Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are a core topic in the IGCSE Mathematics syllabus. In this article, we will explore every method you need to solve them: factorisation, completing the square, the quadratic formula and graphical solutions. Each method is explained step by step with worked examples and exam tips.

二次方程是IGCSE数学教学大纲中的核心主题。在本文中，我们将探讨解二次方程所需的每一种方法：因式分解法、配方法、求根公式和图像法。每种方法都将结合例题和考试技巧逐步讲解。

1. What is a Quadratic Equation? | 什么是二次方程

A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The term ax² is called the quadratic term, bx is the linear term and c is the constant term.

二次方程是任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为实数，且 a ≠ 0。ax² 项称为二次项，bx 项称为一次项，c 项称为常数项。

For example, x² − 5x + 6 = 0 is a quadratic equation, while x² + 2x − 1 = 3 is also quadratic because it can be rearranged to x² + 2x − 4 = 0.

例如，x² − 5x + 6 = 0 是二次方程；而 x² + 2x − 1 = 3 也是二次方程，因为它可以整理为 x² + 2x − 4 = 0。

Crucially, a quadratic equation can have at most two solutions (roots). These roots may be real or non-real, distinct or equal.

关键的是，二次方程最多有两个解（根）。这些根可以是实数也可以是非实数，可以是相异的也可以是相等的。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest way to solve a quadratic when the expression factorises neatly. Always rearrange the equation so that one side is zero first.

当二次表达式可以整齐地因式分解时，因式分解法往往是最快的求解方法。请务必先将方程整理为一边为零的形式。

Worked example: Solve x² − 5x + 6 = 0.

例题：解方程 x² − 5x + 6 = 0。

Find two numbers whose product is 6 and whose sum is −5. These numbers are −2 and −3, so:

找到两个数，使其乘积为 6，和为 −5。这两个数分别是 −2 和 −3，因此：

(x − 2)(x − 3) = 0

Using the zero product property, if the product of two factors is zero, then at least one factor must be zero:

根据零乘积性质，若两个因子的乘积为零，则至少有一个因子为零：

x − 2 = 0 or x − 3 = 0

x = 2 or x = 3

The solutions are x = 2 and x = 3. Always expand your brackets to check the factorisation is correct.

方程的解为 x = 2 和 x = 3。请务必展开括号，以检验因式分解是否正确。

3. Special Factorisation Cases | 特殊因式分解情况

Two special forms appear frequently in IGCSE exams. The first is the difference of two squares: a² − b² = (a − b)(a + b).

在IGCSE考试中，有两种特殊形式经常出现。第一种是平方差公式：a² − b² = (a − b)(a + b)。

For example, x² − 9 = 0 can be written as (x − 3)(x + 3) = 0, giving x = 3 or x = −3.

例如，x² − 9 = 0 可以写成 (x − 3)(x + 3) = 0，得到 x = 3 或 x = −3。

The second special case is a perfect square trinomial: a² ± 2ab + b² = (a ± b)².

第二种特殊情况是完全平方三项式：a² ± 2ab + b² = (a ± b)²。

For example, x² − 6x + 9 = 0 becomes (x − 3)² = 0, so x = 3 is a repeated root (a single solution).

例如，x² − 6x + 9 = 0 可转化为 (x − 3)² = 0，因此 x = 3 是重根（即只有一个解）。

4. Solving by Completing the Square | 配方法

Completing the square rewrites x² + bx + c in the form (x + p)² + q. This method is essential when the quadratic cannot be factorised easily.

配方法将 x² + bx + c

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