$$

Thus, $|q|=\left|x_{1}-x_{2}\right|$
$$
=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{p^{2}-4 r} \text {, }
$$

which means $p^{2}-q^{2}=4 r$.
Also, $s+\frac{1}{s}+2=\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}+2=\frac{\left(x_{1}+x_{2}\right)^{2}}{x_{1} x_{2}}=\frac{p^{2}}{r}$,

which means $\left(\frac{s+1}{p}\right)^{2}=\frac{s}{r}$.
Therefore, $p^{2}=\frac{r(s+1)^{2}}{s}$, and substituting into equation (1) gives
$$
q^{2}=\frac{r(s+1)^{2}}{s}-4 r=\frac{r}{s}(s-1)^{2} \text {. }
$$

From equations (2) and (3), it is easy to see that equation (4) also holds.