📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
Quadratic equations are one of the most important topics in IGCSE and A-Level Mathematics. They connect algebra, graphs, and real-world problem solving, and they appear again and again in later topics such as functions, calculus and coordinate geometry. This guide covers the standard form, expanding brackets, three main solution methods - factorising, completing the square and the quadratic formula - plus the discriminant, quadratic graphs, modelling and exam technique.
二次方程是 IGCSE 与 A-Level 数学中最重要的专题之一。它把代数、图像和实际问题的求解联系在一起，并且在函数、微积分、解析几何等后续内容中反复出现。本指南涵盖标准形式、括号展开、三种主要解法（因式分解、配方法与求根公式），以及判别式、二次函数图像、数学建模与应试技巧。
Every method in this article solves the same equation in a different way. Factorising is fast when the roots are simple integers; completing the square reveals the vertex of the graph; the quadratic formula always works, even when nothing else does. Choosing the right tool is the key exam skill.
本文中的每一种方法都在用不同方式求解同一个方程。当根是简单整数时，因式分解最快；配方法能直接揭示图像的顶点；而求根公式永远有效，即使其他方法都行不通。选择正确的工具，正是考试的关键能力。
1. Recognising Quadratic Equations | 识别二次方程
≠ 0 is essential: if a = 0 the quadratic term vanishes and the equation becomes linear, so the condition a ≠ 0 is exactly what keeps the equation quadratic.
一元二次方程是指可以整理成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。条件 a ≠ 0 至关重要：若 a = 0，二次项消失，方程就退化为一次方程，因此 a ≠ 0 正是保证方程仍为二次的条件。
ax² + bx + c = 0, a ≠ 0
ing. For example, 4x − x² = 5 should become x² − 4x + 5 = 0. It is also worth identifying a, b and c correctly, including negative signs and fractions, because every later method depends on them.
各项的书写顺序可能不同，方程也未必一开始就是标准形式。求解前通常要按 x 的降幂重新排列。例如，4x − x² = 5 应先化为 x² − 4x + 5 = 0。同时要准确识别 a、b、c 的取值，包括负号和分数，因为后续每一种方法都依赖它们。
- a is the coefficient of x² (the quadratic coefficient). / a 是 x² 的系数（二次项系数）。
- b is the coefficient of x (the linear coefficient). / b 是 x 的系数（一次项系数）。
- c is the constant term, including its sign. / c 是常数项，必须带上它的正负号。
2. Expanding Double Brackets | 展开两个括号
Expanding two binomials such as (x + 3)(x − 2) uses the distributive law, often remembered as FOIL: multiply the First, Outer, Inner and Last terms, then collect like terms. Expansion matters because factorising is simply expansion in reverse, so confident expansion makes every factorisation checkable.
展开两个二项式（如 (x + 3)(x − 2)）可以使用分配律，常记作 FOIL 法则：依次相乘首项、外项、内项和末项，然后合并同类项。展开之所以重要，是因为因式分解正是展开的逆过程，因此熟练的展开技巧能让每一次因式分解都可以被检验。
(x + 3)(x − 2) = x² − 2x + 3x − 6 = x² + x − 6
that the middle terms usually combine, which is why the final answer has only three terms.
当系数更大时，仍然是同样的四个乘积。例如 (2x − 5)(3x + 1) = 6x² + 2x − 15x − 5 = 6x² − 13x − 5。注意中间两项通常会合并，这就是最终结果只有三项的原因。
e identity is the whole basis of simple quadratic factorisation, because it tells you that the coefficient of x is the sum of the two numbers and the constant term is their product.
有一个模式值得牢记：(x + p)(x + q) = x² + (p + q)x + pq。这个恒等式就是简单二次三项式因式分解的全部基础，因为它告诉你：x 的系数是两个数之和，常数项是两个数之积。
3. Factorising Simple Monic Quadratics | 因式分解简单首一二次三项式
A monic quadratic has a coefficient of 1 in front of x², such as x² + 7x + 10. To factorise it, find two numbers that multiply to give the constant term and add to give the coefficient of x, then write them inside two brackets.
首一二次三项式是指 x² 项系数为 1 的二次式，例如 x² + 7x + 10。因式分解时，找出两个数，使它们相乘等于常数项、相加等于 x 的系数，然后把它俩写进两个括号里。
x² + 7x + 10 = (x + 2)(x + 5)
he constant term controls the signs of the two numbers: if the constant term is positive, both numbers have the same sign; if it is negative, the two numbers have opposite signs. The sign of the middle coefficient then decides whether that common sign is positive or negative.
对于 x² + 7x + 10，数 2 和 5 符合条件，因为 2 × 5 = 10 且 2 + 5 = 7。常数项的正负决定了两个数的符号：若常数项为正，两个数同号；若常数项为负，两个数异号。而一次项系数的正负，则决定这个共同的符号是正还是负。
| Quadratic | 二次式 | Factorised form | 因式分解形式 | Reason | 理由 |
| x² + 7x + 10 | (x + 2)(x + 5) | 2 × 5 = 10, 2 + 5 = 7 |
| x² − 5x + 6 | (x − 2)(x − 3) | (−2) × (−3) = 6, sum = −5 |
| x² + 2x − 15 | (x + 5)(x − 3) | 5 × (−3) = −15, sum = 2 |
| x² − 9x + 20 | (x − 4)(x − 5) | (−4) × (−5) = 20, sum = −9 |
If no pair of integers satisfies both conditions, the quadratic may not factorise over the integers at all. In that case, do not force an answer: switch to completing the square or the quadratic formula, and state that the expression does not factorise.
如果找不到任何一对整数同时满足两个条件，那么这个二次式在整数范围内可能根本无法因式分解。此时不要强行凑答案，而应改用配方法或求根公式，并说明该式不可因式分解。
4. Factorising Non-Monic Quadratics | 因式分解非首一二次三项式
When the coefficient of x² is not 1, such as 2x² + 7x + 3, use the ac method (also called splitting the middle term). Multiply a and c, find two numbers with that product and with sum equal to b, then split the middle term and factorise in pairs.
当 x² 的系数不是 1 时（例如 2x² + 7x + 3），应使用 ac 法（也叫拆中项法）。先计算 a 与 c 的乘积，找出乘积等于 ac、和等于 b 的两个数，再把中间项拆开，分组提取公因式。
two numbers that multiply to 6 and add to 7 are 1 and 6. Split the middle term and group:
对于 2x² + 7x + 3，有 a × c = 2 × 3 = 6。乘积为 6、和为 7 的两个数是 1 和 6。拆开中间项并分组：
2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)
4x − 10 = 3x(2x − 5) + 2(2x − 5) = (2x − 5)(3x + 2). Always expand mentally to check that the middle term is correct.
第二个带负常数项的例子是 6x² − 11x − 10。此时 a × c = −60，乘积为 −60、和为 −11 的两个数是 −15 和 4。于是 6x² − 15x + 4x − 10 = 3x(2x − 5) + 2(2x − 5) = (2x − 5)(3x + 2)。务必要在脑中展开检验一次，确认中间项正确。
- Step 1: write down a, b, c with their signs. / 第一步：带符号写出 a、b、c。
- Step 2: compute ac and find the factor pair adding to b. / 第二步：计算 ac，找出和为 b 的因数对。
- Step 3: split the middle term into those two terms. / 第三步：把中间项拆成这两项。
- Step 4: group in pairs, take out common factors, then take out the common bracket. / 第四步：两两分组，提取公因式，再提取公共括号。
- Step 5: expand to verify. / 第五步：展开验算。
5. Special Cases: Difference of Two Squares and Perfect Squares | 特殊情形：平方差与完全平方
The second is the perfect square: a² + 2ab + b² = (a + b)², and a² − 2ab + b² = (a − b)².
有两种模式出现频率极高，应当做到一眼识别。第一种是平方差：a² − b² = (a + b)(a − b)。第二种是完全平方：a² + 2ab + b² = (a + b)²，以及 a² − 2ab + b² = (a − b)²。
a² − b² = (a + b)(a − b) a² ± 2ab + b² = (a ± b)²
3)². Note carefully that a sum of squares such as x² + 9 does not factorise over the real numbers, and a common mistake is to write it as (x + 3)², which expands to x² + 6x + 9 instead.
例如，x² − 49 = (x + 7)(x − 7)，4x² − 25 = (2x + 5)(2x − 5)，x² + 6x + 9 = (x + 3)²。要特别注意：像 x² + 9 这样的平方和式在实数范围内不能因式分解，常见错误是把它写成 (x + 3)²，但后者展开是 x² + 6x + 9，并不相等。
These identities also work in reverse to speed up calculations, and they appear in surds, algebraic fractions and later in calculus. Recognising them saves a great deal of time in the exam.
这些恒等式反过来用还能加快运算速度，并且在根式、代数分式以及后续的微积分中都会出现。考试中能识别它们可以节省大量时间。
6. Solving by Factorising and the Zero Product Rule | 因式分解法求解与零积法则
ymbols, if AB = 0 then A = 0 or B = 0. This is why the equation must always be rearranged so that one side equals zero before factorising.
因式分解法依靠零积法则解方程：若两个因式的乘积为零，则其中至少有一个因式为零。用符号表示，若 AB = 0，则 A = 0 或 B = 0。这就是为什么在因式分解前，必须先把方程整理成一边等于零的形式。
AB = 0 ⇒ A = 0 or B = 0
r x = 3. Both answers should be stated, since a quadratic equation normally has two roots.
例如，x² − 5x + 6 = 0 可分解为 (x − 2)(x − 3) = 0，所以 x − 2 = 0 或 x − 3 = 0，得到 x = 2 或 x = 3。两个答案都要写出，因为二次方程通常有两个根。
ividing both sides by x at the start would lose the root x = 0 and is therefore forbidden.
一个经典陷阱是像 2x² = 8x 这样的方程。移项得 2x² − 8x = 0，再提取公因式得 2x(x − 4) = 0，因此 x = 0 或 x = 4。如果一开始就在两边同时除以 x，就会丢掉 x = 0 这个根，因此绝对不可以这样做。
- Rearrange so that the right-hand side is exactly 0. / 先整理，使等式右边恰好为 0。
- Factorise the left-hand side completely. / 把左边彻底因式分解。
- Set each factor equal to zero and solve each linear equation. / 令每个因式等于 0，分别解一次方程。
- Check both roots in the original equation. / 把两个根代回原方程检验。
7. Completing the Square | 配方法
Completing the square rewrites a quadratic in vertex form, which makes the maximum or minimum point visible immediately. For a monic quadratic, halve the coefficient of x, write it inside a bracket squared, then subtract the square of that half-coefficient.
配方法把二次式改写为顶点形式，从而让最大值或最小值点一目了然。对于首一二次式，先取 x 的系数的一半，写成括号平方，再减去这个半数系数的平方。
x² + bx + c = (x + b/2)² − (b/2)² + c
For example, x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8. Similarly, x² − 4x +
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