📚 IGCSE Mathematics: Quadratic Equations and Graphs | IGCSE 数学：二次方程与图像
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from basic factorising to graph sketching and real-life modelling. This revision guide covers the standard form, three main solving methods, the discriminant, graph features, inequalities, and common pitfalls to help you build confidence before the exam.
二次方程贯穿 IGCSE 数学考纲，从基础因式分解到图像绘制和实际建模都有涉及。本复习指南涵盖标准形式、三种主要求解方法、判别式、图像特征、不等式以及常见易错点，帮助你在考试前建立信心。
1. Quadratic Expressions and Standard Form | 二次表达式与标准形式
A quadratic expression in one variable has the form ax² + bx + c, where a, b and c are constants and a ≠ 0.
一元二次表达式形如 ax² + bx + c，其中 a、b、c 是常数，且 a ≠ 0。
The coefficient a cannot be zero because the x² term must be present; otherwise the expression becomes linear.
系数 a 不能为零，因为 x² 项必须存在；否则表达式就变成一次式。
For example, 3x² − 5x + 2 is quadratic with a = 3, b = −5 and c = 2.
例如，3x² − 5x + 2 是二次式，其中 a = 3，b = −5，c = 2。
2. Factorising Quadratics | 因式分解二次式
Factorising is often the fastest method when the quadratic has simple integer roots, especially when a = 1.
当二次式有简单的整数根，尤其是 a = 1 时，因式分解通常是最快的方法。
To factorise x² + bx + c, find two numbers that multiply to c and add to b.
要将 x² + bx + c 因式分解，需要找到两个数，它们的乘积等于 c，和等于 b。
- x² + 5x + 6 = (x + 2)(x + 3) because 2 × 3 = 6 and 2 + 3 = 5.
- x² + 5x + 6 = (x + 2)(x + 3)，因为 2 × 3 = 6，2 + 3 = 5。
- x² − 7x + 12 = (x − 3)(x − 4) because (−3) × (−4) = 12 and (−3) + (−4) = −7.
- x² − 7x + 12 = (x − 3)(x − 4)，因为 (−3) × (−4) = 12，(−3) + (−4) = −7。
When a ≠ 1, use the ac method: multiply a and c, then find two numbers that multiply to ac and add to b, before splitting the middle term.
当 a ≠ 1 时，使用 ac 法：先求 a 与 c 的乘积，再找两个数使它们相乘等于 ac、相加等于 b，然后拆分中间项。
3. Solving Quadratic Equations by Factorising | 因式分解法解二次方程
Once the quadratic expression is factorised, apply the zero product property: if AB = 0, then A = 0 or B = 0.
二次式因式分解后，应用零乘积性质：如果 AB = 0，那么 A = 0 或 B = 0。
Example: Solve x² − x − 6 = 0.
示例：解方程 x² − x − 6 = 0。
Factorise: (x − 3)(x + 2) = 0, so x − 3 = 0 or x + 2 = 0, giving x = 3 or x = −2.
因式分解得 (x − 3)(x + 2) = 0，所以 x − 3 = 0 或 x + 2 = 0，解得 x = 3 或 x = −2。
Always check your solutions by substituting them back into the original equation.
一定要把解代回原方程进行检验。
4. Completing the Square | 配方法
Completing the square is useful when factorising is difficult or when you need the turning point of a quadratic graph.
当因式分解较难或需要求二次图像顶点时，配方法非常有用。
For x² + bx + c, write the first two terms as (x + b/2)² then subtract (b/2)² and add c.
对于 x² + bx + c，先把前两项写成 (x + b/2)²，再减去 (b/2)² 并加上 c。
x² + bx + c = (x + b/2)² − (b/2)² + c
Example: x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4.
示例：x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4。
The completed square form (x + p)² + q directly reveals the minimum or maximum point of the graph.
完全平方形式 (x + p)² + q 能直接揭示图像的最小值点或最大值点。
5. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation ax² + bx + c = 0, where a ≠ 0.
求根公式可以解任意二次方程 ax² + bx + c = 0，其中 a ≠ 0。
x = (−b ± √(b² − 4ac)) ÷ 2a
To use the formula, first identify a, b and c, then substitute carefully and simplify.
使用公式时，先确定 a、b、c，然后小心代入并化简。
Example: Solve 2x² − 3x − 2 = 0. Here a = 2, b = −3, c = −2.
示例：解方程 2x² − 3x − 2 = 0。这里 a = 2，b = −3，c = −2。
Substitute: x = (3 ± √(9 + 16)) ÷ 4 = (3 ± 5) ÷ 4, giving x = 2 or x = −1/2.
代入得：x = (3 ± √(9 + 16)) ÷ 4 = (3 ± 5) ÷ 4，解得 x = 2 或 x = −1/2。
This method is reliable when factorising is not obvious, but watch out for sign errors.
当不容易看出因式分解时，这个方法很可靠，但要留意符号错误。
6. Discriminant and Nature of Roots | 判别式与根的性质
The expression b² − 4ac inside the square root is called the discriminant, often written as Δ.
平方根内的表达式 b² − 4ac 称为判别式，常写作 Δ。
The discriminant tells you how many real solutions the quadratic equation has.
判别式告诉你二次方程有多少个实数解。
| Discriminant | Number of real roots |
| b² − 4ac > 0 | Two distinct real roots |
| b² − 4ac = 0 | One repeated real root |
| b² − 4ac < 0 | No real roots |
如果 Δ > 0，方程有两个不相等的实数根；如果 Δ = 0，方程有一个重根；如果 Δ < 0，方程没有实数根。
Example: For 2x² − 3x − 2 = 0, Δ = (−3)² − 4(2)(−2) = 9 + 16 = 25 > 0, so there are two real roots.
示例：对于 2x² − 3x − 2 = 0，Δ = (−3)² − 4(2)(−2) = 9 + 16 = 25 > 0，所以有两个实数根。
7. Quadratic Graphs: Shape and Key Features | 二次图像：形状与关键特征
The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upwards; if a < 0, it opens downwards.
y = ax² + bx + c 的图像是抛物线。如果 a > 0，开口向上；如果 a < 0，开口向下。
The y-intercept is found by setting x = 0, so the graph crosses the y-axis at (0, c).
y 轴截距通过令 x = 0 求得，因此图像与 y 轴交于 (0, c)。
The x-intercepts are the real roots of ax² + bx + c = 0, if they exist.
x 轴截距是方程 ax² + bx + c = 0 的实数根，如果存在的话。
For example, y = x² − 4 has roots x = ±2 and y-intercept at (0, −4).
例如，y = x² − 4 的根为 x = ±2，y 轴截距为 (0, −4)。
8. Turning Point and Line of Symmetry | 顶点与对称轴
A parabola has one turning point: a minimum if a > 0 and a maximum if a < 0.
抛物线有一个顶点：当 a > 0 时为最小值点，当 a < 0 时为最大值点。
If the quadratic is written as y = (x + p)² + q, the turning point is (−p, q).
如果二次式写成 y = (x + p)² + q，那么顶点是 (−p, q)。
From the standard form, the x-coordinate of the turning point is found using x = −b ÷ 2a.
从标准形式出发，顶点的 x 坐标可以用 x = −b ÷ 2a 求得。
The line of symmetry is the vertical line x = −b ÷ 2a, which passes through the turning point.
对称轴是竖直直线 x = −b ÷ 2a，它经过顶点。
9. Solving Quadratic Inequalities | 解二次不等式
To solve a quadratic inequality such as x² − 5x + 6 > 0, first find the roots of the corresponding equation.
解二次不等式如 x² − 5x + 6 > 0，先求对应方程的根。
Factorise: (x − 2)(x − 3) > 0. The roots are x = 2 and x = 3. These split the number line into three intervals.
因式分解得 (x − 2)(x − 3) > 0。根为 x = 2 和 x = 3。它们把数轴分成三个区间。
Test a value from each interval in the inequality. Here the solution is x < 2 or x > 3.
在每个区间中取一个值代入不等式检验。这里解为 x < 2 或 x > 3。
If the inequality is < 0, the solution would be the interval between the roots: 2 < x < 3.
如果不等式是 < 0，解就是两根之间的区间：2 < x < 3。
10. Applications and Word Problems | 应用与文字题
Quadratic equations often model area, projectile motion, and revenue problems.
二次方程常用于面积、抛体运动和收益问题的建模。
Example: A rectangle has length x + 3 and width x − 1. If its area is 40 cm², form and solve an equation for x.
示例：一个矩形的长为 x + 3，宽为 x − 1。如果它的面积是 40 cm²，列出并解出 x 的方程。
Equation: (x + 3)(x − 1) = 40, so x² + 2x − 3 = 40, giving x² + 2x − 43 = 0. Solve by the quadratic formula if needed.
方程：(x + 3)(x − 1) = 40，所以 x² + 2x − 3 = 40，得 x² + 2x − 43 = 0。必要时用求根公式求解。
In word problems, discard negative solutions that do not make sense for lengths or time.
在文字题中，要舍去对长度或时间无意义的负解。
11. Common Mistakes | 常见错误
Many students forget that a must not be zero in a quadratic equation, or they divide by a term containing x and lose a root.
很多学生忘记二次方程中 a 不能为零，或者除以含 x 的项而丢根。
Sign errors in factorising and in substituting into the quadratic formula are very common, so use brackets for negative values.
因式分解和代入求根公式时的符号错误非常常见，因此负数要加括号。
Do not write √(b² − 4ac) as ±√(b² − 4ac) inside the formula, because the ± already appears before the square root.
在公式中不要把 √(b² − 4ac) 写成 ±√(b² − 4ac)，因为平方根前已经有 ±。
When completing the square, remember to keep the equation balanced by adding and subtracting the same value.
配方时，要记得加减同一个值以保持方程平衡。
12. Exam Tips for Quadratic Questions | 二次方程题考试技巧
Read the question carefully to decide which method is most efficient: factorising for simple roots, the formula for messy decimals, and completing the square for turning points.
仔细审题，选择最有效的方法：简单根用因式分解，小数较复杂用公式，求顶点用配方法。
Show all working clearly, especially when substituting into the quadratic formula, because method marks are often awarded even if the final answer is wrong.
清晰展示所有步骤，尤其是代入求根公式的过程，因为即使最终答案有误，通常也能得到方法分。
If a question asks for answers to a given number of decimal places, calculate the square root carefully and round only at the final step.
如果题目要求答案保留指定小数位，要仔细计算平方根，并且只在最后一步才取近似值。
Use the discriminant to check the nature of roots before solving, and always substitute your solutions back into the original equation as a quick check.
解题前先用判别式判断根的性质，并把解代回原方程进行快速检验。
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