Solving Quadratic Equations | 一元二次方程解题全攻略

📚 Solving Quadratic Equations | 一元二次方程解题全攻略

Quadratic equations form the backbone of IGCSE Mathematics. They appear in Paper 1 and Paper 2, in algebra, geometry and problem-solving contexts, and they prepare students for advanced study in functions, inequalities and calculus. This guide breaks down every method you need, from factorisation to the quadratic formula, with exam-style worked examples and common pitfalls.

一元二次方程是 IGCSE 数学的基石。在试卷一和试卷二中都会出现，涉及代数、几何和应用题，也为今后学习函数、不等式和微积分打下基础。本指南将带你逐个掌握从因式分解到求根公式的每一种解法，并配有考试风格的例题与常见陷阱解析。

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1. The Standard Form | 标准形式

Every quadratic equation can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The coefficient a is attached to x², b to x, and c is the constant term.

任何一个一元二次方程都可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。a 是 x² 的系数，b 是 x 的系数，c 是常数项。

3. Notice that b and c can be negative; this is exactly where sign errors often creep in.

例如在 2x² − 5x + 3 = 0 中，a = 2，b = −5，c = 3。注意 b 和 c 可以是负数；符号错误往往就发生在这里。

quation is given as x² = 9 − 6x, rewrite it as x² + 6x − 9 = 0 before doing anything else.

在使用任何解法之前，必须先整理方程，使所有项在一边、另一边为 0。如果题目给出 x² = 9 − 6x，应先改写为 x² + 6x − 9 = 0，再进入下一步。

Equation 方程	a	b	c
3x² + 4x − 7 = 0	3	4	−7
x² − 9 = 0	1	0	−9
−x² + 5x = 0	−1	5	0

Exam tip: always write down a, b and c clearly before solving. This simple habit eliminates most substitution errors.

考试提示：解题前先清晰写出 a、b、c 的值。这个简单的习惯能避免大部分代入错误。

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2. Solving by Factorisation | 因式分解法

rst find two numbers that multiply to give 6 and add to give 5. Those numbers are 2 and 3.

如果二次三项式可以因式分解，因式分解通常是最快的方法。要解 x² + 5x + 6 = 0，首先找到两个数，相乘得 6 且相加得 5。这两个数就是 2 和 3。

Write the factorised form and apply the zero-product property: if the product of two expressions is zero, then at least one of them must be zero.

写出分解后的形式，并运用零积性质：若两个表达式的乘积为零，则其中至少一个必须为零。

(x + 2)(x + 3) = 0 → x + 2 = 0 or x + 3 = 0

(x + 2)(x + 3) = 0 → x + 2 = 0 或 x + 3 = 0

3. Always state both solutions; some marking schemes award one mark for each correct root.

因此 x = −2 或 x = −3。一定要写出两个解；部分评分标准中每个正确的根各得一分。

at multiply to 2 × 3 = 6 and add to 7; these are 6 and 1. Split the middle term and group:

当 a ≠ 1 时，使用“积和法”，即以 a × c 为积。对于 2x² + 7x + 3 = 0，找到两个数相乘得 2 × 3 = 6 且相加得 7；即 6 和 1。拆分中间项并分组：

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

So (2x + 1)(x + 3) = 0,

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