Solving Quadratic Equations by Completing the Square | 配方法解二次方程

📚 Solving Quadratic Equations by Completing the Square | 配方法解二次方程

A quadratic equation is an equation of the form ax² + bx + c = 0, where a ≠ 0. In IGCSE 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 — a powerful technique that not only 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

A perfect square is a quadratic expression that can be written as (x + p)². For example, x² + 6x + 9 = (x + 3)² and x² − 10x + 25 = (x − 5)². Recognising 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

Notice that the constant term p² is the square of half the coefficient of x. For x² + 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

The coefficient of x is twice a, and the constant is a². So if you are given 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)

This gives x ≈ −0.775 or x ≈ −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

Completing the square is not just a method of solving; it is 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)

The discriminant Δ = b² − 4ac tells us the number of real roots: if Δ > 0 there are 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

When a quadratic function is written as y = a(x − h)² + 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

The vertex is at (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

Since x = −8 gives a 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: 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: When completing the square, the constant 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。

If the question asks for the minimum or maximum value of a quadratic, read it from the completed 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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