Example 3 Consider the equation in $x$
$$
a x^{2}+(a+2) x+9 a=0
$$

has two distinct real roots $x_{1}, x_{2}$, and $x_{1}<\frac{2}{5}<x_{2}$.
(A) $a>-\frac{2}{5}$
(B) $a<-\frac{2}{5}$
(C) $a<-\frac{2}{7}$
(D) $-\frac{2}{11}<a<0$
(2002, National Junior High School Mathematics Competition)
Solution: Let $t=x-1$, then the original equation becomes
$$
a t^{2}+(3 a+2) t+11 a+2=0 \text {. }
$$

Thus, the condition that one root of the original equation is greater than 1 and the other is less than 1 is transformed into the condition that one root of equation (1) is greater than 0 and the other is less than 0. The condition for equation (1) to have one positive and one negative root is
$$
t_{1} t_{2}=\frac{11 a+2}{a}<0 .
$$

Solving this, we get $-\frac{2}{11}<a<0$, so the correct choice is (D).
Note: This problem is solved by transforming the issue through substitution into the sign of the roots, and it can also be solved using the characteristics of the function's graph.

To solve the problem using the characteristics of the function's graph, we first provide an important conclusion:

Let $f(x)=a x^{2}+b x+c(a \neq 0)$, and for constants $m, n (m<n)$, if $f(m) f(n)<0$, then the quadratic equation $f(x)=0$ has exactly one real root in the open interval $(m, n)$.

Using this conclusion to solve the problem, some problems do not require considering the discriminant and the axis of symmetry (such as when the roots are in two different given intervals), while others must consider the discriminant and the axis of symmetry (such as when the roots are in the same interval).