📚 Example 2.3.1: Solving Quadratic Equations | 例题 2.3.1：解二次方程
ng three independent methods: factorisation, completing the square, and the quadratic formula. Each approach not only produces the same roots but also reveals a different aspect of the AQA specification, helping you build a flexible toolkit for the examination.
欢迎来到 TutorHao 为 AQA 开设的 A-Level 数学系列。本文将研究例题 2.3.1——一个关于二次方程的经典问题。我们将用三种独立的方法解方程 x² − 6x + 8 = 0：因式分解、配方法与二次求根公式。每种方法不仅得出相同的根，还从不同角度展示 AQA 考纲的考查方式，帮助你在考试中建立灵活的方法库。
1. Understanding the Quadratic Equation | 理解二次方程
0, where a ≠ 0. In Example 2.3.1 we are given x² − 6x + 8 = 0, so a = 1, b = −6 and c = 8.
二次方程的一般形式为 ax² + bx + c = 0，其中 a ≠ 0。在例题 2.3.1 中，方程为 x² − 6x + 8 = 0，因此 a = 1，b = −6，c = 8。
curve opens upward, forming a U-shape. If a were negative, the curve would open downward.
a 的符号决定抛物线的开口方向。因为 a = 1 > 0，曲线开口向上，呈 U 形。若 a 为负数，则曲线开口向下。
2. Presenting Example 2.3.1 | 呈现例题 2.3.1
Solve the equation x² − 6x + 8 = 0 for real values of x.
解方程 x² − 6x + 8 = 0，求 x 的实数值。
We will apply three methods in turn. The first method, factorisation, is often the quickest when the roots are integers. The second, completing the square, also provides the vertex of the parabola. The third, the quadratic formula, works for every quadratic equation, including those that cannot be factorised.
我们将依次运用三种方法。第一种因式分解在根为整数时通常最快；第二种配方法还能给出抛物线的顶点；第三种二次求根公式适用于所有二次方程，包括无法因式分解的情形。
3. Method 1: Factorisation | 方法一：因式分解
b = −6. The required numbers are −2 and −4, because (−2) × (−4) = 8 and (−2) + (−4) = −6.
我们寻找两个数，使它们的乘积等于 c = 8，和等于 b = −6。这两个数是 −2 和 −4，因为 (−2) × (−4) = 8，且 (−2) + (−4) = −6。
Therefore we may write:
因此可以写成：
x² − 6x + 8 = (x − 2)(x − 4) = 0
hen at least one factor must be zero. Hence x − 2 = 0 or x − 4 = 0, giving x = 2 or x = 4.
根据零乘积性质，若 (x − 2)(x − 4) = 0，则至少有一个因式为零。因此 x − 2 = 0 或 x − 4 = 0，得到 x = 2 或 x = 4。
4. Method 2: Completing the Square | 方法二：配方法
king half of the coefficient of x, which is −6 ÷ 2 = −3, and then squaring it to obtain 9.
配方法将二次式改写为 (x − p)² + q 的形式。首先取 x 系数的一半，即 −6 ÷ 2 = −3，然后平方得到 9。
Add and subtract 9 inside the equation:
在方程中加入 9 再减去 9：
x² − 6x + 9 − 9 + 8 = 0
en (x − 3)² = 1. Taking square roots yields x − 3 = ±1, so x = 3 + 1 = 4 or x = 3 − 1 = 2.
前三项构成完全平方式，得到 (x − 3)² − 1 = 0。于是 (x − 3)² = 1。两边开平方得 x − 3 = ±1，所以 x = 3 + 1 = 4 或 x = 3 − 1 = 2。
5. Method 3: The Quadratic Formula | 方法三：二次求根公式
For any equation of the form ax² + bx + c = 0, the quadratic formula states that x = (−b ± √b² − 4ac) / (2a).
对于任何 ax² + bx + c = 0 形式的方程，二次求根公式为 x = (−b ± √b² − 4ac) / (2a)。
= −6 and c = 8. First compute the discriminant b² − 4ac = (−6)² − 4 × 1 × 8 = 36 − 32 = 4.
代入 a = 1，b = −6，c = 8。先计算判别式 b² − 4ac = (−6)² − 4 × 1 × 8 = 36 − 32 = 4。
x = (6 ± √4) / (2 × 1) = (6 ± 2) / 2
= 2 or x = (6 + 2) / 2 = 4. This confirms the roots obtained by the previous two methods.
因此 x = (6 − 2) / 2 = 2，或 x = (6 + 2) / 2 = 4。这证实了前两种方法求得的根。
6. The Discriminant and the Nature of Roots | 判别式与根的性质
trols how many real roots a quadratic equation has. A summary is shown in the table below.
判别式 Δ = b² − 4ac 决定二次方程实根的个数。下表给出总结。
| Discriminant Δ | Nature of roots | Graph behaviour |
| Δ > 0 | Two distinct real roots | 两个不等实根 | Crosses the x-axis twice | 与 x 轴有两个交点 |
| Δ = 0 | One repeated real root | 一个重根 | Touches the x-axis once | 与 x 轴相切 |
| Δ < 0 | No real roots | 无实根 | Does not meet the x-axis | 与 x 轴无交点 |
> 0, so we expect two distinct real roots. This is exactly what we found: x = 2 and x = 4.
在例题 2.3.1 中，Δ = 4 > 0，因此应有两个不等实根。这正是我们求得的 x = 2 与 x = 4。
7. Linking Roots to the Graph | 将根与图像联系起来
0 are the x-coordinates of the points where the curve y = x² − 6x + 8 crosses the x-axis.
方程 x² − 6x + 8 = 0 的根，就是曲线 y = x² − 6x + 8 与 x 轴交点的横坐标。
0) and (4, 0). Since the parabola opens upward, the curve lies below the x-axis for 2 < x < 4 and above the x-axis for x < 2 or x > 4.
令 y = 0，得到交点 (2, 0) 和 (4, 0)。由于抛物线开口向上，当 2 < x < 4 时曲线位于 x 轴下方；当 x < 2 或 x > 4 时曲线位于 x 轴上方。
This graph interpretation is frequently tested in AQA questions that ask you to solve a quadratic inequality such as x² − 6x + 8 < 0.
这种图像解读在 AQA 试题中经常出现，例如要求解二次不等式 x² − 6x + 8 < 0。
8. Finding the Vertex and Axis of Symmetry | 求顶点与对称轴
From the completed-square form y = (x − 3)² − 1, the vertex of the parabola is at (3, −1).
由配方法得到的 y = (x − 3)² − 1，可知抛物线顶点为 (3, −1)。
passes through the vertex and also through the midpoint of the two roots, (2 + 4) / 2 = 3.
对称轴是竖直线 x = 3，它经过顶点，也穿过两个根的中点，即 (2 + 4) / 2 = 3。
ses the y-axis at (0, 8). With the vertex, the y-intercept and the roots, you can sketch the graph accurately without plotting many points.
令 x = 0 可得 y 截距：y = 0² − 0 + 8 = 8，因此曲线在 (0, 8) 处穿过 y 轴。有了顶点、y 截距和根，你无需描很多点就能准确画出图像。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
quadratic formula, −b means the opposite of b; since b = −6, we must write −b = 6, not −6.
一个常见错误
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