题目
Problem
In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
where is a constant.
Given that is a factor of ,
(a) use the factor theorem to show that
(b) Hence show that the equation has only one real root.
(6)
题目中文翻译
本题必须展示所有解题过程，不能仅依赖计算器。
，其中 是常数。
已知 是 的因式，
(a) 用因式定理证明
(b) 由此证明方程 只有一个实数根。
（6分）
解答
解法一
思路
因式定理：若 是 的因式，则 。代入求 。代入 后用长除法分解，二次因式的判别式 则无实根。
答题过程
(a) Since is a factor of , by the factor theorem .
f\!\left(\frac{3}{2}\right) =&\, 4\left(\frac{3}{2}\right)^3 - 8\left(\frac{3}{2}\right)^2 + 5\left(\frac{3}{2}\right) + a \\[4mm] =&\, 4 \times \frac{27}{8} - 8 \times \frac{9}{4} + 5 \times \frac{3}{2} + a \\[4mm] =&\, \frac{27}{2} - 18 + \frac{15}{2} + a \\[4mm] =&\, \frac{27 + 15}{2} - 18 + a \\[4mm] =&\, 21 - 18 + a \\[4mm] =&\, 3 + a \end{align*}$$ Setting $3 + a = 0$: $$\boxed{a = -3}$$ **(b)** With $a = -3$, $f(x) = 4x^3 - 8x^2 + 5x - 3$. Since $(2x - 3)$ is a factor, dividing $f(x)$ by $(2x - 3)$: $$4x^3 - 8x^2 + 5x - 3 = (2x - 3)(2x^2 - x + 1)$$ For the quadratic $2x^2 - x + 1$, the discriminant is: $$\Delta = (-1)^2 - 4(2)(1) = 1 - 8 = -7 < 0$$ Since $\Delta < 0$, the quadratic $2x^2 - x + 1 = 0$ has no real roots. Therefore $f(x) = 0$ has only one real root, $x = \dfrac{3}{2}$. $\quad\blacksquare$ ## 解法二 #### 思路 (b) 的另一种证法：配方法。将 $2x^2 - x + 1$ 配成完全平方加正常数的形式，说明它恒正，从而无实根。 #### 答题过程 **(b)** With $a = -3$, $f(x) = 4x^3 - 8x^2 + 5x - 3 = (2x - 3)(2x^2 - x + 1)$. Complete the square for $2x^2 - x + 1$: $$\begin{align*} 2x^2 - x + 1 =&\, 2\left(x^2 - \frac{1}{2}x\right) + 1 \\[4mm] =&\, 2\left(x^2 - \frac{1}{2}x + \frac{1}{16} - \frac{1}{16}\right) + 1 \\[4mm] =&\, 2\left(x - \frac{1}{4}\right)^2 - \frac{1}{8} + 1 \\[4mm] =&\, 2\left(x - \frac{1}{4}\right)^2 + \frac{7}{8} \end{align*}$$ Since $2\left(x - \dfrac{1}{4}\right)^2 \geqslant 0$ for all $x$, we have $2x^2 - x + 1 \geqslant \dfrac{7}{8} > 0$. Therefore $2x^2 - x + 1 = 0$ has no real roots, and $f(x) = 0$ has only one real root. $\quad\blacksquare$