📚 Solving Quadratic Equations | 解二次方程的三种方法
stants and a ≠ 0. In IGCSE Mathematics, you must be able to solve quadratic equations using three main methods: factorisation, completing the square, and the quadratic formula. This topic is a cornerstone of the Algebra syllabus, and questions on quadratic equations appear consistently across both Core and Extended papers, often embedded in word problems involving areas, projectile motion, and number puzzles.
二次方程是能够写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。在 IGCSE 数学中，你必须掌握三种主要解法：因式分解法、配方法和公式法。本专题是代数大纲的基石，二次方程相关题目稳定地出现在核心卷和扩展卷中，并且常常嵌入到涉及面积、抛体运动和数字谜题的应用题里。
1. Recognising a Quadratic Equation | 识别二次方程
plying any solution method, because factorising, completing the square, and the formula all require the equation to equal zero with terms arranged in descending powers of x.
若未知数（通常是 x）的最高次数为 2，则该方程为二次方程。例如 3x² − 5x + 1 = 0 是二次方程，但 4x + 7 = 0 是一次方程，x³ − 2x = 0 是三次方程。在运用任何解法之前，必须先整理成标准形式 ax² + bx + c = 0，因为因式分解、配方法和公式法都要求方程右边为 0，且各项按 x 的降幂排列。
2. Method 1 — Factorisation | 方法一：因式分解法
dratics where a = 1, but it also works for a ≠ 1 with careful trial and error or by splitting the middle term.
当二次式 ax² + bx + c 可以写成两个一次因式的乘积时，因式分解法适用。核心思想是：若 p × q = 0，则 p = 0 或 q = 0。因此，一旦将二次式分解为 (mx + n)(px + q) = 0，就分别解 mx + n = 0 和 px + q = 0。对于 a = 1 的简单二次式，此方法最快，但通过仔细试错或裂项法，a ≠ 1 的情况同样适用。
Worked Example 1: Solve x² − 5x + 6 = 0.
例题 1：解方程 x² − 5x + 6 = 0。
- Find two numbers that multiply to +6 and add to −5: they are −2 and −3.
- 找出两个数，乘积为 +6，和为 −5：这两个数是 −2 和 −3。
- Factorise: (x − 2)(x − 3) = 0.
- 因式分解：(x − 2)(x − 3) = 0。
- Set each factor to zero: x − 2 = 0 gives x = 2; x − 3 = 0 gives x = 3.
- 令每个因式为零：x − 2 = 0 得 x = 2；x − 3 = 0 得 x = 3。
Solution: x = 2 or x = 3
解：x = 2 或 x = 3
3. Method 1 — Factorising When a ≠ 1 | 方法一：首项系数不为 1 时的因式分解
x + 3) + 1(x + 3) = (2x + 1)(x + 3). Setting each factor to zero gives x = −1/2 or x = −3.
当 x² 的系数不为 1 时，因式分解需要更加小心。以 2x² + 7x + 3 = 0 为例。我们需要找到两个数，其乘积为 a × c = 2 × 3 = 6，其和为 b = 7。数字 6 和 1 满足条件：将 7x 改写为 6x + x，得到 2x² + 6x + x + 3。然后分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。令每个因式为零，得 x = −1/2 或 x = −3。
nfirms that (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3, so the factors are correct.
始终通过心算展开括号来检验你的因式分解。快速展开可确认 (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3，因此因式正确。
4. Method 2 — Completing the Square | 方法二：配方法
hich can then be solved by taking square roots. This method is essential when the quadratic does not factorise neatly, and it also forms the basis for deriving the quadratic formula and for finding the coordinates of the vertex of a parabola.
配方法将二次式转化为 (x + p)² + q = 0 的形式，然后通过开平方求解。当二次式无法整齐地因式分解时，此方法至关重要，它同时也是推导求根公式以及求抛物线顶点坐标的基础。
Worked Example 2: Solve x² + 6x + 4 = 0 by completing the square.
例题 2：用配方法解方程 x² + 6x + 4 = 0。
- Start with x² + 6x + 4 = 0. Move the constant: x² + 6x = −4.
- 从 x² + 6x + 4 = 0 开始。移常数项：x² + 6x = −4。
- lf of 6, which is 3, and square it: 3² = 9. Add 9 to both sides: x² + 6x + 9 = −4 + 9 = 5.
- 取 6 的一半为 3，再平方：3² = 9。两边同时加 9：x² + 6x + 9 = −4 + 9 = 5。
- Rewrite the left side as a perfect square: (x + 3)² = 5.
- 将左边写成完全平方：(x + 3)² = 5。
- Take square roots: x + 3 = ±√5.
- 开平方：x + 3 = ±√5。
- Solve: x = −3 ± √5.
- 解得：x = −3 ± √5。
Solution: x = −3 + √5 or x = −3 − √5
解：x = −3 + √5 或 x = −3 − √5
5. Method 2 — Completing the Square When a ≠ 1 | 方法二：首项系数不为 1 时的配方法
² = 4. Add 4 to both sides: (x − 2)² = −3/2 + 4 = 5/2. Taking square roots gives x − 2 = ±√5/2, so x = 2 ± √5/2.
当 x² 的系数不为 1 时，必须先将每一项除以该系数。例如，解 2x² − 8x + 3 = 0，先两边同除以 2，得 x² − 4x + 3/2 = 0。然后按之前步骤继续：x² − 4x = −3/2，−4 的一半是 −2，(−2)² = 4。两边加 4：(x − 2)² = −3/2 + 4 = 5/2。开平方得 x − 2 = ±√5/2，所以 x = 2 ± √5/2。
This method is systematic and always works, but it involves fractions more often than the quadratic formula, which is why many students prefer the formula for messy coefficients.
此方法系统性强且始终有效，但相比求根公式，它更常涉及分数运算，这就是许多学生在系数较复杂时更倾向于公式法的原因。
6. Method 3 — The Quadratic Formula | 方法三：求根公式法
ic, regardless of whether it factorises. For ax² + bx + c = 0, the solutions are given by:
求根公式是解二次方程最有力的工具，因为它对每一个二次方程都适用，无论能否因式分解。对于 ax² + bx + c = 0，其解为：
x = (−b ± √b² − 4ac) ÷ (2a)
x = (−b ± √b² − 4ac) ÷ (2a)
s exactly one repeated real root; if D < 0, there are no real roots (the solutions are complex, which is beyond IGCSE).
表达式 D = b² − 4ac 称为判别式。它决定根的性质：若 D > 0，有两个不同的实根；若 D = 0，恰好有一个重根；若 D < 0，没有实根（解为复数，超出 IGCSE 范围）。
7. Worked Example of the Quadratic Formula | 求根公式的完整例题
Worked Example 3: 3x² − 7x − 2 = 0 using the quadratic formula, giving answers correct to 2 decimal places.
例题 3：用求根公式解方程 3x² − 7x − 2 = 0，答案精确到小数点后两位。
- Identify coefficients: a = 3, b = −7, c = −2.
- 确定系数：a = 3，b = −7，c = −2。
- Substitute into the formula: x = (−(−7) ± √((−7)² − 4(3)(−2))) ÷ (2 × 3).
- 代入公式：x = (−(−7) ± √((−7)² − 4(3)(−2))) ÷ (2 × 3)。
- Simplify: x = (7 ± √49 + 24) ÷ 6 = (7 ± √73) ÷ 6.
- 化简：x = (7 ± √49 + 24) ÷ 6 = (7 ± √73) ÷ 6。
- culate: x = (7 + 8.544) ÷ 6 = 2.59 (to 2 d.p.) or x = (7 − 8.544) ÷ 6 = −0.26 (to 2 d.p.).
- 计算：x = (7 + 8.544) ÷ 6 = 2.59（保留两位小数）或 x = (7 − 8.544) ÷ 6 = −0.26（保留两位小数）。
Solution: x ≈ 2.59 or x ≈ −0.26
解：x ≈ 2.59 或 x ≈ −0.26
8. Choosing the Right Method | 选择正确的方法
IGCSE questions sometimes specify the method you must use, but when they do not, choose the most efficient route. Factorisation is fastest when the quadratic is simple and has integer roots. The quadratic formula is best when coefficients are awkward or when the question asks for answers to a given number of decimal places. Completing the square is ideal when the quadratic has a leading coefficient of 1 and an even middle coefficient, or when the question explicitly asks for the vertex form.
IGCSE 考题有时会指定必须使用的方法，但若没有指定，应选择最高效的路径。当二次式简单且有整数根时，因式分解法最快。当系数较复杂或题目要求将答案保留到指定小数位时，求根公式最佳。当二次式的首项系数为 1 且中间项系数为偶数，或者题目明确要求顶点式时，配方法最为理想。
| Method 方法 | Best Used When 最佳使用场景 |
|---|---|
| Factorisation 因式分解法 | Simple quadratics with integer roots 简单二次式且有整数根 |
| Completing the square 配方法 | a = 1 and b is even, or vertex form required a = 1 且 b 为偶数，或需要顶点式 |
| Quadratic formula 求根公式法 | Any quadratic, especially with decimals or non-factorable expressions 任何二次方程，尤其含小数或不可因式分解 |
9. Applications — Word Problems | 实际应用：应用题
torising gives (x + 8)(x − 5) = 0, so x = 5 or x = −8. Since length cannot be negative, the width is 5 m and the length is 8 m.
二次方程经常出现在实际问题中。一个经典例子涉及面积：一个矩形花园的长比宽多 3 米，面积为 40 平方米。设宽为 x 米，则长为 x + 3 米。面积方程为 x(x + 3) = 40，展开得 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 米，长为 8 米。
uadratic. Always reject negative solutions in context, as time and distance cannot be negative.
另一个常见应用是抛体运动，物体在时间 t 后的高度 h 可表示为 h = −5t² + 20t + 15。要求物体何时落地，令 h = 0 并解该二次方程。在具体情境中，始终舍去负解，因为时间和距离不可能为负。
10. Common Mistakes to Avoid | 常见错误与避坑指南
er divide by the unknown variable when solving equations, as this eliminates a valid root.
一个常见错误是在应用解法之前忘记将方程整理成标准形式。例如，解 2x² = 5x 时，若两边同除以 x 会得到 2x = 5，从而丢失解 x = 0。正确的做法是写成 2x² − 5x = 0 并因式分解：x(2x − 5) = 0，得 x = 0 或 x = 5/2。解方程时切勿除以未知量，因为这会消去一个有效根。
ound b and the discriminant to avoid sign errors: x = (−(b) ± √((b)² − 4(a)(c))) ÷ (2(a)).
另一个错误是在求根公式中处理负号不当，尤其是当 b 为负数时。例如，若 b = −4，则 −b = +4。代入时始终用括号将 b 和判别式括起来，以避免符号错误：x = (−(b) ± √((b)² − 4(a)(c))) ÷ (2(a))。
11. Checking Your Answers | 检验你的答案
left side equals the right side, the solution is correct. This habit catches arithmetic errors and sign mistakes, especially in exams where time pressure increases the risk of slips.
始终将解代回原方程进行检验。例如，要检验 x = 3 是 x² − 5x + 6 = 0 的根，将其代入：(3)² − 5(3) + 6 = 9 − 15 + 6 = 0。由于左边等于右边，该解正确。这一习惯能够发现算术错误和符号错误，尤其是在考试时间紧迫、更容易出错的情况下。
For word problems, check that your solutions make sense in context. A negative length, a negative time, or a fractional number of people are all red flags that indicate you should re-examine your equation or reject the inappropriate root.
对于应用题，要检验你的解在情境中是否合理。负长度、负时间或人数为分数都是警示信号，说明你应该重新检查方程或舍去不合理的那一个根。
12. Summary and Exam Tips | 总结与应试技巧
To master quadratic equations, memorise the quadratic formula confidently, practise factorisation until it becomes automatic, and understand completing the square well enough to use it when instructed. In the exam, always show your method step by step, because method marks are awarded even if the final answer is incorrect. Keep your working neat and aligned so examiners can follow your logic easily.
要掌握二次方程，必须熟练记忆求根公式，反复练习因式分解直到熟练自如，并充分理解配方法以便在题目要求时能够运用。考试时务必逐步展示解题过程，因为即使最终答案有误，方法步骤也能得分。保持书写工整、对齐清晰，让阅卷老师能够轻松理解你的逻辑。
Finally, remember that a quadratic equation can have two solutions, one solution, or no real solutions. Do not assume two distinct answers exist; the discriminant tells you how many real roots to expect, so use it strategically in multiple-choice questions to eliminate impossible options quickly.
最后，记住二次方程可以有两个解、一个解或没有实数解。不要想当然地认为一定存在两个不同的答案；判别式能告诉你应该有几个实根，因此在选择题中善用它，可以快速排除不可能的选项。
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