$17 \cdot 73$ Segment $A B$ is $p$ units long, with midpoint $M$ and perpendicular line $M R$; $M R$ is $q$ units long. An arc is drawn from $R$ with a radius equal to $\frac{1}{2} A B$, intersecting $A B$ at $T$. The roots of the following equation are $A T$ and $T B$:
(A) $x^{2}+p x+q^{2}=0$.
(B) $x^{2}-p x+q^{2}=0$.
(C) $x^{2}+p x-q^{2}=0$.
(D) $x^{2}-p x-q^{2}=0$.
(E) $x^{2}-p x+q=0$.
[Solution] By the given and the Pythagorean theorem, we have $M T=\sqrt{\left(\frac{p}{2}\right)^{2}-q^{2}}$, thus
$$
\begin{array}{l}
A T=A M+M T=\frac{p}{2}+\sqrt{\left(\frac{p}{2}\right)^{2}-q^{2}}, \\
B T=B M-M T=\frac{p}{2}-\sqrt{\left(\frac{p}{2}\right)^{2}-q^{2}} .
\end{array}
$$

It is known that $A T, B T$ are the two roots of the equation $x^{2}-p x+q^{2}=0$. Therefore, the correct choice is (B).