Solving Quadratic Equations | IGCSE 数学：解二次方程

📚 Solving Quadratic Equations | IGCSE 数学：解二次方程

Quadratic equations are one of the most important algebra topics in IGCSE Mathematics. They appear in Paper 2 and Paper 4, often linked with graphs, inequalities and real-life problems. This article covers all core methods: factorising, completing the square, using the quadratic formula, and understanding the discriminant.

二次方程是 IGCSE 数学中最重要的代数主题之一。它们常出现在 Paper 2 和 Paper 4 中，经常与图像、不等式和实际问题结合。本文涵盖所有核心方法：因式分解法、配方法、求根公式以及判别式的理解。

1. What is a Quadratic Equation? | 什么是二次方程

A quadratic equation is a polynomial equation of degree 2. Its general form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.

二次方程是一个二次多项式方程。其一般形式为 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。

The term “quadratic” comes from “quadratus”, the Latin word for square. The highest power of the variable x is 2, which is why the graph of y = ax² + bx + c is always a parabola.

“二次”一词来源于拉丁语 “quadratus”，意为平方。变量 x 的最高次数为 2，因此 y = ax² + bx + c 的图像始终是一条抛物线。

In IGCSE exams, you must be able to solve a quadratic equation by at least one method and to choose the most efficient method for a given question.

在 IGCSE 考试中，你必须能够至少用一种方法解二次方程，并能根据给定题目选择最高效的方法。

2. Standard Form and Key Terms | 标准形式与关键术语

Before solving, you should rearrange the equation into standard form: ax² + bx + c = 0. For example, 3x + x² = 4 becomes x² + 3x – 4 = 0.

求解前，你应先将方程整理为标准形式：ax² + bx + c = 0。例如，3x + x² = 4 可化为 x² + 3x – 4 = 0。

Key terms include the coefficient of x² (a), the coefficient of x (b), the constant term (c), the roots (solutions), and the discriminant (b² – 4ac).

关键术语包括 x² 的系数（a）、x 的系数（b）、常数项（c）、根（解）以及判别式（b² – 4ac）。

• Roots are the x-values that make the equation true.
• 根是使方程成立的 x 值。
• A quadratic equation can have two real roots, one repeated real root, or no real roots.
• 二次方程可以有两个实根、一个重根或没有实根。

3. Solving by Factorising | 因式分解法

Factorising is usually the quickest method when the quadratic has simple integer roots. Write the quadratic as a product of two brackets, then set each bracket equal to zero.

当二次方程具有简单的整数根时，因式分解通常是最快的方法。将二次式写成两个括号的乘积，然后令每个括号等于零。

Example: Solve x² – 5x + 6 = 0.

例：解方程 x² – 5x + 6 = 0。

Find two numbers that multiply to +6 and add to -5: they are -2 and -3. So (x – 2)(x – 3) = 0.

找两个数：乘积为 +6，和为 -5：它们是 -2 和 -3。因此 (x – 2)(x – 3) = 0。

Set each bracket to zero: x – 2 = 0 gives x = 2; x – 3 = 0 gives x = 3. The solutions are x = 2 or x = 3.

令每个括号为零：x – 2 = 0 得 x = 2；x – 3 = 0 得 x = 3。解为 x = 2 或 x = 3。

Always check by substituting your answers back into the original equation.

始终通过将答案代回原方程进行检验。

4. Solving by Completing the Square | 配方法

Completing the square is useful when the quadratic cannot be factorised easily, or when you need to find the turning point of a parabola.

当二次方程不易因式分解，或需要求抛物线的顶点时，配方法非常有用。

For x² + bx + c = 0, rewrite as (x + b/2)² – (b/2)² + c = 0.

对于 x² + bx + c = 0，可改写为 (x + b/2)² – (b/2)² + c = 0。

Example: Solve x² + 6x + 2 = 0 by completing the square.

例：用配方法解方程 x² + 6x + 2 = 0。

Half of 6 is 3, and 3² is 9. So x² + 6x +

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