Mastering Quadratic Equations for IGCSE Mathematics | 掌握 IGCSE 数学中的二次方程

📚 Mastering Quadratic Equations for IGCSE Mathematics | 掌握 IGCSE 数学中的二次方程

Quadratic equations are a central topic in IGCSE Mathematics, appearing in algebra, graphs, and real-life problem solving. This revision guide explains the key methods, common pitfalls, and exam strategies you need to master them. Algebraic fluency with quadratics is not only tested directly but also underpins work on functions, inequalities, and coordinate geometry.

二次方程是 IGCSE 数学的核心主题，广泛出现在代数、图像和实际应用题中。本复习指南将讲解关键解法、常见错误以及考试策略，帮助你彻底掌握。二次方程的代数熟练度不仅会直接考查，还会为函数、不等式和坐标几何等内容打下基础。

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

A quadratic equation is any equation that can be rearranged into the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the unknown x is 2, which is why it is called a quadratic equation. The word quadratic comes from the Latin word quadratus, meaning square.

二次方程是任何可以整理为 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。未知数 x 的最高次数是 2，因此称为二次方程。”quadratic” 一词来自拉丁文 quadratus，意为平方。

If a = 0, the equation becomes linear, so the condition a ≠ 0 is essential. For example, 2x² – 5x + 3 = 0 is quadratic, but 0x² + 4x – 7 = 0 is simply the linear equation 4x – 7 = 0.

如果 a = 0，方程就变成一次方程，所以 a ≠ 0 这一条件至关重要。例如，2x² – 5x + 3 = 0 是二次方程，但 0x² + 4x – 7 = 0 只是一次方程 4x – 7 = 0。

The solutions of a quadratic equation are also called roots or zeroes of the quadratic expression. Depending on the discriminant, there may be two distinct real roots, one repeated root, or no real roots at IGCSE level.

二次方程的解也称为二次式的根或零点。根据判别式的不同，在 IGCSE 阶段可能有三个情况：两个不同的实数根、一个重根或没有实数根。

2. Standard Form and Key Terms | 标准形式与关键术语

Before solving, you should always rewrite the equation in standard form ax² + bx + c = 0 and arrange terms in descending powers of x. The values a, b and c are called the quadratic coefficient, linear coefficient and constant term respectively.

解题前，你应始终把方程写成标准形式 ax² + bx + c = 0，并按 x 的降幂排列。a、b、c 分别称为二次项系数、一次项系数和常数项。

For example, 3x² – 5x + 2 = 0 has a = 3, b = -5 and c = 2. Remember to include the sign of each coefficient. If an equation is given as 5 – 3x = 2x², first rearrange it to 2x² + 3x – 5 = 0, giving a = 2, b = 3 and c = -5.

例如，3x² – 5x + 2 = 0 中 a = 3，b = -5，c = 2。注意每个系数都要带符号。如果给出的方程是 5 – 3x = 2x²，应先整理为 2x² + 3x – 5 = 0，于是 a = 2，b = 3，c = -5。

Identifying a, b and c correctly is crucial because the quadratic formula and discriminant depend on these values. A sign error in b or c will lead to an incorrect solution even if the method is correct.

正确确定 a、b、c 至关重要，因为求根公式和判别式都依赖这些值。即使方法正确，b 或 c 的符号错误也会导致解错。

3. Solving by Factorising | 因式分解法

When the quadratic expression can be written as a product of two linear factors, factorising is usually the fastest method. You look for two numbers that multiply to give ac and add to give b, then split the middle term or use direct factorisation.

当二次式可以写成两个一次因式的乘积时，因式分解通常是最快的方法。你需要找两个数，它们的乘积为 ac，和为 b，然后拆中项或直接因式分解。

Example: Solve x² – 5x + 6 = 0. Here a = 1, b = -5 and c = 6, so ac = 6. The numbers -2 and -3 multiply to 6 and add to -5, so the factorised form is (x – 2)(x – 3) = 0. Therefore x = 2 or x = 3.

例题：解 x² – 5x + 6 = 0。这里 a = 1，b = -5，c = 6，所以 ac = 6。数 -2 和 -3 的乘积为 6，和为 -5，因此因式分解为 (x – 2)(x – 3) = 0。所以 x = 2 或 x = 3。

If the coefficient a is not 1, you can still factorise directly or use the ac method. For 2x² + 7x + 3, look for two numbers multiplying to 6 and adding to 7: 6 and 1. Split the middle term: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

如果系数 a 不为 1，你仍然可以直接因式分解或使用 ac 法。对于 2x² + 7x + 3，找两个数乘积为 6 且和为 7：6 和 1。拆分中项：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

Always check your factors by expanding them back to the original expression. This is a quick way to catch sign mistakes and arithmetic slips before you state the final roots.

务必通过展开因式来检验是否与原式一致。这是在你写出最终根之前快速发现符号错误和计算失误的方法。

4. Solving by Completing the Square | 配方法

Completing the square transforms x² + bx into the form (x + p)² + q. Take half of the coefficient of x, square it, and adjust the constant term. This method is extremely useful when the quadratic does not factorise easily and when finding the turning point of a graph.

配方法把 x² + bx 转化为 (x + p)² + q 的形式。取 x 项系数的一半，平方后调整常数项。当二次式不易因式分解以及需要求图像顶点时，这个方法特别有用。

For x² + 6x + 5 = 0, write x² + 6x as (x + 3)² – 9, so the equation becomes (x + 3)² – 9 + 5 = 0, giving (x + 3)² = 4. Hence x + 3 = ±2, so x

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