Completing the Square for Quadratic Equations: Mastering Key Steps | 配方法解二次方程：掌握关键步骤

📚 Completing the Square for Quadratic Equations: Mastering Key Steps | 配方法解二次方程：掌握关键步骤

Completing the square is a fundamental algebraic technique that transforms a quadratic equation into a perfect square trinomial, allowing us to solve equations that cannot be easily factored. This method is essential for IGCSE Edexcel Mathematics, as it appears in both the non-calculator and calculator papers, and forms the basis for understanding the quadratic formula and the vertex form of a parabola.

配方法是一种基础的代数技巧，它将二次方程转化为完全平方三项式，从而解决那些不易因式分解的方程。这一方法在 IGCSE Edexcel 数学考试中至关重要，既出现在非计算器试卷也出现在计算器试卷中，并且是理解二次公式和抛物线顶点形式的基础。

1. Understanding the Structure of a Quadratic | 理解二次方程的结构

A quadratic expression in the form \( ax^2 + bx + c \) can be rewritten using the technique of completing the square. The general result is: \( a(x + \frac{b}{2a})^2 – \frac{b^2}{4a} + c \). For the case where \( a = 1 \), this simplifies to \( (x + \frac{b}{2})^2 – (\frac{b}{2})^2 + c \).

形如 \( ax^2 + bx + c \) 的二次表达式可以通过配方法重新书写。一般结果为：\( a(x + \frac{b}{2a})^2 – \frac{b^2}{4a} + c \)。当 \( a = 1 \) 时，简化为 \( (x + \frac{b}{2})^2 – (\frac{b}{2})^2 + c \)。

Completing the square is particularly useful when solving quadratic equations, finding turning points of parabolas, and simplifying expressions for integration or curve sketching. On the IGCSE exam, questions may ask you to solve equations by completing the square or to express a quadratic in vertex form to find the minimum or maximum value.

配方法在解二次方程、寻找抛物线顶点以及化简表达式时特别有用。在 IGCSE 考试中，题目可能要求你用配方法解方程，或者将二次函数写成顶点形式以求出最小值或最大值。

2. The Basic Procedure: When \( a = 1 \) | 基本步骤：当 \( a = 1 \) 时

When the coefficient of \( x^2 \) is 1, the procedure is straightforward. Start with an equation like \( x^2 + 6x + 5 = 0 \). Take half of the coefficient of \( x \) (which is 6), giving 3. Square this value to get 9. Rewrite the equation as \( (x^2 + 6x + 9) – 9 + 5 = 0 \). This becomes \( (x + 3)^2 – 4 = 0 \).

当 \( x^2 \) 的系数为 1 时，步骤非常直接。例如方程 \( x^2 + 6x + 5 = 0 \)，取 \( x \) 系数（即6）的一半得 3，平方得到 9。将方程重写为 \( (x^2 + 6x + 9) – 9 + 5 = 0 \)，即 \( (x + 3)^2 – 4 = 0 \)。

The golden rule is: add and subtract the square of half the coefficient of \( x \). This keeps the expression equivalent to the original while creating a perfect square. Then solve for \( x \) by isolating the squared term and taking square roots.

黄金法则：加上并减去 \( x \) 系数一半的平方。这样既保持了表达式与原式等价，又构造出完全平方。然后通过移项和开平方来求解 \( x \)。

3. Solving a Quadratic Equation by Completing the Square | 用配方法解二次方程

Let us work through Example 1: Solve \( x^2 + 6x + 5 = 0 \) by completing the square. Follow these steps:

让我们通过例1来练习：用配方法解 \( x^2 + 6x + 5 = 0 \)。按以下步骤进行：

Step 1: Write down the equation: \( x^2 + 6x + 5 = 0 \).
第一步：写出方程：\( x^2 + 6x + 5 = 0 \)。
Step 2: Identify \( b = 6 \). Half of \( b \) is 3. Square it: \( 3^2 = 9 \).
第二步：\( b = 6 \)，取其一半为3，平方得 9。
Step 3: Add and subtract 9 inside the equation: \( x^2 + 6x + 9 – 9 + 5 = 0 \).
第三步：在方程中加入并减去9：\( x^2 + 6x + 9 – 9 + 5 = 0 \)。
Step 4: Factor the perfect square: \( (x + 3)^2 – 4 = 0 \).
第四步：因式分解完全平方：\( (x + 3)^2 – 4 = 0 \)。
Step 5: Isolate the square: \( (x + 3)^2 = 4 \).
第五步：移项得：\( (x + 3)^2 = 4 \)。
Step 6: Take square roots: \( x + 3 = \pm 2 \).
第六步：开平方：\( x + 3 = \pm 2 \)。
Step 7: Solve: \( x = -3 + 2 = -1 \) or \( x = -3 – 2 = -5 \).
第七步：求解：\( x = -1 \) 或 \( x = -5 \)。

The solutions are \( x = -1 \) and \( x = -5 \). You can verify by substituting back into the original equation. On the mark scheme, full marks are awarded for showing each of these key steps clearly.

解为 \( x = -1 \) 和 \( x = -5 \)。你可以代回原方程验证。在评分标准中，清晰展示每一步才能获得满分。

4. When \( a \neq 1 \): Dealing with Coefficients | 当 \( a \neq 1 \)：处理系数

If the coefficient of \( x^2 \) is not 1, factor it out from the first two terms before completing the square. For example, solve \( 2x^2 + 8x + 5 = 0 \). Factor out the 2 from the \( x^2 \) and \( x \) terms: \( 2(x^2 + 4x) + 5 = 0 \). Then complete the square inside the bracket.

如果 \( x^2 \) 的系数不为1，先从前两项中提出该系数再配方。例如解 \( 2x^2 + 8x + 5 = 0 \)，先提出2：\( 2(x^2 + 4x) + 5 = 0 \)，再在括号内配方。

Since half of 4 is 2, square it to get 4. Add and subtract 4 inside the bracket: \( 2[(x^2 + 4x + 4) – 4] + 5 = 0 \). This simplifies to \( 2(x + 2)^2 – 8 + 5 = 0 \), i.e. \( 2(x + 2)^2 – 3 = 0 \).

因为4的一半是2，平方得4。在括号内加4再减4：\( 2[(x^2 + 4x + 4) – 4] + 5 = 0 \)，化简为 \( 2(x + 2)^2 – 8 + 5 = 0 \)，即 \( 2(x + 2)^2 – 3 = 0 \)。

Then solve: \( 2(x + 2)^2 = 3 \), \( (x + 2)^2 = \frac{3}{2} \), \( x + 2 = \pm \sqrt{\frac{3}{2}} \), giving \( x = -2 \pm \sqrt{\frac{3}{2}} \). This is the exact form that examiners expect; converting to decimals may lose marks if the question requires exact answers.

然后求解：\( 2(x + 2)^2 = 3 \)，即 \( (x + 2)^2 = \frac{3}{2} \)，因此 \( x + 2 = \pm \sqrt{\frac{3}{2}} \)，得 \( x = -2 \pm \sqrt{\frac{3}{2}} \)。这正是考官期望的精确形式；如果题目要求精确答案，化为小数可能会丢分。

5. The General Formula: A Shortcut | 一般公式：捷径

From the examples above, we can derive a general algebraic result. For \( ax^2 + bx + c = 0 \), completing the square gives:

从以上例子可以推导出一般代数结论。对于 \( ax^2 + bx + c = 0 \)，配方得到：

\( a\!\left(x + \frac{b}{2a}\right)^2 – \frac{b^2}{4a} + c = 0 \)

This shows that the solutions are \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), which is the quadratic formula. Many students wonder why the formula works; completing the square is the proof behind it. Understanding this derivation helps you remember the formula and its conditions.

由此可见解为 \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)，这正是二次公式。许多学生好奇公式为何成立；配方法就是它的证明基础。理解这一推导有助于记忆公式及其使用条件。

In Edexcel IGCSE, you are given the quadratic formula in the formula booklet, but you must know when and how to apply it. Completing the square is also a standalone skill tested directly, so do not rely solely on the formula.

在 Edexcel IGCSE 中，公式册会提供二次公式，但你必须知道何时以及如何使用。配方法本身也是直接考查的技能，不能只依赖公式。

6. Solving Equations with Irrational Solutions | 解含无理数解的方程

Some quadratic equations have irrational solutions, which cannot be found by simple factoring. For example, solve \( x^2 – 4x – 2 = 0 \) by completing the square. Half of -4 is -2; square to get 4.

有些二次方程包含无理数解，无法通过简单因式分解求出。例如用配方法解 \( x^2 – 4x – 2 = 0 \)。-4的一半是-2，平方得4。

Add and subtract 4: \( x^2 – 4x + 4 – 4 – 2 = 0 \), so \( (x – 2)^2 – 6 = 0 \). Hence \( (x – 2)^2 = 6 \), giving \( x = 2 \pm \sqrt{6} \). These are the exact irrational solutions.

加4再减4：\( x^2 – 4x + 4 – 4 – 2 = 0 \)，即 \( (x – 2)^2 – 6 = 0 \)。所以 \( (x – 2)^2 = 6 \)，得 \( x = 2 \pm \sqrt{6} \)。这就是精确的无理数解。

When the discriminant \( b^2 – 4ac \) is positive but not a perfect square, the solutions are irrational and surds are used. In your final answer, always leave surds in simplified form, e.g. \( \sqrt{8} \) should be written as \( 2\sqrt{2} \), unless the question states otherwise.

当判别式 \( b^2 – 4ac \) 为正但不是完全平方数时，解为无理数并用根式表示。在最终答案中，根式一定要化简，例如 \( \sqrt{8} \) 应写成 \( 2\sqrt{2} \)，除非题目另有要求。

7. Completing the Square with Negative Coefficients | 处理负系数

Negative coefficients can be tricky. Consider the equation \( x^2 – 6x + 2 = 0 \). Half of -6 is -3. Squaring gives +9. Add and subtract 9: \( x^2 – 6x + 9 – 9 + 2 = 0 \). This becomes \( (x – 3)^2 – 7 = 0 \).

负系数容易出错。看方程 \( x^2 – 6x + 2 = 0 \)。-6的一半是-3，平方得9。加9减9：\( x^2 – 6x + 9 – 9 + 2 = 0 \)，即 \( (x – 3)^2 – 7 = 0 \)。

Avoid the common mistake of writing \( (x – 3)^2 + 7 = 0 \). Remember: \( -9 + 2 = -7 \), not \( +7 \). The sign inside the bracket is determined by the sign of \( b \); the constant outside depends on careful arithmetic. Practise several variations to internalise the pattern.

避免常见错误：不要写成 \( (x – 3)^2 + 7 = 0 \)。记住：\( -9 + 2 = -7 \)，而不是 +7。括号内的符号由 \( b \) 的符号决定；外部的常数则取决于细心的运算。多练习不同变式以掌握规律。

8. Applications: Finding Turning Points | 应用：求函数顶点

One major application of completing the square in IGCSE is finding the coordinates of the turning point of a parabola. For \( y = x^2 + 6x + 5 \), completing the square gives \( y = (x + 3)^2 – 4 \). The vertex is at \( (-3, -4) \). The minimum value of the function is -4.

配方法在 IGCSE 中的一个重要应用是求抛物线顶点的坐标。对于 \( y = x^2 + 6x + 5 \)，配方得到 \( y = (x + 3)^2 – 4 \)。顶点坐标为 \( (-3, -4) \)，函数最小值为 -4。

In vertex form \( y = a(x – h)^2 + k \), the turning point is \( (h, k) \). Note the sign: \( (x + 3)^2 \) means \( h = -3 \). If the coefficient \( a \) is positive, the turning point is a minimum; if negative, it is a maximum.

在顶点形式 \( y = a(x – h)^2 + k \) 中，顶点为 \( (h, k) \)。注意符号：\( (x + 3)^2 \) 表示 \( h = -3 \)。如果系数 \( a \) 为正，顶点是最小值；如果为负，则为最大值。

This application is commonly tested in graph sketching questions. You may be asked to state the coordinates of the vertex, the line of symmetry, or the minimum/maximum value, all of which are directly readable from the completed-square form.

这一应用常见于函数图像绘制题。题目可能要求你写出顶点坐标、对称轴或最小值/最大值，这些都可以直接从配方法形式中读出。

9. Worked Example: Exam-Style Question | 例题：考试风格题目

Exam-style problem: Express \( 2x^2 – 12x + 7 \) in the form \( a(x + b)^2 + c \), stating the values of \( a, b, c \).

考试风格题目：将 \( 2x^2 – 12x + 7 \) 化为 \( a(x + b)^2 + c \) 的形式，并写出 \( a、b、c \) 的值。

Step 1: Factor 2 from first two terms: \( 2(x^2 – 6x) + 7 \).
第一步：从前两项提出2：\( 2(x^2 – 6x) + 7 \)。
Step 2: Half of -6 is -3. Square: 9. Add and subtract inside: \( 2(x^2 – 6x + 9 – 9) + 7 \).
第二步：-6的一半是-3，平方得9。在括号内加9减9：\( 2(x^2 – 6x + 9 – 9) + 7 \)。
Step 3: Factor: \( 2[(x – 3)^2 – 9] + 7 \).
第三步：因式分解：\( 2[(x – 3)^2 – 9] + 7 \)。
Step 4: Expand: \( 2(x – 3)^2 – 18 + 7 \).
第四步：展开：\( 2(x – 3)^2 – 18 + 7 \)。
Step 5: Simplify: \( 2(x – 3)^2 – 11 \).
第五步：化简：\( 2(x – 3)^2 – 11 \)。

Thus \( a = 2, b = -3, c = -11 \). Notice that in the requested form, \( b \) is -3 even though the bracket is \( (x – 3)^2 \). Read the question carefully to present your answer correctly.

因此 \( a = 2, b = -3, c = -11 \)。注意在要求的形式中，\( b \) 为 -3，尽管括号中是 \( (x – 3)^2 \)。仔细阅读题目以正确呈现答案。

10. Common Mistakes to Avoid | 常见错误与避坑指南

Students often lose marks in completing-the-square questions due to avoidable errors. The most frequent mistake is forgetting to multiply the subtracted square by the coefficient \( a \) when factoring out \( a \). Always double-check each arithmetic step.

学生在配方法题目中常因可避免的错误而失分。最常见的错误是提出 \( a \) 后忘记将减去的平方项乘上 \( a \)。务必检查每一步的运算。

Mistake 1: Sign errors when \( b \) is negative. Always use brackets carefully.
错误1：当 \( b \) 为负数时出现符号错误。务必小心使用括号。
Mistake 2: Forgetting to add and subtract the square, thereby changing the value of the expression.
错误2：忘记同时加减同一个数，改变了表达式的值。
Mistake 3: Not simplifying surds in the final answer.
错误3：最终答案中的根式未化简。
Mistake 4: Confusing the vertex coordinate sign when reading from \( (x + b)^2 \).
错误4：从 \( (x + b)^2 \) 读取顶点坐标时弄错符号。

Practise with a range of coefficients—fractions, negatives, and values of \( a \) other than 1—to build confidence and avoid these pitfalls on the exam.

用多种系数练习——分数、负数、不等于1的 \( a \) 值——以增强信心并在考试中避开这些陷阱。

11. Practice Problems | 练习题目

Slide 1 of exam revision: attempt these by completing the square. Answers are shown for self-checking, but always write out the full working.

考试复习第一部分：请用配方法完成以下练习。答案供自查，但务必写出完整步骤。

Problem 题目	Answer 答案
1. Solve \( x^2 + 10x + 21 = 0 \)	\( x = -3, -7 \)
2. Solve \( x^2 – 8x + 1 = 0 \)	\( x = 4 \pm \sqrt{15} \)
3. Solve \( 2x^2 + 4x – 3 = 0 \)	\( x = -1 \pm \frac{\sqrt{10}}{2} \)
4. Express \( x^2 – 3x + 1 \) in vertex form	\( (x – \frac{3}{2})^2 – \frac{5}{4} \)
5. Find the minimum value of \( x^2 + 5x + 2 \)	\( -\frac{17}{4} \) at \( x = -\frac{5}{2} \)

Work through these systematically, using the steps outlined in this article. If your answers match, you are on the right track; if not, revisit the relevant section above.

按照本文中的步骤系统练习。如果答案一致，说明掌握良好；如果不一致，请回顾上文对应章节。

12. Summary and Final Tips | 总结与最终建议

Completing the square is a versatile and examinable skill in Edexcel IGCSE Mathematics. Master the steps: identify \( b \), take half, square it, add and subtract, factor, and solve. This method not only solves equations but also reveals the vertex of a parabola, a recurring theme in graph questions.

配方法是 Edexcel IGCSE 数学中的一个多功能且必考的技能。掌握步骤：识别 \( b \)，取一半，平方，加和减，因式分解，求解。该方法不仅能解方程，还能揭示抛物线的顶点，这是图像类题目中的常见考点。

Always present your working clearly, show each line of algebra, and state your final answers in exact form unless told otherwise. Regular practice with different types of coefficients is the surest way to earn full marks.

始终清晰展示计算过程，写出每一步代数变形，并在没有特殊要求时以精确形式给出最终答案。对不同类型的系数进行规律练习，是获得满分的最可靠方法。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Completing the Square for Quadratic Equations: Mastering Key Steps | 配方法解二次方程：掌握关键步骤