Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式

📚 Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式

Quadratic equations are a core part of the IGCSE Mathematics syllabus. They appear in algebra, graph sketching, word problems and even in geometry when working with areas. Mastering three solution methods – factorisation, completing the square and the quadratic formula – will give you flexibility and confidence in the exam. This article explains each method, the discriminant and common pitfalls, with worked examples and clear bilingual notes.

二次方程是 IGCSE 数学课程的核心内容之一。它们出现在代数、函数图像、应用题甚至几何面积计算中。掌握因式分解、配方法和求根公式三种解法，能让你在考试中更加灵活和自信。本文将通过双语讲解、典型例题和常见错误分析，帮助你系统掌握这一考点。

1. Recognising a Quadratic Equation | 识别二次方程

A quadratic equation in one variable is an equation that can be rearranged into the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable is 2, which is why it is called quadratic.

一元二次方程是指可以整理成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。未知数的最高次数是 2，因此称之为二次方程。

For example, x² + 5x + 6 = 0, 3x² – 12 = 0 and 2x² = 7x – 3 are all quadratic equations once rearranged. However, x + 5 = 0 and x³ – 2x = 0 are not quadratic because their highest powers are 1 and 3 respectively.

例如，x² + 5x + 6 = 0、3x² – 12 = 0 和 2x² = 7x – 3 在整理后都是二次方程。然而 x + 5 = 0 和 x³ – 2x = 0 不是二次方程，因为它们的最高次数分别是 1 和 3。

2. The Standard Form and Coefficients | 标准形式与系数

It is useful to identify the coefficients a, b and c before solving. For the equation 2x² – 7x + 3 = 0, we have a = 2, b = -7 and c = 3. If the equation is not in standard form, move all terms to one side first.

在求解之前，先识别系数 a、b 和 c 是很有用的。对于方程 2x² – 7x + 3 = 0，我们有 a = 2、b = -7、c = 3。如果方程不是标准形式，应先将所有项移到一边。

Always remember that the sign belongs to the coefficient. In 5 – 3x – x² = 0, rewriting as -x² – 3x + 5 = 0 gives a = -1, b = -3 and c = 5.

始终记住符号属于系数。在 5 – 3x – x² = 0 中，改写为 -x² – 3x + 5 = 0，可得 a = -1、b = -3、c = 5。

3. Solving by Factorisation | 因式分解法求解

Factorisation is often the fastest method when the quadratic has simple integer factors. The idea is to write ax² + bx + c as a product of two linear brackets, then use the fact that if a product equals zero, at least one factor must be zero.

当二次方程有简单的整数因式时，因式分解法通常是最快的方法。思路是将 ax² + bx + c 写成两个一次括号的乘积，然后利用“乘积为零，则至少一个因式为零”的事实。

For example, x² + 5x + 6 = 0 can be written as (x + 2)(x + 3) = 0. Setting each bracket to zero gives x + 2 = 0 or x + 3 = 0, so x = -2 or x = -3.

例如，x² + 5x + 6 = 0 可以写成 (x + 2)(x + 3) = 0。令每个括号为零，得到 x + 2 = 0 或 x + 3 = 0，所以 x = -2 或 x = -3。

When a ≠ 1, such as 2x² – 7x + 3 = 0, we look for two numbers that multiply to a×c = 2×3 = 6 and add to b = -7. These numbers are -6 and -1, so we split the middle term: 2x² – 6x – x + 3 = 0, then factor by grouping: 2x(x – 3) – 1(x – 3) = 0, giving (2x – 1)(x – 3) = 0. Therefore x = 1/2 or x = 3.

当 a ≠ 1 时，例如 2x² – 7x + 3 = 0，我们寻找两个数，它们相乘等于 a×c = 2×3 = 6，相加等于 b = -7。这两个数是 -6 和 -1，因此我们将中间项拆分：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。因此 x = 1/2 或 x = 3。

4. The Zero Product Property | 零乘积性质

The zero product property states that if p × q = 0, then p = 0 or q = 0 (or both). This principle only works when one side of the equation is exactly zero, so always rearrange the equation before factorising.

零乘积性质指出，如果 p × q = 0，那么 p = 0 或 q = 0（或两者都为 0）。这一性质只有在方程一边恰好为零时才成立，因此在因式分解前一定要先整理方程。

A common mistake is to factorise an expression such as x² – 5x + 6 and then write x = 2 or x = 3 without setting the product equal to zero. Always remember the equation must be in the form (x – 2)(x – 3) = 0 first.

一个常见错误是，对 x² – 5x + 6 进行因式分解后，没有令乘积等于零就直接写出 x = 2 或 x = 3。请记住，方程必须先写成 (x – 2)(x – 3) = 0 的形式。

5. Completing the Square | 配方法

Completing the square transforms a quadratic into the form (x + p)² = q, which can then be solved by taking square roots. The key identity is x² + bx = (x + b/2)² – (b/2)².

配方法将二次方程转化为 (x + p)² = q 的形式，然后可以通过开平方求解。核心恒等式是 x² + bx = (x + b/2)² – (b/2)²。

For example, solve x² + 6x + 2 = 0. Write x² + 6x = -2. Half of 6 is 3, and 3² = 9, so add 9 to both sides: x² + 6x + 9 = -2 + 9, which gives (x + 3)² = 7. Taking square roots gives x + 3 = ±√7, so x = -3 ± √7.

例如，解 x² + 6x + 2 = 0。先写成 x² + 6x = -2。6 的一半是 3，而 3² = 9，因此两边同时加上 9：x² + 6x + 9 = -2 + 9，得到 (x + 3)² = 7。开平方得 x + 3 = ±√7，所以 x = -3 ± √7。

If the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 3 = 0, write 2(x² + 4x) = -3. Half of 4 is 2, and 2² = 4, so 2[(x + 2)² – 4] = -3, leading to 2(x + 2)² = 5 and finally (x + 2)² = 5/2.

如果 x² 的系数不是 1，需要先将其提出。对于 2x² + 8x + 3 = 0，先写成 2(x² + 4x) = -3。4 的一半是 2，而 2² = 4，所以 2[(x + 2)² – 4] = -3，化简得 2(x + 2)² = 5，最终 (x + 2)² = 5/2。

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