📚 IGCSE Maths Lesson 16-9: Quadratic Equations and Problem Solving | IGCSE 数学第16-9课：二次方程与解题方法
In this IGCSE Maths revision lesson we focus on quadratic equations: how to recognise them, how to solve them by factorising, by completing the square and by the quadratic formula, and how to turn a word problem into a quadratic that you can solve. The topic appears in both Core and Extended papers, usually inside the algebra and modelling sections, and it is one of the highest-value skills in the whole IGCSE course because it links algebra, graphs and problem solving together.
在本节 IGCSE 数学复习课中，我们专注于二次方程：如何识别二次方程，如何用因式分解法、配方法以及二次方程求根公式求解，以及如何把一道文字应用题转化为可以求解的二次方程。该主题同时出现在核心（Core）与扩展（Extended）试卷中，通常位于代数与建模部分。由于它把代数、图像与实际解题连接在一起，因此是整个 IGCSE 课程中分值最高的技能之一。
1. What Is a Quadratic Equation? | 什么是二次方程？
what makes the equation quadratic: if a were zero, the x² term would disappear and the equation would become the linear equation bx + c = 0.
二次方程是指可以写成下面标准形式的方程，其中 a、b、c 是常数，且 a ≠ 0。条件 a ≠ 0 正是使方程成为二次方程的关键：若 a 等于 0，x² 项就会消失，方程退化为一次方程 bx + c = 0。
a x² + b x + c = 0, a ≠ 0
. When a is positive the parabola opens upwards and has a minimum point; when a is negative it opens downwards and has a maximum point. In IGCSE questions you will usually meet quadratics with small integer coefficients, so always rearrange the equation so that the right-hand side is 0 before you choose a method.
函数 y = ax² + bx + c 的图像是一条抛物线。当 a 为正时，抛物线开口向上并有一个最低点；当 a 为负时，开口向下并有一个最高点。在 IGCSE 题目中，你通常会遇到系数为小整数的二次方程，所以在选择解法之前，一定要先把方程整理成右边为 0 的形式。
- x² − 4x − 5 = 0 (a = 1, b = −4, c = −5) | x² − 4x − 5 = 0（a = 1，b = −4，c = −5）
- 2x² + 3x + 1 = 0 (a = 2, b = 3, c = 1) | 2x² + 3x + 1 = 0（a = 2，b = 3，c = 1）
- x² − 9 = 0 (b = 0, a difference of two squares) | x² − 9 = 0（b = 0，属于平方差形式）
2. Solving by Factorising | 因式分解法
When a quadratic has simple factors, factorising is the fastest method. Write the expression as a product of two brackets equal to zero, then use the key idea that if two numbers multiply to give zero, at least one of them must be zero. So set each bracket equal to zero and solve the two small linear equations.
当二次式具有简单的因式时，因式分解是最快的方法。把表达式写成两个括号相乘并等于零，然后利用一个关键思想：如果两个数相乘等于零，那么其中至少有一个为零。因此令每个括号分别等于零，解出两个简单的一次方程。
If (p x + q)(r x + s) = 0, then p x + q = 0 or r x + s = 0
− 2)(x − 5). Setting each bracket to zero gives x − 2 = 0 or x − 5 = 0, so x = 2 or x = 5.
例如，解 x² − 7x + 10 = 0。我们需要两个数，使它们的积为 +10、和为 −7，即 −2 和 −5，因此因式为 (x − 2)(x − 5)。令每个括号等于零，得到 x − 2 = 0 或 x − 5 = 0，所以 x = 2 或 x = 5。
0, giving 2x + 1 = 0 or x + 1 = 0, so x = −1/2 or x = −1. Always expand mentally to check the middle term.
当 x² 的系数不是 1 时，思路相同，但必须检查首项。对 2x² + 3x + 1 = 0，因式分解为 (2x + 1)(x + 1) = 0，得到 2x + 1 = 0 或 x + 1 = 0，所以 x = −1/2 或 x = −1。务必在脑中展开检查一次中间项。
- x² − 4x − 5 = 0 → (x − 5)(x + 1) = 0 → x = 5 or x = −1 | x² − 4x − 5 = 0 → (x − 5)(x + 1) = 0 → x = 5 或 x = −1
- x² − 9 = 0 → (x − 3)(x + 3) = 0 → x = 3 or x = −3 | x² − 9 = 0 → (x − 3)(x + 3) = 0 → x = 3 或 x = −3
- x² + 6x + 9 = 0 → (x + 3)² = 0 → x = −3 (a repeated root) | x² + 6x + 9 = 0 → (x + 3)² = 0 → x = −3（一个重根）
3. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q, which makes the turning point of the graph visible and gives exact surd answers. To complete the square for x² + bx + c, take half of b inside the bracket and then subtract the square of that half outside.
配方法把二次式改写成 (x + p)² + q 的形式，这样图像的最高点或最低点就一目了然，也能给出精确的根式答案。对 x² + bx + c 配方时，括号内取 b 的一半，再在括号外减去这一半的平方。
x² + b x + c = (x + b/2)² − (b/2)² + c
)² − 9 + 2 = 0, that is (x − 3)² = 7. Taking the square root of both sides gives x − 3 = ± √7, so x = 3 + √7 or x = 3 − √7. Notice that the answer is exact, which is what examiners want unless the question asks for a rounded decimal.
例如，x² − 6x + 2 = 0 可化为 (x − 3)² − 9 + 2 = 0，即 (x − 3)² = 7。两边开平方得 x − 3 = ± √7，所以 x = 3 + √7 或 x = 3 − √7。注意答案是精确值，除非题目要求保留小数，否则这正是考官想要的形式。
ete the square in the usual way to obtain (x − 2)² = 7/3. The turning point of the curve lies at x = −b/2a, and completing the square proves this result.
当 x² 的系数不是 1 时，先把整个方程除以该系数。对 3x² − 12x + 5 = 0，两边除以 3 得 x² − 4x + 5/3 = 0，再按常规配方得到 (x − 2)² = 7/3。曲线的最低点或最高点位于 x = −b/2a，而配方法正好可以证明这一结论。
4. The Quadratic Formula | 二次方程求根公式
The quadratic formula solves every quadratic equation, including those that do not factorise with integers. You must identify a, b and c correctly from the standard form, then substitute them carefully, keeping the minus signs inside brackets.
二次方程求根公式可以解出任何二次方程，包括那些无法用整数因式分解的方程。你必须从标准形式中正确识别 a、b、c，然后仔细代入，注意把负号放在括号内。
x = ( −b ± √ b² − 4ac ) ÷ ( 2a )
Work through 2x² + 3x + 1 = 0. Here a = 2, b = 3 and c = 1, so b² − 4ac = 9 − 8 = 1 and √1 4 = −1/2 or x = (−3 − 1) ÷ 4 = −1, which agrees with the factorised answer found earlier.
我们用 2x² + 3x + 1 = 0 演练一次。此处 a = 2，b = 3，c = 1，所以 b² − 4ac = 9 − 8 = 1，√1 = 1。于是 x = (−3 + 1) ÷ 4 = −1/2，或 x = (−3 − 1) ÷ 4 = −1，与前面因式分解得到的结果一致。
ommon sources of lost marks. Also remember that if b is negative, −b becomes positive, so for x² − 5x + 3 = 0 you substitute a = 1, b = −5, c = 3 and obtain x = (5 ± √13) ÷ 2.
如果题目要求答案保留两位小数，应先求出精确的根式形式，最后一步才取近似值。过早取近似是最常见的失分原因之一。还要记住：如果 b 为负，则 −b 变为正，因此对 x² − 5x + 3 = 0 应代入 a = 1，b = −5，c = 3，得到 x = (5 ± √13) ÷ 2。
5. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression under the square root, b² − 4ac, is called the discriminant and is often written as Δ. It tells you how many real roots the equation has before you do any further work, which is a very efficient way to check your answer.
根号下的表达式 b² − 4ac 称为判别式，通常记作 Δ。它能在你继续计算之前就告诉你方程有多少个实根，是检查答案的高效方法。
| Value of Δ = b² − 4ac | Nature of roots | 根的性质 | Graph | 图像 |
| Δ > 0 | Two distinct real roots | 两个不同的实根 | Crosses the x-axis twice | 与 x 轴相交两次 |
| Δ = 0 | One repeated root | 一个重根 | Touches the x-axis once | 与 x 轴相切一次 |
| Δ < 0 | No real roots | 无实根 | Does not meet the x-axis | 与 x 轴不相交 |
hile x² + x + 1 = 0 has Δ = 1 − 4 = −3 < 0, so it has no real solutions at IGCSE level.
例如，x² − 4x + 4 = 0 的 Δ = 16 − 16 = 0，因此有一个重根，即 x = 2。而 x² + x + 1 = 0 的 Δ = 1 − 4 = −3 < 0，在 IGCSE 阶段它没有实数解。
6. Graphs, Roots and the Line of Symmetry | 图像、根与对称轴
a quadratic and sketching a parabola are really the same skill viewed from two directions.
方程 ax² + bx + c = 0 的解就是曲线 y = ax² + bx + c 与 x 轴交点的横坐标，因为这些点恰好满足 y = 0。这正是为什么"解二次方程"和"画抛物线草图"其实是同一项技能的两个视角。
Line of symmetry: x = −b ÷ (2a) Turning point: ( −b ÷ 2a , c − b² ÷ 4a )
To sketch the curve you need four pieces of information: the shape (upwards if a > 0, downwards if a < 0), the y-intercept which is simply c, the x-intercepts from solving the equation, and the turning point found by completing the square or by using the line of symmetry. Mark these on the axes and join them with a smooth curve.
要画出抛物线草图，你需要四条信息：开口方向（a > 0 向上，a < 0 向下）、y 轴截距（就是 c）、由解方程得到的 x 轴截距，以及通过配方或利用对称轴求出的顶点。把这些点标在坐标轴上，再用一条光滑曲线连接起来。
the x-axis at x = 5 and x = −1. The line of symmetry is halfway between, at x = 2, and the minimum point is (2, −9).
设 y = x² − 4x − 5。由于 a = 1，曲线开口向上；c = −5，所以它与 y 轴交于 (0, −5)；因式分解形式 (x − 5)(x + 1) 表明它与 x 轴交于 x = 5 和 x = −1。对称轴恰好在两者中间，即 x = 2，最低点为 (2, −9)。
7. Problem Solving Strategy: Setting Up Equations | 解题策略：建立方程
Word problems are the part of this topic that students find hardest, but the method is almost always the same. Follow a fixed sequence of steps every time, and write each step down so that the examiner can award method marks even if the final number is wrong.
应用题是本主题中学生觉得最困难的部分，但方法几乎总是一样的。每次都按固定的步骤顺序进行，并把每一步都写下来，这样即使最终数字算错，考官仍能给步骤分。
- Step 1: Read the question twice and decide what the unknown is. | 第一步：读题两遍，确定未知量是什么。
- Step 2: Define a variable in words, for example "let x be the width in cm". | 第二步：用文字定义变量，例如"设 x 为宽度（单位 cm）"。
- Step 3: Write any other quantities in terms of x. | 第三步：用 x 表示其他所有量。
- Step 4: Use the condition given to form an equation. | 第四步：利用题目给出的条件建立方程。
- Step 5: Expand and rearrange so the right-hand side is zero. | 第五步：展开并整理成右边等于零的形式。
- Step 6: Solve by factorising, completing the square or the formula. | 第六步：用因式分解、配方或公式法求解。
- Step 7: Reject any root that is impossible in context, then answer with units. | 第七步：舍去不符合实际意义的根，然后带单位写出答案。
e cannot be correct, so those roots must be discarded with a short written reason. Writing "x = −8 is rejected because a width cannot be negative" earns credit and shows clear reasoning.
第七步是许多考生容易忘记的。负的长度、负的时间或分数个的人数都不可能正确，因此必须用一句简短的说明把这些根舍去。写出"x = −8 被舍去，因为宽度不能为负"既能得分，也体现了清晰的推理。
8. Area, Perimeter and Consecutive Number Problems | 面积、周长与连续数列问题
gives x(x + 3) = 40. Expanding and rearranging produces x² + 3x − 40 = 0, which factorises as (x + 8)(x − 5) = 0, so x = 5 or x = −8. A width cannot be negative, so x = 5 and the rectangle is 5 cm by 8 cm.
最经典的背景是长方形。一个长方形的长比宽多 3 cm，面积为 40 cm²。设宽为 x，则长为 x + 3，由面积得 x(x + 3) = 40。展开整理得 x² + 3x − 40 = 0，因式分解为 (x + 8)(x − 5) = 0，所以 x = 5 或 x = −8。宽度不能为负，故 x = 5，长方形为 5 cm × 8 cm。
be n + 1, giving n(n + 1) = 156 and n² + n − 156 = 0. This factorises as (n + 13)(n − 12) = 0, so n = 12 and the two numbers are 12 and 13. Check: 12 × 13 = 156.
连续数问题遵循同样的模式。若两个连续正整数的乘积为 156，设较小的为 n，较大的为 n + 1，则 n(n + 1) = 156，即 n² + n − 156 = 0。因式分解得 (n + 13)(n − 12) = 0，所以 n = 12，两个数为 12 和 13。检验：12 × 13 = 156。
For consecutive even or odd numbers the step between them is 2, so you would write n and n + 2. If the problem mentions three consecutive integers, use n, n + 1 and n + 2, and remember that the sum of three consecutive integers is always 3(n + 1).
对于连续偶数或连续奇数，相邻两个数之间差 2，因此应写成 n 与 n + 2。若题目涉及三个连续整数，则用 n、n + 1、n + 2，并记住三个连续整数之和总是 3(n + 1)。
9. Speed, Time and Other Modelling Problems | 速度、时间与其他建模问题
e the quadratic usually comes from. A cyclist travels 60 km. If the speed had been 5 km/h greater, the journey would have taken 1 hour less. Let the actual speed be x km/h.
速度问题使用关系式"时间 = 路程 ÷ 速度"，二次方程通常就是从这里产生的。一名骑行者行驶 60 km。如果速度每小时快 5 km，所需时间会减少 1 小时。设实际速度为 x km/h。
60 ÷ x − 60 ÷ (x + 5) = 1
or x = −20. A speed must be positive, so x = 15 km/h. Check: 60 ÷ 15 = 4 hours and 60 ÷ 20 = 3 hours, a difference of exactly 1 hour.
将每一项乘以 x(x + 5) 得 60(x + 5) − 60x = x(x + 5)，即 300 = x² + 5x，因此 x² + 5x − 300 = 0。因式分解得 (x + 20)(x − 15) = 0，所以 x = 15 或 x = −20。速度必须为正，故 x = 15 km/h。检验：60 ÷ 15 = 4 小时，60 ÷ 20 = 3 小时，正好相差
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