📚 Mastering Quadratic Equations for IGCSE Maths | IGCSE 数学：掌握二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, and real-world modelling, so a confident understanding of how to solve them is essential for a high grade. This article covers the standard form, factorising, the quadratic formula, completing the square, the discriminant, graphs and exam strategy.
二次方程是 IGCSE 数学最重要的专题之一。它们出现在代数、坐标几何和现实建模中，因此熟练掌握其解法是取得高分的关键。本文涵盖标准式、因式分解、求根公式、配方法、判别式、图像以及考试策略。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation that can be arranged into the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The condition a ≠ 0 is essential because if a were 0, the x² term would disappear and the equation would become linear. The graph of a quadratic equation is a parabola, which may open upwards or downwards depending on the sign of a.
二次方程是可以整理为 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。a ≠ 0 这个条件至关重要，因为如果 a = 0，x² 项会消失，方程就变成一次方程。二次方程的图像是一条抛物线，根据 a 的符号可能开口向上或向下。
Examples include 2x² + 3x − 5 = 0, x² − 6x + 9 = 0 and 4x² − 25 = 0. Some equations may need rearranging before they show this standard form, such as x² = 7x − 10, which is equivalent to x² − 7x + 10 = 0.
例如 2x² + 3x − 5 = 0、x² − 6x + 9 = 0 和 4x² − 25 = 0。有些方程需要先整理才能化为标准式，比如 x² = 7x − 10 等价于 x² − 7x + 10 = 0。
2. Standard Form and Identifying a, b, c | 标准式与识别 a、b、c
Before solving a quadratic equation, you should always write it in standard form ax² + bx + c = 0. This means collecting all terms on one side of the equation, usually with the x² term first, then the x term, and finally the constant. The coefficient of x² is called a, the coefficient of x is called b, and the constant term is called c.
在解二次方程之前，你应当始终将其写成标准式 ax² + bx + c = 0。这意味着将所有项移到方程的一边，通常先写 x² 项，再写 x 项，最后是常数项。x² 的系数称为 a，x 的系数称为 b，常数项称为 c。
For example, in 3x² + 5x − 2 = 0, we have a = 3, b = 5 and c = −2. If an equation is written as x² + 6x = 0, then a = 1, b = 6 and c = 0. If it is written as 3x² − 7 = 0, then a = 3, b = 0 and c = −7. Missing terms simply mean that the corresponding coefficient is zero.
例如，在 3x² + 5x − 2 = 0 中，a = 3，b = 5，c = −2。如果方程写成 x² + 6x = 0，那么 a = 1，b = 6，c = 0。如果写成 3x² − 7 = 0，那么 a = 3，b = 0，c = −7。缺少某一项就意味着对应的系数为零。
| Equation | a | b | c |
|---|---|---|---|
| 2x² − 4x + 1 = 0 | 2 | −4 | 1 |
| x² + 6x = 0 | 1 | 6 | 0 |
| 3x² − 7 = 0 | 3 | 0 | −7 |
Identifying a, b and c correctly is especially important when you use the quadratic formula or the discriminant.
正确识别 a、b 和 c 在使用求根公式或判别式时尤为重要。
3. Solving by Factorising | 因式分解法求解
Factorising is often the fastest method when the quadratic has simple integer roots. Write the equation in standard form, factorise the left-hand side into two brackets, set each bracket equal to zero, and solve the resulting linear equations.
当二次方程有简单的整数根时，因式分解通常是最快的方法。先将方程写成标准式，将左边分解为两个括号的乘积，令每个括号等于零，再解所得的一次方程。
For example, solve x² + 7x + 12 = 0. We need two numbers whose product is 12 and whose sum is 7. The numbers are 3 and 4, so x² + 7x + 12 = (x + 3)(x + 4). Therefore (x + 3)(x + 4) = 0, giving x = −3 or x = −4.
例如，解 x² + 7x + 12 = 0。我们需要两个数，它们的乘积为 12，和为 7。这两个数是 3 和 4，因此 x² + 7x + 12 = (x + 3)(x + 4)。所以 (x + 3)(x + 4) = 0，解得 x = −3 或 x = −4。
When a ≠ 1, the factorisation is slightly harder. Solve 2x² − 5x − 3 = 0. We look for two numbers that multiply to 2 × (−3) = −6 and add to −5. The numbers are −6 and 1. Split the middle term: 2x² − 6x + x − 3 = 0, then factorise by grouping: 2x(x − 3) + 1(x − 3) = 0, so (2x + 1)(x − 3) = 0. This gives x = −1/2 or x = 3.
当 a ≠ 1 时，因式分解会稍微复杂一些。解 2x² − 5x − 3 = 0。我们要找两个数，它们相乘等于 2 × (−3) = −6，相加等于 −5。这两个数是 −6 和 1。拆分中间项：2x² − 6x + x − 3 = 0，然后分组分解：2x(x − 3) + 1(x − 3) = 0，得到 (2x + 1)(x − 3) = 0。由此解得 x = −1/2 或 x = 3。
Always expand your brackets to check the factorisation before writing the final answer. Not every quadratic factorises nicely; in that case use the quadratic formula or completing the square.
在写最终答案之前，务必展开括号检查因式分解是否正确。并非每个二次方程都能很好地因式分解；此时可以使用求根公式或配方法。
4. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation in standard form. It is especially useful when factorising is difficult or when the roots are not rational. The formula is:
求根公式可以解任何标准形式的二次方程。当因式分解困难或根不是有理数时，它尤其有用。公式为：
x = (−b ± √(b² − 4ac)) ÷ (2a)
To use the formula, identify a, b and c, substitute them carefully, and simplify. For example, solve 3x² − 7x + 2 = 0. Here a = 3, b = −7 and c = 2. The discriminant is b² − 4ac = (−7)² − 4 × 3 × 2 = 49 − 24 = 25. Then x = (7 ± √25) ÷ 6 = (7 ± 5) ÷ 6, giving x = 12/6 = 2 or x = 2/6 = 1/3.
使用公式时，先确定 a、b、c，仔细代入，然后化简。例如，解 3x² − 7x + 2 = 0。这里 a = 3，b = −7，c = 2。判别式为 b² − 4ac = (−7)² − 4 × 3 × 2 = 49 − 24 = 25。于是 x = (7 ± √25) ÷ 6 = (7 ± 5) ÷ 6，得到 x = 12/6 = 2 或 x = 2/6 = 1/3。
If the discriminant is not a perfect square, leave your answer in surd form unless the question asks for a decimal approximation. For example, x² − 4x − 1 = 0 gives x = (4 ± √20) ÷ 2 = (4 ± 2√5) ÷ 2 = 2 ± √5.
如果判别式不是完全平方数，除非题目要求化为小数，否则答案保留根式形式。例如，x² − 4x − 1 = 0 得到 x = (4 ± √20) ÷ 2 = (4 ± 2√5) ÷ 2 = 2 ± √5。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This is useful for solving equations when factorising is not straightforward, and it also helps to find the vertex of a parabola.
配方法将二次式改写为 a(x + p)² + q 的形式。当因式分解不明显时，这种方法很有用，而且它也有助于求抛物线的顶点。
For a monic quadratic x² + bx + c, you can write x² + bx as (x + b/2)² − (b/2)². For example, to solve x² + 8x + 5 = 0, first move the constant: x² +
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