Solving Quadratic Equations | 二次方程：因式分解、配方法与求根公式

📚 Solving Quadratic Equations | 二次方程：因式分解、配方法与求根公式

A quadratic equation is a polynomial equation of degree 2. In the IGCSE Edexcel syllabus, solving quadratics is essential for algebra, graphs, and real-life problems. This article covers the three main methods, the discriminant, and common pitfalls.

二次方程是次数为 2 的多项式方程。在 IGCSE Edexcel 数学考纲中，解二次方程是代数、函数图像和实际应用题的基础。本文介绍三种主要解法、判别式以及常见易错点。

1. Standard Form of a Quadratic Equation | 二次方程的标准形式

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

二次方程可以写成 ax² + bx + c = 0 的形式，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。

For example, 2x² – 3x + 1 = 0 has a = 2, b = -3, c = 1. If a = 0, the equation becomes linear, so a must not be zero for a quadratic.

例如，2x² – 3x + 1 = 0 中 a = 2，b = -3，c = 1。如果 a = 0，方程就变成了一次方程，所以二次方程中 a 不能为 0。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer factors. We look for two numbers that multiply to ac and add to b, then split the middle term and factor by grouping.

当二次方程含有简单的整数因式时，因式分解是最快捷的方法。我们需要找两个数，它们的乘积等于 ac，和等于 b，然后拆项并分组分解。

For x² + 5x + 6 = 0, we need two numbers that multiply to 6 and add to 5. The numbers are 2 and 3, so (x + 2)(x + 3) = 0. Therefore x = -2 or x = -3.

对于 x² + 5x + 6 = 0，我们需要两个数，乘积为 6，和为 5。这两个数是 2 和 3，所以 (x + 2)(x + 3) = 0。因此 x = -2 或 x = -3。

For harder cases like 2x² + 7x + 3 = 0, multiply a and c: 2 × 3 = 6. Find two numbers with product 6 and sum 7: 1 and 6. Rewrite: 2x² + 1x + 6x + 3 = 0. Group: x(2x + 1) + 3(2x + 1) = 0, so (2x + 1)(x + 3) = 0. Hence x = -½ or x = -3.

对于较复杂的方程如 2x² + 7x + 3 = 0，先算 a × c：2 × 3 = 6。找两个数，乘积为 6，和为 7：1 和 6。改写：2x² + 1x + 6x + 3 = 0。分组：x(2x + 1) + 3(2x + 1) = 0，所以 (2x + 1)(x + 3) = 0。因此 x = -½ 或 x = -3。

3. Difference of Two Squares | 平方差公式

A quadratic of the form x² – a² factorises as (x + a)(x – a). This is called the difference of two squares.

形如 x² – a² 的二次式可以分解为 (x + a)(x – a)，这叫做平方差公式。

For example, x² – 9 = 0 becomes (x + 3)(x – 3) = 0, giving x = 3 or x = -3. Similarly, 4x² – 25 = 0 can be written as (2x)² – 5² = 0, so (2x + 5)(2x – 5) = 0, giving x = ±5/2.

例如，x² – 9 = 0 变为 (x + 3)(x – 3) = 0，得到 x = 3 或 x = -3。类似地，4x² – 25 = 0 可写成 (2x)² – 5² = 0，所以 (2x + 5)(2x – 5) = 0，得到 x = ±5/2。

4. Solving by Completing the Square | 配方法

Completing the square rewrites ax² + bx + c = 0 in the form a(x + p)² + q = 0. This technique is useful when factorisation is difficult and it also helps find the turning point of a quadratic graph.

配方法将 ax² + bx + c = 0 改写为 a(x + p)² + q = 0 的形式。当因式分解困难时，这种方法很有用，而且能帮助求二次函数图像的顶点。

For x² + 6x + 2 = 0, take half of 6, which is 3. Write (x + 3)² – 9 + 2 = 0, so (x + 3)² – 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7, giving x = -3 ± √7.

对于 x² + 6x + 2 = 0，取 6 的一半，即 3。写成 (x + 3)² – 9 + 2 = 0，所以 (x + 3)² – 7 = 0。于是 (x + 3)² = 7，因此 x + 3 = ±√7，即 x = -3 ± √7。

When the coefficient of x² is not 1, for example 2x² + 8x + 5 = 0, first factor out 2: 2(x² + 4x) + 5 = 0. Complete the square for x² + 4x: (x + 2)² – 4. Thus

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