Completing the Square for IGCSE Mathematics | IGCSE 数学配方法完全平方

📚 Completing the Square for IGCSE Mathematics | IGCSE 数学配方法完全平方

Completing the square is an algebraic technique that rewrites a quadratic expression in the form a(x + p)² + q. It is one of the most reliable tools in the IGCSE Mathematics syllabus because it works for every quadratic expression, even when factorising fails.

配方法是一种代数技巧，它把二次式改写成 a(x + p)² + q 的形式。它是 IGCSE 数学课程中最可靠的工具之一，因为即使因式分解失效，配方法也适用于所有二次式。

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1. What Is Completing the Square? | 什么是配方法？

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q, where p and q are constants. This form immediately reveals the turning point of the quadratic graph and makes solving equations more straightforward.

配方法把 ax² + bx + c 改写成 a(x + p)² + q 的形式，其中 p 和 q 是常数。这种形式能直接显示二次函数图像的转折点，也使解方程更加简便。

For example, x² + 6x + 5 can be written as (x + 3)² − 4. The first two terms have been converted into a perfect square, and the constant has been adjusted to keep the expression equal to the original.

例如，x² + 6x + 5 可以写成 (x + 3)² − 4。前两项被转换成了一个完全平方，而常数项经过调整，使表达式与原式保持相等。

In IGCSE questions, you may be asked to express a quadratic in completed square form, find the coordinates of the vertex, or solve an equation by leaving answers in surd form. All of these skills depend on this method.

在 IGCSE 考题中，你可能会被要求把二次式写成完全平方形式、求顶点坐标，或用根式形式解方程。这些技能都依赖于配方法。

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2. Why Do We Need It? | 为什么需要配方法？

Factorising is quick, but it only works when a quadratic has rational roots. Completing the square works for every quadratic, including those whose solutions involve surds or do not factorise neatly.

因式分解很快，但只有当二次方程有有理根时才适用。配方法对所有二次式都有效，包括解为根式或无法整齐因式分解的情况。

IGCSE exam papers often require exact solutions rather than rounded decimals. For example, solving x² + 6x + 4 = 0 gives x = −3 ± √5, which is best obtained by completing the square.

IGCSE 试卷通常要求准确解，而不是四舍五入的小数。例如，解 x² + 6x + 4 = 0 得到 x = −3 ± √5，用配方法最容易得到这个结果。

The method is also essential for finding the minimum or maximum value of a quadratic function. Once the expression is in the form a(x + p)² + q, the maximum or minimum value is simply q.

配方法对于求二次函数的最小值或最大值也很重要。一旦表达式变成 a(x + p)² + q 的形式，最大值或最小值就是 q。

• Find the minimum or maximum value | 求最小值或最大值
• Solve quadratic equations exactly | 准确解二次方程
• Sketch quadratic graphs accurately | 准确绘制二次函数图像
• Identify the line of symmetry | 确定对称轴

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3. The Standard Pattern | 标准模式

For the expression x² + bx, the completed square form is (x + b/2)² − (b/2)². This comes directly from expanding (x + p)² = x² + 2px + p² and comparing coefficients.

对于 x² + bx，完全平方形式是 (x + b/2)² − (b/2)²。这直接来自展开式 (x + p)² = x² + 2px + p²，并进行系数比较。

x² + bx = (x + b/2)² − (b/2)²

When b is positive, the squared bracket uses addition. When b is negative, the bracket uses subtraction. For example, x² − 6x = (x − 3)² − 9.

当 b 为正时，平方括号内用加法。当 b 为负时，括号内用减法。例如，x² − 6x = (x − 3)² − 9。

If b is a fraction, the same rule applies. For x² + 3x, half of 3 is 3/2, and (3/2)² is 9/4, so x² + 3x = (x + 3/2)² − 9/4.

如果 b 是分数，同样规则适用。对于 x² + 3x，3 的一半是 3/2，(3/2)² 是 9/4，所以 x² + 3x = (x + 3/2)² − 9/4。

The constant c can then be added to this result. In general, x² + bx + c becomes (x + b/2)² − (b/2)² + c.

常数项 c 只需要加到这个结果上。一般来说，x² + bx + c 变成 (x + b/2)² − (b/2)² + c。

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4. Steps When a = 1 | 当 a = 1 时的步骤

Follow a clear sequence when completing the square for x² + bx + c. First identify b, the coefficient of x, and ignore the constant c temporarily.

对 x² + bx + c 配方时，按清晰的顺序操作。先确定 x 的系数 b，暂时忽略常数项 c。

• Write the expression as x² + bx + c | 把式子写成 x² + bx + c
• Take half of b, then square it | 取 b 的一半，再平方
• Insert + (b/2)² − (b/2)² before c | 在 c 前插入 + (b/2)² − (b/2)²
• Factor the first three terms into a perfect square | 把前三项因式分解成完全平方
• Simplify the remaining constants | 化简剩余常数

Example: x² + 10x + 7. Half of 10 is 5, and 5² = 25. Rewrite as x² + 10x + 25 − 25 + 7. The first three terms form (x + 5)², so the result is (x + 5)² − 18.

例子：x² + 10x + 7。10 的一半是 5，5² = 25。改写为 x² + 10x + 25 − 25 + 7。前三项构成 (x + 5)²，所以结果是 (x + 5)² − 18。

Always check by expanding the final form. Expanding (x + 5)² − 18 gives x² + 10x + 25 − 18, which simplifies back to x² + 10x + 7.

一定要用展开来检验最终形式。展开 (x + 5)² − 18 得到 x² + 10x + 25 − 18，简化后正好是 x² + 10x + 7。

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5. Handling a Coefficient of x² Not Equal to 1 | 当 x² 的系数不等于 1 时

When the coefficient of x² is not 1, first factor that coefficient from the x² and x terms only. Complete the square inside the bracket, then multiply back carefully.

当 x² 的系数不等于 1 时，先只从 x² 项和 x 项中提出这个系数。在括号内配方，然后小心地乘回去。

Example: 2x² + 8x + 5. Factor 2 from the first two terms: 2(x² + 4x) + 5. For x² + 4x, half of 4 is 2, and 2² = 4, so x² + 4x = (x + 2)² − 4.

例子：2x² + 8x + 5。把前两项中的 2 提出来：2(x² + 4x) + 5。对于 x² + 4x，4 的一半是 2，2² = 4，所以 x² + 4x = (x + 2)² − 4。

Replace inside the bracket: 2[(x + 2)² − 4] + 5. Now multiply the 2 back: 2(x + 2)² − 8 + 5. Simplify to 2(x + 2)² − 3.

代入括号内：2[(x + 2)² − 4] + 5。现在把 2 乘回去：2(x + 2)² − 8 + 5。化简得到 2(x + 2)² − 3。

This form shows a = 2, p = 2, and q = −3. The coefficient a is not

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