Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are a cornerstone of the IGCSE Mathematics syllabus. They connect algebra, graphs, and real-world problem solving, and they appear in almost every examination paper. This guide explains the standard form, the three main solution methods, the discriminant, and common pitfalls, with worked examples in the style of exam questions.

二次方程是 IGCSE 数学课程大纲的核心内容。它将代数、图象与实际应用问题紧密相连，几乎出现在每一份试卷中。本指南将讲解标准形式、三种主要解法、判别式以及常见易错点，并附有贴近考试风格的例题。

1. The Standard Form | 标准形式

A quadratic equation in one variable is any equation that can be rearranged into the standard form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The constant a is called the coefficient of x², b is the coefficient of x, and c is the constant term.

一元二次方程是指可以整理成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。常数 a 称为 x² 的系数，b 称为 x 的系数，c 称为常数项。

Every quadratic equation has at most two solutions, which we call the roots of the equation. For example, x² − 5x + 6 = 0 has two roots, x = 2 and x = 3, because substituting either value into the equation makes the left-hand side equal to zero.

每个二次方程至多有两个解，我们称之为方程的根。例如，x² − 5x + 6 = 0 有两个根 x = 2 和 x = 3，因为将这两个值中的任意一个代入方程，都能使左边等于零。

If a = 0, the equation becomes linear, not quadratic. That is why the condition a ≠ 0 is essential. A quadratic equation is also called a second-degree equation because the highest power of x is 2.

如果 a = 0，方程就变成一次方程而非二次方程，因此条件 a ≠ 0 至关重要。二次方程又称二次方程，因为 x 的最高次数为 2。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic expression factorises neatly into two linear factors. To solve x² + bx + c = 0, we look for two numbers whose product is c and whose sum is b.

当二次表达式能整齐地分解成两个一次因式时，因式分解是最快的方法。要求解 x² + bx + c = 0，我们需要找到两个数，使它们的乘积为 c、和为 b。

For example, to solve x² + 7x + 12 = 0, note that 3 × 4 = 12 and 3 + 4 = 7, so x² + 7x + 12 = (x + 3)(x + 4). Setting each factor to zero gives x = −3 or x = −4.

例如，解 x² + 7x + 12 = 0 时，因为 3 × 4 = 12 且 3 + 4 = 7，所以 x² + 7x + 12 = (x + 3)(x + 4)。令每个因式等于零，得到 x = −3 或 x = −4。

When the coefficient of x² is not 1, we can use the method of splitting the middle term. Solve 2x² + 5x + 2 = 0 by finding two numbers whose product is 2 × 2 = 4 and whose sum is 5; these numbers are 4 and 1. Rewrite the equation as 2x² + 4x + x + 2 = 0 and then factor by grouping:

当 x² 的系数不为 1 时，我们可以使用拆项法。解 2x² + 5x + 2 = 0 时，先找到两个数，使它们的乘积为 2 × 2 = 4、和为 5；这两个数是 4 和 1。将方程改写为 2x² + 4x + x + 2 = 0，再分组分解：

2x² + 4x + x + 2 = 2x(x + 2) + 1(x + 2) = (x + 2)(2x + 1)

Setting (x + 2)(2x + 1) = 0 gives x = −2 or x = −½.

令 (x + 2)(2x + 1) = 0，得到 x = −2 或 x = −½。

Always remember the null factor law: if the product of two expressions is zero, then at least one of them must be zero. This law is the mathematical reason why factorisation works for solving equations.

始终记住零因子定律：如果两个表达式的乘积为零，那么其中至少有一个必须为零。这个定律是因式分解法能够解方程的数学原理。

3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic expression ax² + bx + c in the form a(x + p)² + q. This method is useful for solving equations, for finding the turning point of a parabola, and for proving the quadratic formula.

配方法将二次表达式 ax² + bx + c 改写为 a(x + p)² + q 的形式。这种方法在解方程、求抛物线顶点以及推导二次公式时都非常有用。

For a monic quadratic x² + bx, we add and subtract (b/2)². Take x² + 6x − 7 = 0. Since (6/2)² = 9, we write x² + 6x = (x + 3)² − 9, so the equation becomes (x + 3)² − 16 = 0.

对于首项系数为 1 的二次式 x² + bx，我们在其中加减 (b/2)²。以 x² + 6x − 7 = 0 为例，因为 (6/2)² = 9，所以 x² + 6x = (x + 3)² − 9，于是方程变为 (x + 3)² − 16 = 0。

(x + 3)² = 16 → x + 3 = ±4 → x = 1 or x = −7

When a ≠ 1, first divide the whole equation by a. To solve 3x² − 12x + 6 = 0, divide by 3 to get x² − 4x + 2 = 0, then complete the square:

当 a ≠ 1 时，先将整个方程除以 a。解 3x² − 12x + 6 = 0 时，先除以 3 得到 x² − 4x + 2 = 0，然后配方：

(x − 2)² − 4 + 2 = 0 → (x − 2)² =

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）