📚 Quadratic Equations: Factorising, 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 one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, geometry and real-life modelling. This article explains the standard form, shows three main solving methods, and gives you exam-style practice.
二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、图像、几何和现实生活建模中。本文解释标准形式，展示三种主要解法，并提供考试风格的练习。
1. What Is a Quadratic Equation? | 什么是二次方程？
m ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of x is 2.
二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b 和 c 是常数且 a ≠ 0。x 的最高次数是 2。
If a = 0, the equation becomes linear, not quadratic. For example, x² – 5x + 6 = 0 is quadratic, but 3x + 1 = 0 is linear.
如果 a = 0，方程就变成一次方程，而不是二次方程。例如，x² – 5x + 6 = 0 是二次方程，但 3x + 1 = 0 是一次方程。
2. Standard Form and Key Vocabulary | 标准形式与关键术语
the equation in standard form: ax² + bx + c = 0. Move all terms to one side and simplify.
在求解之前，始终将方程写成标准形式：ax² + bx + c = 0。将所有项移到一边并化简。
ax² + bx + c = 0, a ≠ 0
- a: coefficient of x² — a 是 x² 的系数
- b: coefficient of x — b 是 x 的系数
- c: constant term — c 是常数项
For example, in 4x² – 7x + 2 = 0, we have a = 4, b = –7 and c = 2.
例如，在 4x² – 7x + 2 = 0 中，a = 4，b = –7，c = 2。
3. Solving by Factorising | 因式分解法
Factorising is usually the quickest method when the quadratic expression can be written as a product of two brackets. Set each bracket equal to zero and solve.
当二次式可以写成两个括号的乘积时，因式分解通常是最快的方法。令每个括号等于零并解方程。
For example, solve x² – x – 6 = 0. Find two numbers that multiply to –6 and add to –1: these are –3 and 2. So x² – x – 6 = (x – 3)(x + 2) = 0.
例如，解 x² – x – 6 = 0。找到两个数，它们的乘积为 –6，和为 –1：这两个数是 –3 和 2。因此 x² – x – 6 = (x – 3)(x + 2) = 0。
Then x – 3 = 0 gives x = 3, and x + 2 = 0 gives x = –2. Always check by substituting back.
然后 x – 3 = 0 得 x = 3，x + 2 = 0 得 x = –2。务必代回检验。
When a ≠ 1, try to factorise into (px + q)(rx + s). For 2x² + 5x – 3, factors are (2x – 1)(x + 3).
当 a ≠ 1 时，尝试分解为 (px + q)(rx + s)。对于 2x² + 5x – 3，因式分解为 (2x – 1)(x + 3)。
4. Solving by Completing the Square | 配方法
Completing the square transforms a quadratic expression into the form (x + p)² + q. This method is especially useful for finding turning points of graphs.
配方法将二次式转化为 (x + p)² + q 的形式。该方法对于求图像的顶点尤其有用。
For x² + bx, add and subtract (b/2)²:
对于 x² + bx，加上并减去 (b/2)²：
x² + bx = (x + b/2)² – (b/2)²
Example: solve x² + 6x + 2 = 0. First write x² + 6x = –2. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 7, so (x + 3)² = 7.
例子：解 x² + 6x + 2 = 0。先写成 x² + 6x = –2。两边加上 (6/2)² = 9：x² + 6x + 9 = 7，所以 (x + 3)² = 7。
Then x + 3 = ±√7, giving x = –3 ± √7.
然后 x + 3 = ±√7，得到 x = –3 ± √7。
5. The Quadratic Formula | 求根公式
equation ax² + bx + c = 0. It is derived by completing the square and should be memorised.
求根公式可以解任何二次方程 ax² + bx + c = 0。它由配方法推导而来，应当熟记。
x = [-b ± √b² – 4ac] / (2a)
Example: solve 3x² – 4x – 2 = 0. Here a = 3, b = –4, c = –2. Substitute into the formula.
例子：解 3x² – 4x – 2 = 0。此处 a = 3，b = –4，c = –2。代入公式。
x = [–(–4) ± √((–4)² – 4 × 3 × (–2))] / (2 × 3) = [4 ± √16 + 24] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = [2 ± √10] / 3.
x = [–(–4) ± √((–4)² – 4 × 3 × (–2))] / (2 × 3) = [4 ± √16 + 24] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = [2 ± √10] / 3。
6. The Discriminant and Nature of Roots | 判别式与根的性质
The expression inside the square root, b² – 4ac, is called the discriminant. It tells you how many real roots the equation has without solving it.
平方根内的表达式 b² – 4ac 称为判别式。它告诉你方程有多少个实数根，而无需实际求解。
Δ = b² – 4
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