📚 Quadratic Equations for IGCSE Mathematics: Factorising, Formula and Graphs | IGCSE 数学二次方程：因式分解、求根公式与图像
Quadratic equations are one of the most important algebra topics in IGCSE Mathematics. They appear in Paper 2 and Paper 4, often linked to graphs, inequalities and real-life problems. This guide explains factorising, completing the square, the quadratic formula, the discriminant and graph sketching, with clear worked examples and common exam tips.
二次方程是 IGCSE 数学代数部分最重要的主题之一，在 Paper 2 和 Paper 4 中经常与图像、不等式和实际问题结合考查。本指南讲解因式分解、配方法、求根公式、判别式以及图像草图，并配有清晰的例题和常见考试技巧。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
nts and a ≠ 0. If a = 0, the equation becomes linear, so the condition a ≠ 0 is essential.
二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。若 a = 0，方程就变成一次方程，因此 a ≠ 0 这一条件至关重要。
In IGCSE questions, a, b and c are usually integers or simple fractions. You must be able to identify a, b and c because these values are needed for the quadratic formula and the discriminant.
在 IGCSE 题目中，a、b、c 通常是整数或简单分数。你必须能够识别 a、b、c，因为求根公式和判别式都需要这些值。
2. Solving by Factorising | 因式分解法求解
. Then find two numbers that multiply to give ac and add to give b. Split the middle term, factorise by grouping, and set each bracket equal to zero.
用因式分解法求解时，先把二次方程写成 ax² + bx + c = 0。然后找到两个数，它们的乘积为 ac，和为 b。拆中项、分组分解，再令每个括号等于零。
th product 6 and sum -5; they are -2 and -3. So (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
例题：解 x² - 5x + 6 = 0。需要两个数，乘积为 6，和为 -5；它们是 -2 和 -3。因此 (x - 2)(x - 3) = 0，得到 x = 2 或 x = 3。
Always check your factorised brackets by expanding them mentally. Sign errors are very common when the middle term is negative.
一定要通过心算展开括号来检查因式分解是否正确。当中间项为负时，符号错误非常常见。
3. Solving by Completing the Square | 配方法求解
Completing the square transforms ax² + bx + c into the form a(x + p)² + q. This form immediately shows the turning point and can solve equations that do not factorise neatly.
配方法把 ax² + bx + c 转化为 a(x + p)² + q 的形式。这种形式能直接显示顶点，并能求解不易因式分解的方程。
Example: x² + 6x + 1 = 0 becomes (x + 3)² - 9 + 1 = 0, so (x + 3)² = 8, giving x = -3 ± 2√2.
对于 x² + bx + c，写成 (x + b/2)² - (b/2)² + c。例题：x² + 6x + 1 = 0 变为 (x + 3)² - 9 + 1 = 0，所以 (x + 3)² = 8，得到 x = -3 ± 2√2。
In IGCSE, completing the square is often tested because it links directly to the turning point. You may be asked to express a quadratic in completed square form and then state the minimum or maximum value.
在 IGCSE 中，配方法经常被考查，因为它直接与顶点相关。题目可能要求把二次式写成完全平方形式，然后写出最小值或最大值。
4. The Quadratic Formula | 求根公式
ctorising is difficult, but you must quote it accurately and substitute a, b, c carefully, including negative signs.
求根公式可求解任意二次方程 ax² + bx + c = 0。当因式分解困难时尤其有用，但必须准确写出公式并小心代入 a、b、c，包括负号。
x = (-b ± √b² - 4ac) / 2a
Always check whether the question requires answers in exact surd form or rounded to a given number of decimal places. IGCSE often asks for 3 significant figures unless stated otherwise.
务必检查题目要求答案保留根号形式还是四舍五入到指定小数位。IGCSE 通常要求保留 3 位有效数字，除非另有说明。
Example: Solve 2x² + 3x - 2 = 0 using the formula. Here a = 2, b = 3, c = -2. So x = (-3 ± √3² - 4 × 2 × -2) / (2 × 2) = (-3 ± √9 + 16) / 4 = (-3 ± 5) / 4.
例题：用公式解 2x² + 3x - 2 = 0。这里 a = 2，b = 3，c = -2。因此 x = (-3 ± √3² - 4 × 2 × -2) / (2 × 2) = (-3 ± √9 + 16) / 4 = (-3 ± 5) / 4。
This gives x = 1/2 or x = -2. Both roots are correct because the discriminant is positive.
得到 x = 1/2 或 x = -2。这两个根都正确，因为判别式为正。
5. The Discriminant and Number of Roots | 判别式与根的个数
The discriminant is the expression b² - 4ac under the square root in the quadratic formula. It tells you how many real roots the equation has before you solve it.
判别式是求根公式中平方根下的表达式 b² - 4ac。它能在求解前告诉你方程有多少个实数根。
| Discriminant | Number of real roots | 根的个数 |
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