📚 Mastering Quadratic Equations | 掌握二次方程
Quadratic equations appear in virtually every IGCSE Mathematics examination paper. Whether you sit the Core or Extended paper, the ability to solve and interpret quadratics is a non-negotiable skill. This revision guide breaks down every method — factorisation, completing the square and the quadratic formula — into clear, exam-focused steps.
二次方程几乎出现在每份 IGCSE 数学考卷中。无论你考 Core（核心）还是 Extended（扩展）卷，求解并理解二次函数都是必备技能。本复习指南将因式分解、配方法与二次公式这三种解法拆解为清晰、紧扣考点的步骤。
1. What Is a Quadratic Equation? | 什么是二次方程？
stants and a ≠ 0. The coefficient a cannot be zero, otherwise the equation would reduce to a linear equation.
二次方程是指能写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。系数 a 不能为零，否则方程就会退化为一次方程。
x² + bx + c is a smooth curve called a parabola. When a > 0 the parabola opens upwards; when a < 0 it opens downwards.
变量的最高次数为 2，因此二次方程也被称为二次方程。函数 y = ax² + bx + c 的图像是一条平滑的曲线，称为抛物线。当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。
= 0 and 9x² − 16 = 0. Note that 9x² = 16 must first be rearranged into the standard form.
常见例子包括 x² − 4x + 3 = 0、2x² + 5x − 7 = 0 以及 9x² − 16 = 0。注意 9x² = 16 需要先移项整理成标准形式。
x − 3)(x + 5) = 0. You must still recognise that expanding it would give x² + 2x − 15 = 0.
有些题目中二次方程以因式形式出现，例如 (x − 3)(x + 5) = 0。你应该能看出展开后就是 x² + 2x − 15 = 0。
2. Review: Expanding and Factorising | 复习：展开与因式分解
Before solving quadratics, you must be fluent in expanding brackets and factorising trinomials. These are the inverse operations that let you move between factored form and standard form.
在求解二次方程之前，你必须熟练展开括号和分解因式。这两种运算互为逆运算，帮助你在这两种形式之间自由转换。
actorise x² + bx + c you simply look for two numbers whose product is c and whose sum is b.
一般的展开公式为 (x + p)(x + q) = x² + (p + q)x + pq。要将 x² + bx + c 因式分解，只需找到两个数，使其乘积为 c、和为 b。
Three special products are tested frequently and are worth memorising:
三个特殊公式经常被考查，值得牢记：
(a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b², (a + b)(a − b) = a² − b²
(a + b)² = a² + 2ab + b²， (a − b)² = a² − 2ab + b²， (a + b)(a − b) = a² − b²
m. Here a × c = −6, so look for two numbers with product −6 and sum 5; those are 6 and −1.
当 x² 的系数不为 1 时，例如 2x² + 5x − 3，可先计算 a × c 再拆中项。此处 a × c = −6，所以找乘积为 −6、和为 5 的两个数，即 6 和 −1。
3. Solving by Factorisation | 用因式分解求解
Factoring is the fastest method when the quadratic has rational roots. The key principle is the zero product property: if the product of two expressions is zero, then at least one of them must be zero.
当二次方程有有理数根时，因式分解是最快的方法。核心原则是“零乘积性质”：若两个表达式的乘积为零，则其中至少有一个必须为零。
Worked example | 例： Solve x² − 5x + 6 = 0.
例题：解方程 x² − 5x + 6 = 0。
Write down the equation in standard form. | 写出标准形式。
x² − 5x + 6 = 0
Factorise the left-hand side. | 对左边因式分解。
(x − 2)(x − 3) = 0
Apply the zero product property. | 应用零乘积性质。
x − 2 = 0 or x − 3 = 0
所以 x − 2 = 0 或 x − 3 = 0
State the solution set. | 写出解集。
x = 2 or x = 3
x = 2 或 x = 3
fore factorising. If you try to factorise x² − 5x = −6 directly, you may make sign errors.
因式分解前务必把方程整理成一侧为零。若直接对 x² − 5x = −6 分解，很容易出现符号错误。
4. Solving by Completing the Square | 配方法
. This method works for any quadratic and is essential when factorisation is not possible.
配方法把二次式改写成一个完全平方加一个常数：x² + bx = (x + b/2)² − (b/2)²。这种方法适用于任何二次方程，尤其在无法因式分解时必不可少。
Worked example | 例： Solve x² + 6x + 2 = 0 by completing the square.
例题：用配方法解方程 x² + 6x + 2 = 0。
Start from x² + 6x + 2 = 0. Move the constant to the other side. | 从 x² + 6x + 2 = 0 出发，把常数项移到另一边。
x² + 6x = −2
, which is 3, and form (x + 3)². Since (x + 3)² = x² + 6x + 9, subtract 9 to keep balance. | 取 6 的一半得 3，组成 (x + 3)²。因为 (x + 3)² = x² + 6x + 9，所以减去 9 以保持等式平衡。
(x + 3)² − 9 = −2
Add 9 to both sides. | 两边同时加 9。
(x + 3)² = 7
Take square roots and remember to include ±. | 两边开平方根，记得加 ±。
x + 3 = ±√7
Subtract 3 to obtain the final answers. | 移项得最终答案。
x = −3 ± √7
Answers written in this form are perfectly acceptable in IGCSE unless the question asks for a decimal approximation. You may be asked to give answers to 3 significant figures or 2 decimal places.
除非题目要求小数近似值，否则以这种形式书写答案在 IGCSE 中是完全可以的。题目有时会要求保留 3 位有效数字或 2 位小数。
5. The Quadratic Formula | 二次公式
eting the square on ax² + bx + c = 0, but in the exam you only need to apply it correctly.
二次公式可以一次性解出所有二次方程。它由对 ax² + bx + c = 0 配方推导而来，但在考试中你只需正确应用它。
x = (−b ± √b² − 4ac) / (2a)
Worked example | 例： Solve 2x² − 4x − 3 = 0, giving your answer to 2 decimal places.
例题：解方程 2x² − 4x − 3 = 0，答案保留 2 位小数。
Identify a = 2, b = −4, c = −3. | 确定 a = 2、b = −4、c = −3。
Substitute into the formula. | 代入公式。
x = (4 ± √((−4)² − 4 × 2 × (−3))) / (2 × 2)
Simplify inside the square root. | 化简根号内的部分。
x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4
Simplify the surd: √40 = 2√10. | 化简根式：√40 = 2√10。
x = (2 ± √10) / 2
Use a calculator to get decimal answers. | 用计算器得到小数答案。
x = 2.58 or x = −0.58 (to 2 d.p.)
x = 2.58 或 x = −0.58（保留 2 位小数）
The most common mistake is confusing the sign of b. Always write brackets around a negative b, as shown above.
最常见的错误是搞错 b 的符号。代入负数 b 时一定要加括号，如上所示。
6. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant, usually written as Δ. It tells you how many real roots a quadratic equation has without solving it.
根号内的表达式 b² − 4ac 称为判别式，通常记为 Δ。它能在不解方程的情况下告诉我们二次方程有多少个实数根。
| Discriminant Δ | Nature of roots | 根的性质 | Graph meaning | 图像意义 |
| Δ > 0 | Two distinct real roots | 两个不相等的实数根 | Parabola crosses the x-axis twice | 抛物线与 x 轴有两个交点 |
| Δ = 0 | One repeated real root | 一个相等的实数根（二重根） | Parabola touches the x-axis at one point | 抛物线与 x 轴相切于一点 |
| Δ < 0 | No real roots
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