97彰化藝術高中
這是教師甄選的其中一題目而已。是跟瑋岳老師一起討論出來的。
97彰化藝術高中教師甄選
解方程式:\(\left( {x^2 - 3x - 12} \right)^3 + 6\left( {x^2 - 3x - 12} \right) = x^3 + 6x\) 答案: \(6,-2\)
解答: 令 \(\left( {x^2 - 3x - 12} \right) = t\)
\(\begin{array}{l}
\left( {x^2 - 3x - 12} \right)^3 + 6\left( {x^2 - 3x - 12} \right) = x^3 + 6x \\
\Rightarrow t^3 + 6t = x^3 + 6x \\
\Rightarrow \left( {t^3 - x^3 } \right) + 6\left( {t - x} \right) = 0 \\
\Rightarrow \left( {t - x} \right)\left( {t^2 + tx + x^2 } \right) + 6\left( {t - x} \right) = 0 \\
\Rightarrow \left( {t - x} \right)\left( {t^2 + tx + x^2 + 6} \right) = 0 \\
\Rightarrow t - x = 0 \vee \left( {t^2 + tx + x^2 + 6} \right) = 0 \\
\end{array}\)
(1) 當 \(t-x=0\)
\(\begin{array}{l}
x^2 - 3x - 12 - x = 0 \\
\Rightarrow x^2 - 4x - 12 = 0 \\
\Rightarrow \left( {x - 6} \right)\left( {x + 2} \right) = 0 \\
x = 6 \vee - 2 \\
\end{array}\)
(2) 當 \(t^2 + tx + x^2 + 6 = 0\)
\(
\begin{array}{l}
t^2 + tx + x^2 + 6 = 0 \\
\Rightarrow \left\{ {t^2 + tx + \left( {\frac{x}{2}} \right)^2 } \right\} + \frac{3}{4}x^2 + 6 = \left( {t + \frac{x}{2}} \right)^2 + \frac{3}{4}x^2 + 6 \ge 6 \\
\end{array}
\)
所以 \(t^2 + tx + x^2 + 6 \) 恆正,因此可知 \(t^2 + tx + x^2 + 6 = 0\) 無實數解。
由以上可知,方程式的解為 \(x=6,x=-2\)