Four. (25 points) Given the quadratic function $f(x)=x^{2}+p x+q$, and the equation $f(x)=0$ has the same non-zero real roots as $f(2 x)=0$.
(1) Find the value of $\frac{q}{p^{2}}$;
(2) If $f(1)=28$, solve the equation $f(x)=0$.
(1) Let the two roots of $f(x)=0$ be $x_{1}$ and $x_{2}$, and $x_{1} \leqslant x_{2}$,
then
$$
\begin{array}{l}
f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right), \\
f(2 x)=\left(2 x-x_{1}\right)\left(2 x-x_{2}\right) \\
=4\left(x-\frac{x_{1}}{2}\right)\left(x-\frac{x_{2}}{2}\right) .
\end{array}
$$

Thus, the two roots of $f(2 x)=0$ are $\frac{x_{1}}{2}$ and $\frac{x_{2}}{2}$, and $\frac{x_{1}}{2} \leqslant \frac{x_{2}}{2}$.
Given that $f(x)=0$ and $f(2 x)=0$ have the same non-zero real roots,
so, $x_{1}=\frac{x_{2}}{2}$, which means $x_{2}=2 x_{1}$.
$$
\begin{aligned}
\text { Hence } \frac{q}{p^{2}} & =\frac{x_{1} x_{2}}{\left(x_{1}+x_{2}\right)^{2}}=\frac{1}{\left(\frac{x_{1}}{x_{2}}+1\right)\left(\frac{x_{2}}{x_{1}}+1\right)} \\
= & \frac{1}{\left(\frac{1}{2}+1\right)(2+1)}=\frac{2}{9} .
\end{aligned}
$$
(2) From (1), we have $f(x)=\left(x-x_{1}\right)\left(x-2 x_{1}\right)$.

Also, $f(1)=28$, so $\left(1-x_{1}\right)\left(1-2 x_{1}\right)=28$.
Solving this, we get $x_{1}=-3$ or $\frac{9}{2}$.
Thus, $x_{2}=-6$ or 9.
Therefore, the two sets of solutions for $f(x)=0$ are
$$
\left\{\begin{array}{l}
x_{1}=-3, \\
x_{2}=-6
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}=\frac{9}{2}, \\
x_{2}=9 .
\end{array}\right.\right.
$$