📚 Solving Quadratic Equations | 解二次方程
where a, b and c are constants and a ≠ 0. This topic is one of the most frequently tested areas in IGCSE Mathematics, appearing in both Core and Extended papers through factorisation, formula, graphs and word problems.
二次方程是最高次数为二的整式方程，标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。这是 IGCSE 数学中考查频率最高的内容之一，在 Core 卷和 Extended 卷中都有涉及，包括因式分解、公式法、图像法和应用题等题型。
1. Identifying a Quadratic Equation | 识别二次方程
An equation is quadratic when the highest power of the variable is 2. In standard form, terms must be arranged in descending order: first the x² term, then the x term, then the constant term, with the whole expression set equal to zero.
判断一个方程是否为二次方程，只需看变量的最高次数是否为 2。标准形式要求按降幂排列：先是 x² 项，然后是 x 项，最后是常数项，且整个式子等于零。
a ≠ 0 is essential: if a = 0, the equation becomes linear because the x² term disappears.
例如，3x² − 5x + 2 = 0 是二次方程，而 2x + 1 = 0 是一次方程，x³ − 1 = 0 是三次方程。条件 a ≠ 0 非常重要：若 a = 0，x² 项消失，方程就退化为一次方程了。
l terms to the left side. Always simplify and rearrange before choosing a solution method.
有些方程一开始并不是标准形式，需要先移项整理。例如 4x² = 9 − 2x 移项后得到 4x² + 2x − 9 = 0。在决定解法之前，务必先化简整理。
Similarly, x(x − 5) = 0 expands to x² − 5x = 0 and has the two solutions x = 0 and x = 5.
要警惕特殊形式。x² = 16 本质上是二次方程：两边同减 16 得 x² − 16 = 0，因式分解为 (x + 4)(x − 4) = 0。同样，x(x − 5) = 0 展开为 x² − 5x = 0，其两个解为 x = 0 和 x = 5。
2. Solving by Factorisation | 用因式分解法求解
Factorisation is often the quickest method for solving a quadratic equation. The key idea is that if two expressions multiply to zero, then at least one of them must be zero.
因式分解法是解二次方程最快的方法之一。核心思想是：若两个因式的乘积为零，则至少有一个因式为零。
If (px + q)(rx + s) = 0, then px + q = 0 or rx + s = 0
若 (px + q)(rx + s) = 0，则 px + q = 0 或 rx + s = 0
two numbers that multiply to 12 and add to −7. The pair is −3 and −4, so the factorisation is:
例：解方程 x² − 7x + 12 = 0。寻找两个数，使其乘积为 12、和为 −7。这个数对是 −3 和 −4，因此因式分解为：
x² − 7x + 12 = (x − 3)(x − 4) = 0
4. Always expand your factorisation mentally to check it returns the original expression.
令每个因式分别为零，得 x − 3 = 0 或 x − 4 = 0，所以 x = 3 或 x = 4。建议在脑中展开检验，确认能否还原原式。
umbers whose product is 6 and sum is b = 5, which are 2 and 3. Then split the middle term:
当 x² 系数不是 1 时，过程需要更仔细。解 2x² + 5x + 3 = 0。先计算 a × c = 2 × 3 = 6，再找两个数使乘积为 6、和为 b = 5，即 2 和 3。然后拆中项：
2x² + 2x + 3x + 3 = 0
oup the terms: 2x(x + 1) + 3(x + 1) = 0, so (2x + 3)(x + 1) = 0. Hence x = −3/2 or x = −1.
分组提取公因式：2x(x + 1) + 3(x + 1) =
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