Example 2 When $x$ is a rational number, why is the value of the algebraic expression $9 x^{2}+$ $23 x-2$ exactly the product of two consecutive even numbers.



Preserve the original text's line breaks and format, and output the translation result directly.
Analysis: Let two consecutive even numbers be $n, n+2$. The problem is transformed into: For what value of $n$ does the equation $9 x^{2}+23 x-2$ $=n(n+2)$ have rational roots.

The problem of rational roots is essentially also a problem of integer roots, requiring the discriminant of the equation to be a perfect square of an integer or a rational number. Consider the discriminant
$$
\begin{aligned}
\Delta & =23^{2}+36\left(n^{2}+2 n+2\right) \\
& =36(n+1)^{2}+565 .
\end{aligned}
$$

Since $n$ is an integer, the discriminant must be a perfect square of an integer. Let
$$
36(n+1)^{2}+565=m^{2}(m \text { is a natural number greater than } \sqrt{565} \text { }
$$

Factoring by moving terms, we get
$$
(m+6 n+6)(m-6 n-6)=1 \times 5 \times 113 \text {. }
$$

Only $\left\{\begin{array}{l}m+6 n+6=113 \\ m-6 n-6=5\end{array}\right.$. or $\left\{\begin{array}{l}m+6 n+6=565, \\ m-6 n-6=1 .\end{array}\right.$
Solving gives $n=8$, or $n=46$.
Substituting these values back into the original equation, the rational solutions are
$$
-\frac{41}{9}, 2 \text { or } \frac{130}{9},-17 \text {. }
$$

Note: When the discriminant is a quadratic expression in terms of a certain parameter, and the coefficient of the quadratic term is positive, the method of completing the square can be used to transform it into: $f^{2}(\alpha)+$ constant ( $\alpha$ is an integer). Then, the method used in Example 1 can be applied to find the solution.