Example 7 When is $x$ a rational number such that the algebraic expression $9 x^{2}$ $+23 x-2$ is exactly the product of two consecutive positive even numbers?



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Solution: Let the two consecutive positive even numbers be $k, k+2$. Then
$$
9 x^{2}+23 x-2=k(k+2) \text {, }
$$

which means $9 x^{2}+23 x-\left(k^{2}+2 k+2\right)=0$.
Since $x$ is a rational number, the discriminant must be a perfect square, i.e.,
$$
\begin{aligned}
\Delta & =23^{2}+4 \times 9\left(k^{2}+2 k+2\right) \\
& =565+[6(k+1)]^{2} .
\end{aligned}
$$

Let $\Delta=p^{2}(p \geqslant 0)$, then
$$
\begin{array}{l}
p^{2}-[6(k+1)]^{2} \\
=565=113 \times 5=565 \times 1 . \\
\text { The left side }=[p+6(k+1)][p-6(k+1)],
\end{array}
$$
with $p \geqslant 0, k>0$, we get
$$
\left\{\begin{array}{l}
p+6(k+1)=113, \\
p-6(k+1)=5,
\end{array}\right.
$$

or $\left\{\begin{array}{l}p+6(k+1)=565, \\ p-6(k+1)=1,\end{array}\right.$.
Solving (1) yields $k=8$, thus, $x=2$ or $-\frac{41}{9}$;
Solving (2) yields $k=46$, thus, $x=-17$ or $\frac{130}{9}$.
In summary, when $x=2, -\frac{41}{9}$ or $x=-17, \frac{130}{9}$, $9 x^{2}+23 x-2$ is exactly the product of the two positive even numbers 8 and 10, or 46 and 48.