$2 \cdot 60$ The larger root minus the smaller root of the equation $(7+4 \sqrt{3}) x^{2}+(2+\sqrt{3}) x-2=0$ is
(A) $-2+3 \sqrt{3}$.
(B) $2-\sqrt{3}$.
(C) $6+3 \sqrt{3}$.
(D) $6-3 \sqrt{3}$.
(E) $3 \sqrt{3}+2$.
(15th American High School Mathematics Examination, 1964)
［Solution］Let the two roots of the equation be $r, s$ and $r \geqslant s$.
By
$$
\begin{aligned}
r-s & =\sqrt{(r-s)^{2}}=\sqrt{(r+s)^{2}-4 r s} \\
& =\frac{\sqrt{(2+\sqrt{3})^{2}+8(7+4 \sqrt{3})}}{7+4 \sqrt{3}} \\
& =\frac{\sqrt{9(2+\sqrt{3})^{2}}}{(2+\sqrt{3})^{2}}=\frac{3}{2+\sqrt{3}} \\
& =6-3 \sqrt{3} .
\end{aligned}
$$

Therefore, the answer is $(D)$.