📚 Solving Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 解二次方程：因式分解、配方法与公式法
Quadratic equations are among the most important algebraic skills in IGCSE and A Level Mathematics. They appear in pure algebra questions, coordinate geometry, quadratic graph work and real-life applications such as area optimisation and projectile motion. This article explains the three core methods - factorising, completing the square and the quadratic formula - and shows you how to choose the best one under timed exam conditions, with worked examples and a full list of the mistakes examiners see every year.
二次方程是 IGCSE 与 A Level 数学中最重要的代数技能之一。它出现在纯代数题、坐标几何、二次函数图像题，以及面积优化、抛体运动等实际应用问题中。本文系统讲解三种核心解法——因式分解法、配方法与公式法，并说明在限时考试中如何选择最快的方法，同时配有完整例题与考试中最常见的错误清单。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation in which the highest power of the unknown, usually x, is 2. It can be written in the general form below, where a, b and c are constants and a does not equal zero. The constant a is called the leading coefficient, b is the coefficient of the linear term, and c is the constant term. The word "quadratic" comes from the Latin quadratus, meaning "made square".
二次方程是指未知数（通常为 x）的最高次数为 2 的方程。它可以写成下面的一般形式，其中 a、b、c 为常数，且 a 不等于 0。a 称为二次项系数，b 称为一次项系数，c 称为常数项。"二次"一词源自拉丁语 quadratus，意为"平方"。
a x² + b x + c = 0, a ≠ 0
utions, often called roots. If a = 0 the x² term vanishes, the equation becomes linear, for example b x + c = 0, and it is no longer quadratic. For instance, 2x² + 3x − 5 = 0 is quadratic, while 3x − 5 = 0 is linear.
方程的解就是使等式成立的 x 的取值。由于最高次数为 2，一个二次方程最多有两个不同的解，通常称为根。如果 a = 0，x² 项消失，方程变成一次方程，例如 b x + c = 0，就不再是二次方程。例如 2x² + 3x − 5 = 0 是二次方程，而 3x − 5 = 0 是一次方程。
2. Writing the Equation in Standard Form | 将方程整理为标准形式
Before you use any method, move every term to one side so that the other side equals zero, and write the terms in descending powers of x. This is called standard form. Problems are usually set in a disguised form precisely to test whether you do this step correctly.
在运用任何方法之前，应先把所有项移到一边，使另一边等于 0，并把各项按 x 的降幂排列，这称为标准形式。题目常常故意把方程写成"伪装形式"，正是为了考查你是否能正确完成这一步。
a x² + b x + c = 0
+ 3) = 0, so x = 6 or x = −3. Example with a fraction: 2 / x + x = 3, where x ≠ 0, becomes 2 + x² = 3x after multiplying both sides by x, so x² − 3x + 2 = 0, giving x = 1 or x = 2.
例如：3x + x² = 7 应整理为 x² + 3x − 7 = 0。再如：(x + 2)(x − 5) = 8 必须先展开，得到 x² − 3x − 10 = 8，再整理为 x² − 3x − 18 = 0，于是 (x − 6)(x + 3) = 0，所以 x = 6 或 x = −3。含分式的例子：2 / x + x = 3（x ≠ 0），两边同乘 x 得 2 + x² = 3x，即 x² − 3x + 2 = 0，解得 x = 1 或 x = 2。
le only works when the product equals zero, so you must always rearrange to ... = 0 first.
一个非常常见的陷阱是把 (x + 2)(x − 5) = 8 拆成 x + 2 = 8 和 x − 5 = 8 两个等式。这是错的。零积法则只在乘积等于 0 时才成立，所以务必先整理成"…= 0"的形式。
3. Method 1: Factorising | 方法一：因式分解法
ers whose product is c and whose sum is b. Then write (x + p)(x + q) = 0 and use the zero product rule: if a product of two factors is zero, at least one factor must be zero, so x = −p or x = −q.
当二次式能干净地分解成两个有理系数的线性因式时，因式分解是最快的方法。对于 x² + b x + c = 0，你需要找到两个数，其乘积等于 c、和等于 b。然后写成 (x + p)(x + q) = 0，并运用零积法则：若两个因式的乘积为 0，则至少有一个因式为 0，所以 x = −p 或 x = −q。
since (−2) × (−3) = 6 and (−2) + (−3) = −5. So (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
例 A：x² − 5x + 6 = 0。两个数为 −2 与 −3，因为 (−2) × (−3) = 6，(−2) + (−3) = −5。所以 (x − 2)(x − 3) = 0，解得 x = 2 或 x = 3。
When the leading coefficient is not 1, use the "ac method" 7, so the numbers are 6 and 1. Split: 2x² + 6x + x + 3 = 0, then 2x(x + 3) + 1(x + 3) = 0, so (2x + 1)(x + 3) = 0 and x = −1/2 or x = −3.
当二次项系数不为 1 时，可用"ac 法"：找两个数，其乘积为 a × c、和为 b，把中间项拆开，再分组分解。例 B：2x² + 7x + 3 = 0。此处 a × c = 6，b = 7，两个数为 6 与 1。拆项得 2x² + 6x + x + 3 = 0，即 2x(x + 3) + 1(x + 3) = 0，所以 (2x + 1)(x + 3) = 0，解得 x = −1/2 或 x = −3。
- Difference of two squares: x² − 9 = (x − 3)(x + 3). | 平方差：x² − 9 = (x − 3)(x + 3)。
- Perfect square: x² + 6x + 9 = (x + 3)². | 完全平方：x² + 6x + 9 = (x + 3)²。
- Always expand your brackets mentally to check the middle term. | 一定要在脑中展开括号，检验一次项是否正确。
4. Method 2: Completing the Square | 方法二：配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. It is the method examiners ask for when they want the vertex of a parabola, the minimum or maximum value, or a proof of the quadratic formula. The rule is simple: halve the coefficient of x, square it, add it and subtract it.
配方法把二次式改写为 a(x + p)² + q 的形式。当题目要求抛物线的顶点、最大值或最小值，或要求证明求根公式时，考官通常指定使用这种方法。做法很简单：取一次项系数的一半，平方，加上再减去。
x² + b x + c = (x + b/2)² + c − b²/4
e expression x² + 6x − 7 also equals (x + 3)² − 16, so the minimum value is −16 when x = −3.
例：x² + 6x − 7 = 0。6 的一半是 3，3² = 9，所以 x² + 6x = (x + 3)² − 9。方程变为 (x + 3)² − 9 − 7 = 0，即 (x + 3)² = 16，于是 x + 3 = ±4，解得 x = 1 或 x = −7。同时 x² + 6x − 7 = (x + 3)² − 16，因此当 x = −3 时取得最小值 −16。
4] + 3 = 2(x − 2)² − 8 + 3 = 2(x − 2)² − 5. So 2(x − 2)² = 5, (x − 2)² = 5/2, and x = 2 ± √5/2 = 2 ± √10 / 2. The curve has a minimum point (2, −5) and its line of symmetry is x = 2.
当 a ≠ 1 时的例子：2x² − 8x + 3 = 0。先把 x 的项提出因子 2：2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 8 + 3 = 2(x − 2)² − 5。于是 2(x − 2)² = 5，(x − 2)² = 5/2，x = 2 ± √5/2 = 2 ± √10 / 2。该曲线有最小值点 (2, −5)，对称轴为 x = 2。
5. Method 3: The Quadratic Formula | 方法三：公式法
The quadratic formula solves any quadratic equation, even when the roots are irrational or the quadratic refuses to factorise. Learn it by heart; it is not given in most IGCSE papers.
求根公式可以解任何二次方程，即使根是无理数、或因式分解根本做不出来也能用。请务必背熟它——大多数 IGCSE 试卷不会给出这个公式。
x = ( −b ± √b² − 4ac ) / (2a)
= −2. First compute the discriminant: b² − 4ac = (−5)² − 4 × 3 × (−2) = 25 + 24 = 49. Then √49 he minus sign gives x = −2 / 6 = −1/3. Check: 3(2)² − 5(2) − 2 = 12 − 10 − 2 = 0, correct.
例：3x² − 5x − 2 = 0，所以 a = 3，b = −5，c = −2。先算判别式：b² − 4ac = (−5)² − 4 × 3 × (−2) = 25 + 24 = 49。再取 √49 = 7，得 x = (5 ± 7) / (2 × 3) = (5 ± 7) / 6。取加号得 x = 12 / 6 = 2；取减号得 x = −2 / 6 = −1/3。检验：3(2)² − 5(2) − 2 = 12 − 10 − 2 = 0，正确。
Two practical tips. First, always write down a, b and c explicitly before substituting, so you do not confuse them. Second, when b is negative, use brackets: −(−5) becomes +5. Also remember that the whole numerator is divided by 2a, not just the −b part.
两个实用技巧。第一，代入前先把 a、b、c 明确写出来，避免张冠李戴。第二，当 b 为负数时要用括号：−(−5) 变成 +5。还要记住，分母 2a 是除整个分子，而不仅仅是 −b 那一项。
6. The Discriminant and the Number of Roots | 判别式与根的个数
nt because it discriminates between the possible cases. Its sign alone tells you how many real roots the equation has, without solving it.
根号下的表达式 Δ = b² − 4ac 称为判别式，因为它能"判别"各种情况。只凭它的符号，就能在不求解的情况下判断方程有多少个实根。
| Value of Δ | Δ 的取值 | Number of real roots | 实根个数 | Graph behaviour | 图像表现 |
| Δ > 0 | Two distinct real roots | 两个不同实根 | Crosses the x-axis twice | 与 x 轴交于两点 |
| Δ = 0 | One repeated root | 一个重根 | Touches the x-axis | 与 x 轴相切 |
| Δ < 0 | No real roots | 无实根 | Does not meet the x-axis | 与 x 轴不相交 |
4, which is negative, so there are no real roots. If Δ is a perfect square, the roots are rational and the quadratic factorises over the integers; if Δ is not a perfect square, the roots are irrational surds.
例：x² + 4x + 5 = 0，Δ = 16 − 20 = −4，为负数，所以没有实根。若 Δ 是完全平方数，则根为有理数，二次式可在整数范围内因式分解；若 Δ 不是完全平方数，则根为无理数（根式）。
7. Choosing the Best Method Under Exam Pressure | 考场中如何选择最佳方法
Spend ten seconds checking whether the quadratic factorises before you commit to a long method. If two integers with the right sum and product exist, factorising is by far the fastest. If the question says "by completing the square" or "express in the form a(x + p)² + q", you must use that method even if factorising would work.
先用十秒判断二次式能否因式分解，再决定用哪种方法。如果确实存在满足"积与和"的两个整数，因式分解明显最快。如果题目明确要求"用配方法"或"表示为 a(x + p)² + q 的形式"，即使能因式分解也必须按要求作答。
- Factorising: quickest for integer roots; also gives the x-intercepts directly. | 因式分解：整数根时最快，还能直接读出 x 轴截距。
- Completing the square: required for vertex, minimum or maximum values, and for proving the formula. | 配方法：求顶点、最值，以及证明公式时的指定方法。
- Quadratic formula: the universal fallback, ideal when coefficients are large, decimal or surd roots appear, or when you can use a calculator. | 公式法：万能后备方案，系数较大、出现小数或根式根，或允许使用计算器时最合适。
Whichever method you use, finish with a quick substitution of your answers into the original equation. Thirty seconds of checking can save several marks.
无论用哪种方法，最后都要把答案代回原方程快速检验。三十秒的检查往往能救回好几分。
8. Quadratic Graphs, Roots and the Vertex | 二次函数图像、根与顶点
0 are exactly the x-coordinates where the parabola meets the x-axis, which is why the discriminant counts intersections. If a > 0 the parabola opens upwards and has a minimum point; if a < 0 it opens downwards and has a maximum point.
y = a x² + b x + c 的图像是一条抛物线。方程 a x² + b x + c = 0 的根，正是抛物线与 x 轴交点的横坐标——这也解释了为什么判别式能"数出"交点个数。若 a > 0，抛物线开口向上，有最小值点；若 a < 0，开口向下，有最大值点。
Line of symmetry: x = −b / (2a) Vertex: ( −b / (2a), c − b² / (4a) )
e vertex instantly. Factorised form (x − p)(x − q) shows the intercepts p and q, while completed-square form shows the vertex; exam questions often ask you to move between these three forms.
例：y = x² − 4x + 1。对称轴为 x = 4 / 2 = 2，顶点为 (2, 1 − 4) = (2, −3)。写成配方式 y = (x − 2)² − 3，可立刻看出同一顶点。因式分解式 (x − p)(x − q) 直接给出 x 轴截距 p 与 q，配方式给出顶点；考题经常要求在这三种形式
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