📚 IGCSE Maths: 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 modelling projectile motion. Understanding how to solve them and how their graphs behave is essential for both Core and Extended candidates.
二次方程贯穿 IGCSE 数学课程，从基础因式分解到抛体运动建模都有涉及。掌握二次方程的解法及其图像行为，对 Core 和 Extended 考生都至关重要。
1. What Is a Quadratic Equation? | 什么是二次方程？
c are constants with a ≠ 0. The highest power of x is 2, which gives the equation its characteristic curved graph.
二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 x 是变量，a、b、c 是常数且 a ≠ 0。x 的最高次数为 2，这使得它的图像具有特有的曲线形状。
−x² + 6x = 0. If a = 0, the equation becomes linear, so the condition a ≠ 0 is essential.
例子包括 x² − 4x + 3 = 0、2x² + 5x − 7 = 0 和 −x² + 6x = 0。如果 a = 0，方程就变成了一次方程，因此 a ≠ 0 的条件非常关键。
2. Standard Form and Coefficients | 标准形式与系数
, where a is the coefficient of x², b is the coefficient of x, and c is the constant term.
求解前，应先将方程的所有项移到一边，使另一边为 0。标准形式为 ax² + bx + c = 0，其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。
example, 3x + 2 = x² can be rearranged to x² − 3x − 2 = 0. Here a = 1, b = −3 and c = −2.
例如，3x + 2 = x² 可以整理为 x² − 3x − 2 = 0。此时 a = 1，b = −3，c = −2。
- Always write the x² term first, then the x term, then the constant.
- 始终先写 x² 项，再写 x 项，最后写常数项。
3. Solving by Factorising | 因式分解法求解
least one factor must be zero. Solve x² − 5x + 6 = 0 by writing it as (x − 2)(x − 3) = 0.
如果二次式可以因式分解为两个一次因式，我们可以利用“乘积为零，则至少一个因式为零”的性质。将 x² − 5x + 6 = 0 写成 (x − 2)(x − 3) = 0 来求解。
(x − 2)(x − 3) = 0 ⇒ x = 2 or x = 3
he roots of the equation. Always check by expanding the brackets to make sure the factorisation is correct.
解 x = 2 和 x = 3 被称为方程的根。记得展开括号验证因式分解是否正确。
middle term: 2x² + 2x + 3x + 3 = 0, then factorise by grouping to get (2x + 3)(x + 1) = 0.
当系数 a 不等于 1 时，例如 2x² + 5x + 3 = 0，需要找两个数，它们相乘等于 2 × 3 = 6，相加等于 5。这两个数是 2 和 3，因此拆分中间项：2x² + 2x + 3x + 3 = 0，然后分组因式分解得到 (2x + 3)(x + 1) = 0。
4. The Quadratic Formula | 二次公式
factorising is difficult or impossible. For ax² + bx + c = 0, the solutions are given by:
二次公式适用于所有二次方程，即使因式分解很困难或不可能。对于 ax² + bx + c = 0，解为：
x = (−b ± √b² − 4ac) ÷ 2a
2 = 0, a = 2, b = 3, c = −2, so b² − 4ac = 3² − 4 × 2 × (−2) = 9 + 16 = 25, and x = (−3 ± √25) ÷ 4 = (−3 ± 5) ÷ 4, giving x = 0.5 or x = −2.
代入负数时要小心使用括号。对于 2x² + 3x − 2 = 0，a = 2，b = 3，c = −2，因此 b² − 4ac = 3² − 4 × 2 × (−2) = 9 + 16 = 25，x = (−3 ± √25) ÷ 4 = (−3 ± 5) ÷ 4，得到 x = 0.5 或 x = −2。
In exams, state the formula first, then show the substitution step clearly. This earns method marks even if a small arithmetic error is made later.
考试时，应先写出公式，再清楚地展示代入步骤。这样即使后面出现小的计算错误，也能得到方法分。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)
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