📚 Solving Quadratic Equations by Factorising, Completing the Square and the Quadratic Formula | 用因式分解、配方法和求根公式解二次方程
Quadratic equations are a central topic in IGCSE Mathematics. This revision guide covers the three main algebraic methods for solving them: factorising, completing the square, and using the quadratic formula. Each method is explained with step-by-step examples, common errors, and exam tips to help you choose the most efficient approach in a test.
二次方程是 IGCSE 数学的核心主题。本复习指南涵盖解二次方程的三种主要代数方法：因式分解、配方法和求根公式。每种方法都配有分步示例、常见错误和考试技巧，帮助你在考试中选择最高效的解题路径。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the unknown x is 2. For example, x² – 5x + 6 = 0 and 3x² + 2x – 1 = 0 are quadratic equations, but 2x – 3 = 0 is linear because the highest power of x is 1.
二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b 和 c 是常数，且 a ≠ 0。未知数 x 的最高次数是 2。例如，x² – 5x + 6 = 0 和 3x² + 2x – 1 = 0 都是二次方程，但 2x – 3 = 0 是一次方程，因为 x 的最高次数是 1。
The graph of y = ax² + bx + c is a parabola. The solutions of the quadratic equation are the x-values where this parabola crosses the x-axis, so they are also called roots, zeros, or x-intercepts. An IGCSE quadratic equation can have two real solutions, one repeated solution, or no real solutions.
函数 y = ax² + bx + c 的图像是一条抛物线。二次方程的解就是抛物线与 x 轴交点的横坐标，因此解也称为根、零点或 x 轴截距。IGCSE 中的二次方程可以有两个实数解、一个重根，或没有实数解。
2. Standard Form and Key Vocabulary | 标准形式与关键术语
Before solving, always rearrange the equation into standard form ax² + bx + c = 0. The number a is the coefficient of x², b is the coefficient of x, and c is the constant term. For 3x² – 7x + 2 = 0, we have a = 3, b = -7 and c = 2. If a = 1, the quadratic is called monic, which is the easiest case for factorising.
在求解之前，一定要先把方程整理成标准形式 ax² + bx + c = 0。数字 a 是 x² 的系数，b 是 x 的系数，c 是常数项。对于 3x² – 7x + 2 = 0，我们有 a = 3，b = -7，c = 2。如果 a = 1，这个二次式就称为首一二次式，这是最容易因式分解的情况。
The solutions are also called roots. The expression b² – 4ac is called the discriminant, and it tells you how many real roots the equation has. You do not need to solve the equation fully to use the discriminant, which makes it a common short-answer exam question.
解也称为根。表达式 b² – 4ac 叫做判别式，它可以告诉你方程有多少个实数根。使用判别式时不需要完整解方程，因此它是考试中常见的简答题类型。
3. The Zero Product Property | 零积性质
Factorising relies on a simple rule: if the product of two expressions is zero, then at least one of the expressions must be zero. In symbols, if A × B = 0, then A = 0 or B = 0. This is called the zero product property.
因式分解法依赖一条简单规则：如果两个式子的乘积为零，那么其中至少一个式子必须为零。用符号表示就是：如果 A × B = 0，那么 A = 0 或 B = 0。这叫做零积性质。
For example, if (x – 2)(x + 3) = 0, then either x – 2 = 0 or x + 3 = 0. Solving these two linear equations gives x = 2 or x = -3. This is why factorising converts a quadratic equation into two simple linear equations.
例如，如果 (x – 2)(x + 3) = 0，那么要么 x – 2 = 0，要么 x + 3 = 0。解这两个一次方程得到 x = 2 或 x = -3。这就是因式分解能把二次方程转化为两个简单一次方程的原因。
4. Factorising When a = 1 | a = 1 时的因式分解
When the quadratic is monic, factorising means finding two numbers whose product is c and whose sum is b. For x² + 7x + 12 = 0, we need two numbers that multiply to 12 and add to 7. The numbers 3 and 4 work, so x² + 7x + 12 = (x + 3)(x + 4). Setting each bracket to zero gives x = -3 or x = -4.
当二次式是首一形式时，因式分解就是找到两个数，使它们的乘积等于 c，和等于 b。对于 x² + 7x + 12 = 0，我们需要两个数，乘积为 12，和为 7。数字 3 和 4 符合条件，因此 x² + 7x + 12 = (x + 3)(x + 4)。令每个括号等于零，得到 x = -3 或 x = -4。
When the constant term is negative, the two numbers have opposite signs. For x² – x – 6 = 0, we need two numbers that multiply to -6 and add to -1. The numbers -3 and 2 work, so (x – 3)(x + 2) = 0, giving x = 3 or x = -2. Always expand the brackets mentally to check your factorisation.
当常数项为负数时，这两个数异号。对于 x² – x – 6 = 0，我们需要两个数，乘积为 -6，和为 -1。数字 -3 和 2 符合条件，因此 (x – 3)(x + 2) = 0，得到 x = 3 或 x = -2。一定要在心里展开括号来检验你的因式分解是否正确。
5. Factorising When a ≠ 1 | a ≠ 1 时的因式分解
When a is not 1, we often use the ac method. First multiply a and c, then find two numbers that multiply to ac and add to b. For 2x² + 7x + 3 = 0, we have ac = 2 × 3 = 6. We need two numbers that multiply to 6 and add to 7, namely 6 and 1.
当 a 不等于 1 时，我们常用 ac 法。先用 a 乘以 c，然后找到两个数，使它们乘积等于 ac，和等于 b。对于 2x² + 7x + 3 = 0，我们有 ac = 2 × 3 = 6。我们需要两个数，乘积为 6，和为 7，即 6 和 1。
Split the middle term using these numbers: 2x² + 6x + x + 3. Now factorise in pairs: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Setting each bracket to zero gives 2x + 1 = 0, so x = -1/2, or x + 3 = 0, so x = -3. The solutions are x = -1/2 or x = -3.
用这两个数拆开中间项：2x² + 6x + x + 3。然后两两分组因式分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。令每个括号等于零，得到 2x + 1 = 0，所以 x = -1/2；或者 x + 3 = 0，所以 x = -3。解为 x = -1/2 或 x = -3。
6. Completing the Square: The Key Idea | 配方法的基本思想
Completing the square rewrites a quadratic expression x² + bx in the form (x + p)² + q. The essential identity is x² + bx = (x + b/2)² – (b/2)². For example, x² + 8x can be written as (x + 4)² – 16, because (x + 4)² = x² + 8x + 16, so we subtract 16 to keep the expression unchanged.
配方法把二次式 x² + bx 改写成 (x + p)² + q 的形式。核心恒等式是 x² + bx = (x + b/2)² – (b/2)²。例如，x² + 8x 可以写成 (x + 4)² – 16，因为 (x + 4)² = x² + 8x + 16，所以要减去 16 才能保持原式不变。
This method is useful when factorising is difficult or impossible with integers, and it also reveals the vertex of the parabola. The vertex of y = x² + 8x + 3 is found from y = (x + 4)² – 13, so the vertex is (-4, -13).
当因式分解困难或无法用整数分解时，配方法非常有用，而且它还能揭示抛物线的顶点。由 y = x² + 8x + 3 可以配方为 y = (x + 4)² – 13，因此顶点是 (-4, -13)。
7. Completing the Square: Worked Example | 配方法解题示例
Solve x² + 6x – 7 = 0 by completing the square. First move the constant term to the right: x² + 6x = 7. Take half of the coefficient of x, which is 6/2 = 3, and square it to get 9. Add 9 to both sides: x² + 6x + 9 = 16.
用配方法解 x² + 6x – 7 = 0。首先把常数项移到右边：x² + 6x = 7。取 x 系数的一半，即 6/2 = 3，再平方得到 9。两边同时加上 9：x² + 6x + 9 = 16。
The left side is now a perfect square: (x + 3)² = 16. Take the square root of both sides, remembering the ± sign: x + 3 = ±4. This gives two cases: x + 3 = 4, so x = 1, or x + 3 = -4, so x = -7. The solutions are x = 1 or x = -7.
左边现在是一个完全平方：(x + 3)² = 16。两边开平方，记得要有正负号：x + 3 = ±4。由此得到两种情况：x + 3 = 4，所以 x = 1；或者 x + 3 = -4，所以 x = -7。解为 x = 1 或 x = -7。
8. Deriving the Quadratic Formula | 求根公式的推导
The quadratic formula comes from completing the square on the general quadratic ax² + bx + c = 0. First divide every term by a: x² + (b/a)x + c/a = 0. Move c/a to the right: x² + (b/a)x = -c/a.
求根公式来自对一般二次方程 ax² + bx + c = 0 进行配方。首先每一项都除以 a：x² + (b/a)x + c/a = 0。把 c/a 移到右边：x² + (b/a)x = -c/a。
Add (b/2a)² to both sides. The left side becomes a perfect square:
两边同时加上 (b/2a)²。左边变成一个完全平方：
(x + b/2a)² = (b² – 4ac) / 4a²
Take the square root and rearrange to isolate x:
两边开平方并移项求出 x：
x = (-b ± √(b² – 4ac)) / 2a
This formula works for every quadratic equation, including those that cannot be factorised using integers. You should memorise it exactly, including the denominator 2a and the ± sign.
这个公式适用于所有二次方程，包括不能用整数因式分解的方程。你必须准确记住它，包括分母 2a 和正负号。
9. Using the Quadratic Formula | 使用求根公式
To use the quadratic formula, first identify a, b and c from standard form. For 2x² – 4x – 3 = 0, we have a = 2, b = -4 and c = -3. Substitute these values carefully into x = (-b ± √(b² – 4ac)) / 2a.
使用求根公式时，首先从标准形式中确定 a、b 和 c。对于 2x² – 4x – 3 = 0，我们有 a = 2，b = -4，c = -3。把这些值小心地代入 x = (-b ± √(b² – 4ac)) / 2a。
The discriminant is b² – 4ac = (-4)² – 4 × 2 × (-3) = 16 + 24 = 40. Therefore x = (4 ± √40) / 4. Since √40 = √(4 × 10) = 2√10, this simplifies to x = (4 ± 2√10) / 4 = 1 ± √10/2.
判别式为 b² – 4ac = (-4)² – 4 × 2 × (-3) = 16 + 24 = 40。因此 x = (4 ± √40) / 4。因为 √40 = √(4 × 10) = 2√10，所以可以化简为 x = (4 ± 2√10) / 4 = 1 ± √10/2。
Always simplify surds and fractions in your final answer unless the question says otherwise. Leaving an unsimplified answer can lose method or accuracy marks in IGCSE exams.
除非题目另有说明，否则最终答案一定要化简根式和分数。在 IGCSE 考试中，答案没有化简可能会丢掉方法分或准确度分。
10. The Discriminant: What It Tells You | 判别式的作用
The discriminant is D = b² – 4ac. It determines the nature of the roots without solving the equation. If D > 0, the quadratic has two distinct real roots. If D = 0, it has one repeated real root. If D < 0, it has no real roots, though IGCSE Higher students may meet complex numbers later.
判别式是 D = b² – 4ac。它可以在不解方程的情况下判断根的性质。如果 D > 0，二次方程有两个不同的实数根；如果 D = 0，它有一个重根；如果 D < 0，它没有实数根，不过 IGCSE 高阶学生在以后可能会学到复数。
For example, in 3x² + 2x + 5 = 0, we have a = 3, b = 2, c = 5, so D = 2² – 4 × 3 × 5 = 4 – 60 = -56. Since D is negative, the equation has no real solutions. Its graph does not cross the x-axis.
例如，对于 3x² + 2x + 5 = 0，我们有 a = 3，b = 2，c = 5，所以 D = 2² – 4 × 3 × 5 = 4 – 60 = -56。因为 D 为负数，该方程没有实数解。它的图像不与 x 轴相交。
11. Choosing the Best Method and Checking Answers | 选择最佳方法与检验答案
Use factorising when the quadratic has small integer coefficients and you can spot the two numbers quickly. Use completing the square when a = 1 and b is even, or when the question asks for the vertex form. Use the quadratic formula when factorising is difficult, when coefficients are decimals or fractions, or when you are asked to give answers correct to a given number of decimal places.
当二次式的系数为较小的整数，并且你能快速找到两个关键数字时，使用因式分解法。当 a = 1 且 b 为偶数，或者题目要求写出顶点式时，使用配方法。当因式分解困难、系数为小数或分数，或者题目要求把答案保留到指定小数位时，使用求根公式。
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