📚 IGCSE Mathematics: Quadratic Equations & Simultaneous Equations | IGCSE 数学复习：二次方程与联立方程
Welcome to this TutorHao revision article, part of our IGCSE Mathematics Revision Series. This article focuses on two essential topics: quadratic equations and simultaneous equations. You will learn the standard methods of solution, how to apply the discriminant, and how to solve mixed systems involving one linear and one quadratic equation. Worked examples and common exam pitfalls are included to help you prepare effectively.
欢迎阅读 TutorHao 复习系列文章之一，本文属于 IGCSE 数学复习板块。本文重点讲解两大核心主题：二次方程与联立方程。你将学习标准的解法、判别式的应用，以及如何求解含有一个线性方程和一个二次方程的混合方程组。文中包含典型例题和常见考试陷阱，帮助你高效备考。
1. Forms of a Quadratic Equation | 二次方程的形式
0. The values a, b and c are real numbers, and x is the unknown. The highest power of x is 2, which gives the equation its name 'quadratic'.
一元二次方程是指可以写成 ax² + bx + c = 0 的方程，其中 a ≠ 0。a、b、c 是实数，x 是未知数。x 的最高次数为 2，因此称为“二次”方程。
- General form: ax² + bx + c = 0
- 一般形式：ax² + bx + c = 0
- If b = 0, the equation becomes ax² + c = 0, a simple quadratic.
- 若 b = 0，方程变为 ax² + c = 0，即简单二次方程。
- If c = 0, the equation becomes ax² + bx = 0, which can be factorised by taking x out.
- 若 c = 0，方程变为 ax² + bx = 0，可通过提取 x 进行因式分解。
Before solving, always rearrange the equation so that one side equals 0. This is the first and most important step in any quadratic solution method.
在求解之前，务必先整理方程，使一边等于 0。这是所有二次方程解法的第一步，也是最关键的一步。
2. Solving by Factorisation | 因式分解法
product property: if the product of two expressions is 0, at least one of them must be 0.
当二次方程具有简单的整数根时，因式分解是最快捷的方法。我们将二次式写成两个一次式的乘积：ax² + bx + c = (px + q)(rx + s)。然后利用零积性质：若两个表达式的乘积为 0，则其中至少一个必须为 0。
Worked Example: Solve x² - 5x + 6 = 0
例题：解方程 x² - 5x + 6 = 0
We look for two numbers that multiply to 6 and add to -5. These numbers are -2 and -3. Therefore:
我们需要找到两个数，它们的乘积为 6，和为 -5。这两个数是 -2 和 -3。因此：
x² - 5x + 6 = (x - 2)(x - 3) = 0
So x - 2 = 0 or x - 3 = 0, which gives x = 2 or x = 3.
因此 x - 2 = 0 或 x - 3 = 0，所以 x = 2 或 x = 3。
Always check your factorisation by expanding the brackets. This takes only a few seconds and prevents careless errors in the exam.
务必通过展开括号来检验你的因式分解是否正确。这只需几秒钟，可以有效避免考试中的粗心错误。
3. Solving by Completing the Square | 配方法
nd the vertex of a parabola. The method is based on the identity x² + 2px + p² = (x + p)².
配方法将二次方程转化为 a(x + p)² + q = 0 的形式。当方程不易因式分解，或需要求抛物线的顶点时，这种方法尤其有用。该方法基于恒等式 x² + 2px + p² = (x + p)²。
Worked Example: Solve x² + 6x + 2 = 0 by completing the square.
例题：用配方法解方程 x² + 6x + 2 = 0。
Step 1: Halve the coefficient of x, which is 6, to get 3. Write (x + 3)² = x² + 6x + 9.
步骤 1：将 x 的系数 6 取半，得到 3。写出 (x + 3)² = x² + 6x + 9。
tep 2: Rewrite the equation as (x + 3)² - 9 + 2 = 0, which simplifies to (x + 3)² - 7 = 0.
步骤 2：将原方程改写为 (x + 3)² - 9 + 2 = 0，即 (x + 3)² - 7 = 0。
Step 3: Solve: (x + 3)² = 7, so x + 3 = ±√7, hence x = -3 ± √7.
步骤 3：解方程：(x + 3)² = 7，因此 x + 3 = ±√7，所以 x = -3 ± √7。
For a quadratic where the coefficient of x² is not 1, first divide the whole equation by a before completing the square.
若二次项系数不为 1，应先将整个方程除以 a，再配方。
4. The Quadratic Formula | 二次公式
od that solves every quadratic equation. For ax² + bx + c = 0, the solutions are given by:
二次公式是一种通用解法，适用于所有二次方程。对于 ax² + bx + c = 0，解为：
x = (−b ± √b² − 4ac) ⁄ 2a
This formula is derived by completing the square on the general form. You are expected to memorise it for the IGCSE examination.
该公式是由一般形式配方推导而来。在 IGCSE 考试中，你需要熟记此公式。
Worked Example: Solve 2x² - 4x - 3 = 0 using the quadratic formula.
例题：用二次公式解方程 2x² - 4x - 3 = 0。
Here a = 2, b = -4, c = -3. Substitute into the formula:
这里 a = 2，b = -4，c = -3。代入公式：
x = (−(−4) ± √((−4)² − 4×2×(−3))) ⁄ (2×2) = (4 ± √16 + 24) ⁄ 4 = (4 ± √40) ⁄ 4
Simplify √40 = 2√10, so x = (4 ± 2√10) ⁄ 4 = 1 ± (√10) ⁄ 2. These are the exact solutions.
化简 √40 = 2√10，所以 x = (4 ± 2√10) ⁄ 4 = 1 ± (√10) ⁄ 2。这就是精确解。
When a calculator is allowed, you may give decimal approximations, but always show the substitution clearly to earn method marks.
若允许使用计算器，你可以给出小数近似值，但一定要清晰写出代入过程以获取方法分。
5. The Discriminant | 判别式
The discriminant is the expression b² - 4ac inside the square root of the quadratic formula. It tells us how many and what type of roots a quadratic equation has, without solving the equation completely.
判别式是二次公式根号内的表达式 b² - 4ac。它告诉我们二次方程有多少个根、是什么类型的根，而无需完全解方程。
| Discriminant 判别式 | Nature of roots 根的性质 |
| b² - 4ac > 0 | Two distinct real roots 两个不相等的实数根 |
| b² - 4ac = 0 | One repeated real root (two equal roots) 两个相等的实数根（重根） |
| b² - 4ac < 0 | No real roots (two complex roots) 没有实数根（两个复数根） |
Worked Example: Find the value of k for which x² + kx + 9 = 0 has one repeated root.
例题：求 k 的值，使得方程 x² + kx + 9 = 0 有一个重根。
For one repeated root, the discriminant must be 0. Here a = 1, b = k, c = 9:
要有一个重根，判别式必须为 0。这里 a = 1，b = k，c = 9：
k² - 4×1×9 = 0 → k² = 36 → k = ±6
So k can be 6 or -6. Both values give a perfect square trinomial.
因此 k 可以是 6 或 -6。两个值都能使三项式成为完全平方。
6. Simultaneous Equations: One Linear, One Quadratic | 联立方程：一个线性、一个二次
A system of simultaneous equations often contains one linear equation and one quadratic equation. The solution involves substitution: solve the linear equation for one variable, substitute into the quadratic equation, then solve the resulting quadratic.
联立方程组通常包含一个线性方程和一个二次方程。解法采用代入法：先由线性方程解出一个变量，代入二次方程，再解所得的一元二次方程。
Worked Example: Solve y = x + 1 and y = x² - 3x + 2.
例题：解方程组 y = x + 1 和 y = x² - 3x + 2。
Since both expressions equal y, we can equate them:
由于两个表达式都等于 y，可以将它们相等：
x + 1 = x² - 3x + 2
Rearrange to form a quadratic: x² - 4x + 1 = 0.
整理为二次方程：x² - 4x + 1 = 0。
Use the quadratic formula with a = 1, b = -4, c = 1:
使用二次公式，其中 a = 1，b = -4，c = 1：
x = (4 ± √16 - 4) ⁄ 2 = (4 ± √12) ⁄ 2 = 2 ± √3
Then substitute each x into y = x + 1 to find the corresponding y values:
然后将每个 x 代入 y = x + 1 求出对应的 y 值：
- If x = 2 + √3, then y = 3 + √3
- 若 x = 2 + √3，则 y = 3 + √3
- If x = 2 - √3, then y = 3 - √3
- 若 x = 2 - √3，则 y = 3 - √3
The solutions are the coordinate pairs (2 + √3, 3 + √3) and (2 - √3, 3 - √3). Graphically, these are the intersection points of a line and a parabola.
解为坐标对 (2 + √3, 3 + √3) 和 (2 - √3, 3 - √3)。在图形上，它们是直线与抛物线的交点。
7. Solving Word Problems | 应用题解法
Exam questions often present quadratic and simultaneous equations in real-world contexts. The key is to define variables clearly, translate the conditions into equations, then solve using the appropriate method.
考试题目通常将二次方程和联立方程放在实际情境中。关键是要清晰地定义变量，把条件转化为方程，然后用适当方法求解。
Worked Example: The product of two consecutive positive integers is 240. Find the integers.
例题：两个连续正整数的乘积是 240。求这两个整数。
Let the smaller integer be n. Then the larger is n + 1. The equation is:
设较小的整数为 n，则较大的为 n + 1。方程为：
n(n + 1) = 240 → n² + n - 240 = 0
Factorise: (n + 16)(n - 15) = 0. Since n is positive, n = 15. The integers are 15 and 16.
因式分解：(n + 16)(n - 15) = 0。由于 n 为正整数，n = 15。所以这两个整数是 15 和 16。
Always check that your answers make sense in the original context. Negative or fraction answers may be invalid if the problem demands positive integers.
务必检查答案是否符合原题情境。如果题目要求正整数，负数或分数答案可能无效。
8. Common Exam Mistakes and Tips | 常见考试错误与提示
Students often lose marks in IGCSE exams due to small but avoidable errors. Here are the most common mistakes and how to avoid them.
学生在 IGCSE 考试中常因一些细小但可避免的错误而失分。以下是最常见的错误及避免方法。
- to rearrange the equation to ax² + bx + c = 0 before applying the formula or factorising.
- 忘记在套用公式或因式分解前，先将方程整理为 ax² + bx + c = 0。
- Sign errors when substituting negative values into the quadratic formula, especially with b.
- 将负值代入二次公式时出现符号错误，尤其是 b 项。
- Dropping the ± symbol when taking square roots in solving or completing the square.
- 在求解或配方时遗漏 ± 符号。
- When solving simultaneous equations, not substituting both x values to find both y values.
- 解联立方程时，没有将两个 x 值都代入求出对应的两个 y 值。
- Rounding too early: keep exact values until the final step.
- 过早四舍五入：直到最后一步前请保留精确值。
Tip: After finding roots, substitute them back into the original equation to verify your answer. This simple check can save many marks.
提示：求出根后，将根代回原方程验证答案。这个简单检查可以帮你保住很多分数。
9. Practice Questions | 练习题目
Try these questions yourself before looking at the answers. They cover all methods discussed in this article.
请先独立尝试以下题目，再查看答案。这些题目覆盖了本文讨论的所有方法。
- Solve by factorisation: x² - 7x + 10 = 0.
- 用因式分解法解：x² - 7x + 10 = 0。
- Solve by completing the square: x² + 8x + 5 = 0.
- 用配方法解：x² + 8x + 5 = 0。
- ow many real roots each equation has: 3x² + 2x + 1 = 0; x² - 6x + 9 = 0; 2x² - 5x + 1 = 0.
- 用判别式判断每个方程有幾個实数根：3x² + 2x + 1 = 0；x² - 6x + 9 = 0；2x² - 5x + 1 = 0。
- Solve the simultaneous equations: y = 2x - 3 and y = x² - 2x - 1.
- 解联立方程组：y = 2x - 3 和 y = x² - 2x - 1。
Answers: 1) x = 2 or x = 5. 2) x = -4 ± √11. 3) No real roots; one repeated real root; two distinct real roots. 4) (2, 1) and (-2, -7).
答案：1) x = 2 或 x = 5。2) x = -4 ± √11。3) 无实数根；一个重根；两个不相等的实数根。4) (2, 1) 和 (-2, -7)。
Review any question you answered incorrectly and redo it without notes. Practice is the key to mastering quadratic and simultaneous equations.
请认真复习做错的题目，并在不看笔记的情况下重做。练习是掌握二次方程和联立方程的关键。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图