2.65 Given the quadratic equation $x^{2}-4 a x+5 a^{2}-6 a=0$ has two real roots, and the absolute value of the difference between these two roots is 6. Then, the value of $a$ is
(A) -1.
(B) 1.
(C) 3.
(D) 5.
(China Beijing Junior High School Mathematics Competition, 1991)
［Solution］Let the two roots of the equation be $x_{1}, x_{2}$. By Vieta's formulas, we have
$$
x_{1}+x_{2}=4 a \text {, and } x_{1} x_{2}=5 a^{2}-6 a \text {. }
$$

Then $\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}$
$$
\begin{array}{l}
=\sqrt{16 a^{2}-4\left(5 a^{2}-6 a\right)} \\
=2 \sqrt{6 a-a^{2}}=6,
\end{array}
$$
$\therefore \sqrt{6 a-a^{2}}=3$, which means $a^{2}-6 a+9=0$. Solving this, we get $a=3$, which satisfies $\quad 6 a-a^{2}>0$. Therefore, the answer is $(C)$.