📚 Worked Example 1.3.1: Solving Quadratic Equations by Completing the Square | 例题 1.3.1：配方法解二次方程
ple, we use the equation x² + 6x + 5 = 0 to show every step clearly, including how the same process links to the graph of the function.
配方法是 AQA A-Level 数学中最重要的代数技巧之一。它可以把任意二次式改写成 a(x + p)² + q 的形式，从而直接显示抛物线的顶点，并提供一种可靠的二次方程解法。在本例题中，我们用方程 x² + 6x + 5 = 0 来逐步展示完整过程，并说明这一过程与函数图像之间的联系。
1. What Completing the Square Means | 什么是配方法
The phrase 'completing the square' example, x² + 6x becomes a perfect square when you add 9, because x² + 6x + 9 = (x + 3)².
'配方法' 就是把 x² + bx + c 这样的二次式改写成一个完全平方加上或减去一个常数。例如，x² + 6x 加上 9 后就变成完全平方，因为 x² + 6x + 9 = (x + 3)²。
This technique is not just a trick for solving equations. It is also the key to finding the minimum or maximum value of a quadratic function and understanding transformations of graphs.
这一技巧不只是解方程的窍门。它也是求二次函数最小值或最大值、理解图像变换的关键工具。
2. The Standard Quadratic Form | 二次方程的标准形式
s and a ≠ 0. In Example 1.3.1, the equation is x² + 6x + 5 = 0, so a = 1, b = 6 and c = 5.
一元二次方程的一般形式是 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。在例题 1.3.1 中，方程为 x² + 6x + 5 = 0，因此 a = 1，b = 6，c = 5。
re straightforward because the leading coefficient is already 1. If a is not 1, we must first divide every term by a before adding the square.
当 a = 1 时，配方法会更加直接，因为首项系数已经是 1。如果 a 不等于 1，我们必须在加上平方之前先用 a 去除每一项。
3. Setting Up the Worked Example | 建立本例题
. The goal is to isolate the x² and x terms on one side, then add a carefully chosen constant so that the left-hand side becomes a perfect square.
我们从方程 x² + 6x + 5 = 0 开始。目标是把 x² 项和 x 项单独放在一边，然后加上一个精心选择的常数，使左边成为一个完全平方。
x² + 6x + 5 = 0
The constant term 5 is moved to the right-hand side first. This makes the left-hand side ready for the square-completing step.
先把常数项 5 移到右边。这样左边就为配方步骤做好了准备。
4. Step 1: Move the Constant Term | 第一步：移项常数项
. It is important to keep the equation balanced by doing the same operation on both sides.
从方程两边同时减去 5。得到 x² + 6x = −5。必须对方程两边进行相同的运算以保持平衡，这一点很重要。
x² + 6x = −5
Now the left-hand side contains only the x² and x terms. We are ready to find the number that completes the square.
现在左边只含有 x² 项和 x 项。我们接下来找出能够配成完全平方的那个数。
5. Step 2: Find Half of the Coefficient of x | 第二步：求 x 系数的一半
he result: 3² = 9. This number 9 is the constant that turns x² + 6x into a perfect square.
观察 x 的系数，它是 6。取这个系数的一半：6 ÷ 2 = 3。然后把结果平方：3² = 9。这个数 9 就是使 x² + 6x 变成完全平方的常数。
half of 6 = 3, and 3² = 9
For any expression of the form x² + bx, adding (b/2)² gives x² + bx + (b/2)² = (x + b/2)².
这一步背后的公式是 (b/2)²。对于任何形如 x² + bx 的式子，加上 (b/2)² 就得到 x² + bx + (b/2)² = (x + b/2)²。
6. Step 3: Add the Square to Both Sides | 第三步：两边同时加上该平方
e left side becomes x² + 6x + 9, and the right side becomes −5 + 9, which simplifies to 4.
为了保持方程平衡，在 x² + 6x = −5 的两边同时加上 9。左边变为 x² + 6x + 9，右边变为 −5 + 9，化简后为 4。
x² + 6x + 9 = 4
A very common mistake is to add 9 only to the left-hand side. Always add the same value to both sides when solving an equation.
一个非常常见的错误是只在左边加上 9。解方程时必须始终在两边加上相同的值。
7. Step 4: Factor the Perfect Square | 第四步：因式分解完全平方
en as (x + 3)². This is because (x + 3)² = x² + 6x + 9. The equation becomes (x + 3)² = 4.
左边 x² + 6x + 9 现在可以写成 (x + 3)²。这是因为 (x + 3)² = x² + 6x + 9。方程变为 (x + 3)² = 4。
(x + 3)² = 4
This is the completed square form. It is much easier to solve because the unknown x now appears only once, inside the squared bracket.
这就是配方法的完成形式。它更容易求解，因为未知数 x 现在只出现一次，就在平方括号内。
8. Step 5: Solve Using Square Roots | 第五步：开平方求解
a positive number has two square roots: one positive and one negative. Therefore x + 3 = ±√4, which gives x + 3 = ±2.
对 (x + 3)² = 4 的两边同时开平方。记住正数有两个平方根：一个正根和一个负根。因此 x + 3 = ±√4，即 x + 3 = ±2。
x + 3 = ±2
separate linear equations: x + 3 = 2 or x + 3 = −2. Solving these gives x = −1 or x = −5.
现在把它拆成两个独立的一次方程：x + 3 = 2 或 x + 3 = −2。解这两个方程得到 x = −1 或 x = −5。
x = −1 or x = −5
These are the two solutions of the original quadratic equation. In set notation, the solution set is {−5, −1}.
这就是原二次方程的两个解。用集合表示，解集为 {−5, −1}。
9. Verifying the Solutions | 验证所得解
orrect. Now substitute x = −5: (−5)² + 6(−5) + 5 = 25 − 30 + 5 = 0, which is also correct.
把 x = −1 代入原方程：(−1)² + 6(−1) + 5 = 1 − 6 + 5 = 0，结果正确。再把 x = −5 代入：(−5)² + 6(−5) + 5 = 25 − 30 + 5 = 0，结果也正确。
Verification is a good habit in A-Level exams because it can catch sign errors and accidental slips, especially when the coefficient of x is negative or when a is not equal to 1.
在 A-Level 考试中养成验证的习惯是很好的，因为它能发现符号错误和意外失误，尤其是当 x 的系数为负或者 a 不等于 1 时。
10. Linking to the Vertex Form and Graph | 与顶点式和图像的联系
= (x + 3)² − 4. This is the vertex form y = a(x − h)² + k, with a = 1, h = −3 and k = −4.
配方法得到的 (x + 3)² = 4 可以改写为 (x + 3)² − 4 = 0。如果定义 y = x² + 6x + 5，那么 y = (x + 3)² − 4。这就是顶点式 y = a(x − h)² + k，其中 a = 1，h = −3，k = −4。
Therefore the parabola has its vertex at (−3, −4). Since a is positive, the parabola opens upwards and the vertex is the minimum point of the graph.
因此抛物线的顶点为 (−3, −4)。由于 a 为正，抛物线开口向上，顶点是图像的最低点。
vertex = (−3, −4)
ly the points where the graph crosses the x-axis. Completing the square therefore connects the algebraic solution to the geometric picture of the curve.
解 x = −5 和 x = −1 正是图像与 x 轴交点的横坐标。因此配方法把代数解与曲线的几何图像联系了起来。
11. Common Mistakes and How to Avoid Them | 常见错误及避免方法
y when b is negative. For example, if b = −6, then half of b is −3, and (−3)² = 9, not −9.
一个常见错误是忘记在方程两边同时加上平方数。另一个错误是当 b 为负数时计算 (b/2)² 出错。例如，如果 b = −6，那么 b 的一半是 −3，而 (−3)² = 9，不是 −9。
8x + 3 = 0, you must first write x² + 4x + 3/2 = 0, then complete the square for x² + 4x.
学生们有时在首项系数不为 1 时直接配方，却没有先除以 a。对于 2x² + 8x + 3 = 0，你必须先写成 x² + 4x + 3/2 = 0，然后再对 x² + 4x 进行配方。
9, which is missing the middle term 6x. Always check by expanding: (x + 3)² = x² + 6x + 9.
另一个错误是把 (x + 3)² 写成 x² + 9，这漏掉了中间项 6x。务必通过展开来检查：(x + 3)² = x² + 6x + 9。
12. AQA Exam Tips and Variations | AQA 考试技巧与变形
In AQA exam papers, you may be asked to solve a quadratic by completing the square, to express a quadratic in the form (x + p)² + q, or to state the coordinates of the vertex. The same method handles all of these tasks.
在 AQA 考试试卷中，题目可能要求你用配方法解二次方程，把二次式写成 (x + p)² + q 的形式，或者写出顶点坐标。同样的方法可以处理所有这些任务。
+ 5/3 = 0, and you then complete the square for x² − 4x by adding (−2)² = 4 to both sides.
如果 x² 的系数不是 1，在配方之前要先把这个系数除到每一项。例如，3x² − 12x + 5 = 0 变为 x² − 4x + 5/3 = 0，然后你对 x² − 4x 进行配方，在两边加上 (−2)² = 4。
Also remember that the discriminant b² − 4ac tells you how many real roots a quadratic has. Completing the square gives you the actual roots and also reveals the vertex, so it is often more informative than using the quadratic formula alone.
还要记住判别式 b² − 4ac 可以告诉你二次方程有多少个实根。配方法不仅能求出具体的根，还能显示顶点，因此通常比单独使用求根公式提供更多信息。
Summary | 小结
then solved to find x = −1 or x = −5. The same expression also gives the vertex (−3, −4).
配方法把 x² + bx + c = 0 转化为 (x + p)² = q 的形式。对于 x² + 6x + 5 = 0，我们在两边加上 9 得到 (x + 3)² = 4，然后解得 x = −1 或 x = −5。同样的表达式也给出了顶点 (−3, −4)。
Mastering this method gives you a powerful tool for solving quadratics, analysing graphs, and answering a wide range of AQA Pure Mathematics questions.
掌握这一方法可以为你提供强大的工具，用于解二次方程、分析图像，并解答 AQA Pure Mathematics 中各种类型的题目。
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