📚 182. Quadratic Equations – Solving by Factorising, Formula and Graphs | 182. 二次方程：因式分解、公式与图像解法
Quadratic equations lie at the very heart of IGCSE Mathematics, and Edexcel examiners love to test them in both papers. Whether you are factorising a simple trinomial, applying the quadratic formula, or interpreting a parabola on a graph, a solid understanding of quadratics can unlock a large number of marks. This article covers every method you need to solve quadratic equations at IGCSE level, linking each technique to the demands of the Edexcel specification. From spotting the standard form to avoiding common sign errors, you will find clear explanations supported by step‑by‑step examples.
二次方程是 IGCSE 数学的核心内容，Edexcel 考试经常在卷一和卷二中出现相关题目。无论是因式分解一个简单的三项式、套用二次公式，还是通过图像解读抛物线，扎实掌握二次方程都能为你稳稳拿下大量分数。本文涵盖 IGCSE 阶段解二次方程的所有方法，并将每种技巧与 Edexcel 考纲的要求紧密挂钩。从识别标准形式到避开常见的符号错误，你将读到清晰的讲解，并配有循序渐进的示例。
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 general form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The exponent of the highest‑degree term is 2, which is why we call it 'quadratic' — from the Latin word 'quadratus' meaning 'square'. If a were allowed to be zero, the equation would collapse to a linear one, losing its characteristic curved shape when graphed.
任何可以写成一般形式 ax² + bx + c = 0 的方程都叫二次方程，其中 a、b、c 是常数且 a ≠ 0。最高次项的次数为 2，这也是 'quadratic' 一词的来源——拉丁语 'quadratus' 意为‘平方’。如果允许 a 为零，方程就会退化成一次方程，画出来的图形也不再具备抛物线特有的弯曲形状。
in standard form, but occasionally you must rearrange terms first. For instance, 3x² = 5x – 2 should be rewritten as 3x² – 5x + 2 = 0 before you begin solving.
在 Edexcel IGCSE 的题目中，二次方程通常已经写成标准形式，但有时你需要先进行整理。例如，3x² = 5x – 2 应先移项为 3x² – 5x + 2 = 0，再着手求解。
2. The Standard Form ax² + bx + c = 0 | 标准式 ax² + bx + c = 0
ons or brackets, expand and simplify first. For example, (x + 3)² = 16 becomes x² + 6x + 9 – 16 = 0, leading to x² + 6x – 7 = 0.
求解二次方程时，务必把所有项移到等号同一侧，使另一边等于零。标准形式 ax² + bx + c = 0 不仅让你直接读出 a、b、c 的值，更是使用二次公式或判断判别式的前提。如果方程含有分数或括号，请先展开并化简。例如，(x + 3)² = 16 可化为 x² + 6x + 9 – 16 = 0，得到 x² + 6x – 7 = 0。
Identifying a, b and c correctly is crucial. In the equation 5 – 2x – x² = 0, it is helpful to multiply through by –1 and reorder as x² + 2x – 5 = 0, so a = 1, b = 2, c = –5. Many marks are lost simply because students misread the sign of c.
准确识别 a、b、c 至关重要。在方程 5 – 2x – x² = 0 中，不妨两边同乘 –1 并整理为 x² + 2x – 5 = 0，这样 a = 1，b = 2，c = –5。许多考生仅仅因为看错 c 的符号而白白丢分。
3. Solving by Factorising – Simple Monic Quadratics | 因式分解法——首项系数为 1 的二次方程
o expressions equals zero, at least one of them must be zero. This gives the solutions x = –p or x = –q.
当 a = 1 时，二次方程称作首一多项式。因式分解 x² + bx + c = 0 时，要寻找两个数，它们的积等于 c，和等于 b。将方程因式分解为 (x + p)(x + q) = 0 后，利用零乘积性质：若两个式子的乘积为零，则至少有一个式子为零。由此得到解 x = –p 或 x = –q。
For example, to solve x² – 5x + 6 = 0, we need two numbers whose product is 6 and sum is –5. The pair (–2) and (–3) works, giving (x – 2)(x – herefore x = 2 or x = 3. Edexcel often asks for both solutions, so never stop at just one.
例如，解方程 x² – 5x + 6 = 0，需要找两个数，积为 6，和为 –5。–2 和 –3 满足要求，于是得到 (x – 2)(x – 3) = 0，从而 x = 2 或 x = 3。Edexcel 常要求写出两个解，千万不要只写一个就结束。
Always check your solutions by substituting them back into the original equation. This habit catches sign errors and builds confidence before moving on to more complex problems.
养成把解代回原方程检验的习惯。这能及时发现符号错误，也能在进入更复杂的题目之前增强你的信心。
4. Factorising When a > 1 | 首项系数大于 1 时的因式分解
o 7: they are 1 and 6. Rewrite 7x as 1x + 6x, giving 2x² + 1x + 6x + 3. Group as (2x² + 1x) + (6x + 3) and factor each group: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
当 x² 的系数不为 1 时，因式分解会稍难一些。你需要找到两个数，它们的积为 a × c，和为 b，然后拆中项并分组提取公因式。以 2x² + 7x + 3 为例，先计算 a × c = 2 × 3 = 6。需要两个数，积为 6，和为 7：正是 1 和 6。把 7x 拆成 1x + 6x，式子变为 2x² + 1x + 6x + 3。分组为 (2x² + 1x) + (6x + 3)，分别提取公因式：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。
Practise the grouping method until it feels automatic. Many Edexcel higher‑tier questions feature quadratics with a > 1, and correctly factorising a non‑monic quadratic can save you the trouble of using the formula or completing the square in a later part of the question.
反复练习分组分解法，直到变成条件反射。Edexcel 进阶卷的题目中经常出现 a > 1 的二次方程，能够熟练分解一个非首一二次式，可以省去你在后续小问中使用公式或配方法的麻烦。
5. Factorising Special Cases – Difference of Two Squares | 特殊情况——平方差公式
Recognise the pattern a² – b² = (a + b)(a – of quadratics, this appears when b = 0 and c is a negative perfect square. For example, x² – 25 = 0 factorises as (x + 5)(x – 5), giving solutions x = –5, 5. Similarly, 4x² – 9 = (2x + 3)(2x – 3). These questions often appear as quick wins in the non‑calculator paper, so spotting them instantly saves time.
你要能识别平方差公式 a² – b² = (a + b)(a – b)。在二次方程中，当 b = 0 且 c 是一个负的完全平方数时，就会出现这种形式。例如，x² – 25 = 0 可分解为 (x + 5)(x – 5)，解为 x = –5，5。类似地，4x² – 9 = (2x + 3)(2x – 3)。这类题在非计算器试卷中经常作为送分题出现，一眼看穿可以节约大量时间。
A common pitfall is to treat x² + 25 as a difference of squares — er the real numbers at IGCSE level, so a quadratic like x² + 25 = 0 has no real solutions.
一个常见误区是把 x² + 25 也当作平方差来处理——这是错误的，因为常数项是正数。在 IGCSE 范围内，平方和在实数范围内无法因式分解，所以像 x² + 25 = 0 这样的方程没有实数解。
6. Completing the Square | 配方法
Completing the square transforms any quadratic into the form (x + p)² – square it, and add and subtract that value inside the expression. The result is (x + b/2)² – (b/2)² + c = 0. For example, with x² + 6x – 7 = 0, half of 6 is 3, and 3² = 9. Rewrite as (x² + 6x + 9 – 9) – 7 = 0, which simplifies to (x + 3)² – 16 = 0. Hence (x + 3)² = 16, giving x + 3 = ±4, so x = 1 or x = –7.
配方法可以将任何二次方程转化为 (x + p)² – q = 0 的形式，再通过移项和开方求出解。从 x² + bx + c 开始，取 b 的一半并平方，在式子中加上并减去这个平方数。最终形式为 (x + b/2)² – (b/2)² + c = 0。例如，x² + 6x – 7 = 0，6 的一半是 3，3² = 9。变形为 (x² + 6x + 9 – 9) – 7 = 0，化简得 (x + 3)² – 16 = 0。于是 (x + 3)² = 16，x + 3 = ±4，解得 x = 1 或 x = –7。
This method is particularly powerful because it works even when the quadratic does not factorise nicely, and it is the stepping stone to deriving the quadratic formula. It is also the go‑to technique when the question specifically asks for answers in surd form or requires the coordinates of a turning point.
配方法的威力在于，即使二次方程不能漂亮地因式分解，它依然有效，同时也是推导二次公式的基石。当题目明确要求以根式形式给出答案，或需要找出抛物线顶点坐标时，配方法也是首选技术。
7. The Quadratic Formula | 二次公式
The quadratic formula provides the solutions of ax² + bx + c = 0 directly:
x = (–b ± √b² – 4ac) / (2a)
You must memorise this formula precisely — la sheet but you cannot afford to waste time looking it up under exam pressure. The symbol '±' indicates there are normally two solutions, corresponding to the '+' and '–' cases.
二次公式直接给出 ax² + bx + c = 0 的解：
x = (–b ± √b² – 4ac) / (2a)
你必须准确记住这个公式——虽然 Edexcel IGCSE 公式表上提供了它，但在考试压力下你可能来不及反复翻看。符号 '±' 表示通常有两个解，分别对应取 '+' 和 '–' 的情况。
b and c from the standard form, substitute them carefully, and compute the discriminant b² – 4ac first. For 2x² – 4x – 6 = 0, a = 2, b = –4, c = –6. Then b² – 4ac = (–4)² – 4(2)(–6) = 16 + 48 = 64. The formula gives x = (4 ± √64) / (4) = (4 ± 8) / 4, yielding x = 3 or x = –1. Always simplify the fraction completely.
使用公式时，先从标准形式中识别出 a、b、c，小心代入，优先计算判别式 b² – 4ac。对于 2x² – 4x – 6 = 0，a = 2，b = –4，c = –6。b² – 4ac = (–4)² – 4(2)(–6) = 16 + 48 = 64。代入公式得 x = (4 ± √64) / (4) = (4 ± 8) / 4，解得 x = 3 或 x = –1。最后一定要把分数彻底化简。
8. Discriminant and Nature of Roots | 判别式与根的性质
The expression b² – 4ac, called the discriminant and often denoted by Δ, determines how many real solutions a quadratic has. The following table summarises the rules:
表达式 b² – 4ac 称为判别式，常记为 Δ，它决定了二次方程有多少个实数解。下表总结了判别规则：
| Condition | Nature of Roots |
|---|---|
| b² – 4ac > 0 | Two distinct real roots |
| b² – 4ac = 0 | One repeated real root (equal roots) |
| b² – 4ac < 0 | No real roots |
| 条件 | 根的性质 |
|---|---|
| b² – 4ac > 0 | 两个不相等的实数根 |
| b² – 4ac = 0 | 一个二重实数根（相等根） |
| b² – 4ac < 0 | 没有实数根 |
Edexcel frequently asks questions such as 'Find the value of k for which the equation has equal roots.' In that case, set b² – 4ac = 0 and solve for k. For example, if x² + kx + 9 = 0 has equal roots, then k² – 4(1)(9) = k² – 36 = 0, so k = ±6.
Edexcel 经常出这样的题：'求 k 的值，使得该方程有重根。' 这时，只需令 b² – 4ac = 0，然后解出 k。例如，若 x² + kx + 9 = 0 有重根，则 k² – 4(1)(9) = k² – 36 = 0，因此 k = ±6。
9. Solving Quadratic Equations Graphically | 图像法解二次方程
The graph of y = ax² + bx + c is a parabola. The x‑ aph are the real solutions of the equation ax² + bx + c = 0. If the parabola touches the x‑axis at one point, the equation has a repeated root; if it does not cross the x‑ on helps when a question provides a sketch or asks you to estimate solutions from a graph.
y = ax² + bx + c 的图像是一条抛物线。图像与 x 轴的交点横坐标就是方程 ax² + bx + c = 0 的实数解。如果抛物线刚好与 x 轴相切于一点，方程就有重根；如果抛物线完全不穿过 x 轴，则二次方程没有实数解。理解这种联系，在题目给出草图或要求你通过图像估算解时会极有帮助。
phically by drawing a suitable straight line on a given parabola. For example, to solve x² – 3x – points. Recognising how to rearrange the target equation to match a graph is a key skill.
Edexcel IGCSE 可能会要求你通过在给定的抛物线上添画一条适当的直线，来图解一个二次方程。例如，为解 x² – 3x – 4 = 0，可先画出 y = x²，再画直线 y = 3x + 4，然后找出交点横坐标。能够将目标方程变形，使其与已给图像匹配，是一项关键技能。
10. Applications – Word Problems and Real‑Life Contexts | 应用题与实际背景
equations often model area, projectile motion, or number puzzles. A classic IGCSE problem: 'The product of two consecutive positive integers is 132. Find the integers.' x, then the next is x + 1. The equation becomes x(x + 1) = 132, which simplifies to x² + x – 132 = 0. Factorising gives (x – 11)(x + 12) = 0. Since x must be positive, x = 11, so the integers are 11 and 12.
二次方程常用来建模面积、投掷运动或数字谜题。一道经典的 IGCSE 题目是：'两个连续正整数的乘积是 132，求这两个数。' 设第一个整数为 x，则下一个为 x + 1。列出方程 x(x + 1) = 132，化简为 x² + x – 132 = 0。因式分解得 (x – 11)(x + 12) = 0。由于 x 必须为正，取 x = 11，因此这两个数是 11 和 12。
When tackling word problems, always define your variable clearly, write the equation in standard form, solve it, and then check whether both mathematical solutions make sense in the real‑world context. Negative lengths or times, for instance, should be rejected.
解应用题时，一定要先明确定义变量，将方程写成标准形式，求解后再检验两个数学解在实际情况中是否都有意义。例如，长度或时间为负数的解应当舍去。
11. Common Mistakes and How to Avoid Them | 常见错误与避坑指南
- Forgetting to set the equation to zero before factorising. Always rearrange first.
- Losing a sign when moving terms across the equals sign.
- Dividing by x when x could be zero, thereby losing the solution x = 0.
- Misapplying the quadratic formula by plugging in wrong a, b or c values, especially when coefficients are negative.
- Stopping after finding only one solution when the discriminant is positive.
- 因式分解前忘记将方程设为零，务必先移项。
- 移项时抄错符号。
- 在 x 可能为零的情况下两边除以 x，从而丢失 x = 0 这个解。
- 代入二次公式时弄错 a、b、c 的值，尤其是系数为负时。
- 当判别式为正时，只求出一个解就停笔。
A disciplined approach — write out a, b, c clearly; check discriminant sign; verify solutions by substitution — will eliminate most of these errors. Practice with past‑paper questions where these traps are deliberately set, and you will learn to spot them instantly.
养成规范化的解题习惯——清晰写出 a、b、c 的值；检查判别式的符号；把解代回原方程验证——就能消除绝大多数错误。多做那些故意设置这些陷阱的真题，你会很快学会一眼识破。
12. Summary and Edexcel Exam Tips | 小结与 Edexcel 考试策略
In Edexcel IGCSE Mathematics, quadratic equations can appear in both the foundation and higher tiers, but higher‑tier papers demand fluency in factorising non‑monic quadratics, completing the square, and using the discriminant. The four main solution methods — factorising, completing the square, quadratic formula, and graphical — are not independent; they reinforce one another. Choose the method that best fits the structure of the equation and the wording of the question.
在 Edexcel IGCSE 数学中，二次方程既出现在基础卷也出现在进阶卷，但进阶卷要求你熟练分解非首一二次式、配方法以及运用判别式。四种主要解法——因式分解、配方法、二次公式和图像法——彼此并不孤立，而是相互强化。要根据方程的结构和题目的措辞，选择最合适的方法。
Always
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