Mastering Quadratic Equations: Factorising, Solving and Graphing | 掌握二次方程：因式分解、求解与作图

📚 Mastering Quadratic Equations: Factorising, Solving and Graphing | 掌握二次方程：因式分解、求解与作图

Quadratic equations appear in almost every IGCSE Mathematics paper. Whether you are taking the Extended or Core syllabus, you will be asked to factorise, solve, sketch and apply quadratic expressions in real-world problems. This revision guide covers every essential skill you need, with worked examples and examiner tips that will help you turn a weak topic into a reliable source of marks.

二次方程几乎出现在每一份 IGCSE 数学试卷中。无论你学习的是 Extended（扩展）还是 Core（核心）大纲，你都需要对二次表达式进行因式分解、求解、作图，并将其应用于现实问题。本复习指南涵盖了你所需的全部核心技能，配有带解答的例题和考官建议，帮助你把这个薄弱主题转化为稳定的得分来源。

1. The Standard Form ax² + bx + c = 0 | 二次方程的标准形式 ax² + bx + c = 0

Every quadratic equation can be written in the standard form ax² + bx + c = 0, where a, b and c are real numbers and a is not zero. The coefficient a multiplies x², b multiplies x, and c is the constant term. If a = 0, the equation becomes linear, so the condition a ≠ 0 is what makes an equation truly quadratic.

每一个二次方程都可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 是实数，且 a 不等于零。系数 a 乘以 x²，b 乘以 x，c 是常数项。如果 a = 0，方程就会退化为一次方程，因此 a ≠ 0 这一条件正是方程被称为”二次”的根本原因。

For example, in 3x² − 4x + 5 = 0, we have a = 3, b = −4 and c = 5. Before solving any quadratic, you must rearrange the equation so that every term is on one side and the other side equals zero.

例如，在 3x² − 4x + 5 = 0 中，a = 3，b = −4，c = 5。在求解任何二次方程之前，你必须先整理方程，把所有项移到一边，使另一边等于 0。

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2. Expanding Two Brackets: The Reverse Process | 展开两个括号：因式分解的逆运算

Factorising is the reverse process of expanding. To expand (x + 2)(x + 3), multiply each term in the first bracket by each term in the second. The result is x² + 3x + 2x + 6 = x² + 5x + 6. Notice that 2 + 3 = 5 (the coefficient of x) and 2 × 3 = 6 (the constant term).

因式分解是展开的逆运算。要展开 (x + 2)(x + 3)，需要用第一个括号中的每一项分别乘第二个括号中的每一项，结果为 x² + 3x + 2x + 6 = x² + 5x + 6。注意 2 + 3 = 5（即 x 的系数），而 2 × 3 = 6（即常数项）。

To expand two brackets quickly, use the FOIL method: First, Outer, Inner, Last. For (x + 4)(x − 2): First gives x², Outer gives −2x, Inner gives 4x, Last gives −8. Combining the middle terms gives x² + 2x − 8.

要快速展开两个括号，可以使用 FOIL 法：First（首项）、Outer（外项）、Inner（内项）、Last（末项）。对于 (x + 4)(x − 2)：First 得到 x²，Outer 得到 −2x，Inner 得到 4x，Last 得到 −8。合并中间两项后得到 x² + 2x − 8。

The table below summarises the signs inside the brackets for the factorisation of x² + bx + c.

下表总结了 x² + bx + c 因式分解时括号内符号的判断规律。

Condition 条件	Example 例子	Factorised form 分解形式
c > 0, b > 0	x² + 5x + 6	(x + 2)(x + 3)
c > 0, b < 0	x² − 5x + 6	(x − 2)(x − 3)
c < 0	x² − x − 6	(x − 3)(x + 2)

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3. Factorising x² + bx + c When a = 1 | 因式分解 a = 1 的二次式

When the coefficient of x² is 1, look for a factor pair of c whose sum is b. Follow these four steps: first, write down all factor pairs of c; second, find the pair that adds to b; third, write the answer in the form (x ± p)(x ± q); fourth, expand your answer to check it works.

当 x² 的系数为 1 时，你需要找到 c 的一对因数，使它们的和等于 b。按以下四步操作：第一步，写出 c 的所有因数对；第二步，找出相加等于 b 的那一对；第三步，把答案写成 (x ± p)(x ± q) 的形式；第四步，展开答案进行检验。

Worked example: factorise x² − 7x + 12. The factor pairs of 12 are (1, 12), (2, 6) and (3, 4). Only 3 and 4 add to give 7. Since b = −7, both brackets must contain a minus sign.

典型例题：分解 x² − 7x + 12。12 的因数对有 (1, 12)、(2, 6) 和 (3, 4)。只有 3 和 4 相加等于 7。因为 b = −7，所以两个括号中都必须带负号。

x² − 7x + 12 = (x − 3)(x − 4)

Expanding (x − 3)(x − 4) gives x² − 4x − 3x + 12 = x² − 7x + 12, which confirms the factorisation is correct. Always remember: expanding is your built-in checking method.

展开 (x − 3)(x − 4) 得到 x² − 4x − 3x + 12 = x² − 7x + 12，从而确认因式分解正确。永远记住：展开就是你自带的检验工具。

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4. Factorising ax² + bx + c When a ≠ 1 | 当 a ≠ 1 时的因式分解

Harder quadratics have a coefficient of x² greater than 1, such as 2x² + 7x + 3. The grouping method works well here. First multiply a and c: 2 × 3 = 6. Then find a factor pair of 6 that adds to b = 7: that pair is 1 and 6. Rewrite the middle term using these numbers, then factorise in pairs.

较难的二次式中，x² 的系数大于 1，例如 2x² + 7x + 3。此时使用分组法非常有效。先用 a 乘 c：2 × 3 = 6。然后找到 6 的一对因数，使它们的和等于 b = 7：这一对数是 1 和 6。用这两个数重写中间项，然后两两分组提取公因式。

2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)

A special case appears when the expression is a difference of two squares. Any expression of the form a² − b² equals (a − b)(a + b). For example, 9x² − 16 = (3x)² − 4² = (3x − 4)(3x + 4). The middle term disappears because the inner and outer terms cancel exactly.

有一种特殊情况需要注意：平方差公式。任何形如 a² − b² 的表达式都等于 (a − b)(a + b)。例如，9x² − 16 = (3x)² − 4² = (3x − 4)(3x + 4)。由于内项与外项完全抵消，所以展开后没有中间项。

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5. Solving Quadratic Equations by Factorising | 用因式分解法求解二次方程

The fundamental principle is the zero product property: if A × B = 0, then A = 0 or B = 0. Once a quadratic is factorised into two brackets, you set each bracket equal to zero and solve the two resulting linear equations.

这里的基本原理是零积性质：如果 A × B = 0，那么 A = 0 或 B = 0。当二次式被分解为两个括号后，你只需令每个括号分别等于 0，然后解两个一次方程即可。

Worked example: solve x² − 5x − 14 = 0. The numbers −7 and 2 multiply to −14 and add to −5, so the factorisation is (x − 7)(x + 2) = 0. Therefore x − 7 = 0 or x + 2 = 0.

典型例题：解方程 x² − 5x − 14 = 0。−7 和 2 相乘得 −14，相加得 −5，所以分解为 (x − 7)(x + 2) = 0。因此 x − 7 = 0 或 x + 2 = 0。

x = 7 或 x = −2

Write your final answer as ‘x = 7 or x = −2’. Never write ‘x = 7 and x = −2’, because a quadratic equation may have two different solutions, not one value that is both numbers at once.Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

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