Example 3 If the two roots of the equation $(x-1)\left(x^{2}-2 x+m\right)=0$ can serve as the lengths of the three sides of a triangle, then the range of the real number $m$ is ( ).
(A) $0 \leqslant m \leqslant 1$
(B) $m \geqslant \frac{3}{4}$
(C) $\frac{3}{4}<m \leqslant 1$
(D) $\frac{3}{4} \leqslant m \leqslant 1$
Solution: It is obvious that the original equation has a root of 1. Let $\alpha, \beta$ be the two roots of the equation $x^{2}-2 x+m=0$. By Vieta's formulas, we have
$$
\alpha+\beta=2, \alpha \beta=m .
$$

Since the three sides of the triangle are $1, \alpha, \beta$, then
$$
\begin{aligned}
& |\alpha-\beta|<1 \\
& 1+\alpha>\beta, 1+\beta>\alpha, \alpha+\beta>1 .
\end{aligned}
$$
From the above, we get
$$
\frac{3}{4}<m<1 .
$$
Therefore, the correct choice is (C).