📚 Completing the Square: A Step-by-Step Guide | 配方法：分步指南
Completing the square is a central algebraic technique in the IGCSE Mathematics syllabus. It allows you to rewrite quadratic expressions in a form that reveals the turning point of a parabola, makes solving equations more elegant, and provides a bridge to the quadratic formula. This article gives a clear, methodical walkthrough of the technique with worked examples and common pitfalls.
配方法是 IGCSE 数学课程中的核心代数技巧。它能把二次表达式改写为直接显示抛物线顶点的形式，使方程求解更加优美，并为二次公式提供桥梁。本文将分步骤清晰讲解这一技巧，配合典型例题和常见错误分析。
1. What Is Completing the Square? | 什么是配方法？
Completing the square means rewriting a quadratic expression of the form ax² + bx + c into the form a(x + p)² + q. The new form contains a perfect square term, which makes key features of the quadratic graph immediately visible. The values p and q are determined by the coefficients of the original expression.
配方法是指将形如 ax² + bx + c 的二次表达式改写为 a(x + p)² + q 的形式。改写后的形式包含一个完全平方项，使我们能直接看出二次函数图像的关键特征。p 和 q 的值由原表达式的系数决定。
For example, the expression x² + 6x + 5 can be rewritten as (x + 3)² − 4. The turning point of the graph y = x² + 6x + 5 is therefore at (−3, −4). This simple transformation is extremely powerful in graphing, equation solving, and evaluating maxima or minima.
例如，表达式 x² + 6x + 5 可改写为 (x + 3)² − 4。于是函数 y = x² + 6x + 5 的顶点坐标为 (−3, −4)。这一简单的变形在画图、解方程以及求最大值或最小值中都非常有力。
2. The Basic Formula | 基本公式
For a simple quadratic where the coefficient of x² is 1, the formula is based on the identity (x + p)² = x² + 2px + p². To complete the square for x² + bx, take half of b, square it, and then adjust by subtracting the square outside the bracket.
对于 x² 系数为 1 的简单二次表达式，配方基于恒等式 (x + p)² = x² + 2px + p²。对于 x² + bx，取 b 的一半，平方后放入完全平方项，再在括号外减去这个平方以保持相等。
(x + b/2)² − (b/2)² = x² + bx
In general, if you have x² + bx + c, the completed square form is:
一般情况下，对于 x² + bx + c，配方法结果为：
(x + b/2)² − (b/2)² + c
This identity is the foundation for all completing-the-square problems. Always note that b/2 may be negative if b is negative.
这个恒等式是所有配方问题的基础。注意当 b 为负数时，b/2 也可能是负数。
3. Worked Example 1: Simple Quadratic | 例题一：简单二次表达式
Complete the square for x² + 8x + 7.
请为 x² + 8x + 7 配方。
Step 1: Take half of 8, which is 4. Write (x + 4)².
第一步：取 8 的一半，即 4，写出 (x + 4)²。
Step 2: Expand (x + 4)² to get x² + 8x + 16. This has an extra 16 compared with x² + 8x, so subtract 16.
第二步：展开 (x + 4)² 得到 x² + 8x + 16。这比 x² + 8x 多出 16，因此要减去 16。
Step 3: Add the constant +7 from the original expression.
第三步：加上原表达式中的常数项 +7。
x² + 8x + 7 = (x + 4)² − 16 + 7 = (x + 4)² − 9
Thus the turning point of y = x² + 8x + 7 is (−4, −9).
因此，y = x² + 8x + 7 的顶点坐标为 (−4, −9)。
4. Worked Example 2: Coefficient of x² Not 1 | 例题二：x² 系数不为 1
When the coefficient of x² is not 1, factor it out from the first two terms before completing the square. For example, complete the square for 2x² + 12x + 5.
当 x² 系数不为 1 时，先将该系数从前两项中提出来，再配方。例如，请为 2x² + 12x + 5 配方。
Step 1: Factor 2 from 2x² + 12x to get 2(x² + 6x). Keep the constant +5 outside for now.
第一步：从 2x² + 12x 中提出 2，得到 2(x² + 6x)，暂时将常数 +5 留在外面。
Step 2: Complete the square inside the bracket: x² + 6x = (x + 3)² − 9.
第二步：在括号内配方：x² + 6x = (x + 3)² − 9。
Step 3: Multiply through by 2 and add 5.
第三步：乘以 2 后再加 5。
2x² + 12x + 5 = 2[(x + 3)² − 9] + 5 = 2(x + 3)² − 18 + 5 = 2(x + 3)² − 13
The vertex is at (−3, −13), and since a = 2 > 0, the parabola opens upward.
顶点为 (−3, −13)，由于 a = 2 > 0，抛物线开口向上。
5. Solving Equations by Completing the Square | 用配方法解方程
To solve a quadratic equation, set the completed square expression equal to 0 and then isolate the square before taking square roots. Consider the equation x² − 6x − 7 = 0.
要解二次方程，先配方得到等于 0 的形式，然后通过开平方求出 x。考虑方程 x² − 6x − 7 = 0。
Complete the square on the left side:
对方程左边配方：
x² − 6x = (x − 3)² − 9
(x − 3)² − 9 − 7 = 0 → (x − 3)² − 16 = 0
Now add 16 to both sides and take the square root:
两边加 16，再开平方：
(x − 3)² = 16 → x − 3 = ±4
Therefore x = 3 + 4 = 7 or x = 3 − 4 = −1.
所以 x = 3 + 4 = 7 或 x = 3 − 4 = −1。
This method always works, even when the quadratic does not factor easily.
这种解法总是有效，即使二次式不容易因式分解也能使用。
6. Connection to the Quadratic Formula | 与二次公式的联系
The quadratic formula x = (−b ± √(b² − 4ac)) / (2a) can be derived directly by completing the square on the general equation ax² + bx + c = 0.
二次公式 x = (−b ± √(b² − 4ac)) / (2a) 可以直接通过对一般式 ax² + bx + c = 0 配方导出。
Start by dividing by a and moving the constant term to the other side:
先将两边除以 a，并将常数项移到等号另一边：
x² + (b/a)x = −c/a
Complete the square on the left:
对左边配方：
(x + b/(2a))² − b²/(4a²) = −c/a
Then rearrange and take square roots to obtain the familiar formula.
然后整理并开平方，即可得到我们熟悉的二次公式。
This connection helps you remember the formula and understand where the discriminant b² − 4ac comes from.
这个联系有助于记忆公式，并理解判别式 b² − 4ac 的来源。
7. Sketching Graphs Using the Completed Square Form | 利用配方形式画函数图像
For a quadratic in completed square form y = a(x + p)² + q, the vertex is (−p, q). The line of symmetry is x = −p. The value of a determines whether the graph opens upward (a > 0) or downward (a < 0).
对于配方法形式 y = a(x + p)² + q，顶点坐标为 (−p, q)，对称轴为 x = −p。a 的正负决定抛物线开口向上 (a > 0) 或向下 (a < 0)。
To sketch the graph, start by plotting the vertex. Then find the y-intercept by substituting x = 0. You can also find x-intercepts by solving y = 0, if real roots exist.
画图时，先标出顶点，然后代入 x = 0 求 y 轴截距。若存在实数根，还可以解 y = 0 求 x 轴截距。
Example: y = 2(x + 3)² − 13 has vertex (−3, −13), axis of symmetry x = −3, and y-intercept 2(3)² − 13 = 5. Since a = 2 > 0, the parabola opens upward.
例如：y = 2(x + 3)² − 13 的顶点为 (−3, −13)，对称轴为 x = −3，y 轴截距为 2(3)² − 13 = 5。由于 a = 2 > 0，抛物线开口向上。
8. Common Mistakes | 常见错误
One frequent error is forgetting to subtract the square when the coefficient of x² is not 1. Always multiply the subtracted square by the factor you took out. For instance, in 2(x + 3)² − 9, the −9 must be multiplied by 2 to give −18.
常见错误之一是在 x² 系数不为 1 时忘记减去平方项。记住提取公因数后，括号中减去的平方必须乘以该因数。例如，2(x + 3)² − 9 中的 −9 必须乘以 2，得到 −18。
Another mistake is mixing up signs when b is negative. For x² − 10x, half of b is −5, so the bracket is (x − 5)², and the subtracted square is 25, not −25.
另一个错误是当 b 为负数时弄错符号。比如 x² − 10x，b 的一半是 −5，所以括号为 (x − 5)²，减去的平方是 25，而不是 −25。
A third error involves solving equations: forgetting to take both positive and negative square roots. Always write x + p = ±√k when (x + p)² = k.
第三个错误是解方程时忘记取正负平方根。当 (x + p)² = k 时，一定要写 x + p = ±√k。
9. Practice Questions | 练习题目
Try these questions to master the technique.
请尝试以下题目以掌握配方法。
-
1. Complete the square for x² + 10x + 21.
1. 为 x² + 10x + 21 配方。
-
2. Complete the square for 3x² − 6x + 2.
2. 为 3x² − 6x + 2 配方。
-
3. Solve x² + 4x − 3 = 0 by completing the square, giving answers in surd form.
3. 用配方法解 x² + 4x − 3 = 0，答案保留根号形式。
Answers: 1. (x + 5)² − 4; 2. 3(x − 1)² − 1; 3. x = −2 ± √7.
答案：1. (x + 5)² − 4；2. 3(x − 1)² − 1；3. x = −2 ± √7。
10. Applications in Problem Solving | 在解题中的应用
Completing the square is often used to find the maximum or minimum value of a quadratic model. For example, the height h(t) of a projectile may be modelled by h(t) = −5t² + 40t + 2. Complete the square to find the maximum height and the time at which it is reached.
配方法常用于求二次模型的最大值或最小值。例如，抛射物的高度 h(t) 可用 h(t) = −5t² + 40t + 2 建模。通过配方可求出最大高度及达到该高度的时刻。
h(t) = −5(t² − 8t) + 2 = −5[(t − 4)² − 16] + 2 = −5(t − 4)² + 80 + 2 = −5(t − 4)² + 82
Therefore the maximum height is 82, reached at t = 4. The completed square form immediately gives the vertex of the parabola.
因此最大高度为 82，在 t = 4 时达到。配方形式直接给出了抛物线的顶点。
11. Summary | 总结
Completing the square is a versatile tool in IGCSE Mathematics. Practice the basic identity, learn to factor out non‑1 coefficients, solve equations carefully, and use the final form to sketch graphs. This skill also deepens your understanding of the quadratic formula and discriminants.
配方法是 IGCSE 数学中的多功能工具。熟练基本恒等式，学会提出非 1 系数，仔细求解方程，并利用最终形式画图。这项技能还能加深你对二次公式和判别式的理解。
With consistent practice, you will be able to recognise when completing the square is the fastest route and avoid the common errors that cost marks.
通过持续练习，你将能够判断何时使用配方法最快捷，并避免那些失分的常见错误。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导