48 The roots of the equation $x^{2}-2 x+k=0$ are $\alpha$, $\beta$. And $|\alpha-\beta|=2 \sqrt{2}$, then the value of $k$ is $(\quad)$
A. $-1,-3$
B. 3,1
C. $-1,3$
D. $1,-3$
48 C. $(x+1)^{2}=1-k$. If $k \leqslant 1$ then $2 \sqrt{2}=|\alpha-\beta|=2 \sqrt{1-k}$, thus $k=-1$. If $k>1$, then $2 \sqrt{2}=|\alpha-\beta|=2|i \sqrt{k-1}|=2 \sqrt{k-1}$, thus $k=3$.