📚 Solving Quadratic Equations | 解一元二次方程
b and c are constants and a ≠ 0. Quadratics appear in nearly every IGCSE Mathematics paper — factorisation, the three standard solving methods, the discriminant and exam-style applications, with bilingual explanations to make every step clear.
一元二次方程是指形如 ax² + bx + c = 0（其中 a、b、c 为常数且 a ≠ 0）的方程。二次函数几乎出现在每份 IGCSE 数学试卷中——无论是直接考查，还是作为几何、数列和实际问题中的隐含步骤。本指南将带你复习展开与因式分解，掌握三种标准解法、判别式以及考试风格的应用题，并配以中英双语讲解，确保每一步都清晰易懂。
1. Expanding Double Brackets | 展开二项式括号
Before solving quadratics, you must be comfortable expanding and factorising. To expand (x + p)(x + q), multiply each term in the first bracket by each term in the second. The acronym FOIL (First, Outer, Inner, Last) helps you remember the order. For example:
在解一元二次方程之前，你必须熟练掌握展开与因式分解。要展开 (x + p)(x + q)，需要用第一个括号中的每一项去乘第二个括号中的每一项。首外内尾（FOIL）口诀可以帮助你记住顺序。例如：
(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15
Notice that the coefficient of x is the sum of 3 and 5, and the constant term is their product. This pattern leads directly to factorisation: to factorise x² + bx + c, find two numbers that multiply to c and add to b.
注意：x 的系数是 3 和 5 的和，常数项是 3 和 5 的积。这一规律直接引出因式分解：要把 x² + bx + c 分解因式，只需找到两个数，使它们的乘积为 c、和为 b。
2. Factorising Quadratic Expressions | 二次多项式的因式分解
When the coefficient of x² is 1, factorising is straightforward. For x² − 7x + 12, we need two numbers with product 12 and sum −7. These are −3 and −4, so:
当 x² 的系数为 1 时，因式分解非常简单。对于 x² − 7x + 12，我们需要找到乘积为 12、和为 −7 的两个数。这两个数是 −3 和 −4，因此：
x² − 7x + 12 = (x − 3)(x − 4)
When the coefficient of x² is not 1, for example 2x² + 7x + 3, you can use the 'grouping' method. Multiply a and c (2 × 3 = 6), then find two numbers whose product is 6 and whose sum is 7 — that is 1 and 6. Split the middle term and factor by grouping:
当 x² 的系数不为 1 时，例如 2x² + 7x + 3，可以使用“分组分解”法。先求出 a 与 c
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