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. It appears in almost every examination paper, either as a direct solving question, or as part of a graph-sketching task, a word problem, or an algebraic manipulation exercise. Mastering the three core methods — factorisation, completing the square, and the quadratic formula — is essential for success at both Core and Extended levels.

一元二次方程是 IGCSE 数学中最重要的考点之一。几乎每一份试卷都会考查这一内容，无论是直接解方程，还是作为画图题、应用题或代数变形的一部分。掌握三种核心方法——因式分解法、配方法和求根公式法——对于 Core 和 Extended 两个级别的考试都至关重要。

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

An equation is called quadratic, or of degree two, when the highest power of the unknown variable is 2. The general form is:

当一个方程中未知数的最高次数为 2 时，它被称为二次方程。其一般形式为：

ax² + bx + c = 0, where a ≠ 0

Here, a is called the leading coefficient, b is the linear coefficient, and c is the constant term. The condition a ≠ 0 is essential: if a were zero, the equation would be linear, not quadratic. In IGCSE examinations, you must be able to identify a, b and c quickly, since these values are required when applying the quadratic formula or finding the discriminant.

其中，a 称为二次项系数，b 称为一次项系数，c 称为常数项。条件 a ≠ 0 是必要的：若 a = 0，方程就变成一次方程而不是二次方程。在 IGCSE 考试中，你必须能够快速识别 a、b、c 的值，因为使用求根公式或计算判别式时需要它们。

2. Solving by Factorisation | 用因式分解法解方程

Factorisation is the fastest method when it works. We rewrite the quadratic expression as a product of two linear brackets, then use the zero-product property: if the product of two factors is zero, then at least one of the factors must be zero.

因式分解法在所有方法中速度最快。我们将二次表达式改写为两个一次括号的乘积，然后利用零积性质：若两个因式的乘积为零，则至少有一个因式必须为零。

Example 1: Solve x² − 5x + 6 = 0.

例 1：解方程 x² − 5x + 6 = 0。

Find two numbers whose product is 6 and whose sum is −5. The numbers are −2 and −3, so:

寻找两个数，使其乘积为 6，和为 −5。这两个数是 −2 和 −3，因此：

(x − 2)(x − 3) = 0

Hence x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3. Both values satisfy the original equation, which can be confirmed by substitution.

因此 x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。将这两个值代回原方程均可验证成立。

When the coefficient of x² is not 1, such as 2x² − 7x + 3 = 0, look for factors in the form (2x ± p)(x ± q). In this case, 2x² − 7x + 3 = (2x − 1)(x − 3), so x = ½ or x = 3. Always expand your brackets to check that no sign error has been made.

当 x² 的系数不为 1 时，例如 2x² − 7x + 3 = 0，应寻找形如 (2x ± p)(x ± q) 的因式。本例中，2x² − 7x + 3 = (2x − 1)(x − 3)，所以 x = ½ 或 x = 3。务必展开括号以确认没有出现符号错误。

3. The Difference of Two Squares | 平方差公式

An important special case of factorisation is the difference of two squares. A quadratic of this form has no linear term:

因式分解中一个重要的特例是平方差公式。这种二次方程不含一次项：

x² − a² = (x + a)(x − a)

Example 2:Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

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