📚 Solving Quadratic Equations | 解一元二次方程
Quadratic equations appear in nearly every IGCSE Mathematics paper, typically worth 4–8 marks across Paper 2 and Paper 4. Knowing when to factorise, when to complete the square, and when to reach for the quadratic formula is an essential exam skill. This revision guide breaks down each method step by step, with worked examples and common pitfalls highlighted throughout.
一元二次方程几乎出现在每一份 IGCSE 数学试卷中，通常在 Paper 2 和 Paper 4 中占 4–8 分。知道何时用因式分解法、何时用配方法、何时使用求根公式，是至关重要的考试技能。本复习指南将逐步拆解每一种方法，配有完整例题，并全程标注常见易错点。
1. Standard Form of a Quadratic Equation | 一元二次方程的标准形式
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 reason we require a ≠ 0 is that if a were 0, the equation would become linear rather than quadratic.
一元二次方程是指可以写成一般形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。要求 a ≠ 0 的原因在于：如果 a 等于 0，方程就会变成一次方程（线性方程），而不是二次方程。
= 1, b = 4 and c = 0. Notice that b or c (or both) may be zero, but a must never be zero.
例如，在方程 3x² − 5x + 2 = 0 中，a = 3，b = −5，c = 2。在方程 x² + 4x = 0 中，a = 1，b = 4，c = 0。注意：b 或 c（或两者同时）可以为 0，但 a 绝对不能为 0。
s a frequent source of lost marks when students solve 2x² = 8x by dividing both sides by x — a step that loses the root x = 0.
在运用任何解法之前，务必先将方程整理为所有项在等号一侧、另一侧等于 0 的形式。这是常见的失分点：例如解 2x² = 8x 时，如果两边直接除以 x，就会丢失 x = 0 这个根。
2. Solving by Factorisation | 因式分解法
Factorisation is the fastest method when the quadratic has simple integer factors. The method relies on the zero product property: if the product of two expressions is zero, then at least one of them must be zero.
当二次式具有简单的整数因子时，因式分解法是最快的解法。该方法基于“零乘积性质”：如果两个表达式的乘积为零，那么其中至少有一个表达式必须为零。
Example 1: Solve x² − 7x + 12 = 0.
例 1：解方程 x² − 7x + 12 = 0。
ce (−3) × (−4) = 12 and (−3) + (−4) = −7. Hence the factorised form is (x − 3)(x − 4) = 0.
找到两个数，它们相乘得 12，相加得 −7。这两个数是 −3 和 −4，因为 (−3) × (−4) = 12，且 (−3) + (−4) = −7。因此因式分解形式为 (x − 3)(x − 4) = 0。
(x − 3)(x − 4) = 0, so x = 3 or x = 4
(x − 3)(x − 4) = 0，所以 x = 3 或 x = 4
Example 2: ing in pairs: 2x(x + 3) − 1(x + 3) = 0, so (2x − 1)(x + 3) = 0. Therefore x = ½ or x = −3.
例 2：解方程 2x² + 5x − 3 = 0。这里我们需要找到两个数，相乘得 2 × (−3) = −6，相加得 5。这一对数是 6 和 −1。将中间项拆分得 2x² + 6x − x − 3 = 0，然后分组因式分解：2x(x + 3) − 1(x + 3) = 0，所以 (2x − 1)(x + 3) = 0。因此 x = ½ 或 x = −3。
Always expand your factorised answer mentally to check it. If the expansion does not return to the original expression, your factorisation is wrong.
完成后务必在脑中展开因式后的结果进行检验。如果展开后不能回到原表达式，说明因式分解有误。
3. Solving by Completing the Square | 配方法
Completing the square rewrites the quadratic in the form (x + p)² + q. It is especially useful when the equation cannot be factorised easily, and it also reveals the turning point of the graph directly.
配方法将二次式改写为 (x + p)² + q 的形式。当方程不易因式分解时特别有用，同时它也能直接揭示图像的顶点坐标。
Example: Solve x² − 6x + 4 = 0 by completing the square.
例：用配方法解方程 x² − 6x + 4 = 0。
f x, which is −6 ÷ 2 = −3, and square it: (−3)² = 9. Then rewrite the equation as follows:
取 x 项系数的一半，即 −6 ÷ 2 = −3，然后平方：(−3)² = 9。将方程改写如下：
(x − 3)² − 9 + 4 = 0, so (x − 3)² = 5
(x − 3)² − 9 + 4 = 0，所以 (x − 3)² = 5
es. Remember that square roots can be positive or negative, so we must include the ± sign:
现在对两边同时开平方。记住平方根可正可负，因此必须加上 ± 号：
x − 3 = ±√5, so x = 3 ± √5
x − 3 = ±√5，所以 x = 3 ± √5
The two solutions are x = 3 + √5 and x = 3 − √5, which are approximately 5.24 and 0.76. Leave the answers in exact surd form unless the question asks for decimals.
两个解为 x = 3 + √5 和 x = 3 − √5，分别约为 5.24 和 0.76。除非题目要求小数，否则答案应保留精确的根式形式。
2x² + 8x + 5 = 0 becomes 2(x² + 4x) + 5 = 0, then complete the square inside the bracket.
如果 x² 的系数不是 1，先将其提取出来。例如，2x² + 8x + 5 = 0 可化为 2(x² + 4x) + 5 = 0，再在括号内配方。
4. The Quadratic Formula | 求根公式
olutions directly. It works for every quadratic, whether factorisable or not, and is the most reliable method under exam pressure.
对于任意一元二次方程 ax² + bx + c = 0，求根公式可以直接给出解。它对所有二次方程都适用，无论能否因式分解，也是考试压力下最可靠的方法。
x = (−b ± √b² − 4ac) / (2a)
x = (−b ± √b² − 4ac) / (2a)
Example: Solve 2x² + 3x − 5 = 0 using the quadratic formula.
例：用求根公式解方程 2x² + 3x − 5 = 0。
= −5. Substitute these values carefully, paying close attention to the negative sign on c:
这里 a = 2，b = 3，c = −5。小心代入这些数值，特别注意 c 为负数：
x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2)
x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2)
x = (−3 ± √9 + 40) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4
x = (−3 ± √9 + 40) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4
4 = 4/4 = 1, or x = (−3 − 7) / 4 = −10/4 = −2.5. The two solutions are x = 1 and x = −2.5.
因此 x = (−3 + 7) / 4 = 4/4 = 1，或 x = (−3 − 7) / 4 = −10/4 = −2.5。两个解为 x = 1 和 x = −2.5。
When using the formula, write down the substitution line first. This earns method marks even if your final arithmetic goes wrong, and it helps you spot sign errors before you continue.
使用求根公式时，先写出代入步骤。即使最后的计算出错，这一步也能获得方法分，同时帮助你在继续之前发现符号错误。
5. The Discriminant | 判别式
The expression under the square root in the quadratic formula, b² − 4ac, is called the discriminant, often denoted Δ. It tells us the number of real roots without actually solving the equation.
求根公式中根号内的表达式 b² − 4ac 称为判别式，通常记为 Δ。它能在不解方程的情况下告诉我们方程有多少个实数根。
- If Δ > 0: the equation has two distinct real roots. The graph crosses the x-axis at two points.
- 若 Δ > 0：方程有两个不相等的实数根，图像与 x 轴有两个交点。
- If Δ = 0: the equation has exactly one repeated real root. The graph touches the x-axis at one point (the vertex).
- 若 Δ = 0：方程有两个相等的实数根（一个重根），图像与 x 轴相切于一点（即顶点处）。
- If Δ < 0: the equation has no real roots. The graph never crosses the x-axis.
- 若 Δ < 0：方程没有实数根，图像与 x 轴无交点。
Example: tion has one repeated root. Indeed, x² − 4x + 4 factorises as (x − 2)², giving x = 2 only.
例：判断方程 x² − 4x + 4 = 0 的根的性质。这里 a = 1，b = −4，c = 4，所以 Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0。因为 Δ = 0，方程有一个重根。事实上，x² − 4x + 4 可因式分解为 (x − 2)²，解得 x = 2。
as two distinct real roots. Set Δ > 0 and solve the resulting inequality: 4 − 12k > 0, giving k < ⅓.
考试还可能要求你求 k 的范围，使得方程 kx² + 2x + 3 = 0 有两个不相等的实数根。令 Δ > 0 并解所得不等式：4 − 12k > 0，即 k < ⅓。
6. Solving Quadratics Graphically | 图像法解二次方程
bx + c crosses the x-axis. These points are called the roots or x-intercepts of the curve.
方程 ax² + bx + c = 0 的解恰好对应于函数 y = ax² + bx + c 的图像与 x 轴交点的横坐标。这些点称为曲线与 x 轴的交点（即根）。
Example: are the solutions to x² − 4x + 3 = 0, which matches the factorisation (x − 1)(x − 3) = 0.
例：对于图像 y = x² − 4x + 3，曲线在 x = 1 和 x = 3 处与 x 轴相交。这两个值就是方程 x² − 4x + 3 = 0 的解，与因式分解 (x − 1)(x − 3) = 0 的结果一致。
oint) is at (2, −1). Since the coefficient of x² is positive, the vertex is a minimum point.
抛物线的对称轴位于两个根的正中间，即 x = 2。将 x = 2 代入 y = x² − 4x + 3 得 y = −1，所以顶点（驻点）坐标为 (2, −1)。由于 x² 的系数为正，该顶点是最小值点。
solve equations like x² − 4x + 3 = k graphically. This is a common Paper 4 question style.
当试卷上已给出图像时，可以用它来估算难以精确因式分解的根。画出直线 y = k 并读取交点，即可用图像法解方程 x² − 4x + 3 = k。这是 Paper 4 中常见的题型。
7. Word Problems Involving Quadratics | 二次方程应用题
Many real-world problems in the IGCSE syllabus lead to quadratic equations, especially those involving areas, motion, and number relationships. The key is to define a variable clearly, translate the information into an equation, solve it, and then interpret the answers in context.
IGCSE 考纲中的许多实际问题都会引出一元二次方程，尤其是涉及面积、运动和数字关系的问题。关键是清晰地设定变量，将信息转化为方程，解方程，然后在具体情境中解释答案。
Example: A rectangle has length 3 cm greater than its width. Its area is 40 cm². Find the width of the rectangle.
例：一个长方形的长比宽多 3 cm，面积为 40 cm²。求长方形的宽。
e a width cannot be negative, we reject x = −8. The width is 5 cm, and the length is 8 cm.
设宽为 x cm，则长为 (x + 3) cm。面积条件给出 x(x + 3) = 40，展开得 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于宽度不能为负数，舍去 x = −8。因此宽为 5 cm，长为 8 cm。
Example 2: gives n = 11 or n = −13, so the pairs are 11 and 13, or −13 and −11. Both pairs are valid.
例 2：两个连续奇数的乘积为 143。设这两个数为 n 和 n + 2，则 n(n + 2) = 143，得 n² + 2n − 143 = 0。用求根公式或因式分解得 n = 11 或 n = −13，因此这两组数为 11 和 13，或 −13 和 −11。两组答案均有效。
Always check whether the question describes a physical context that requires a positive answer. If in doubt, substitute your answers back into the original information to confirm they work.
始终检查题目所描述的实际情境是否要求正数答案。如果有疑问，将答案代回原题信息中进行验证。
8. Exam Tips and Common Mistakes | 考试技巧与常见错误
The table below summarises the most common mistakes students make in quadratic equation questions and how to avoid them. Reviewing this list before the exam can save several marks.
下表总结了学生在解一元二次方程时最常见的错误以及避免方法。考前复习这份清单可以帮你多拿几分。
| Common mistake | Correct approach |
| Dividing both sides by x, losing a root | Rearrange to the form ax² + bx + c = 0, then factorise fully |
| Forgetting the ± sign when taking square roots | Always write ± before the square root: x = −p ± √q |
| Misidentifying a, b and c when b or c is negative | Write the equation in standard form first, then list a, b, c separately |
| Rounding surds too early | Keep exact values such as √5 until the final answer |
| 常见错误 | 正确做法 |
| 两边除以 x，丢失一个根 | 先整理成 ax² + bx + c = 0，然后完整因式分解 |
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