📚 Mastering Quadratic Equations for IGCSE | 掌握一元二次方程：IGCSE 数学核心攻略
Quadratic equations are one of the most heavily tested topics in the IGCSE Mathematics syllabus. They appear in algebra, coordinate geometry, graphs and problem-solving questions. This article provides a complete, exam-focused revision guide to quadratic equations, from factorisation to the quadratic formula, with bilingual explanations for every step.
一元二次方程是 IGCSE 数学考纲中的高频考点，广泛出现在代数、坐标几何、函数图象与应用题中。本文提供一套完整、紧扣考点的复习指南：从因式分解到求根公式，每一步都配有中英双语讲解，帮助你系统掌握。
1. What Is a Quadratic Equation? | 什么是一元二次方程
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. Its standard form is ax² + bx + c = 0, where a, b and c are real constants and a ≠ 0. If a = 0, the equation becomes linear, not quadratic. The values of x that satisfy the equation are called roots or solutions.
一元二次方程是最高次数为 2 的整式方程，即变量的最高幂次为 2。其标准形式为 ax² + bx + c = 0，其中 a、b、c 为实数常数，且 a ≠ 0。若 a = 0，方程就退化为一次方程，不再是二次方程。满足方程的 x 值称为方程的根或解。
In the IGCSE examination, you must be able to solve quadratic equations using three main methods: factorisation, completing the square and the quadratic formula. You should also be able to sketch the graph of y = ax² + bx + c and interpret its turning point, intercepts and symmetry. This article covers all these skills in detail.
在 IGCSE 考试中，你必须掌握解一元二次方程的三种主要方法：因式分解法、配方法和求根公式法。同时，你还需要会画 y = ax² + bx + c 的图象，并能解读其顶点、交点与对称性。本文将对这些技能逐一详述。
2. The Standard Form and Coefficients | 标准形式与相关术语
Before solving any quadratic equation, always check that it is written in standard form: ax² + bx + c = 0. If it is not, expand brackets and collect all terms on one side so that the other side is zero. For example, (x + 3)(x − 2) = 4 must first be expanded and rearranged.
求解任何二次方程之前，务必先检查它是否写成标准形式 ax² + bx + c = 0。若未写成标准形式，要先展开括号、移项合并，使方程右边为零。例如，(x + 3)(x − 2) = 4 必须先展开并整理。
The table below shows how to identify the coefficients a, b and c from different-looking equations. When a term is missing, its coefficient is zero.
下表展示了如何从不同形式的方程中提取系数 a、b、c。注意：当某项缺失时，其对应的系数为 0。
| Equation | a | b | c |
| 2x² + 3x − 5 = 0 | 2 | 3 | −5 |
| x² − 4x = 0 | 1 | −4 | 0 |
| 3x² = 12 | 3 | 0 | −12 |
In the last example, rewriting 3x² = 12 as 3x² − 12 = 0 shows that b = 0. This equation can be solved quickly by taking square roots: x² = 4, so x = ±2.
在最后一个例子中，将 3x² = 12 改写为 3x² − 12 = 0，可知 b = 0。这类方程可直接开平方求解：x² = 4，所以 x = ±2。
3. Solving by Factorisation | 因式分解法
Factorisation is usually the fastest method when the equation has simple integer coefficients. The aim is to write ax² + bx + c as a product of two linear factors, such as (px + q)(rx + s). For example, x² + 5x + 6 = (x + 2)(x + 3).
当方程系数为简单整数时，因式分解法通常最快。目标是将 ax² + bx + c 写成两个一次因式的乘积，如 (x + 2)(x + 3)。例如 x² + 5x + 6 = (x + 2)(x + 3)。
After factorising, apply the zero product property: if AB = 0, then A = 0 or B = 0. Consider the equation x² − 7x + 12 = 0. We need two numbers that multiply to 12 and add to −7; these numbers are −3 and −4. Hence (x − 3)(x − 4) = 0, so x = 3 or x = 4.
分解后，利用“零积性质”求解：若 AB = 0，则 A = 0 或 B = 0。例如 x² − 7x + 12 = 0，需要找两个数，使它们的乘积为 12、和为 −7，这两个数就是 −3 与 −4。因此 (x − 3)(x − 4) = 0，解得 x = 3 或 x = 4。
(x − 3)(x − 4) = 0 → x = 3 or x = 4
You should also recognise the difference of two squares: x² − 9 = (x − 3)(x + 3), and the perfect square: x² + 6x + 9 = (x + 3)². These special forms save valuable time in the exam.
还应掌握平方差公式：x² − 9 = (x − 3)(x + 3)，以及完全平方式：x² + 6x + 9 = (x + 3)²。这些特殊形式能在考试中为你节省宝贵时间。
4. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x − h)² + k. This is a powerful technique because it reveals the turning point directly and can solve any quadratic equation, including those that cannot be factorised.
配方法将二次式化为 a(x − h)² + k 的形式。该法功能强大，因为它能直接显示顶点坐标，并且可以求解任何一元二次方程，包括无法因式分解的方程。
Consider x² + 8x + 13 = 0. Half of the coefficient of x is 4, and 4² = 16. We therefore write (x + 4)² − 16 + 13 = 0, which simplifies to (x + 4)² − 3 = 0. Then (x + 4)² = 3, so x + 4 = ±√3 and x = −4 ± √3.
以 x² + 8x + 13 = 0 为例。一次项系数 8 的一半为 4，而 4² = 16。于是将方程改写为 (x + 4)² − 16 + 13 = 0，即 (x + 4)² − 3 = 0。由此 (x + 4)² = 3，所以 x + 4 = ±√3，故 x = −4 ± √3。
(x + 4)² − 3 = 0 → x = −4 ± √3
When a ≠ 1, first factor out a from the first two terms before completing the square. For example, 2x² + 4x − 1 = 2(x² + 2x) − 1 = 2[(x + 1)² − 1] − 1 = 2(x + 1)² − 3.
当 a ≠ 1 时，先从前两项提出 a，再进行配方。例如 2x² + 4x − 1 = 2(x² + 2x) − 1 = 2[(x + 1)² − 1] − 1 = 2(x + 1)² − 3。
5. The Quadratic Formula | 求根公式
The quadratic formula is a universal method that works for any quadratic equation. It is derived by completing the square on the general form ax² + bx + c = 0, and it is especially useful when factorisation is difficult or impossible.
求根公式是适用于一切一元二次方程的通用方法。它由对一般式 ax² + bx + c = 0 配方推导而来，在因式分解困难或无法分解时尤为有用。
x = (−b ± √(b² − 4ac)) / 2a
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