📚 Quadratic Equations | 二次方程
Quadratic equations are among the most frequently tested topics in IGCSE Mathematics. Whether you are sitting the Core or Extended paper, you will meet questions on solving, factorising, completing the square, and applying quadratics to real-life situations. This guide consolidates the essential knowledge, step-by-step methods, and exam strategies you need to secure full marks.
二次方程是IGCSE数学中最高频的考点之一。无论你报考核心（Core）还是扩展（Extended）试卷，都会遇到求解、因式分解、配方，以及将二次方程应用于实际情境的题目。本指南整合了核心知识、逐步方法和考试策略，助你稳拿满分。
1. The Standard Form | 标准形式
Every quadratic equation can be written in the standard form:
每一个二次方程都可以写成标准形式：
ax² + bx + c = 0, a ≠ 0
Here, a, b and c are real numbers, and the coefficient a cannot be zero; otherwise the equation would be linear. The highest power of x is 2, which gives the equation its name — “quadratic”.
其中 a、b、c 为实数，且系数 a 不能为零，否则方程就变成了一次方程。x 的最高次数为 2，这就是“二次”（quadratic）名称的来源。
For example, in 2x² – 3x + 5 = 0, we have a = 2, b = -3, c = 5. Identifying these values correctly is the essential first step before choosing a solution method.
例如，在 2x² – 3x + 5 = 0 中，a = 2，b = -3，c = 5。正确识别这些数值是选择求解方法前最关键的第一步。
2. Solving by Factorisation | 因式分解法求解
Factorisation is the fastest method when a quadratic has simple integer roots. The goal is to rewrite ax² + bx + c as a product of two linear expressions.
当二次方程具有简单的整数根时，因式分解是最快捷的方法。目标是将 ax² + bx + c 写成两个一次表达式的乘积。
Consider the equation x² – 5x + 6 = 0. We look for two numbers that multiply to give 6 and add to give -5. These numbers are -2 and -3, so we factorise as (x – 2)(x – 3) = 0.
考虑方程 x² – 5x + 6 = 0。我们寻找两个数，其乘积为 6、和为 -5。这两个数是 -2 和 -3，因此因式分解为 (x – 2)(x – 3) = 0。
Using the zero-product property, if the product of two expressions is zero, then at least one of them must be zero. Hence x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.
根据零乘积性质，若两个表达式的乘积为零，则至少有一个表达式为零。因此 x – 2 = 0 或 x – 3 = 0，解得 x = 2 或 x = 3。
When the coefficient of x² is not 1, such as in 2x² + 7x + 3 = 0, first find factor pairs of ac = 6 that add to b = 7: the pair is 1 and 6. Split the middle term: 2x² + 6x + x + 3, then factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 x² 的系数不为 1 时，例如 2x² + 7x + 3 = 0，先找出 ac = 6 的因数对，使其和为 b = 7：该因数对为 1 和 6。拆分中间项：2x² + 6x + x + 3，再分组因式分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
Always expand your result to check your work: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3. ✓
务必展开因式分解结果进行验算：(2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3。✓
3. Special Patterns | 特殊形式
Some quadratics have a common factor or follow a special product pattern. For example, 3x² – 12x = 0 can be solved by factoring out the common factor 3x: 3x(x – 4) = 0, giving x = 0 or x = 4.
有些二次方程含有公因式，或者符合特殊的乘积模式。例如，3x² – 12x = 0 可以先提取公因式 3x：3x(x – 4) = 0，得到 x = 0 或 x = 4。
The difference of two squares is another important pattern: x² – 9 = (x – 3)(x + 3) = 0, so x = 3 or x = -3. The general rule is:
平方差公式是另一个重要模式：x² – 9 = (x – 3)(x + 3) = 0，所以 x = 3 或 x = -3。一般规律为：
a² – b² = (a – b)(a + b)
A perfect square trinomial such as x² + 6x + 9 can be written as (x + 3)². Recognising these patterns quickly saves valuable time in the exam.
完全平方式如 x² + 6x + 9 可以写成 (x + 3)²。在考试中快速识别这些模式可以节省宝贵的时间。
4. The Quadratic Formula | 求根公式
When factorisation is difficult or impossible, use the quadratic formula. This formula works for every quadratic equation:
当因式分解困难或无法进行时，可以使用求根公式。该公式适用于一切二次方程：
x = (-b ± √(b² – 4ac)) / (2a)
Let us solve 3x² – 5x – 2 = 0. Here a = 3, b = -5, c = -2. Substitute into the formula:
让我们求解 3x² – 5x – 2 = 0。此时 a = 3，b = -5，c = -2。代入公式：
x = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6
This gives two solutions: x = (5 + 7)/6 = 2, and x = (5 – 7)/6 = -1/3.
由此得到两个解：x = (5 + 7)/6 = 2，以及 x = (5 – 7)/6 = -1/3。
In the IGCSE exam, always write down the substitution clearly. A correct substitution earns you method marks even if you make a small arithmetic error later.
在IGCSE考试中，务必清楚地写出代入过程。即使后续出现细微的计算失误，正确的代入也能为你赢得方法分。
5. The Discriminant | 判别式
The expression under the square root, b² – 4ac, is called the discriminant, denoted by Δ. It tells us how many real roots the equation has without solving it:
根号内的表达式 b² – 4ac 称为判别式，记作 Δ。无需解方程，它就能告诉我们方程实数根的个数：
Δ = b² – 4ac
-
If Δ > 0: two
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