Solving Quadratic Equations for IGCSE Mathematics | IGCSE 数学：解二次方程

📚 Solving Quadratic Equations for IGCSE Mathematics | IGCSE 数学：解二次方程

In IGCSE Mathematics, quadratic equations appear in both the Core and Extended tiers. A quadratic equation involves an unknown raised to the second power, and it can model many real-life situations such as projectile motion, area problems, and business profit. This article explains the main methods for solving quadratics and gives practical tips for exams.

在 IGCSE 数学中，二次方程在核心和扩展层级都会出现。二次方程包含一个未知数的二次方，它可以描述许多实际情境，例如抛体运动、面积问题和商业利润。本文解释解二次方程的主要方法，并提供实用考试技巧。

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

A quadratic equation is an equation of degree 2. In one variable x, its standard form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The condition a ≠ 0 is essential because if a = 0, the equation becomes linear. The highest power of x is 2, so a quadratic equation has at most two solutions, also called roots.

二次方程是次数为 2 的方程。对于变量 x，其标准形式为 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。a ≠ 0 这个条件很关键，因为如果 a = 0，方程就变成了一次方程。x 的最高次数是 2，因此二次方程最多有两个解，也称根。

Quadratic equations can be written in different forms, such as expanded form, factorised form, or completed-square form. For example, x² + 5x + 6 = 0 can also be written as (x + 2)(x + 3) = 0. Each form reveals different information about the roots and the graph.

二次方程可以写成不同形式，例如展开式、因式分解形式或配方形式。例如，x² + 5x + 6 = 0 也可以写成 (x + 2)(x + 3) = 0。每种形式都能揭示根和图像的不同信息。

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

Before solving, always rearrange the equation into the standard form ax² + bx + c = 0. Identify a, b and c carefully, including their signs. For instance, in -2x² + 3x – 7 = 0, we have a = -2, b = 3 and c = -7. Missing terms mean the coefficient is zero: in 4x² – 5 = 0, b = 0.

解题前，要先把方程整理成标准形式 ax² + bx + c = 0。仔细识别 a、b、c 及其符号。例如，在 -2x² + 3x – 7 = 0 中，a = -2，b = 3，c = -7。缺少项表示系数为零：在 4x² – 5 = 0 中，b = 0。

Rearranging often involves expanding brackets, collecting like terms, and moving all terms to one side. If a is negative, you may multiply both sides by -1 to make the coefficient positive, but this is not compulsory. The key is to keep the equation balanced.

整理方程通常包括展开括号、合并同类项，并把所有项移到一边。如果 a 是负数，可以将两边同时乘以 -1，使二次项系数变为正数，但这并不是必须的。关键是保持方程平衡。

3. Solving by Factorising | 因式分解法

Factorising is the fastest method when the quadratic factorises easily with integer roots. After writing ax² + bx + c = 0, find two numbers that multiply to give ac and add to give b. Then split the middle term, factorise by grouping, and set each factor equal to zero.

当二次方程能方便地用整数根因式分解时，因式分解法是最快的方法。写出 ax² + bx + c = 0 后，找到两个数，使它们的乘积为 ac，和为 b。然后拆项、分组分解，再让每个因式等于零。

Example: Solve x² + 7x + 12 = 0. We need two numbers with product 12 and sum 7. The numbers are 3 and 4, so (x + 3)(x + 4) = 0. Therefore x + 3 = 0 or x + 4 = 0, giving x = -3 or x = -4.

例子：解 x² + 7x + 12 = 0。我们需要两个数，乘积为 12，和为 7。这两个数是 3 和 4，所以 (x + 3)(x + 4) = 0。因此 x + 3 = 0 或 x + 4 = 0，得到 x = -3 或 x = -4。

When a ≠ 1, the process requires more care. For 2x² + 5x + 3 = 0, we need two numbers with product 2 × 3 = 6 and sum 5, namely 2 and 3. Split 5x into 2x + 3x, then factorise: 2x² + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1) = 0. The roots are x = -3/2 and x = -1.

当 a ≠ 1 时，需要更加小心。对于 2x² + 5x + 3 = 0，我们需要两个数乘积为 2 × 3 = 6，和为 5，即 2 和 3。将 5x 拆成 2x + 3x，再分解：2x² + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1) = 0。根为 x = -3/2 和 x = -1。

4. Solving by the Quadratic Formula | 二次公式法

The quadratic formula solves any quadratic equation, even when factorising is difficult or roots are irrational. For ax² + bx + c = 0, the solutions are given by:

二次公式可以解任何二次方程，即使因式分解困难或根是无理数。对于 ax² + bx + c = 0，解为：

x = (-b ± √(b² – 4ac)) / 2a

This formula comes from completing the square on the general quadratic. It is given in the IGCSE formula sheet, but you must practise substituting values accurately. Pay attention to negative signs, especially when b is negative or c is negative.

这个公式源于对一般二次方程进行配方。它在 IGCSE 公式表中给出，但你必须练习准确代入数值。注意负号，特别是当 b 为负或 c 为负时。

Example: Solve 3x² – 4x – 2 = 0. Here a = 3, b = -4, c = -2. Substituting: x = ( -(-4) ± √((-4)² – 4 × 3 × (-2)) ) / (2 × 3) = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3.

例子：解 3x² – 4x – 2 = 0。这里 a = 3，b = -4，c = -2。代入：x = ( -(-4) ± √((-4)² – 4 × 3 × (-2)) ) / (2 × 3) = (4 ± √(16 + 24)) /

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