11. If $1 \leqslant p \leqslant 20,1 \leqslant q \leqslant 10$, and the roots of the equation $4 x^{2}-p x+q$ $=0$ are both odd numbers. Then the roots of this equation are $\qquad$ .
11. $x_{1}=x_{2}=1$.

Let $x_{1}$ and $x_{2}$ be the two roots of the equation, then
$$
x_{1}+x_{2}=\frac{p}{4}, x_{1} x_{2}=\frac{q}{4} \text {. }
$$

Since $x_{1}$ and $x_{2}$ are both odd numbers, $x_{1}+x_{2}$ is even, and $x_{1} x_{2}$ is odd.
$$
\begin{array}{l}
\text { Also, } 1 \leqslant p \leqslant 20,1 \leqslant q \leqslant 10, \text { then } \\
\frac{1}{4} \leqslant \frac{p}{4} \leqslant 5, \frac{1}{4} \leqslant \frac{q}{4} \leqslant \frac{5}{2} . \\
\text { Therefore, } \frac{q}{4}=1, q=4 .
\end{array}
$$

By $\Delta=p^{2}-16 q \geqslant 0$, solving gives $p \geqslant 8$. Thus, $\frac{p}{4} \geqslant 2$.
So, $\frac{p}{4}=2$ or 4, i.e., $p=8$ or $p=16$.
When $p=8$, $x_{1}=x_{2}=1$, which meets the condition;
When $p=16$, $x_{1}$ and $x_{2}$ are both irrational numbers, which does not meet the condition, so it is discarded.
Therefore, the roots of the original equation are $x_{1}=x_{2}=1$.