Quadratic Equations and Functions | 二次方程与函数

📚 Quadratic Equations and Functions | 二次方程与函数

A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in almost every paper, from simple factorisation to complex problem-solving. This guide covers everything you need to know: expanding, factorising, solving, graphing, and applying quadratics in real-world contexts.

二次方程是IGCSE数学中最重要的考点之一。从简单的因式分解到复杂的综合应用题，它几乎出现在每一份试卷中。本指南将涵盖你所需的一切：展开、因式分解、求解、作图以及在实际情境中的应用。

1. Understanding Quadratic Expressions | 理解二次表达式

A quadratic expression is any expression of the form ax² + bx + c, where a, b and c are constants and a ≠ 0. The highest power of x is 2, which is why it is called ‘quadratic’ — from the Latin ‘quadratus’, meaning ‘square’.

二次表达式是形如 ax² + bx + c 的表达式，其中 a、b 和 c 是常数，且 a ≠ 0。x 的最高次数为 2，这就是它被称为”二次”的原因——源自拉丁语 ‘quadratus’，意为”平方”。

Common examples include x² + 5x + 6, 2x² − 3x + 1, and −x² + 4x − 7. If a = 0, the expression becomes linear, not quadratic. The coefficient a is called the leading coefficient, and it determines the shape and direction of the graph.

常见例子包括 x² + 5x + 6、2x² − 3x + 1 和 −x² + 4x − 7。如果 a = 0，表达式就变成线性而非二次。系数 a 称为首项系数，它决定图像的形状和开口方向。

2. Expanding and Factorising | 展开与因式分解

Expanding means removing brackets by multiplying terms. The most common pattern is the double-bracket expansion: (x + p)(x + q) = x² + (p + q)x + pq.

展开是指通过乘法运算去掉括号。最常见的模式是双括号展开：(x + p)(x + q) = x² + (p + q)x + pq。

(x + 3)(x + 5) = x² + 8x + 15

Here we multiply x by x, x by 5, 3 by x, and 3 by 5 — the famous FOIL method (First, Outer, Inner, Last).

这里我们依次将 x 乘 x、x 乘 5、3 乘 x、3 乘 5——即著名的 FOIL 方法（首项、外项、内项、末项）。

Factorising is the reverse process. To factorise x² + 8x + 15, find two numbers that multiply to 15 and add to 8. The numbers 3 and 5 satisfy both conditions, so x² + 8x + 15 = (x + 3)(x + 5).

因式分解是相反的过程。要对 x² + 8x + 15 因式分解，需要找到两个数相乘得 15、相加得 8。数字 3 和 5 同时满足这两个条件，因此 x² + 8x + 15 = (x + 3)(x + 5)。

When the coefficient of x² is not 1, we use the ‘ac method’. For 2x² + 7x + 3, multiply a by c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: they are 1 and 6. Split the middle term: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).

当 x² 的系数不为 1 时，我们使用”ac 方法”。对于 2x² + 7x + 3，先计算 a × c：2 × 3 = 6。找两个数相乘得 6、相加得 7：它们是 1 和 6。然后拆分中间项：2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。

3. Solving by Factorisation | 用因式分解求解

To solve a quadratic equation, first rearrange it so one side equals zero. Then factorise and apply the zero-product property: if ab = 0, then a = 0 or b = 0.

要解二次方程，首先将其整理为一侧等于零的形式。然后因式分解并利用零积性质：如果 ab = 0，那么 a = 0 或 b = 0。

x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3

Always check your solutions by substituting them back into the original equation. This verifies that no

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