答案：(a) $f(x)=13x+x^2$ (b) $x=5$ or $x=-18$
- 設 $f(x) = k_1 x + k_2 x^2$，其中 $k_1$ 為 $k_2$ 常數。
$\begin{array}{rcl}
f(3) & = & 48 \\
k_1 ( 3) + k_2 (3)^2 & = & 48 \\
k_1 + 3k_2 & = & 16 \ \ldots \unicode{x2460}
\end{array}$$\begin{array}{rcl}
f(9) & = & 198 \\
k_1(9) +k_2(9)^2 & = & 198 \\
k_1 + 9k_2 & = & 22 \ \ldots \unicode{x2461}
\end{array}$$ \unicode{x2461} – \unicode{x2460}$，可得
$\begin{array}{rcl}
6k_2 & = & 6 \\
k_2 & = & 1
\end{array}$把 $k_2=1$ 代入 $\unicode{x2460}$，可得
$\begin{array}{rcl}
k_1 + 3(1) & = & 16 \\
k_1 & = & 13
\end{array}$$\therefore f(x) =13x +x^2$.
- 根據 (a) 的結果，可得
$\begin{array}{rcl}
13x + x^2 & = & 90 \\
x^2 + 13x -90 & = & 0 \\
(x-5)(x+18) & = & 0 \\
\end{array}$$\therefore x=5 \ \text{或} \ x = -18$。
