📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与二次公式
ng, in inequalities and in word problems about areas, speeds and profits. This revision guide explains the standard form ax² + bx + c = 0 and then works through the three core solution methods: factorising, completing the square and the quadratic formula. You will also learn how the discriminant predicts the number of roots, how completing the square reveals the turning point of a parabola, and how to choose the fastest method under exam pressure.
二次方程是 IGCSE 与 A-Level 代数的核心内容，广泛出现在纯代数题、坐标几何、图像绘制、不等式以及面积、速度、利润等应用题中。本复习指南先讲解标准形式 ax² + bx + c = 0，再逐一梳理三种核心解法：因式分解法、配方法和二次公式法。你还将学习判别式如何预测根的个数、配方法如何揭示抛物线的顶点，以及在考试压力下如何选择最快的方法。
1. What is a Quadratic Equation? | 什么是二次方程？
, b and c are real constants and a ≠ 0. The condition a ≠ 0 is essential: if a were 0 the x² term would vanish and the equation would collapse into a linear equation such as bx + c = 0.
二次方程是次数为 2 的多项式方程，即未知数的最高次数为 2。它的标准形式是 ax² + bx + c = 0，其中 a、b、c 为实数常数且 a ≠ 0。条件 a ≠ 0 至关重要：如果 a = 0，x² 项就会消失，方程退化为 bx + c = 0 这样的一次方程。
(x − 2) = 7 needs expanding and rearranging to 3x² − 6x − 7 = 0 before any method can be applied. The values of x that satisfy the equation are called the roots, solutions or zeros.
典型例子包括 x² − 5x + 6 = 0、2x² + 3x − 2 = 0 和 x² − 4 = 0。有些方程初看并不像二次方程，必须先整理，例如 3x(x − 2) = 7 需要展开并移项为 3x² − 6x − 7 = 0，才能使用任何解法。使方程成立的 x 值称为根、解或零点。
ax² + bx + c = 0, a ≠ 0
s are exactly the x-coordinates where the parabola meets the x-axis. If a > 0 the parabola opens upwards, and if a < 0 it opens downwards.
二次方程可以有两个不同的实根、一个重根，或者没有实根。y = ax² + bx + c 的图像是一条抛物线，方程的根恰好是抛物线与 x 轴交点的横坐标。当 a > 0 时抛物线开口向上，当 a < 0 时开口向下。
2. Standard Form and Coefficients | 标准形式与系数
de and zero on the other. Then identify the coefficients a, b and c carefully, including their signs. Sign errors at this stage are the single most common cause of lost marks in quadratic questions.
使用任何方法之前，都要先把方程整理成标准形式 ax² + bx + c = 0，即把所有项移到一边、另一边为零。然后仔细确定系数 a、b、c，包括它们的正负号。这一步的符号错误是二次方程题目失分最常见的原因。
des gives 5x² − 2x − 1 = 0, so a = 5, b = −2 and c = −1. A careless student writes b = +2 or c = +1 and then produces two wrong roots.
以 3x² − 7x + 2 = 0 为例，a = 3，b = −7，c = 2。注意 b 是负 7 而不是正 7。再假设题目给出 5x² = 2x + 1，两边同时减去 2x 和 1，得到 5x² − 2x − 1 = 0，因此 a = 5，b = −2，c = −1。粗心的学生会写成 b = +2 或 c = +1，最终算出两个错误的根。
| Equation | 方程 | a | b | c |
|---|---|---|---|
| 3x² − 7x + 2 = 0 | 3 | −7 | 2 |
| x² + 4x = 0 | 1 | 4 | 0 |
| 5x² = 2x + 1 | 5 | −2 | −1 |
| x² − 9 = 0 | 1 | 0 | −9 |
| 2x(x − 3) = 5 | 2 | −6 | −5 |
3. Method 1: Factorising (when a = 1) | 方法一：因式分解（a = 1 时）
). So to factorise x² + bx + c you must find two numbers that multiply together to give c and add together to give b.
因式分解是展开括号的逆过程。对于二次项系数为 1 的方程，我们使用恒等式 x² + (p + q)x + pq = (x + p)(x + q)。因此要分解 x² + bx + c，只需找到两个数：它们的乘积等于 c，它们的和等于 b。
ve: they are −2 and −3. Hence x² − 5x + 6 = (x − 2)(x − 3). Expanding to check gives x² − 3x − 2x + 6 = x² − 5x + 6, which is correct.
考虑 x² − 5x + 6 = 0。我们需要两个数，乘积为 +6，和为 −5。因为乘积为正而和为负，两数都必须是负数，即 −2 和 −3。因此 x² − 5x + 6 = (x − 2)(x − 3)。展开检验：x² − 3x − 2x + 6 = x² − 5x + 6，结果正确。
Two special patterns save a great deal of time in the exam.
有两个特殊模式可以在考试中节省大量时间。
- Difference of two squares | 平方差：a² − b² = (a + b)(a − b), so x² − 4 = (x + 2)(x − 2) and 9x² − 25 = (3x + 5)(3x − 5).
- Perfect square trinomials | 完全平方式：x² + 6x + 9 = (x + 3)² and x² − 10x + 25 = (x − 5)².
e, 2x² − 8 = 2(x² − 4) = 2(x + 2)(x − 2), and 3x² + 12x + 12 = 3(x² + 4x + 4) = 3(x + 2)².
永远先检查是否有公因数。例如 2x² − 8 = 2(x² − 4) = 2(x + 2)(x − 2)，3x² + 12x + 12 = 3(x² + 4x + 4) = 3(x + 2)²。
4. Factorising when a ≠ 1 | 当 a ≠ 1 时的因式分解
When the coefficient of x² is not 1, the method of finding two numbers that multiply to c and add to b no longer works directly. The most reliable technique is the ac method, also called splitting the middle term.
当 x² 的系数不是 1 时，直接寻找"乘积为 c、和为 b"的两个数不再适用。最可靠的方法是 ac 法，也叫拆中间项法。
Step 1: compute the product ac. Step 2: find two numbers whose product is ac and whose sum is b. Step 3: split the middle term using those numbers. Step 4: factorise by grouping in pairs.
第一步：计算乘积 ac。第二步：找到两个数，其乘积为 ac、和为 b。第三步：用这两个数把中间项拆开。第四步：分组提取公因式。
We need two numbers with product −4 and sum 3: these are +4 and −1. Split the middle term:
以 2x² + 3x − 2 为例。这里 a = 2，b = 3，c = −2，所以 ac = 2 × (−2) = −4。需要两个数乘积为 −4、和为 3，即 +4 和 −1。拆开中间项：
2x² + 4x − x − 2
ays expect to spend a moment on trial and improvement, and always check by expanding at the end.
现在分组：2x(x + 2) − 1(x + 2) = (x + 2)(2x − 1)。在脑中展开可验证 2x² + 4x − x − 2 = 2x² + 3x − 2。要习惯用试错法寻找这对数，并且最后一定要展开检验。
If no integer pair exists, the quadratic does not factorise over the rational numbers, and you should switch to completing the square or the quadratic formula rather than wasting time.
如果找不到整数对，说明该二次式在有理数范围内不能分解，此时应改用配方法或二次公式，而不是继续浪费时间。
5. Solving by Factorising and the Zero Product Rule | 用因式分解与零积法则求解
0 then A = 0 or B = 0. This turns one quadratic equation into two simple linear equations.
一旦二次式被写成两个一次因式的乘积等于零，就可以使用零积法则：若 A × B = 0，则 A = 0 或 B = 0。这样就把一个二次方程转化为两个简单的一次方程。
giving x = 2 or x = 3. Notice the sign flip: the factor (x − 2) produces the root x = +2.
解 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0，所以 x − 2 = 0 或 x − 3 = 0，即 x = 2 或 x = 3。注意符号反转：因式 (x − 2) 对应根 x = +2。
. From the previous section, (x + 2)(2x − 1) = 0, so x = −2 or x = ½. Substitute both back
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