2.70 If $\sqrt{28-10 \sqrt{3}}$ is a root of the equation $x^{2}+a x+b=0$ (where $a$, $b$ are rational numbers), then $a b$ is
(A) $-220+\sqrt{3}$.
(B) $220-\sqrt{3}$.
(C) -220 .
(D) 220 .
(China Sichuan Province Junior High School Mathematics Competition, 1990)
[Solution] From $\sqrt{28-10 \sqrt{3}}=\sqrt{5^{2}-10 \sqrt{3}+(\sqrt{3})^{2}}=5-\sqrt{3}$, and $x^{2}+a x+b=0$ is a rational coefficient equation, then $5+\sqrt{3}$ is its other root.
Thus, $a=-[(5+\sqrt{3})+(5-\sqrt{3})]=-10$,
and $b=(5+\sqrt{3})(5-\sqrt{3})=25-3=22$.
Therefore,
$$
a b=-220 \text {. }
$$

Hence, the answer is $(C)$.