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。

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

c are constants and a ≠ 0. The largest exponent of x is 2, which is why it is described as "quadratic" d 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。

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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。

roduct 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。

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3. The Quadratic Formula | 求根公式法

mula 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。

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4. Completing the Square | 配方法

. 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。

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。

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

the square root sign in the quadratic formula and determines the number and type of roots.

判别式

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