IGCSE Maths: Quadratic Equations - Factorising, Completing the Square and the Quadratic Formula | IGCSE 数学：二次方程 - 因式分解、配方法与求根公式

📚 IGCSE Maths: Quadratic Equations - Factorising, Completing the Square and the Quadratic Formula | IGCSE 数学：二次方程 - 因式分解、配方法与求根公式

Quadratic equations sit at the heart of the IGCSE Mathematics syllabus. In the Core tier they appear as straightforward factorising questions; in the Extended tier they grow into completing the square, the quadratic formula, the discriminant and the analysis of quadratic graphs. Whatever tier you are sitting, the same small set of tools solves almost every quadratic you will ever meet, and examiners reward clear, structured working far more than a lucky final answer. This guide works through every method in turn, keeps the algebra honest, and finishes with exam-style questions and a revision checklist you can use the night before the paper.

二次方程是 IGCSE 数学课程的核心内容之一。在核心层（Core），它表现为直接的因式分解题；在扩展层（Extended），则延伸为配方法、求根公式、判别式以及二次函数图像的分析。无论你考哪一层，同一套工具几乎可以解决你遇到的所有二次方程，而考官真正给分的是清晰、有条理的解题过程，而不是碰巧正确的最终答案。本文逐一讲解每种方法，保证代数推导严谨，并在最后附上考试型例题与考前复习清单。

1. What is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are real constants and a ≠ 0. The highest power of x is 2, so a quadratic equation can have up to two real solutions, called roots.

二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是实数常数且 a ≠ 0。x 的最高次数是 2，因此二次方程最多可以有两个实数解，称为根。

ax² + bx + c = 0, a ≠ 0

− 4 = 0 are both quadratic equations, but 4x + 7 = 0 is linear because it has no x² term.

左边的表达式 ax² + bx + c 称为二次式。例如 2x² − 5x + 3 = 0 和 x² − 4 = 0 都是二次方程，而 4x + 7 = 0 是一次方程，因为它没有 x² 项。

b and c, however, may be zero, which is why x² − 4 = 0 and x² − 5x = 0 are still perfectly good quadratics.

a ≠ 0 这一条件非常重要。如果 a 等于 0，x² 项就消失了，方程会退化为只有一个解的一次方程。而 b 和 c 可以为零，所以 x² − 4 = 0 和 x² − 5x = 0 依然是标准的二次方程。

on true. On a graph, those same values are the x-intercepts of the curve y = ax² + bx + c.

不同考试局对同一概念使用的术语略有不同：解（solutions）、根（roots）、零点（zeros）指的都是使方程成立的 x 值。在图像上，这些值就是曲线 y = ax² + bx + c 与 x 轴交点的横坐标。

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2. Solving by Factorising | 因式分解法求解

x² + 5x + 6 = (x + 2)(x + 3), so the equation x² + 5x + 6 = 0 becomes (x + 2)(x + 3) = 0.

当二次式可以写成两个一次因式的乘积时，我们就能快速解方程。例如 x² + 5x + 6 = (x + 2)(x + 3)，因此方程 x² + 5x + 6 = 0 就变成 (x + 2)(x + 3) = 0。

d 3 work because 2 × 3 = 6 and 2 + 3 = 5. The factorised form is therefore (x + 2)(x + 3).

要对 x² + bx + c 形式的二次式进行因式分解，需要找到两个数，它们的乘积等于 c，和等于 b。对于 x² + 5x + 6，2 和 3 符合条件，因为 2 × 3 = 6 且 2 + 3 = 5。因此因式分解形式为 (x + 2)(x + 3)。

The same rule handles negative numbers. For x² − 7x + 12 you need a product of +12 and a sum of −7, so both numbers must be negative: −3 and −4. For x² + 2x − 15 you need a product of −15 and a sum of +2, so the numbers have opposite signs: +5 and −3.

这条规则同样适用于负数。对于 x² − 7x + 12，需要乘积为 +12、和为 −7，因此两个数都必须是负数：−3 和 −4。对于 x² + 2x − 15，需要乘积为 −15、和为 +2，因此两数异号：+5 和 −3。

3, the factorised form is (2x + 1)(x + 3) because expanding gives 2x² + 6x + x + 3 = 2x² + 7x + 3.

如果 x² 的系数不是 1，则需要使用更系统的方法，例如拆分中项或合理的尝试法。对于 2x² + 7x + 3，因式分解形式为 (2x + 1)(x + 3)，因为展开后得到 2x² + 6x + x + 3 = 2x² + 7x + 3。

to x + 6x, giving 2x² + x + 6x + 3, then factorise in pairs: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).

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对 2x² + 7x + 3 使用拆分中项法：先算 a × c = 2 × 3 = 6，再找乘积为 6、和为 7 的两个数，即 1 和 6。把 7x 拆成 x + 6x，得到 2x² + x + 6x + 3，然后分组提取公因式：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。

tic with no constant term gives x² − 5x = x(x − 5). Always look for a common factor first.

两种特殊结构能在考场上节省时间。平方差给出 x² − 4 = (x − 2)(x + 2)；没有常数项的二次式给出 x² − 5x = x(x − 5)。拿到题目时，永远先看有没有公因式。

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3. The Zero Product Property | 零乘积性质

e zero. From (x + 2)(x + 3) = 0, we set each factor equal to zero: x + 2 = 0 or x + 3 = 0.

因式分解之所以有效，是因为零乘积性质：如果两个表达式的乘积为零，那么其中至少有一个必须为零。由 (x + 2)(x + 3) = 0，我们令每个因式等于零：x + 2 = 0 或 x + 3 = 0。

d practice to substitute each answer back into the original equation to confirm the result, because a sign slip in the factorising step is easy to make and easy to catch.

解这两个一次方程就得到根 x = −2 和 x = −3。写出答案后最好把每个根代回原方程检验，因为因式分解时很容易出现符号错误，而检验能迅速发现问题。

range to (x + 2)(x + 3) − 6 = 0, that is x² + 5x = 0, and only then factorise: x(x + 5) = 0, so x = 0 or x = −5.

关键条件是乘积必须等于零。如果题目给出 (x + 2)(x + 3) = 6，绝不能把因式直接等于 6。必须先展开、整理成 (x + 2)(x + 3) − 6 = 0，即 x² + 5x = 0，这时才能因式分解：x(x + 5) = 0，所以 x = 0 或 x = −5。

Write the solutions clearly, and remember that a quadratic equation can also have one repeated root or no real roots depending on the discriminant, which is discussed later in Section 6.

解答要写清楚，同时要记住：根据判别式的不同，二次方程也可能只有一个重根，或者没有实数根，这将在第 6 节讨论。

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4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. It always works, even when the expression refuses to factorise with integers, and it reveals the vertex of the graph directly.

配方法把二次式改写成 (x + p)² + q 的形式。它总是可行，即使该二次式无法用整数分解，而且能直接揭示图像的顶点。

x² + bx + c = (x + b ÷ 2)² − (b ÷ 2)² + c

The recipe is: take half of the coefficient of x, write it inside the bracket, then subtract the square of that half outside the bracket. Nothing has been added or lost, so the expression is unchanged in value.

口诀是：取 x 的系数的一半，写进括号里，再在括号外减去这个一半的平方。整个过程没有增加也没有减少任何量，所以表达式的值保持不变。

the expression is −8, reached when x = −3. The vertex of the curve is therefore (−3, −8).

例如 x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。由于平方永远非负，该表达式的最小值就是 −8，在 x = −3 时取到。因此曲线顶点为 (−3, −8)。

form equal to zero and undo the square: (x + 3)² − 8 = 0 gives (x + 3)² = 8, so x + 3 = ±√8 and x = −3 ± √8. In exact form this is x = −3 ± 2√2, which is approximately −0.17 or −5.83.

用配方法解方程时，把配方后的形式设为 0，再开平方：(x + 3)² − 8 = 0 得 (x + 3)² = 8，于是 x + 3 = ±√8，x = −3 ± √8。精确形式为 x = −3 ± 2√2，约等于 −0.17 或 −5.83。

When the coefficient of x² is not 1, take it out as a common factor first. For 2x

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