Solving Quadratic Equations: Methods and Applications | 二次方程的解法与应用

📚 Solving Quadratic Equations: Methods and Applications | 二次方程的解法与应用

Quadratic equations are a cornerstone of IGCSE Mathematics. This revision article explores standard forms, solution methods, discriminant analysis, and exam-oriented practice, using a worked example throughout.

二次方程是 IGCSE 数学的重要组成部分。本复习文章围绕标准形式、求解方法、判别式分析以及应试练习展开，并始终以一道典型例题贯穿全文。

1. Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation in one variable x can be written in the standard form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0.

一元二次方程的标准形式为 ax² + bx + c = 0，其中 a、b、c 为实数，且 a ≠ 0。

For example, 2x² – 5x + 3 = 0 has a = 2, b = -5, c = 3. This equation will be used as the worked example in this article.

例如，2x² – 5x + 3 = 0 中，a = 2，b = -5，c = 3。本文后续将以该方程作为示范例题。

It is important to rearrange any given equation into this standard form before applying solution methods.

在应用解法之前，务必先将题目给出的方程整理为标准形式。

2. Solving by Factoring | 因式分解法

Factoring uses the null-factor law: if p × q = 0, then p = 0 or q = 0. To apply this to a quadratic, we express ax² + bx + c as a product of two linear factors.

因式分解法利用零因子定律：若 p × q = 0，则 p = 0 或 q = 0。我们将 ax² + bx + c 写成两个一次因式的乘积，再令每个因式等于零。

For 2x² – 5x + 3 = 0, we look for two factors of 2 × 3 = 6 that add to -5. These numbers are -2 and -3, so we rewrite: 2x² – 2x – 3x + 3 = 0, then group: 2x(x – 1) – 3(x – 1) = 0, giving (2x – 3)(x – 1) = 0.

对于 2x² – 5x + 3 = 0，我们需要找出两个数，乘积为 2 × 3 = 6，且和为 -5。这两个数是 -2 和 -3，于是改写为 2x² – 2x – 3x + 3 = 0，再分组：2x(x – 1) – 3(x – 1) = 0，得到 (2x – 3)(x – 1) = 0。

Therefore, x = 3/2 or x = 1. Always check your factors by expanding them back.

因此，x = 3/2 或 x = 1。务必通过展开来检验因式是否正确。

3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + h)² + k = 0, which can then be solved by taking square roots.

配方法将二次式改写为 a(x + h)² + k = 0 的形式，然后通过开平方求解。

First divide by 2: x² – (5/2)x + 3/2 = 0. Move the constant: x² – (5/2)x = -3/2. Add (5/4)² = 25/16 to both sides: (x – 5/4)² = -3/2 + 25/16 = 1/16.

首先两边除以 2：x² – (5/2)x + 3/2 = 0。移常数项：x² – (5/2)x = -3/2。两边加上 (5/4)² = 25/16：(x – 5/4)² = -3/2 + 25/16 = 1/16。

Taking square roots gives x – 5/4 = ±1/4, so x = 5/4 + 1/4 = 3/2 or x = 5/4 – 1/4 = 1.

开平方得 x – 5/4 = ±1/4，所以 x = 5/4 + 1/4 = 3/2，或 x = 5/4 – 1/4 = 1。

Note: When the coefficient of x² is not 1, divide the entire equation by that coefficient first.

提示：当 x² 的系数不为 1 时，先对方程两边同时除以该系数。

4. Solving by the Quadratic Formula | 公式法

The quadratic formula solves any quadratic equation directly:

二次公式可以直接求解任何二次方程：

x = (-b ± √(b² – 4ac)) / (2a)

For 2x² – 5x + 3 = 0, substitute

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