📚 Mastering Quadratic Equations for IGCSE Mathematics | 精通 IGCSE 数学二次方程
Quadratic equations sit at the centre of IGCSE Mathematics. They connect algebra, graphs and real-world modelling, and they appear in both Paper 2 (non-calculator) and Paper 4 (calculator). This revision guide walks through the definition, the three standard forms, the four main solving methods, the discriminant, graph sketching, word problems, inequalities, simultaneous equations and the mistakes that cost the most marks.
二次方程是 IGCSE 数学的核心内容。它把代数、图像和实际建模连接在一起，在 Paper 2（不可用计算器）和 Paper 4（可用计算器）中都会出现。本复习指南将系统讲解定义、三种标准形式、四种主要解法、判别式、图像画法、应用题、不等式、联立方程，以及最容易丢分的错误。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation in which the highest power of the unknown is 2. Its general form is shown below, where a, b and c are constants and a cannot be zero. If a were zero, the equation would become linear rather than quadratic.
二次方程是未知数最高次数为 2 的方程。其一般式如下，其中 a、b、c 是常数，且 a 不能为零。若 a 为零，方程就退化为一次方程，而不是二次方程。
a x² + b x + c = 0, a ≠ 0
The values of x that satisfy the equation are called roots or solutions. Depending on the discriminant, a quadratic equation can have two distinct real roots, one repeated real root, or no real roots at all. On a graph, the roots are the x-coordinates where the curve crosses the x-axis.
满足方程的 x 值称为根或解。根据判别式的不同，二次方程可以有两个不同的实根、一个重根，或没有实根。在图像上，根就是曲线与 x 轴交点的横坐标。
be rearranged into general form, for example 3x² = 5x - 2 becomes 3x² - 5x + 2 = 0. Others are disguised quadratics, such as x⁴ - 5x² + 4 = 0, which becomes a normal quadratic when you substitute u = x².
在 IGCSE 试题中，你可能会被要求解方程、画图像、列方程或解释二次方程。有些题看起来难，其实只需先整理成一般式，例如 3x² = 5x - 2 应写成 3x² - 5x + 2 = 0。还有一些是"伪二次方程"，如 x⁴ - 5x² + 4 = 0，令 u = x² 后就变成普通二次方程。
2. Recognising the Three Standard Forms | 识别三种标准形式
g the quadratic formula and for reading off the y-intercept, which is c. It also tells you immediately whether the curve opens upwards (a > 0) or downwards (a < 0).
二次表达式有三种重要形式。一般式 y = a x² + b x + c 最适合使用二次公式，也便于直接读出 y 轴截距 c。它还能立刻告诉你抛物线开口向上（a > 0）还是向下（a < 0）。
alfway between the two roots, at x = (p + q) ÷ 2, and substituting this value back gives the turning point.
因式分解式 y = a(x - p)(x - q) 可直接给出根 x = p 和 x = q，也就是 x 轴截距。对称轴正好位于两个根的正中间，即 x = (p + q) ÷ 2，把这个值代回原式就得到顶点坐标。
therefore the fastest route to maximum and minimum problems and to describing transformations of the basic curve y = x².
配方式 y = a(x - h)² + k 可直接给出顶点 (h, k)。若 a > 0，这是最低点；若 a < 0，这是最高点。因此配方式解决最大值、最小值问题最快，也最容易看出它相对基本曲线 y = x² 的变换。
| Form | 形式 | Best for | 最适合 |
| y = a x² + b x + c | Quadratic formula, y-intercept | 二次公式、y 轴截距 |
| y = a(x - p)(x - q) | Roots, x-intercepts | 根、x 轴截距 |
| y = a(x - h)² + k | Vertex, max or min value | 顶点、最大或最小值 |
3. Solving by Factorisation | 因式分解法求解
product of two factors is zero, then at least one factor must be zero. So once you have (x - p)(x - q) = 0, you can write x = p or x = q.
因式分解是最快的方法，也是不可用计算器的试卷中最可靠的方法。它的依据是零乘积性质：若两个因式的乘积为零，则至少有一个因式为零。所以一旦得到 (x - p)(x - q) = 0，就可以写出 x = p 或 x = q。
2x - 5)(x + 3) without checking by expanding, which wastes time and loses the answer mark.
因式分解之前，务必先把方程整理成一边为零。常见错误是把 2x² - 5x + 3 = 0 错误地分解为 (2x - 5)(x + 3) 而不展开检验，这样既浪费时间又会丢掉答案分。
- Common factor | 提公因式: 4x² - 12x = 0 → 4x(x - 3) = 0 → x = 0 or x = 3.
- Difference of two squares | 平方差: 9x² - 16 = 0 → (3x - 4)(3x + 4) = 0 → x = 4/3 or x = -4/3.
- Simple trinomial | 简单三项式: x² - 7x + 12 = 0 → (x - 3)(x - 4) = 0 → x = 3 or x = 4.
- Trinomial with a ≠ 1 | 首项系数不为 1: 2x² + 5x - 3 = 0 → (2x - 1)(x + 3) = 0 → x = 1/2 or x = -3.
To factorise a x² + b x + c when a ≠ 1, find two numbers that multiply to a c and add to b, then split the middle term and factorise in pairs. Always expand mentally to check your result before using the zero product property.
当 a ≠ 1 时，分解 a x² + b x + c 的方法是：找两个数，使它们的积为 a c、和为 b，然后拆分中间项并分组提取公因式。使用零乘积性质前，先在脑中展开检验结果。
4. Completing the Square | 配方法
e coefficient of x and square it, then add and subtract that value. The standard result is shown below.
配方法把二次式改写成一个完全平方加上一个常数。当 a = 1 时，取 x 的系数的一半再平方，然后加上并减去这个值。标准结果如下。
x² + b x + c = (x + b/2)² - (b/2)² + c
x + 3)² - 9 + 1 = (x + 3)² - 8. Setting this equal to zero gives (x + 3)² = 8, so x = -3 ± √8 = -3 ± 2√2. This is the exact surd form, which is what a non-calculator paper expects.
例如，x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8。令其等于零，得 (x + 3)² = 8，所以 x = -3 ± √8 = -3 ± 2√2。这就是精确的根式形式，也是不可用计算器的试卷所要求的形式。
r instance, 2x² - 8x + 5 = 2(x² - 4x) + 5 = 2[(x - 2)² - 4] + 5 = 2(x - 2)² - 3. The vertex is therefore (2, -3) and the minimum value is -3.
当 a ≠ 1 时，先从含 x 的前两项中提出 a，在括号内配方，再乘开并调整常数项。例如 2x² - 8x + 5 = 2(x² - 4x) + 5 = 2[(x - 2)² - 4] + 5 = 2(x - 2)² - 3。因此顶点是 (2, -3)，最小值是 -3。
Completing the square is the method of choice for three question types: finding the vertex or the maximum/minimum value, solving equations and leaving answers in surd form, and proving that an expression is always positive or always negative.
配方法是以下三类题型的首选：求顶点或最大最小值；解方程并保留根式答案；以及证明某个表达式恒为正或恒为负。
5. The Quadratic Formula | 二次公式
pleting the square on the general form a x² + b x + c = 0. You should memorise it exactly.
二次公式可以解任何二次方程，包括无法因式分解的方程。它的推导过程就是对一般式 a x² + b x + c = 0 配方。你必须一字不差地记住它。
x = ( -b ± √b² - 4ac ) ÷ (2a)
we have a = 2, b = -4 and c = -5. The discriminant is b² - 4ac = 16 + 40 = 56. So x = (4 ± √56) ÷ 4 = (4 ± 2√14) ÷ 4 = (2 ± √14) ÷ 2, giving x ≈ 2.87 or x ≈ -0.87 to 3 significant figures.
仔细看这个例子。对 2x² - 4x - 5 = 0，有 a = 2、b = -4、c = -5。判别式 b² - 4ac = 16 + 40 = 56。所以 x = (4 ± √56) ÷ 4 = (4 ± 2√14) ÷ 4 = (2 ± √14) ÷ 2，即 x ≈ 2.87 或 x ≈ -0.87（保留 3 位有效数字）。
n brackets. Do not simplify the surd unless the question asks for a decimal answer or says "give your answer in the form p ± q√r".
代入正确是有分的，所以一定要先写出公式，再把 a、b、c 的值带括号代入。除非题目要求小数答案，或要求写成 p ± q√r 的形式，否则不要随意化简根式。
6. The Discriminant | 判别式
The discriminant is the expression under the square root sign in the quadratic formula, and it tells you how many real roots the equation has before you solve it.
判别式就是二次公式中根号下的表达式，它能在你求解之前告诉你方程有几个实根。
Δ = b² - 4ac
| Discriminant | 判别式 | Roots | 根的个数 | Graph | 图像 |
| Δ > 0 | Two distinct real roots | 两个不同实根 | Crosses x-axis twice | 与 x 轴相交两次 |
| Δ = 0 | One repeated real root | 一个重根 | Touches x-axis at vertex | 顶点与 x 轴相切 |
| Δ < 0 | No real roots | 无实根 | Does not meet x-axis | 与 x 轴不相交 |
0, giving k = 6 or k = -6. Always consider both signs, because k² = 36 has two solutions.
典型考题：求使 x² + k x + 9 = 0 有等根的 k 值。等根即 Δ = 0，所以 k² - 36 = 0，得 k = 6 或 k = -6。要考虑正负两种情况，因为 k² = 36 有两个解。
product is c ÷ a. These are helpful for checking answers and for a small number of extension questions.
两个相关的有用结论是根的和与积。对 a x² + b x + c = 0，两根之和为 -b ÷ a，两根之积为 c ÷ a。它们可以用来检验答案，也能应对少量拓展题。
7. Graphs of Quadratic Functions | 二次函数的图像
curve is symmetric about the vertical line x = -b ÷ (2a), which passes through the vertex.
二次函数的图像是一条抛物线。若 a > 0，开口向上，有最小值；若 a < 0，开口向下，有最大值。曲线关于竖直线 x = -b ÷ (2a) 对称，这条线经过顶点。
he vertex from the completed square form or from x = -b ÷ (2a); decide the direction of opening from the sign of a; then join the points with a smooth symmetrical curve. Label the intercepts and the vertex, because these usually carry marks.
要准确画出二次函数图像，按五步走：令 x = 0 求 y 轴截距；用因式分解或公式求根；由配方式或 x = -b ÷ (2a) 求顶点；由 a 的符号判断开口方向；最后用平滑对称的曲线连接各点。要标出截距和顶点，这些通常都有分。
s steeper than y = x² when a > 1 and flatter when 0 < a < 1. A negative a reflects the curve in the x-axis.
你还要能识别图像变换。曲线 y = (x - p)² + q 就是 y = x² 向右平移 p 个单位、向上平移 q 个单位。当 a > 1 时，y = a x² 的图像比 y = x² 更陡；当 0 < a < 1 时更平缓。a 为负则把曲线沿 x 轴翻折。
8. Forming Quadratic Equations from Word Problems | 由应用题列二次方程
Word problems usually describe a geometric figure, an age, a consecutive number, or a rate such as speed. Your first job is to define a variable clearly, then translate each sentence into algebra before solving.
应用题通常描述几何图形、年龄、连续整数或速度等。第一步是清楚设定变量，然后在求解之前把每一句话翻译成代数式。
arranging gives x² + 3x - 40 = 0, which factorises as (x + 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.
看这个例子：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，于是 x(x + 3) = 40。整理得 x² + 3x - 40 = 0，分解为 (x + 8)(x - 5) = 0，所以 x = 5 或 x = -8。由于长度不可能为负，宽为 5 cm，长为 8 cm。
Always interpret your solutions in context and reject the ones that do not make sense. When the question involves consecutive integers, use x and x + 1; for consecutive even or odd integers, use x and x + 2. Check your final answer against the original wording, not just against your own equation.
一定要结合情境解释你的解，并舍去不合理的答案。若题目涉及连续整数，用 x 和 x + 1；若涉及连续偶数或奇数，用 x 和 x + 2。最后要用原题的文字核对答案，而不仅仅核对自己列的方程。
9. Quadratic Inequalities and Simultaneous Equations | 二次不等式与联立方程
To solve a quadratic inequality, first solve the corresponding equation to find the critical values, then sketch the parabola to decide which region satisfies the inequality.
解二次不等式时，先解对应的方程求出临界值，再画出抛物线，判断哪个区间满足不等式。
wards, the curve is below the x-axis between the roots and above it outside the roots. Therefore x² - x - 6 < 0 gives -2 < x < 3, while x² - x - 6 > 0 gives x < -2 or x > 3. The word "or" is essential in the second case.
例如，x² - x - 6 = 0 得 (x - 3)(x + 2) = 0，临界值为 x = 3 和 x = -2。因为抛物线开口向上，曲线在两根之间位于 x 轴下方，在两根之外位于 x 轴上方。因此 x² - x - 6 < 0 的解为 -2 < x < 3；而 x² - x - 6 > 0 的解为 x < -2 或 x > 3。第二种情况中的"或"字必不可少。
ance, y = x² + 1 and y = 2x + 4 give x² + 1 = 2x + 4, so x² - 2x - 3 = 0, hence (x - 3)(x + 1) = 0 and x = 3 or x = -1. Substituting back gives the points (3, 10) and (-1, 2).
解"一个一次、一个二次"的联立方程时，把一次方程代入二次方程以消去一个变量。例如 y = x² + 1 与 y = 2x + 4，得 x² + 1 = 2x + 4，即 x² - 2x - 3 = 0，分解为 (x - 3)(x + 1) = 0，x = 3 或 x = -1。代回得交点 (3, 10) 和 (-1, 2)。
discriminant: two intersections mean Δ > 0, tangency means Δ = 0, and no intersection means Δ < 0.
如果题目只问直线与曲线有几个交点，你不必完全解出方程。只需组成合并后的二次方程，观察判别式：两个交点对应 Δ > 0，相切对应 Δ = 0，无交点对应 Δ < 0。
10. Common Pitfalls in the Exam | 考试中的常见陷阱
Most lost marks in quadratic questions come from a small number of repeated errors. Knowing them in advance is the cheapest way to raise your grade.
二次方程题目丢的分，多数来自少数几个反复出现的错误。提前了解这些错误，是提高分数最省力的办法。
- Not rearranging to zero | 没有整理成等于零: You can only use the zero product property when one side is exactly zero.
- Dividing by x | 两边同时除以 x: If x² = 5x, dividing by x loses the root x = 0. Instead write x² - 5x = 0 and factorise.
- Sign errors | 符号错误: Take special care when b or c is negative; write the substitution in brackets, for example ( -3 )².
- Forgetting the ± | 忘记 ±: A square root gives two values, so a quadratic usually gives two answers.
- Rounding too early | 过早取近似值: Keep full accuracy through the working and round only at the final step.
- Ignoring the context | 忽略实际情境: A negative length, time or age must be rejected even if the algebra is correct.
Another frequent problem is a careless error when expanding brackets, which makes every later step wrong. If you have time, substitute your final root back into the original equation to confirm that both sides are equal.
另一个常见问题是展开括号时粗心出错，导致后面每一步都错。如果时间允许，把求出的根代回原方程，确认两边相等。
11. Choosing the Right Method | 选择正确的方法
Choosing an efficient method saves time and reduces errors. Use the table below as a quick decision guide during revision.
选择高效的方法能节省时间、减少错误。复习时可以把下表当作快速判断指南。
| Situation | 情况 | Best method | 最佳方法 |
| Simple integer roots, no calculator | 整数根，无计算器 | Factorisation | 因式分解 |
| Need the vertex or max
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