📚 Worked Example 2.6.1: Completing the Square for x² + 6x + 1 = 0 | 例题 2.6.1：配方法解 x² + 6x + 1 = 0
The method is especially valuable when a quadratic does not factorise over integers, because it turns the expression into a squared bracket plus or minus a constant.
在 AQA A-Level 数学中，配方法是解二次方程和确定顶点最可靠的方法之一。本文详细解析例题 2.6.1，求解二次方程 x² + 6x + 1 = 0。当二次式不能进行整数因式分解时，该方法尤为重要，因为它能把表达式转化为一个完全平方项加或减一个常数。
1. Why Completing the Square Matters in AQA | 配方法在 AQA 考试中的重要性
AQA frequently tests completing the square in both Pure Mathematics and applied contexts. It is not only used to solve equations but also to find the vertex of a parabola, to integrate certain rational functions, and to derive the quadratic formula. Mastering the technique in Example 2.6.1 gives you a strong foundation for these topics.
AQA 在纯数学和应用数学中频繁考查配方法。它不仅用于解方程，还用于求抛物线的顶点、对某些有理函数积分以及推导二次公式。掌握例题 2.6.1 中的技巧将为这些主题打下坚实基础。
Completing the square rewrites a quadratic in the form a(x + p)² + q. This completed-square form immediately reveals the minimum or maximum point of the graph, which is often asked in AQA exam questions. Understanding the algebra behind the method helps you avoid careless errors when questions become more complex.
配方法将二次式写成 a(x + p)² + q 的形式。这种完全平方形式能立即显示图像的最小值或最大值点，这在 AQA 试题中经常出现。理解该方法背后的代数原理有助于在题目变复杂时避免粗心错误。
2. The Given Problem: x² + 6x + 1 = 0 | 题目呈现：x² + 6x + 1 = 0
e completing-the-square process is straightforward. The first step is to isolate the x terms on one side, leaving the constant on the other.
我们需要求解的方程是 x² + 6x + 1 = 0。x² 的系数为 1，因此配方法过程较为直接。第一步是把含 x 的项移到一边，常数项留在另一边。
This quadratic cannot be factorised using integer factors because no pair of integers has product 1 and sum 6. The quadratic formula would work, but completing the square gives a neat exact answer in surd form and is often preferred by examiners for method marks.
这个二次式不能用整数因数分解，因为没有一对整数的积为 1 且和为 6。二次公式可以解决，但配方法能给出简洁的根式精确答案，并且阅卷老师通常会给方法分。
3. Step 1: Move the Constant Term | 第一步：移项
quation to separate the constant term from the x terms. The equation becomes x² + 6x = -1.
将方程两边同时减去 1，把常数项与含 x 的项分离。方程变为 x² + 6x = -1。
x² + 6x = -1
Always keep the equation balanced. Moving the constant term to the right creates space on the left for adding the special number that completes the square. If you forget this step and try to factorise immediately, the process will not work.
务必保持方程平衡。把常数项移到右边后，左边就为加上配平方数留出了空间。如果忘记这一步并试图立即因式分解，配方法就无法进行。
4. Step 2: Calculate the Magic Number | 第二步：计算配平方数
To complete the square, take half of the coefficient of x, which is 6, and square it. Half of 6 is 3, and 3² is 9. This number, 9, is added to both sides of the equation so that the left-hand side becomes a perfect square.
为了配平方，取 x 的系数 6 的一半再平方。6 的一半是 3，3² 为 9。这个数 9 要加到方程两边，使左边成为完全平方。
(6 ÷ 2)² = 3² = 9
uation balanced. The left side becomes x² + 6x + 9, and the right side becomes -1 + 9 = 8.
两边同时加上 9，注意保持方程平衡。左边变为 x² + 6x + 9，右边变为 -1 + 9 = 8。
x² + 6x + 9 = 8
o a = 3, which gives a² = 9. This relationship is the core logic of completing the square.
我们之所以取系数的一半，是因为完全平方展开式 (x + a)² = x² + 2ax + a²。这里 2a = 6，所以 a = 3，从而 a² = 9。这一关系正是配方法的核心逻辑。
5. Step 3: Factorise to a Perfect Square | 第三步：因式分解为完全平方
as (x + 3)², since x² + 6x + 9 = (x + 3)(x + 3). The equation simplifies to (x + 3)² = 8.
此时左边是一个完全平方三项式，可以写成 (x + 3)²，因为 x² + 6x + 9 = (x + 3)(x + 3)。方程简化为 (x + 3)² =
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