📚 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
A quadratic equation is one of the most important algebraic skills in the IGCSE Mathematics syllabus. It appears in algebra, coordinate geometry, mensuration and word problems, so a confident method for solving quadratics is essential for high marks. This guide covers the standard form, the zero product property, factorising, completing the square, the quadratic formula, the discriminant, method selection, modelling questions and the mistakes that cost candidates the most marks.
二次方程是 IGCSE 数学大纲中最重要的代数技能之一。它出现在代数、坐标几何、测量和应用题中，因此熟练掌握解二次方程的方法对取得高分至关重要。本文涵盖标准形式、零积性质、因式分解、配方法、求根公式、判别式、方法选择、建模应用题，以及最容易失分的常见错误。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation that can be written in the standard form below, where a, b and c are constants and a ≠ 0. The highest power of the unknown is 2, and that single condition is what makes the equation quadratic rather than linear.
二次方程是可以写成下面标准形式的方程，其中 a、b、c 为常数且 a ≠ 0。未知数的最高次数是 2，正是这个条件使方程成为二次方程而不是一次方程。
a x² + b x + c = 0, a ≠ 0
uation becomes linear, so it would be solved by simple rearrangement instead. Here a is the coefficient of x², b is the coefficient of x, and c is the constant term.
如果 a = 0，x² 项消失，方程就变成一次方程，需要用简单移项来解。这里 a 是 x² 的系数，b 是 x 的系数，c 是常数项。
ns. The solutions of a quadratic equation are also called roots, or x-intercepts, because on a graph they are the points where the curve cuts the x-axis.
例如 x² - 5x + 6 = 0、2x² + 3x - 2 = 0 和 x² - 9 = 0 都是二次方程。二次方程的解也称为根或 x 轴截距，因为在图像上它们就是曲线与 x 轴相交的点。
In IGCSE questions, you are usually asked to solve a quadratic equation by factorising, completing the square or using the quadratic formula. Choosing the fastest method saves time and reduces errors. Quadratic equations are also used for modelling real-life situations, such as finding the area of a shape, calculating compound interest or predicting the path of a projectile.
在 IGCSE 题目中，通常会要求你用因式分解、配方法或求根公式来解二次方程。选择最快的方法可以节省时间并减少错误。二次方程在建立现实情境模型时也非常有用，例如求图形面积、计算复利或预测抛体运动轨迹。
2. The Zero Product Property | 零积性质
e of those numbers must be zero. Algebraically, if p q = 0, then p = 0 or q = 0 (or both).
零积性质是因式分解法的逻辑基础：如果两个数的乘积为零，那么其中至少有一个数为零。用代数表示，若 p q = 0，则 p = 0 或 q = 0（或两者同时为 0）。
nnot conclude that x - 2 = 6 or x - 3 = 6. Nothing useful follows, so the first step of every factorising solution is to rearrange the equation to equal zero.
这个性质之所以成立，只因为右边是零。如果你把 x² - 5x + 6 分解为 (x - 2)(x - 3)，但方程等于 6 而不是 0，你无法得出 x - 2 = 6 或 x - 3 = 6。这样推不出任何有用结论，所以因式分解解法的第一步永远是先把方程整理成等于零的形式。
(x - 2)(x - 3) = 0 ⇒ x - 2 = 0 or x - 3 = 0 ⇒ x = 2 or x = 3
Once the quadratic is written as a product of two brackets equal to zero, each bracket can be set to zero in turn and solved independently. This turns one difficult quadratic into two easy linear equations, which is exactly why factorising is the quickest route whenever the factors are obvious.
一旦二次式写成两个因式之积等于零的形式，就可以依次令每个因式等于零并分别求解。这样就把一个困难的二次方程转化为两个简单的一次方程，这也正是当因式容易看出时因式分解法最快的原因。
3. Solving by Factorising (a = 1) | 因式分解法（a = 1）
Factorising relies on the zero product property: once the quadratic is written as a product of two brackets equal to zero, each bracket can be set to zero. When the coefficient of x² is 1, you need two numbers that multiply to give c and add to give b.
因式分解法依赖于零积性质：一旦二次式写成两个因式之积等于零，就可以令每个因式等于零。当 x² 的系数为 1 时，你需要找两个数，使它们的乘积等于 c，和等于 b。
ratic factorises as (x - 2)(x - 3) = 0. Setting each bracket to zero gives x = 2 or x = 3.
以 x² - 5x + 6 = 0 为例。这里 b = -5，c = 6。数对 -2 与 -3 的乘积是 +6，和是 -5，因此该二次式分解为 (x - 2)(x - 3) = 0。令每个因式等于零，得到 x = 2 或 x = 3。
- If c is positive, both numbers have the same sign, and it matches the sign of b. | 如果 c 为正，两个数同号，且符号与 b 相同。
- If c is negative, the two numbers have opposite signs. | 如果 c 为负，两个数异号。
- If b = 0, the numbers are equal in size but opposite in sign. | 如果 b = 0，两个数大小相同、符号相反。
6, which matches the original. A five-second check prevents most lost marks in this topic.
一定要在脑中展开检验：(x - 2)(x - 3) = x² - 3x - 2x + 6 = x² - 5x + 6，与原式一致。花五秒钟检验，可以避免这个专题里的大部分失分。
4. Factorising When a ≠ 1 and Special Cases | a ≠ 1 的因式分解与特殊情形
to -4 and add to b = 3: these are +4 and -1. Split the middle term and factorise in pairs.
当 x² 的系数不为 1 时，寻找过程稍微长一些。对于 2x² + 3x - 2 = 0，先算 a 与 c 的乘积得 -4，再找两个数使其乘积为 -4、和为 b = 3：即 +4 与 -1。拆分中间项后分组分解。
2x² + 4x - x - 2 = 2x(x + 2) - 1(x + 2) = (2x - 1)(x + 2)
eger roots are perfectly acceptable in IGCSE answers, and they are a strong hint that the quadratic formula would also work well here.
令每个因式等于零，得 2x - 1 = 0 或 x + 2 = 0，所以 x = ½ 或 x = -2。在 IGCSE 中，非整数根完全可接受，而且这往往提示求根公式在这里也同样好用。
, because x³ - 4x = x(x² - 4) = x(x - 2)(x + 2) turns a cubic into three simple linear statements.
有两种特殊结构反复出现。平方差没有一次项：x² - 9 = (x - 3)(x + 3)，所以 x = ±3。永远要先看有没有公因式，因为 x³ - 4x = x(x² - 4) = x(x - 2)(x + 2)，把一个三次式化成三个简单的一次式结论。
5. Completing the Square: The Technique | 配方法：技巧
Completing the square rewrites a quadratic in the form (x + p)² + q. It is the method of choice when the question asks for the completed-square form, the turning point of a graph, or when factorising is not possible with integers.
配方法把二次式改写为 (x + p)² + q 的形式。当题目要求配方式、要求图像的顶点，或无法用整数因式分解时，这是首选方法。
For x² + bx, take half of b and square it. Add and subtract that value so the expression is unchanged in value:
对于 x² + b x，取 b 的一半并平方。加上再减去这个值，表达式的值保持不变：
x² + b x = (x + b ÷ 2)² - (b ÷ 2)²
(x + 3)² - 8. The completed-square form is therefore (x + 3)² - 8, giving a minimum turning point at (-3, -8).
例题：x² + 6x + 1。6 的一半是 3，3² = 9，所以 x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8。因此配方式为 (x + 3)² - 8，最小顶点坐标为 (-3, -8)。
, write 2(x² - 4x) + 5 = 2[(x - 2)² - 4] + 5 = 2(x - 2)² - 3, so the vertex is at (2, -3).
如果 a 不为 1，先把 a 从前两项中提出来再配方。对于 2x² - 8x + 5，写成 2(x² - 4x) + 5 = 2[(x - 2)² - 4] + 5 = 2(x - 2)² - 3，顶点为 (2, -3)。
6. Solving by Completing the Square | 用配方法解方程
. Remember that every square root has two signs, so the ± symbol must appear at that step.
用配方法解二次方程时，要把平方因式单独留在一边，然后对两边开平方。注意每个平方根都有两个符号，因此 ± 符号必须在这一步出现。
pleting the square gives (x + 3)² - 8 = 0, so (x + 3)² = 8. Taking square roots, x + 3 = ±√8, hence x = -3 ± √8 = -3 ± 2√2, which is approximately 1.83 or -4.83 to three significant figures.
解 x² + 6x + 1 = 0。配方得 (x + 3)² - 8 = 0，所以 (x + 3)² = 8。开平方得 x + 3 = ±√8，因此 x = -3 ± √8 = -3 ± 2√2，约等于 1.83 或 -4.83（保留三位有效数字）。
Simplify surds fully unless the question asks for decimals. √8 = 2√2 because 8 = 4 × 2 and √4 = 2. Leaving the answer as -3 ± √8 is usually acceptable, but -3 ± 2√2 is the cleaner form examiners prefer when the instruction says 'leave your answer in surd form'.
除非题目要求小数，否则要把根式化到最简。√8 = 2√2，因为 8 = 4 × 2 且 √4 = 2。把答案写成 -3 ± √8 通常可以接受，但当题目要求"以根式形式作答"时，-3 ± 2√2 是考官更喜欢的简洁形式。
rmula is nothing more than completing the square on a x² + b x + c = 0 in full generality.
这个方法也是通往求根公式的桥梁，因为推导公式本质上就是对方程 a x² + b x + c = 0 一般地进行配方。
7. Deriving the Quadratic Formula | 推导求根公式
+ (c ÷ a) = 0. Move the constant term to the right, then complete the square on the left.
从 a x² + b x + c = 0（a ≠ 0）出发，每项除以 a：x² + (b ÷ a)x + (c ÷ a) = 0。把常数项移到右边，再对左边配方。
(x + b ÷ 2a)² = (b² - 4ac) ÷ 4a²
Taking the square root of both sides gives x + b ÷ 2a = ±√b² - 4ac ÷ 2a, and subtracting b ÷ 2a from both sides produces the quadratic formula.
两边开平方得 x + b ÷ 2a = ±√b² - 4ac ÷ 2a，再从两边减去 b ÷ 2a，就得到求根公式。
x = (-b ± √b² - 4ac) ÷ 2a
comes from, why the denominator is 2a, and why the expression under the root, b² - 4ac, controls the number of solutions.
除非题目明确要求展示推导过程，否则考试中不需要复现这一推导，但理解它能解释 ± 从哪里来、为什么分母是 2a，以及为什么根号下的表达式 b² - 4ac 决定了解的数量。
8. Using the Quadratic Formula | 使用求根公式
The quadratic formula solves any quadratic equation, which makes it the most reliable fallback. Write the equation in standard form first, then identify a, b and c carefully, including their signs.
求根公式能解任何二次方程，因此是最可靠的保底方法。先写成标准形式，再仔细确定 a、b、c 的值，包括它们的符号。
2x² + 3x - 2 = 0. Here a = 2, b = 3, c = -2, so b² - 4ac = 9 - 4(2)(-2) = 9 + 16 = 25, and √25 Both roots match the factorised answer from Section 4, which is a good consistency check.
解 2x² + 3x - 2 = 0。这里 a = 2，b = 3，c = -2，所以 b² - 4ac = 9 - 4(2)(-2) = 9 + 16 = 25，√25 = 5。代入得 x = (-3 ± 5) ÷ 4，所以 x = 2 ÷ 4 = ½ 或 x = -8 ÷ 4 = -2。两个根与第 4 节因式分解的结果一致，这是一个很好的自检。
- Always write down a, b and c before substituting, and bracket negative values. | 代入前先写出 a、b、c，并把负值加括号。
- Do the discriminant b² - 4ac as a separate line to avoid arithmetic slips. | 把判别式 b² - 4ac 单独写一行，避免算术失误。
- Give answers to the required accuracy, usually 3 significant figures or 2 decimal places. | 按题目要求的精度作答，通常是三位有效数字或两位小数。
- When the discriminant is not a perfect square, simplify the surd or give decimals as instructed. | 当判别式不是完全平方数时，按指示化简根式或给出小数。
9. The Discriminant b² - 4ac | 判别式 b² - 4ac
c. Because you cannot take the square root of a negative number in real numbers, its sign tells you exactly how many real solutions the equation has.
判别式是公式中根号下的部分，Δ = b² - 4ac。由于在实数范围内不能对负数开平方，它的符号直接告诉你方程有多少个实数解。
| Discriminant | 判别式 | Number of real roots | 实数根个数 | Graph behaviour | 图像特征 |
|---|---|---|
| b² - 4ac > 0 | Two distinct real roots | 两个不同实根 | Curve crosses the x-axis twice | 曲线与 x 轴相交两次 |
| b² - 4ac = 0 | One repeated real root | 一个重根 | Curve touches the x-axis | 曲线与 x 轴相切 |
| b² - 4ac < 0 | No real roots | 无实数根 | Curve never meets the x-axis | 曲线不与 x 轴相交 |
olve the resulting equation, which is usually linear or a simple quadratic in the unknown.
考试中非常常见的一种题给出一个未知系数，要求你求出使方程有等根的值。令 b² - 4ac = 0，再解得到的方程，通常是关于该未知数的一次方程或简单二次方程。
hat both roots of the discriminant equation are valid answers, and questions worth several marks expect you to find both.
例如，当 k² - 36 = 0 时，x² + k x + 9 = 0 有等根，所以 k = 6 或 k = -6。注意判别式方程的两个解都是有效答案，分值较高的题目会期望你找全两者。
10. Choosing the Best Method | 选择最优方法
All three methods give the same roots, so the choice is about speed, accuracy and what the question asks for. Read the command word before you start: 'solve by completing the square' and 'find the coordinates of the turning point' both point to completing the square.
三种方法给出相同的根，所以选择的关键是速度、准确性以及题目的要求。动笔前先看题干指令："用配方法解"和"求顶点坐标"都指向配方法。
- Factorising: fastest when the roots are integers or simple fractions. | 因式分解：当根是整数或简单分数时最快。
- Completing the square: use when the question asks for the completed-square form or the maximum or minimum point. | 配方法：当题目要求配方式或最大、最小值点时使用。
- Quadratic formula: use when factorising fails, when coefficients are awkward, or when the discriminant is needed. | 求根公式：因式分解行不通、系数复杂或需要判别式时使用。
A quick check before factorising: calculate b² - 4ac. If it is a perfect square such as 1, 4, 9, 16, 25, 36, then the roots are rational and factorising will work. If it is negative or not a perfect square, go straight to the formula.
因式分解前的快速判断：先算 b² - 4ac。如果它是完全平方数，如 1、4、9、16、25、36，那么根是有理数，因式分解可行。如果它为负或不是完全平方数，直接使用求根公式。
11. Word Problems and Modelling | 应用题与建模
Applied quadratic questions always follow the same pattern: form the equation from the information given, solve it, then interpret the roots in context. Forming the equation is normally worth more marks than solving it, so show that step clearly.
二次方程应用题遵循同样的套路：根据已知信息列出方程、解方程、再把根放回情境中解释。列方程通常比解方程分值更高，因此要把这一步写清楚。
8)(x - 5) = 0, so x = 5 or x = -8. Since a length cannot be negative, the width is 5 cm and the length is 8 cm.
例题：一个矩形的长比宽多 3 cm，面积为 40 cm²。设宽为 x，则 x(x + 3) = 40，得 x² + 3x - 40 = 0。因式分解得 (x + 8)(x - 5) = 0，所以 x = 5 或 x = -8。由于长度不能为负，宽为 5 cm，长为 8 cm。
0: -5t² + 20t = 0, so -5t(t - 4) = 0 and t = 4 (rejecting t = 0, the launch moment). The ball reaches 15 m when -5t² + 20t - 15 = 0, that is t² - 4t + 3 = 0, so t = 1 or t = 3.
抛体问题同样常见。若一个球在 t 秒后的高度（米）由 h = -5t² + 20t 给出，则当 h = 0 时球落地：-5t² + 20t = 0，即 -5t(t - 4) = 0，得 t = 4（舍去 t = 0，即抛出时刻）。当 -5t² + 20t - 15 = 0，即 t² - 4t + 3 = 0 时，球达到 15 m，所以 t = 1 或 t = 3。
Always reject roots that are physically impossible: negative lengths, negative times, or answers exceeding a stated domain. Examiners award a mark for correct interpretation, and 'the ball is at 15 m at 1 s and again at 3 s' shows you understand that the ball passes the same height twice.
一定要舍去物理上不可能的根：负的长度、负的时间，或超出题目给定范围的答案。考官会为正确解释给
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