Completing the Square | 配方法

📚 Completing the Square | 配方法

Completing the square is a fundamental algebraic technique in IGCSE Mathematics. It rewrites a quadratic expression ax² + bx + c into the form a(x − h)² + k, revealing the turning point of a parabola and enabling the solution of any quadratic equation. This technique appears across all examination boards, from simple factorisation questions to complex graph-sketching problems.

配方法是IGCSE数学中的一项基础代数技巧。它把一个二次表达式 ax² + bx + c 改写为 a(x − h)² + k 的形式，从而揭示抛物线的顶点，并能够解出任何二次方程。这一技巧在各考试局的试卷中都会出现，从简单的因式分解题到复杂的函数图像作图题，应用非常广泛。

1. The Perfect Square Pattern | 完全平方模式

Before mastering the technique, you must recognise the pattern of a perfect square. When we expand (x + p)², we obtain:

在掌握配方法之前，你必须先认识完全平方的展开模式。当我们展开 (x + p)² 时，得到：

(x + p)² = x² + 2px + p²

The key observation is that the constant term p² is the square of half the coefficient of x, namely (2p ÷ 2)² = p². This relationship is the essence of completing the square.

关键的观察在于：常数项 p² 是 x 系数一半的平方，即 (2p ÷ 2)² = p²。这个关系正是配方法的核心本质。

For example, (x + 5)² = x² + 10x + 25. Notice that 25 = (10 ÷ 2)². Any quadratic of the form x² + bx + c can be rewritten using this perfect-square identity by manipulating the constant term.

例如，(x + 5)² = x² + 10x + 25。注意 25 = (10 ÷ 2)²。任何形如 x² + bx + c 的二次表达式都可以通过调整常数项，利用这个完全平方恒等式进行改写。

2. Basic Technique: Coefficient of x² is 1 | 基本技巧：x² 的系数为 1

When the coefficient of x² is 1, completing the square follows a straightforward three-step procedure. Consider the expression x² + 6x + 5.

当 x² 的系数为 1 时，配方法遵循一个简单的三步流程。以表达式 x² + 6x + 5 为例。

• Step 1: Take half of the coefficient of x. Here, 6 ÷ 2 = 3.

  第一步：取 x 系数的一半。这里，6 ÷ 2 = 3。

• Step 2: Square this value: 3² = 9. Write the expression as x² + 6x + 9 − 9 + 5.

  第二步：将这个值平方：3² = 9。把表达式写成 x² + 6x + 9 − 9 + 5。

• Step 3: Factorise the first three terms: (x + 3)² − 9 + 5 = (x + 3)² − 4.

  第三步：将前三项因式分解：(x + 3)² − 9 + 5 = (x + 3)² − 4。

Thus, x² + 6x + 5 = (x + 3)² − 4. The +9 introduced by squaring 3 is immediately cancelled by the extra −9, preserving the original value of the expression.

因此，x² + 6x + 5 = (x + 3)² − 4。由 3 的平方产生的 +9 被额外添加的 −9 立即抵消，从而保持原表达式的值不变。

3. Solving Quadratic Equations | 解二次方程

Completing the square provides a reliable method for solving quadratic equations when factorisation is not obvious. Starting from the completed-square form, we isolate the squared term and take square roots.

当因式分解不明显时，配方法提供了求解二次方程的可靠方法。从配方的形式出发，我们分离出平方项，然后两边开平方。

Solve x² + 6x + 5 = 0 using the form found above:

利用上面得到的形式解方程 x² + 6x + 5 = 0：

(x + 3)² − 4 = 0

(x + 3)² = 4

x + 3 = ± 2

The symbol ± is crucial: the equation (x + 3)² = 4 has two solutions because both 2 and −2 square to give 4. Solving the two linear equations gives x = −1 and x = −5.

± 这个符号至关重要：方程 (x + 3)² = 4 有两个解，因为 2 和 −2 的平方都等于 4。分别解这两个一次方程，得到 x = −1 和 x = −5。

4. When a ≠ 1: Factoring Out the Leading Coefficient | 当 a ≠ 1 时：提取首项系数

If the coefficient a of x² is not 1, you must factorise it out of the x² and x terms before completing the square. Consider 2x² + 8x + 3.

如果 x² 的系数 a 不等于 1，你必须先把 a 从 x² 项和 x 项中提取出来，然后再配方。以 2x² + 8x + 3 为例。

2x² + 8x + 3 = 2(x² + 4x) + 3

Inside the bracket, the coefficient of x is 4. Half of

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