Solving Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程解法全解析

📚 Solving Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程解法全解析

Quadratic equations are one of the most important algebra topics in the IGCSE Mathematics syllabus. They appear in Paper 2 and Paper 4, often linked with graphs, areas and motion problems. This article explains the key methods you need to master: factorising, completing the square, the quadratic formula, and using the discriminant.

二次方程是 IGCSE 数学大纲中最重要的代数专题之一。它们在 Paper 2 和 Paper 4 中经常出现，常与图像、面积和运动问题结合。本文讲解你需要掌握的核心方法：因式分解法、配方法、公式法以及判别式的使用。

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1. What Is a Quadratic Equation? | 什么是二次方程？

nown is 2. Its general form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.

二次方程是一个次数为 2 的多项式方程，也就是说未知数的最高次数是 2。它的一般形式是 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。

ax² + bx + c = 0, a ≠ 0

and 3x² = 6x − 1. The variable can be any letter, but x is most common in IGCSE questions.

如果 a = 0，方程就变成一次方程而不是二次方程。例如 2x² + 3x − 5 = 0、x² − 4 = 0、3x² = 6x − 1 都是二次方程。未知数可以是任意字母，但在 IGCSE 题目中 x 最常用。

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2. Standard Form and Coefficient Identification | 标准形式与系数识别

refully, including their signs. For example, in 2x² − 5x + 3 = 0, a = 2, b = −5 and c = 3.

解题前要先把方程整理成标准形式 ax² + bx + c = 0，然后认真识别 a、b、c，包括符号。例如 2x² − 5x + 3 = 0 中 a = 2、b = −5、c = 3。

coefficients. Write down a, b and c separately before substituting them into any formula.

对 4x − x² = 1，应整理为 −x² + 4x − 1 = 0 或 x² − 4x + 1 = 0，再识别系数。在代入任何公式之前，要单独写下 a、b、c 的值。

Equation	a	b	c
2x² − 5x + 3 = 0	2	−5	3
x² − 4 = 0	1	0	−4
4x − x² = 1 → x² − 4x + 1 = 0	1	−4	1
3x² = 6x − 1 → 3x² − 6x + 1 = 0	3	−6	1

This step is essential for the quadratic formula and the discriminant, because sign errors here lead to wrong answers later.

这一步对公式法和判别式非常关键，因为这里符号出错会导致后面答案错误。

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3. Solving by Factorising | 因式分解法

When a quadratic expression can be written as the product of two linear factors, factorising is often the fastest method. For a simple quadratic x² + bx + c, find two numbers that multiply to c and add to b.

当二次式可以写成两个一次因式的乘积时，因式分解法通常是最快的方法。对于简单的二次式 x² + bx + c，要找两个数，使它们相乘得 c、相加得 b。

= 12 and 3 + 4 = 7. By the zero product rule, x + 3 = 0 or x + 4 = 0, so x = −3 or x = −4.

例如 x² + 7x + 12 = 0 可写成 (x + 3)(x + 4) = 0，因为 3 × 4 = 12 且 3 + 4 = 7。根据零积法则，x + 3 = 0 或 x + 4 = 0，所以 x = −3 或 x = −4。

o watch for common factors and difference of two squares, such as x² − 4 = (x − 2)(x + 2).

对于 a ≠ 1 的二次式，如 2x² + 5x − 3，可以拆分中间项或尝试因式组合。2x² + 5x − 3 = (2x − 1)(x + 3)。一定要展开检验因式分解是否正确。还要注意公因式和平方差，例如 x² − 4 = (x − 2)(x + 2)。

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4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. It is useful when factorising is difficult and when you need the vertex of a quadratic graph. Start by making the coefficient of x² equal to 1.

配方法把二次式改写成 (x + p)² + q 的形式。当因式分解不容易时，或者需要求二次函数图像的顶点时，它非常有用。第一步是让 x² 的系数变为 1。

x² + bx + c = (x + b/2)² − (b/2)² + c

For example, x² + 6x + 1 = 0 becomes (x + 3)² − 9 + 1 = 0, so (x + 3)² = 8. Then x + 3 = ±√8 = ±2√2, giving x = −3 ± 2√2.

例如 x² + 6x + 1 = 0 可写成 (x + 3)² − 9 + 1 = 0，所以 (x + 3)² = 8。于是 x + 3 = ±√8 = ±2√2，得到 x = −3 ± 2√2。

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by 2 to get x² + 4x + 3/2 = 0, then (x + 2)² − 4 + 3/2 = 0, so (x + 2)² = 5/2 and x = −2 ± √5/2.

如果 a ≠ 1，先整式除以 a，或把 a 提取出来。对于 2x² + 8x + 3 = 0，除以 2 得 x² + 4x + 3/2 = 0，再配方得 (x + 2)² − 4 + 3/2 = 0，所以 (x + 2)² = 5/2，x = −2 ± √5/2。

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5. Deriving and Using the Quadratic Formula | 公式法推导与使用

x² + bx + c = 0. It solves any quadratic equation, even when the factors are not integers.

二次公式来自对一般式 ax² + bx + c = 0 进行配方。它可以解任何二次方程，即使因式不是整数也可以使用。

x = (−b ± √b² − 4ac) / (2a)

discriminant part is b² − 4ac = 25 − 24 = 1, so x = (5 ± 1) / 4, giving x = 3/2 or x = 1.

使用时先写下 a、b、c，再仔细代入。对于 2x² − 5x + 3 = 0：a = 2，b = −5，c = 3。判别式部分 b² − 4ac = 25 − 24 = 1，所以 x = (5 ± 1) / 4，得到 x = 3/2 或 x = 1。

Another example is x² + 4x + 1 = 0. Here x = (−4 ± √16 − 4) / 2 = (−4 ± √12) / 2 = −2 ± √3. IGCSE answers may be required in exact surd form or corrected to a given number of decimal places.

再看 x² + 4x + 1 = 0。这里 x = (−4 ± √16 − 4) / 2 = (−4 ± √12) / 2 = −2 ± √3。IGCSE 答案可能要求保留精确根式，也可能要求按题目指定的小数位数四舍五入。

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6. The Discriminant and Number of Roots | 判别式与根的个数

Δ = b² − 4ac. It tells you how many real roots the equation has without solving it fully.

判别式是二次公式中根号下的表达式：Δ = b² − 4ac。它可以在不完全求解的情况下告诉你方程有多少个实根。

• If Δ > 0, there are two distinct real roots.

  如果 Δ > 0，方程有两个不同的实根。

• If Δ = 0, there is one repeated real root.

  如果 Δ = 0，方程有一个重根。

• If Δ < 0, there are no real roots.

  如果 Δ < 0，方程没有实根。

, so one repeated root x = −3. x² + x + 1 = 0 has Δ = 1 − 4 = −3 < 0, so no real roots.

例如 x² + 7x + 12 = 0 的 Δ = 49 − 48 = 1 > 0，所以有两个根。x² + 6x + 9 = 0 的 Δ = 36 − 36 = 0，所以有一个重根 x = −3。x² + x + 1 = 0 的 Δ = 1 − 4 = −3 < 0，所以没有实根。

o find unknown constants. If x² + kx + 9 = 0 has equal roots, then k² − 36 = 0, so k = ±6.

判别式也常用来求未知常数。如果 x² + kx + 9 = 0 有等根，则 k² − 36 = 0，所以 k = ±6。

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7. Solving Quadratic Equations by Graphs | 图像法解二次方程

x-coordinates where the graph crosses the x-axis. These are called the x-intercepts or roots.

y = ax² + bx + c 的图像是一条抛物线。ax² + bx + c = 0 的解就是图像与 x 轴交点的横坐标。这些交点称为 x 轴截距或根。

-axis, there are no real roots. This matches the discriminant: two roots when Δ > 0, one root when Δ = 0, no real roots when Δ < 0.

如果抛物线与 x 轴相交两次，就有两个实根。如果只接触 x 轴一次，就有一个重根。如果不与 x

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