📚 Solving Quadratic Equations | 解二次方程
A quadratic equation is one of the most frequently tested topics in IGCSE Mathematics. Whether you sit for Cambridge IGCSE, Edexcel International GCSE, or other major boards, questions on factorising, solving, and interpreting quadratics appear in almost every examination paper. Mastering this topic unlocks access to higher-level work in algebra, graphs, and problem solving, and it forms the foundation for many calculus topics studied later.
二次方程是 IGCSE 数学中最常考查的内容之一。无论你参加 Cambridge IGCSE、Edexcel International GCSE 还是其他主流考试局，几乎所有试卷都会出现因式分解、求解和解读二次方程的问题。掌握这一主题不仅有助于你进一步学习代数、函数图像和实际应用问题，也为日后学习微积分等更高阶内容打下坚实基础。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is an equation that takes the general form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The term quadratic comes from a Latin word meaning square, because the highest power of the unknown is 2. The degree of the equation is therefore always 2.
二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。术语 "quadratic" 源自拉丁语中意为"正方形"的词，因为未知数的最高次数为 2，因此方程的次数总是 2。
otice that b and c may be zero, but the coefficient a of the x² term can never be zero; otherwise the equation would reduce to a linear equation. Constant terms such as c can be positive, negative, or zero.
例如，在方程 3x² − 5x + 2 = 0 中，a = 3，b = −5，c = 2。注意 b 和 c 可以为零，但二次项系数 a 绝不能为零，否则方程就退化成了线性方程。常数项 c 可以是正数、负数或零。
fficient of x² positive. For instance, x² = 4x + 5 becomes x² − 4x − 5 = 0 before solving.
在求解之前，你需要重整方程，使所有非零项都在等号一侧，右侧等于零。先去括号、合并同类项，并在可能的情况下使 x² 项系数为正。例如 x² = 4x + 5 应先行移项得到 x² − 4x − 5 = 0 再求解。
ax² + bx + c = 0, where a ≠ 0
2. Solving by Factorisation | 因式分解法
a product of two binomial factors set equal to zero, then apply the zero product property.
因式分解是当二次方程具有简单整数系数时最直接的方法。其原理是将 ax² + bx + c = 0 改写为两个二项式因子的乘积等于零，然后运用零积性质求解。
tiply to 6 and sum to 5: these are 2 and 3. Hence the equation becomes (x + 2)(x + 3) = 0.
对于首项系数为 1 的二次方程，寻找两个数使其乘积等于 c、和等于 b。例如解 x² + 5x + 6 = 0 时，寻找乘积为 6、和为 5 的两个数：即 2 和 3。因此方程改写为 (x + 2)(x + 3) = 0。
r x = −3. Always present both solutions clearly on separate lines or separated by the word "or".
接着运用零积性质：若两个因式的乘积为零，则至少有一个因式为零。因此 x + 2 = 0 或 x + 3 = 0，解得 x = −2 或 x = −3。务必把两个解分别写在同一行的两端，并用"或"分隔。
(2x + 1) + 3(2x + 1) = 0, giving (2x + 1)(x + 3) = 0. The solutions are x = −½ and x = −3.
当首项系数 a 不为 1 时，例如 2x² + 7x + 3 = 0，将 a 与 c 相乘得到 2 × 3 = 6，然后寻找乘积为 6、和为 7 的两个数：即 1 和 6。将中间项拆分得 2x² + x + 6x + 3 = 0，再分组因式分解：x(2x + 1) + 3(2x + 1) = 0，即 (2x + 1)(x + 3) = 0。解为 x = −½ 和 x = −3。
3. Difference of Two Squares | 平方差公式
A special and very important case of factorisation is the difference of two squares: a² − b² = (a + b)(a − b). This identity appears constantly in IGCSE papers, both in pure algebra and embedded within larger problems. It applies whenever a quadratic consists only of an x² term and a constant term which is a perfect square.
因式分解中一个特殊且非常重要的情形是平方差公式：a² − b² = (a + b)(a − b)。这个恒等式在 IGCSE 试卷中出现频率极高，既出现在纯代数题中，也融入更复杂的综合题中。只要二次式仅含 x² 项和一个完全平方的常数项，就可以运用该公式。
ation could also be solved by moving 9 across and taking square roots, but factorisation reveals both roots in one step and avoids sign errors.
例如解 x² − 9 = 0 时，确认 9 = 3²，因此 x² − 9 = (x + 3)(x − 3) = 0。解得 x = −3 或 x = 3。注意此
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