106 Let the two imaginary roots of the equation $x^{2}-2 \sqrt{2} x+m=0$ be $\alpha$ and $\beta$, and $|\alpha-\beta|=3$, then the value of the real number $m$ is
A. $-\frac{1}{4}$
B. $\frac{1}{4}$
C. $-\frac{17}{4}$
D. $\frac{17}{4}$
106 D. $\alpha+\beta=2 \sqrt{2},|\alpha-\beta|=3$, and $\alpha-\beta$ is a pure imaginary number, so $\alpha-\beta= \pm 3 i$.
$$
\begin{array}{c}
4 a \beta=(\alpha+\beta)^{2}-(\alpha-\beta)^{2}=(2 \sqrt{2})^{2}+3^{2}=17 \\
m=\alpha \beta=\frac{17}{4}
\end{array}
$$

Solving $3=|\alpha-\beta|=\sqrt{4 m-(2 \sqrt{2})^{2}}$, so
$$
m=\frac{1}{4}\left(3^{2}+(2 \sqrt{2})^{2}\right)=\frac{17}{4} .
$$