3. The equation $x^{2}-4[x]+3=0, x \in \mathbf{R}$ has
A. 2 solutions
B. 3 solutions
C. 1 solution
D. 4 solutions
3. B From $x \leqslant[x]<x+1$, we know $2-\sqrt{5}<x<2+\sqrt{5}$. Also, $x^{2}=4[x]-3 \in \mathbf{Z}, x^{2} \leqslant 17, -1 \leqslant[x] \leqslant$ 4, substituting into the equation, we get $x=3, \sqrt{5}$ or 1.