Solving Quadratic Equations | 二次方程解法

📚 Solving Quadratic Equations | 二次方程解法

In the IGCSE Mathematics syllabus, quadratic equations appear in almost every examination paper. Whether you are aiming for a grade C or a grade A*, mastering this topic is non-negotiable. This article explains the four main methods of solving quadratics, the role of the discriminant, how to sketch the graph of a quadratic function, and how to apply these skills to examination questions.

在 IGCSE 数学考纲中，二次方程几乎出现在每一份试卷中。无论你的目标是 C 等还是 A*，掌握这一知识点都是必不可少的。本文将讲解解二次方程的四种主要方法、判别式的作用、如何绘制二次函数图像，以及如何将这些技能应用于考试题目。

1. The Standard Form | 标准形式

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The coefficient a is the coefficient of x², b is the coefficient of x, and c is the constant term. If a = 0, the equation becomes linear, so the condition a ≠ 0 is what makes the equation genuinely quadratic.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 是实数，且 a ≠ 0。系数 a 是 x² 的系数，b 是 x 的系数，c 是常数项。如果 a = 0，方程就变成了一次方程，因此 a ≠ 0 这一条件保证了方程真正具有”二次”性质。

Example: 2x² − 5x + 3 = 0 is a quadratic equation with a = 2, b = −5 and c = 3. In an examination, you may first need to rearrange an equation into this form. For instance, 3x² + 2 = 7x must be rewritten as 3x² − 7x + 2 = 0 before any method can be applied.

例如：2x² − 5x + 3 = 0 是二次方程，其中 a = 2，b = −5，c = 3。在考试中，你可能首先需要将方程整理成这种形式。例如，3x² + 2 = 7x 必须先改写为 3x² − 7x + 2 = 0，之后才能应用任何解法。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic has simple integer roots. The idea is to write the left-hand side as a product of two linear factors: (px + q)(rx + s) = 0. Then, because the product is zero, at least one of the factors must be zero, which gives two simple linear equations to solve.

当二次方程具有简单的整数根时，因式分解往往是最快捷的方法。其核心思路是将等号左边写成两个一次因式的乘积：(px + q)(rx + s) = 0。由于乘积为零，至少有一个因式必须为零，由此得到两个简单的一次方程。

Example: Solve x² + 5x + 6 = 0. We look for two numbers whose sum is 5 and whose product is 6. These numbers are 2 and 3. Therefore x² + 5x + 6 = (x + 2)(x + 3) = 0, so x = −2 or x = −3.

例：解 x² + 5x + 6 = 0。我们需要找两个数，它们的和为 5，乘积为 6。这两个数是 2 和 3。因此 x² + 5x + 6 = (x + 2)(x + 3) = 0，故 x = −2 或 x = −3。

When the coefficient of x² is not 1, the factorisation requires more care. For 2x² − 5x + 3 = 0, we try (2x − 3)(x − 1) = 0. Expanding gives 2x² − 2x − 3x + 3 = 2x² − 5x + 3, which confirms the factorisation. Hence x = 3/2 or x = 1.

当 x² 的系数不为 1 时，因式分解需要更加细心。对于 2x² − 5x + 3 = 0，我们尝试 (2x − 3)(x − 1) =

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