📚 Solving Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 解二次方程：因式分解、配方法与求根公式
Quadratic equations lie at the heart of IGCSE Mathematics. They appear in algebra, coordinate geometry, measurement problems and even basic optimisation. This revision guide explains three powerful methods for solving any quadratic equation: factorising, completing the square and using the quadratic formula. You will learn when to use each method, how to interpret the discriminant and how to avoid common errors.
二次方程是 IGCSE 数学的核心内容。它们出现在代数、坐标几何、测量问题甚至基础优化中。本复习指南将讲解解任何二次方程的三种有效方法：因式分解法、配方法和求根公式法。你将学会何时使用每种方法、如何解释判别式以及如何避免常见错误。
1. What Is a Quadratic Equation? | 什么是二次方程？
raph called a parabola. Examples include x² - 5x + 6 = 0, 2x² + 3x - 2 = 0 and x² - 4 = 0.
一元二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数且 a ≠ 0。未知数 x 的最高次数是 2，这使得它的图像是一条特征性的 U 形曲线，称为抛物线。例子包括 x² - 5x + 6 = 0、2x² + 3x - 2 = 0 和 x² - 4 = 0。
In IGCSE exams, a quadratic equation may already be in standard form, or you may need to expand brackets, collect like terms and rearrange it before solving. Always check that the right-hand side is zero before you start solving.
在 IGCSE 考试中，二次方程可能已经是标准形式，也可能需要你先展开括号、合并同类项并移项才能求解。开始求解前，务必检查右边是否为零。
2. Why Do We Need Several Methods? | 为什么需要多种方法？
Not every quadratic expression factorises nicely. Some equations have irrational roots, repeated roots or no real roots at all. Having three methods allows you to choose the quickest route for each problem and provides a built-in way to check your answers.
并非每个二次式都能很好地因式分解。有些方程有无理根、重根或根本没有实根。掌握三种方法可以让你针对每道题选择最快的路径，同时提供一种内置的答案检查方式。
- Factorising is fast when the quadratic has simple integer roots. — 因式分解法在二次式有简单整数根时最快。
- Completing the square is useful for finding the vertex and handling quadratics that do not factorise. — 配方法对求顶点和处理不能因式分解的二次式很有用。
- The quadratic formula always works, even for messy decimals or irrational coefficients. — 求根公式始终有效，即使系数是复杂的小数或无理数也可用。
Examiners often specify a method, so you must be able to use all three confidently and show full working.
考官通常会指定某种方法，因此你必须能够自信地使用全部三种方法，并展示完整的解题过程。
3. Factorising: The Basics | 因式分解法：基础
en apply the zero product property: if the product of two expressions is zero, at least one of the expressions must be zero. This means you can set each bracket equal to zero and solve two simple linear equations.
使用因式分解法时，把 ax² + bx + c = 0 写成 (px + q)(rx + s) = 0 的形式。然后应用零积性质：如果两个式子的乘积为零，那么至少有一个式子为零。这意味着你可以令每个括号分别等于零，再解两个简单的一次方程。
)(x - 3) = 0. Setting each factor to zero gives x - 2 = 0 or x - 3 = 0, so x = 2 or x = 3.
例如，x² - 5x + 6 = 0 可以分解为 (x - 2)(x - 3) = 0。令每个因式为零，得到 x - 2 = 0 或 x - 3 = 0，所以 x = 2 或 x = 3。
Always write down the two linear equations separately. This shows your method clearly and helps you avoid missing one root.
务必分别写出这两个一次方程。这样可以清晰地展示你的方法，并帮助你避免漏掉一个根。
4. Factorising Method: Worked Examples | 因式分解法：例题详解
x, then factor by grouping: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0.
解 2x² + 7x + 3 = 0。首先找到两个数，它们的乘积是 a × c = 2 × 3 = 6，和是 b = 7。这两个数是 6 和 1。把 7x 改写成 6x + x，然后分组分解：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。
s 2x + 1 = 0, so x = -1/2, and x + 3 = 0, so x = -3. The solutions are x = -1/2 or x = -3.
令每个括号分别为零，得到 2x + 1 = 0，即 x = -1/2；以及 x + 3 = 0，即 x = -3。解为 x = -1/2 或 x = -3。
= 0, so x = 3 or x = -3. Notice that the two roots are equal in size but opposite in sign.
另一种常见类型是平方差。解 x² - 9 = 0。由于 9 = 3²，方程变为 (x - 3)(x + 3) = 0，所以 x = 3 或 x = -3。注意这两个根大小相等但符号相反。
| Type 类型 | Example 例子 | Factorised form 因式分解形式 |
| Simple trinomial 简单三项式 | x² + 5x + 6 | (x + 2)(x + 3) |
| Difference of squares 平方差 | x² - 16 | (x - 4)(x + 4) |
| Perfect square 完全平方 | x² + 6x + 9 | (x + 3)² |
5. Completing the Square: The Idea | 配方法：思路
Completing the square rewrites a quadratic expression in the form a(x - h)² + k. This form immediately shows the turning point or vertex of the parabola, which is at (h, k). It is also extremely useful for solving equations that cannot be factorised easily.
配方法把二次式改写成 a(x - h)² + k 的形式。这种形式能立即显示抛物线的转折点或顶点，即 (h, k)。它在求解不容易因式分解的方程时也非常有用。
(b/2)². This identity says that to make a perfect square from x² + bx, you take half the coefficient of x, square it, and add it to both sides.
关键的恒等式是 x² + bx = (x + b/2)² - (b/2)²。这个恒等式说明，要把 x² + bx 配成完全平方，你需要取 x 系数的一半，把它平方，然后在两边加上它。
6. Completing the Square: Step-by-Step | 配方法：步骤
= 3, and square it: 3² = 9. Add 9 to both sides: x² + 6x + 9 = -5 + 9, so x² + 6x + 9 = 4.
解 x² + 6x + 5 = 0。首先把常数移到右边：x² + 6x = -5。取 x 系数的一半，即 6/2 = 3，并平方：3² = 9。两边同时加 9：x² + 6x + 9 = -5 + 9，得到 x² + 6x + 9 = 4。
left side as a perfect square: (x + 3)² = 4. Take the square root of both sides: x + 3 = ±√4, so x + 3 = ±2. Therefore x = -3 ± 2, giving x = -1 or x = -5.
现在把左边写成完全平方：(x + 3)² = 4。两边开平方：x + 3 = ±√4，所以 x + 3 = ±2。因此 x = -3 ± 2，得到 x = -1 或 x = -5。
cient. For example, 2x² + 8x + 6 = 0 becomes x² + 4x + 3 = 0 before completing the square.
如果 x² 的系数不是 1，先让整个方程除以该系数。例如，2x² + 8x + 6 = 0 先变成 x² + 4x + 3 = 0，然后再配方。
so lets you write down the vertex directly: (2, -5). This is a common IGCSE question type.
配成完全平方后的形式如 (x - 2)² - 5 = 0 还能让你直接写出顶点：(2, -5)。这是 IGCSE 常见的题型。
7. Quadratic Formula: Derivation and Memory | 求根公式：推导与记忆
ssion. It is derived by completing the square on the general quadratic equation. You should memorise it and be able to identify a, b and c correctly from the given equation.
求根公式用一个式子给出 ax² + bx + c = 0 的解。它是由一般二次方程配方法推导出来的。你应当牢记该公式，并能从给定方程中正确识别 a、b 和 c。
x = (-b ± √b² - 4ac) ÷ (2a)
two values: one with the plus sign and one with the minus sign. The term inside the square root, b² - 4ac, determines how many real solutions exist.
这里的符号 ± 表示你要计算两个值：一个取加号，另一个取减号。平方根内的 b² - 4ac 决定了存在多少个实数解。
For 3x² - 5x - 2 = 0, we have a = 3, b = -5 and c = -2. Substitute carefully and simplify.
使用该公式时，先把方程写成标准形式，然后列出 a、b 和 c。对于 3x² - 5x - 2 = 0，有 a = 3、b = -5、c = -2。仔细代入并化简。
8. The Discriminant and Nature of Roots | 判别式与根的性质
e discriminant. It tells you how many real solutions a quadratic equation has without actually solving it. This is a very common IGCSE examination question.
表达式 D = b² - 4ac 称为判别式。它能告诉你在不实际求解的情况下，二次方程有多少个实数解。这是 IGCSE 考试中非常常见的题目。
- If D > 0, the equation has two distinct real roots. — 如果 D > 0，方程有两个不相等的实根。
- If D = 0, the equation has one repeated real root. — 如果 D = 0，方程有一个重根。
- If D < 0, the equation has no real roots. — 如果 D < 0，方程没有实根。
hen D = (-4)² - 4 × 2 × 3 = 16 - 24 = -8. Since D < 0, the equation has no real solutions.
例题：对于 2x² - 4x + 3 = 0，a = 2，b = -4，c = 3。那么 D = (-4)² - 4 × 2 × 3 = 16 - 24 = -8。由于 D < 0，该方程没有实数解。
Being able to calculate the discriminant quickly also helps you predict whether factorising is possible using integers. If D is not a perfect square, the roots will be irrational and the quadratic formula is usually the best choice.
能快速计算判别式还能帮助你预判是否可能用整数因式分解。如果 D 不是完全平方数，根将是无理数，此时求根公式通常是最佳选择。
9. Choosing the Right Method | 选择正确的方法
Your choice of method depends on the form of the quadratic and the demands of the question. If the question says 'solve by factorising' or 'complete the square', follow that instruction. If no method is specified, use the one that is most efficient.
方法的选择取决于二次式的形式以及题目要求。如果题目说 '用因式分解法求解' 或 '配方'，请遵循该指令。如果没有指定方法，请使用最有效的一种。
| Situation 情况 | Best method 最佳方法 |
| Simple integer factors are obvious 简单的整数因式明显
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