Example 1. If the equation $m x^{2}-2(m+2) x+m$ $+5=0$ has no real roots with respect to $x$, then the number of real roots of the equation $(m-5) x^{2}$ $-2(m+2) x+m=0$ with respect to $x$ is ( ).
(A) 2
(B) 1
(C) 0
(D) Uncertain
(1989, National Junior High School Mathematics Competition)
Solve: For $m x^{2}-2(m+2) x+m+5=0$ to have no real roots, then
$$
\Delta_{1}=4(m+2)^{2}-4 m(m+5)4$.
Also, the discriminant of $(m-5) x^{2}-2(m+2) x+m=0$ is
$$
\Delta_{2}=4(m+2)^{2}-4 m(m-5)=36 m+16 \text {. }
$$

When $m>4$ and $m \neq 5$, $\Delta_{2}>0$, the equation has two real roots; when $m=5$, the equation is a linear equation, with only one real root. Therefore, the number of real roots of the equation is uncertain. Hence, the answer is (D).