📚 Mastering Quadratic Equations for IGCSE Mathematics | 精通 IGCSE 数学二次方程
Quadratic equations are a central topic in IGCSE Mathematics. They appear in algebra, coordinate geometry, and real-life modelling, so a solid understanding of how to solve and interpret them is essential for success in the examination. This guide covers every method on the syllabus, shows you how to choose the fastest route to the answer, and highlights the mistakes that cost candidates the most marks.
二次方程是 IGCSE 数学的核心主题。它们出现在代数、坐标几何和现实建模中，因此扎实掌握如何求解并解读二次方程对考试成功至关重要。本指南覆盖考纲中的全部解法，教你如何选择最快的解题路径，并指出最容易让考生丢分的错误。
You will meet quadratics in Paper 2 and Paper 4 in the 0580 syllabus, and heavily in 0606 Additional Mathematics. The skills build on each other: factorising leads to completing the square, completing the square derives the quadratic formula, and the formula leads to the discriminant. Work through the sections in order if you are revising from scratch, or jump straight to the method you need.
在 0580 考纲中，你会在 Paper 2 和 Paper 4 遇到二次方程；在 0606 进阶数学中则会大量出现。这些技能层层递进：因式分解通向配方法，配方法推导出求根公式，求根公式又引出判别式。如果你是零基础复习，请按顺序阅读；如果只想查漏补缺，可直接跳到你需要的方法。
1. What Is a Quadratic Equation? | 什么是二次方程？
can usually be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是一个最高次数为 2 的多项式方程。在单变量情况下，它的最高次幂是 x²，通常可以写成 ax² + bx + c = 0 的形式，其中 a、b、c 是常数且 a ≠ 0。
ather than quadratic. The presence of the x² term is what gives the equation its curved graph and its two possible solutions.
如果 a = 0，方程就变成了一次方程，而不是二次方程。x² 项的存在使方程图像呈曲线，并使得方程可能有两个解。
The word "quadratic" comes from the Latin word for square, and the defining feature of every quadratic is that the unknown appears squared at least once. The equation may also contain a linear term, a constant term, or both - but it must contain the x² term.
"quadratic" 一词源自拉丁语中"平方"的意思，而每一个二次方程的决定性特征就是未知量至少出现一次平方。方程中可能同时含有一次项、常数项，或两者兼有——但必须含有 x² 项。
ax² + bx + c = 0, a ≠ 0
2. Standard Form and Coefficients | 标准形式与系数
coefficients a, b and c can be positive, negative, or zero, except that a cannot be zero.
在求解之前，要先把方程化为标准形式：ax² + bx + c = 0。系数 a、b、c 可以是正数、负数或零，但 a 不能为零。
ation 3x - 5 = 2x² should be rearranged to 2x² - 3x + 5 = 0. Here a = 2, b = -3 and c = 5.
例如，方程 3x - 5 = 2x² 应重新整理为 2x² - 3x + 5 = 0。这里 a = 2，b = -3，c = 5。
A reliable habit is to move every term to the side where the x² coefficient is positive, then write down a, b and c explicitly before substituting them into any formula. Writing them out separately dramatically reduces sign errors later.
一个可靠的习惯是：把所有项移到 x² 系数为正的一侧，然后在代入任何公式之前明确写出 a、b、c。把它们单独写出来能大幅减少后续的符号错误。
= x + 2 becomes x² + 2x - 3 = 0 after multiplying through by x, with the condition x ≠ 0.
有些二次方程隐藏在其他形式之中。方程 x(x - 4) = 5 在展开括号前并不是标准形式，展开后得到 x² - 4x - 5 = 0；而 3/x = x + 2 在两边同乘 x（并注明 x ≠ 0）后变为 x² + 2x - 3 = 0。
| Equation 原方程 | Standard form 标准形式 | a | b | c |
|---|---|---|---|---|
| 3x - 5 = 2x² | 2x² - 3x + 5 = 0 | 2 | -3 | 5 |
| x² = 9 | x² - 9 = 0 | 1 | 0 | -9 |
| (x + 1)² = 4x | x² - 2x + 1 = 0 | 1 | -2 | 1 |
3. Solving by Factorisation | 因式分解法求解
or x + q = 0, because a product can only be zero when at least one of its factors is zero.
当二次方程可以表示为两个一次因式的乘积时，因式分解法是最快的方法。如果 (x + p)(x + q) = 0，那么 x + p = 0 或 x + q = 0，因为只有当至少一个因式为零时，乘积才为零。
o 6 and add to 5: they are 2 and 3. Therefore (x + 2)(x + 3) = 0, giving x = -2 or x = -3.
当 a = 1 时，寻找两个数，它们相乘得 c、相加得 b。例如，解 x² + 5x + 6 = 0。我们寻找两个数，它们相乘得 6、相加得 5：即 2 和 3。因此 (x + 2)(x + 3) = 0，得到 x = -2 或 x = -3。
1)(x + 3) = 0, so x = -1/2 or x = -3. Check by expanding: 2x² + 6x + x + 3 = 2x² + 7x + 3.
当 a ≠ 1 时，寻找过程稍长，但逻辑完全相同。对于 2x² + 7x + 3 = 0，因式分解为 (2x + 1)(x + 3) = 0，所以 x = -1/2 或 x = -3。展开检验：2x² + 6x + x + 3 = 2x² + 7x + 3。
e such as x² - 10x + 25 = 0 becomes (x - 5)² = 0, which has only one repeated root, x = 5.
留意特殊模式。平方差形式如 x² - 49 = 0 可立即分解为 (x - 7)(x + 7) = 0；完全平方形式如 x² - 10x + 25 = 0 变为 (x - 5)² = 0，只有一个重根 x = 5。
Always check by expanding the brackets. This method works best when the roots are integers or simple fractions; if they are not, switch to completing the square or the formula.
始终要通过展开括号来检验。当根是整数或简单分数时，这种方法最有效；如果不是，请改用配方法或求根公式。
4. Solving by Completing the Square | 配方法求解
ays works, even when factorisation fails, and it is the standard way to find the turning point of a curve.
配方法把二次式改写为 (x + p)² + q = 0 的形式。它总是可行，即使因式分解失败时也一样，而且它是求曲线转折点的标准方法。
= 0, half of 6 is 3 and 3² = 9, so we write x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8.
技巧是把 x 的系数除以 2，再平方，然后加上并减去这个值。对于 x² + 6x + 1 = 0，6 的一半是 3，3² = 9，所以写成 x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8
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