📚 Solving Quadratic Equations: Factorising, Completing the Square and the Formula | 解二次方程：因式分解、配方法与求根公式
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, problem solving and even coordinate geometry. This guide covers every method you need to solve a quadratic equation, with worked examples and exam tips for Cambridge IGCSE and other major boards.
二次方程是 IGCSE 数学中最重要的话题之一。它出现在代数、图像、应用题乃至坐标几何中。本指南覆盖解二次方程所需的全部方法，并配有典型例题和适合各大考试局的考场技巧。
1. The General Form | 一般形式
A quadratic equation is any equation that can be written in the general form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The value of a is never zero, because if a = 0 the equation becomes linear and no longer contains an x² term.
二次方程是指可以写成一般形式 ax² + bx + c = 0 的方程，其中 a、b、c 是实数且 a ≠ 0。a 的值绝不能为零，因为如果 a = 0，方程就变成一次方程，不再含有 x² 项。
The coefficient a also controls the shape of the graph. If a > 0, the parabola opens upwards and looks like a ‘U’. If a < 0, it opens downwards and looks like an inverted 'U'.
系数 a 还决定图像的开口方向。当 a > 0 时，抛物线开口朝上，形如字母 U；当 a < 0 时，抛物线开口朝下，形如倒置的 U。
ax² + bx + c = 0, a ≠ 0
2. Expanding Double Brackets | 展开双括号
Before you can factorise, you must be able to expand two linear brackets. Use the FOIL rule: First, Outer, Inner, Last. For example, expand (x + 5)(x – 2):
在学习因式分解之前，你必须先会展开两个一次括号。使用 FOIL 法则：First（首项）、Outer（外项）、Inner（内项）、Last（末项）。例如展开 (x + 5)(x – 2)：
- First: x × x = x²
- Outer: x × (-2) = -2x
- Inner: 5 × x = 5x
- Last: 5 × (-2) = -10
- 首项：x × x = x²
- 外项：x × (-2) = -2x
- 内项：5 × x = 5x
- 末项：5 × (-2) = -10
Now combine the like terms: -2x + 5x = 3x. So (x + 5)(x – 2) = x² + 3x – 10. The general result is worth memorising: (x + p)(x + q) = x² + (p + q)x + pq.
现在合并同类项：-2x + 5x = 3x。所以 (x + 5)(x – 2) = x² + 3x – 10。一个值得记忆的通用结论是：(x + p)(x + q) = x² + (p + q)x + pq。
3. Factorising When a = 1 | a = 1 时的因式分解
Factorising is the reverse of expanding. When the coefficient of x² is 1, we look for two numbers m and n such that m + n = b and m × n = c. Then the factorised form is (x + m)(x + n).
因式分解是展开的逆运算。当 x² 的系数为 1 时，我们要找两个数 m 和 n，使得 m + n = b 且 m × n = c。那么分解结果为 (x + m)(x + n)。
Example: factorise x² – 7x + 12. We need two numbers whose product is 12 and whose sum is -7. Since (-3) × (-4) = 12 and (-3) + (-4) = -7, the answer is (x – 3)(x – 4). Always expand to check your result.
例如：分解 x² – 7x + 12。我们需要两个数，其乘积为 12，和为 -7。因为 (-3) × (-4) = 12，且 (-3) + (-4) = -7，所以答案为 (x – 3)(x – 4)。务必展开检验你的结果。
x² + bx + c = (x + m)(x + n), where m + n = b and mn = c
4. Special Products and Difference of Two Squares | 特殊因式与平方差
Two special patterns appear frequently in exams. The first is the difference of two squares: a² – b² = (a – b)(a + b). For example, x² – 81 = (x – 9)(x + 9). The second is a perfect square: a² ± 2ab + b² = (a ± b)². For example, x² – 10x + 25 = (x – 5)².
有两种常见于考试的特殊模式。第一种是平方差公式：a² – b² = (a – b)(a + b)。例如，x² – 81 = (x – 9)(x + 9)。第二种是完全平方：a² ± 2ab + b² = (a ± b)²。例如，x² – 10x + 25 = (x – 5)²。
a² – b² = (a – b)(a + b)
a² ± 2ab + b² = (a ± b)²
Watch out for expressions such as x² – 16, 4x² – 9 or 25x² – 1. Each of them is a difference of two squares: (x – 4)(x + 4), (2x – 3)(2x + 3) and (5x – 1)(5x + 1) respectively.
注意形如 x² – 16、4x² – 9 或 25x² – 1 的表达式，它们都是平方差：分别分解为 (x – 4)(x + 4)、(2x – 3)(2x + 3) 和 (5x – 1)(5x + 1)。
5. Factorising When a ≠ 1 | a ≠ 1 时的因式分解
When the coefficient of x² is not 1, use the product-sum method with grouping. Take the quadratic 3x² + 10x + 8 as an example.
当 x² 的系数不为 1 时，使用“乘积-和”配合分组的方法。以二次式 3x² + 10x + 8 为例。
- Multiply a by c: 3 × 8 = 24.
- Find two factors of 24 whose sum is b = 10: 4 and 6.
- Split the middle term:
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