📚 Solving Quadratic Equations by Completing the Square | 配方法解二次方程
GCSE Mathematics, you are expected to solve such equations by three methods: factorisation, the quadratic formula, and completing the square. This article focuses on completing the square — solves equations but also helps you sketch quadratic graphs and solve real-life problems.
二次方程是形如 ax² + bx + c = 0 的方程，其中 a ≠ 0。在 IGCSE 数学中，你应当掌握三种解法：因式分解法、求根公式法和配方法。本文重点讲解配方法——它不仅能够解方程，还能帮助你绘制二次函数图象并解决实际生活中的问题。
1. What Is a Perfect Square | 什么是完全平方式
gnising perfect squares quickly is the first step towards mastering completing the square.
完全平方式是可以写成 (x + p)² 的二次表达式。例如，x² + 6x + 9 = (x + 3)²，x² − 10x + 25 = (x − 5)²。快速识别完全平方式是掌握配方法的第一步。
The expansion rule gives us a crucial pattern:
(x + p)² = x² + 2px + p²
6x + 9, half of 6 is 3, and 3² = 9. For x² − 10x + 25, half of −10 is −5, and (−5)² = 25.
注意，常数项 p² 等于 x 系数一半的平方。对于 x² + 6x + 9，6 的一半是 3，3² = 9。对于 x² − 10x + 25，−10 的一半是 −5，(−5)² = 25。
2. Expanding (x + a)² — The Foundation | 展开 (x + a)² —— 基础
Before completing the square, you must be fluent in expanding (x + a)². The table below shows several examples that illustrate the pattern.
在开始配方法之前，你必须熟练展开 (x + a)²。下表展示了多个例子，帮助你理解其中的规律。
| Expanded form | Completed square |
| x² + 4x + 4 | (x + 2)² |
| x² − 6x + 9 | (x − 3)² |
| x² + 14x + 49 | (x + 7)² |
| x² − 20x + 100 | (x − 10)² |
To expand (x + a)², use the formula
(x + a)² = x² + 2ax + a²
x² + 12x + 36, you can reverse the process: half of 12 is 6, and 6² = 36, giving (x + 6)².
展开 (x + a)² 时使用公式
(x + a)² = x² + 2ax + a²
x 的系数是 2a，常数项是 a²。所以如果你看到 x² + 12x + 36，可以逆向操作：12 的一半是 6，6² = 36，因此结果是 (x + 6)²。
3. The Core Method: Completing the Square | 配方法的核心步骤
Many quadratic expressions are not perfect squares, but we can force them to be one. For an expression x² + bx + c, the method works as follows:
许多二次表达式并不是完全平方式，但我们可以把它改造成完全平方式。对于形如 x² + bx + c 的表达式，配方法的步骤如下：
- Take half of the coefficient of x, that is b/2.
- Square it to obtain (b/2)².
- Rewrite the expression as (x + b/2)² − (b/2)² + c.
- 取 x 系数的一半，即 b/2。
- 将其平方，得到 (b/2)²。
- 把原式改写为 (x + b/2)² − (b/2)² + c。
Let us illustrate with x² + 8x + 7. Here b = 8, so half of b is 4 and 4² = 16. Therefore:
我们用 x² + 8x + 7 来示范。这里 b = 8，b 的一半是 4，4² = 16。因此：
x² + 8x + 7 = (x + 4)² − 16 + 7 = (x + 4)² − 9
The expression is now in completed square form. The term −9 appears because we added 16 inside the bracket, so we must subtract 16 outside to keep the expression unchanged.
现在表达式已化为配方法形式。出现 −9 是因为我们在括号内加上了 16，所以必须在括号外减去 16，以保持原式的值不变。
4. Solving Equations with Leading Coefficient 1 | 解首项系数为 1 的方程
We now use completing the square to solve x² + 6x − 7 = 0. Follow the steps carefully.
现在我们用配方法来解 x² + 6x − 7 = 0。请仔细按照以下步骤操作。
Step 1: Complete the square on the left-hand side.
第 1 步：对左边进行配方。
x² + 6x − 7 = (x + 3)² − 9 − 7 = (x + 3)² − 16
The equation becomes
方程变为
(x + 3)² − 16 = 0
Step 2: Isolate the perfect square.
第 2 步：将完全平方式单独放在等式一边。
(x + 3)² = 16
Step 3: Take the square root of both sides. Remember the ± sign.
第 3 步：两边开平方。记住要加 ± 号。
x + 3 = ± 4
Step 4: Solve for x.
第 4 步：求出 x。
x = −3 + 4 = 1 or x = −3 − 4 = −7
The solution set is x = 1 或 x = −7. You can verify both answers by substituting them back into the original equation.
解集为 x = 1 或 x = −7。你可以把两个答案代回原方程进行验证。
5. Example: Solving x² − 10x + 21 = 0 | 示例：解 x² − 10x + 21 = 0
Let us work through another example with a negative coefficient of x.
我们再来看一个 x 系数为负的完整例子。
Half of −10 is −5, and (−5)² = 25. Therefore:
−10 的一半是 −5，(−5)² = 25。因此：
x² − 10x + 21 = (x − 5)² − 25 + 21 = (x − 5)² − 4
Now set the equation to zero:
现在令方程等于零：
(x − 5)² − 4 = 0 ⇒ (x − 5)² = 4
x − 5 = ± 2 ⇒ x = 7 or x = 3
Both answers check: 7² − 70 + 21 = 49 − 70 + 21 = 0, and 3² − 30 + 21 = 9 − 30 + 21 = 0.
两个答案都成立：7² − 70 + 21 = 49 − 70 + 21 = 0，3² − 30 + 21 = 9 − 30 + 21 = 0。
6. Leading Coefficient Not Equal to 1 | 首项系数不为 1 的情况
When the coefficient of x² is not 1, factor it out from the first two terms before completing the square.
当 x² 的系数不是 1 时，先从前两项中提出该系数，再进行配方。
Solve 2x² + 8x + 5 = 0.
解 2x² + 8x + 5 = 0。
Factor out 2 from the quadratic and linear terms:
从二次项和一次项中提出 2：
2x² + 8x + 5 = 2(x² + 4x) + 5
Complete the square inside the bracket: half of 4 is 2, and 2² = 4.
对括号内配方：4 的一半是 2，2² = 4。
2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5
2x² + 8x + 5 = 2(x + 2)² − 3
Now solve the equation:
现在解方程：
2(x + 2)² − 3 = 0 ⇒ (x + 2)² = 3/2
x + 2 = ± √3/2 ⇒ x = −2 ± √3/2
≈ −3.225. Always keep the exact form unless the question asks for a decimal approximation.
因此 x ≈ −0.775 或 x ≈ −3.225。除非题目要求近似值，否则应始终保留精确形式。
7. Connection to the Quadratic Formula | 与求根公式的联系
s also the derivation of the quadratic formula. Starting from ax² + bx + c = 0 with a ≠ 0:
配方法不仅是一种解法，它还是求根公式的推导基础。从 ax² + bx + c = 0（a ≠ 0）出发：
a(x² + (b/a)x) + c = 0
a[(x + b/(2a))² − b²/(4a²)] + c = 0
a(x + b/(2a))² − b²/(4a) + c = 0
(x + b/(2a))² = (b² − 4ac)/(4a²)
x = (−b ± √b² − 4ac) / (2a)
two distinct roots; if Δ = 0 there is one repeated root; if Δ < 0 there are no real roots.
判别式 Δ = b² − 4ac 决定实根的个数：若 Δ > 0，有两个不相等实根；若 Δ = 0，有一个重根；若 Δ < 0，则没有实根。
8. Sketching Graphs Using Completed Square Form | 利用配方法绘制函数图象
+ k, the vertex of the parabola is at (h, k). This is the most useful form for sketching.
当二次函数写成 y = a(x − h)² + k 时，抛物线的顶点坐标为 (h, k)。这是绘制图象时最有用的形式。
Sketch y = x² − 4x + 1. Complete the square:
绘制 y = x² − 4x + 1 的图象。先配方：
y = (x − 2)² − 4 + 1 = (x − 2)² − 3
). Since a = 1 > 0, the parabola opens upwards. The y-intercept is found by setting x = 0:
顶点为 (2, −3)。因为 a = 1 > 0，抛物线开口向上。令 x = 0 可求 y 轴截距：
y = 0² − 0 + 1 = 1
To find the x-intercepts, set y = 0 and solve using completing the square:
令 y = 0 并配方求解，可求出 x 轴截距：
(x − 2)² = 3 ⇒ x = 2 ± √3 ≈ 3.73 or 0.27
With the vertex, intercepts, and direction of opening, you can produce an accurate sketch.
有了顶点、截距和开口方向，你就能画出准确的图象。
9. Word Problem Application | 实际应用题
A rectangular garden has length (x + 4) m and width (x − 2) m. Its area is 40 m². Find the value of x.
一块长方形花园的长为 (x + 4) m，宽为 (x − 2) m，面积为 40 m²。求 x 的值。
The area gives the equation:
由面积可得方程：
(x + 4)(x − 2) = 40
x² + 2x − 8 = 40 ⇒ x² + 2x − 48 = 0
Complete the square:
配方：
(x + 1)² − 1 − 48 = 0 ⇒ (x + 1)² = 49
x + 1 = ± 7 ⇒ x = 6 or x = −8
negative width, we discard it. The valid solution is x = 6. The rectangle is 10 m by 4 m.
因为 x = −8 会使宽为负数，所以舍去。有效解为 x = 6。此时长方形为长 10 m、宽 4 m。
10. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Students often lose marks on completing the square due to a few repeated errors. Pay special attention to the following points.
学生在配方法上丢分，往往是因为几个反复出现的错误。请特别注意以下几点。
- Forgetting the ± sign: th : When taking the square root of both sides, always write ±. For example, (x + 3)² = 16 gives x + 3 = ±4, not just x + 3 = 4.
- Half before square: You must halve the coefficient of x before squaring, not square first. For x² + 12x, use half of 12, which is 6, then square to get 36. Squaring first gives 144, which is wrong.
- Ignoring the constant term: tant c must be adjusted correctly. For x² + 6x + 2, write (x + 3)² − 9 + 2 = (x + 3)² − 7.
- Forgetting to factor when a ≠ 1: If the coefficient of x² is not 1, factor it out from the first two terms first. Only then complete the square.
- 忘记 ± 号：等式两边开平方时，一定要写 ±。例如，(x + 3)² = 16 得到 x + 3 = ±4，而不是只写 x + 3 = 4。
- 先平方再取半：必须先取 x 系数的一半再平方。对于 x² + 12x，应先取 12 的一半得 6，再平方得 36。先平方会得到 144，那就错了。
- 忽略常数项：配方时必须正确处理常数 c。对于 x² + 6x + 2，应写成 (x + 3)² − 9 + 2 = (x + 3)² − 7。
- a ≠ 1 时忘记提取系数：当 x² 的系数不是 1 时，应先将该系数从前两项中提出，然后再配方。
Always check your final answers by substituting them into the original equation. This takes only a few seconds and prevents careless errors.
完成解答后，务必把答案代回原方程验证。这只需几秒钟，却能避免粗心造成的错误。
11. Exam Tips and Quick Checklist | 考试技巧与快速检查清单
In an IGCSE exam, completing the square may appear in algebra, calculus (differentiation), and graph-sketching questions. The following checklist will help you answer with confidence.
在 IGCSE 考试中，配方法常出现在代数、微分以及函数图象等题型中。以下清单能帮助你自信作答。
| Step | Action | Action (中文) |
| 1 | Divide by the coefficient of x² if needed. | 如需，先除以 x² 的系数。 |
| 2 | Halve the coefficient of x and square it. | 取 x 系数的一半并平方。 |
| 3 | Rewrite as (x + p)² − p² + c. | 改写为 (x + p)² − p² + c。 |
| 4 | Move the constant to the right side. | 把常数项移到等号右边。 |
| 5 | Take ± square roots. | 开平方并加上 ±。 |
| 6 | Solve the linear equations and verify. | 解一次方程并验证。 |
ed square form. For y = a(x − h)² + k, the minimum/maximum value is k, occurring at x = h.
如果题目要求二次函数的最小值或最大值，可以直接从配方法形式中读出。对于 y = a(x − h)² + k，最小值或最大值为 k，在 x = h 处取得。
12. Practice Questions | 巩固练习
Try these questions on your own, then check the answers below.
请先独立完成以下题目，再对照参考答案。
- Q1. Solve x² + 4x − 12 = 0 by completing the square.
- Q2. Solve x² − 6x + 4 = 0, giving your answers in surd form.
- Q3. Express 2x² − 12x + 7 in the form a(x − h)² + k.
- Q4. Find the vertex and y-intercept of y = x² + 8x + 15.
- 第 1 题：用配方法解 x² + 4x − 12 = 0。
- 第 2 题：解 x² − 6x + 4 = 0，答案保留根号形式。
- 第 3 题：将 2x² − 12x + 7 写成 a(x − h)² + k 的形式。
- 第 4 题：求 y = x² + 8x + 15 的顶点坐标与 y 轴截距。
Answers | 参考答案
Q1. (x + 2)² − 16 = 0 → (x + 2)² = 16 → x = 2 or x = −6.
第 1 题：(x + 2)² − 16 = 0 → (x + 2)² = 16 → x = 2 或 x = −6。
Q2. (x − 3)² − 5 = 0 → x = 3 ± √5.
第 2 题：(x − 3)² − 5 = 0 → x = 3 ± √5。
Q3. 2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11.
第 3 题：2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11。
Q4. y = (x + 4)² − 1, so the vertex is (−4, −1) and the y-intercept is 15.
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