Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

A quadratic equation is one of the most important topics in IGCSE Mathematics. This article is designed for teachers, providing a step-by-step teaching guide with worked examples and common pitfalls. All methods are aligned with the IGCSE syllabus.

一元二次方程是 IGCSE 数学中最重要的主题之一。本文专为教师编写，提供分步教学指南、典型例题和常见错误分析，所有方法均对应 IGCSE 大纲要求。

1. What Is a Quadratic Equation? | 什么是一元二次方程？

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The largest exponent of x is 2, which is why it is described as “quadratic”. The equation has degree 2, and it may have zero, one or two real solutions, depending on the value of the discriminant.

凡可写成 ax² + bx + c = 0 形式的方程都是一元二次方程，其中 a、b、c 为常数，且 a ≠ 0。由于 x 的最高次数为 2，故称为“二次”。该方程的根可能为零个、一个或两个实根，具体取决于判别式的值。

If a = 0, the equation becomes linear, not quadratic. / 如果 a = 0，方程退化为一次方程，而非二次方程。
Standard form: ax² + bx + c = 0. / 标准形式为 ax² + bx + c = 0。

Before solving, always rewrite the given equation in standard form, gathering all terms on one side with zero on the other.

求解之前，应先将给定方程化为标准形式，将所有项移到等号一侧，使另一侧为零。

2. The Factorisation Method | 因式分解法

Factorisation is often the quickest method when the quadratic can be written as a product of two linear factors. The steps are: list the factor pairs of ac, find the pair whose sum is b, split the middle term, group and factor, then set each factor equal to zero.

当二次式可以写成两个一次因式的乘积时，因式分解往往是最快的方法。步骤为：列出 ac 的因数对，找出和为 b 的一组，拆中项，分组分解，再令每个因式等于零。

For example, solve x² − 5x + 6 = 0. We need two numbers with product 6 and sum −5: those are −2 and −3. Hence (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

例如，解 x² − 5x + 6 = 0。需找两个数，其乘积为 6、和为 −5，即 −2 与 −3。因此 (x − 2)(x − 3) = 0，得 x = 2 或 x = 3。

(x − p)(x − q) = 0 ⇒ x = p or x = q

The zero-product property states that if two factors multiply to zero, at least one factor must be zero. This is the key logical step behind the factorisation method.

零积性质表明：若两个因式相乘为零，则至少有一个因式为零。这是因式分解法背后关键的逻辑依据。

3. The Quadratic Formula | 求根公式法

When factorisation is difficult or impossible, the quadratic formula always works. For any quadratic equation ax² + bx + c = 0, the solutions are given by

当因式分解困难或无法进行时，求根公式始终有效。对任意一元二次方程 ax² + bx + c = 0，其解为

x = (−b ± √(b² − 4ac)) / 2a

This formula is printed on the IGCSE formula sheet, but students must know how to substitute values correctly. Evaluate the discriminant b² − 4ac first, then simplify the root.

该公式印在 IGCSE 公式表中，但学生必须会正确代入求值。先计算判别式 b² − 4ac，再化简根式。

Example: solve 2x² + 4x − 3 = 0. Here a = 2, b = 4, c = −3. Then

例：解 2x² + 4x − 3 = 0。此时 a = 2，b = 4，c = −3。则

x = (−4 ± √(16 + 24)) / 4 = (−4 ± √40) / 4 = −1 ± √10/2

Students should be reminded to round only at the final step if an exact value is not required.

应提醒学生，若题目不要求精确值，只在最后一步四舍五入。

4. Completing the Square | 配方法

Completing the square rewrites x² + bx + c = 0 in the form (x + b/2)² = (b/2)² − c. Then solve by taking the square root of both sides. This method is especially useful for finding the turning point of a quadratic graph.

配方法将 x² + bx + c = 0 改写为 (x + b/2)² = (b/2)² − c，然后两边开平方求解。该方法尤其适用于求二次函数图象的顶点坐标。

Example: solve x² + 6x + 1 = 0. Rewrite as (x + 3)² − 9 + 1 = 0, so (x + 3)² = 8. Taking square roots gives x = −3 ± 2√2.

例：解 x² + 6x + 1 = 0。改写为 (x + 3)² − 9 + 1 = 0，即 (x + 3)² = 8。两边开平方得 x = −3 ± 2√2。

For a non-monic quadratic ax² + bx + c = 0, first divide by a, then complete the square.

对于非首一的二次方程 ax² + bx + c = 0，先两边除以 a，再进行配方。

5. The Discriminant and the Nature of Roots | 判别式与根的性质

The discriminant is the expression Δ = b² − 4ac. It appears under the square root sign in the quadratic formula and determines the number and type of roots.

判别式

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Solving Quadratic Equations | 解一元二次方程