Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程完全掌握

📚 Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程完全掌握

A quadratic equation is one of the most important topics in the IGCSE Mathematics syllabus. It appears in algebra, graphs, problem-solving and even in topics such as geometry and sequences. Mastering quadratics means you can confidently factorise, use the quadratic formula, interpret graphs and apply these skills in exam questions.

二次方程是 IGCSE 数学大纲中最重要的主题之一。它出现在代数、图像、实际问题解决中，甚至出现在几何和数列等主题里。掌握二次方程意味着你能够自信地进行因式分解、使用求根公式、理解函数图像，并在考试题中灵活运用这些技能。

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

Its standard form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The word 'quadratic' comes from the Latin word 'quadratus', meaning square, because the highest power of the variable is 2.

二次方程是指次数为 2 的多项式方程。其标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。“二次”一词来源于拉丁语“quadratus”，意为“平方”，因为变量的最高次数为 2。

2 = 0 and x² − 4 = 0. Notice that every example can be written with the highest power x².

二次方程的例子包括 x² − 5x + 6 = 0、2x² + 3x − 2 = 0 和 x² − 4 = 0。注意每一个例子都可以写成含有最高次项 x² 的形式。

ratic but linear. This condition is frequently tested, so always check that the coefficient of x² is not zero before using quadratic techniques.

如果 a = 0，方程就不再是二次方程，而是一次方程。这个条件经常在考试中出现，因此在使用二次方程方法前，一定要检查 x² 的系数是否不为零。

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2. Standard Form and Key Terms | 标准形式与关键术语

. The term ax² is the quadratic term, bx is the linear term, and c is the constant term. The coefficients a, b and c may be positive, negative or zero, except a cannot be zero.

在求解之前，应始终把方程整理成标准形式 ax² + bx + c = 0。ax² 是二次项，bx 是一次项，c 是常数项。系数 a、b、c 可以是正数、负数或零，但 a 不能为零。

The solutions of a quadratic equation are also called roots, zeros or x-intercepts. These names describe the same values from different viewpoints: algebraically they are roots, graphically they are x-intercepts.

二次方程的解也称为根、零点或 x 轴截距。这些名称从不同角度描述同样的值：从代数角度看它们是根，从图像角度看它们是 x 轴截距。

ts. Being able to identify these coefficients quickly is essential for using the quadratic formula.

例如，在 x² − 5x + 6 = 0 中，a = 1、b = −5、c = 6 就是各项系数。快速识别这些系数是使用求根公式的基础。

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3. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic expression factorises easily. The steps are: write the equation in standard form, factorise the left-hand side, set each factor equal to zero, then solve each simple linear equation.

当二次式容易分解时，因式分解法通常是最快的方法。步骤是：把方程写成标准形式；将左边分解因式；令每个因式等于零；然后解每个简单的一次方程。

6 and add to −5. These numbers are −2 and −3, so the factorisation is (x − 2)(x − 3) = 0.

例如，要解 x² − 5x + 6 = 0，需要找到两个数，它们相乘得 6，相加得 −5。这两个数是 −2 和 −3，所以因式分解为 (x − 2)(x − 3) = 0。

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

x² − 5x + 6 = 0 ⇒ (x − 2)(x − 3) = 0 ⇒ x = 2 或 x = 3

iginal equation. Here, 2² − 5(2) + 6 = 0 and 3² − 5(3) + 6 = 0, so both roots are correct.

一定要把解代回原方程进行检验。这里 2² − 5(2) + 6 = 0，3² − 5(3) + 6 = 0，所以两个根都是正确的。

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4. Solving by Completing the Square | 配方法

Completing the square is a powerful method because it works for every quadratic equation and leads directly to the vertex form of a quadratic function. The idea is to rewrite x² + bx as part of a perfect square.

配方法是一种非常强大的方法，因为它适用于所有二次方程，并且可以直接转化为二次函数的顶点式。其核心思想是把 x² + bx 改写为一个完全平方的一部分。

f of 6 is 3, and 3² = 9, so add 9 to both sides: x² + 6x + 9 = 4. This gives (x + 3)² = 4.

例如，要解 x² + 6x + 5 = 0，首先把常数项移项：x² + 6x = −5。6 的一半是 3，3² = 9，所以两边同时加 9：x² + 6x + 9 = 4。于是得到 (x + 3)² = 4。

(x + 3)² = 4 ⇒ x + 3 = ±2 ⇒ x = −1 or x = −5

(x + 3)² = 4 ⇒ x + 3 = ±2 ⇒ x = −1 或 x = −5

first. For example, 2x² + 8x + 6 = 0 becomes x² + 4x + 3 = 0 before completing the square.

当 x² 的系数不是 1 时，应先把整个方程除以 a。例如 2x² + 8x + 6 = 0 先化为 x² + 4x + 3 = 0，再进行配方。

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5. The Quadratic Formula | 求根公式

ny quadratic equation. For ax² + bx + c = 0, the solutions are given by the formula below.

求根公式提供了一种直接求解任意二次方程的方法。对于 ax² + bx + c = 0，其解由以下公式给出。

x = (−b ± √b² − 4ac) ÷ 2a

x = (−b ± √b² − 4ac) ÷ 2a

4 and c = −6. Substituting into the formula gives x = (4 ± √((−4)² − 4(2)(−6))) ÷ 4 = (4 ± √16 + 48) ÷ 4 = (4 ± 8) ÷ 4.

例如，解 2x² − 4x − 6 = 0。这里 a = 2，b = −4，c = −6。代入公式得 x = (4 ± √((−4)² − 4(2)(−6))) ÷ 4 = (4 ± √16 + 48) ÷ 4 = (4 ± 8) ÷ 4。

x = (4 + 8) ÷ 4 = 3 or x = (4 − 8) ÷ 4 = −1

x = (4 + 8) ÷ 4 = 3 或 x = (4 − 8) ÷ 4 = −1

This method is especially useful when factorisation is difficult or when the coefficients are decimals or fractions. Remember to write the equation in standard form before identifying a, b and c.

当因式分解困难或系数是小数、分数时，此方法特别有用。请记住，在确定 a、b、c 之前，要先把方程写成标准形式。

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6. The Discriminant | 判别式

s given by Δ = b² − 4ac and tells us the nature of the roots without solving the equation.

判别式是求根公式中根号内的表达式。它由 Δ = b² − 4ac 给出，可以在不求解方程的情况下判断根的性质。

Discriminant	Number of real roots	Graph behaviour
Δ > 0	2 distinct real roots	Parabola cuts x-axis twice
Δ = 0	1 repeated real root	Parabola touches x-axis once
Δ < 0	No real roots	Parabola does not cut x-axis

判别式	实根个数	图像表现
Δ > 0	2 个不同实根	抛物线与 x 轴相交两次
Δ = 0	1 个重复实根	抛物线与 x 轴相切一次
Δ < 0	无实根	抛物线不与 x 轴相交

, so it has no real roots. Its graph lies entirely above the x-axis because a is positive.

例如，x² + 4x + 5 = 0 的判别式 Δ = 4² − 4(1)(5) = 16 − 20 = −4，因此它没有实数根。由于 a 为正，其图像完全位于 x 轴上方。

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7. Graphs of Quadratic Functions | 二次函数图像

x² + bx + c and its graph is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.

二次函数的形式为 y = ax² + bx + c，其图像是一条抛物线。如果 a > 0，抛物线开口向上，有最低点。如果 a < 0，抛物线开口向下，有最高点。

ning point. Its equation is x = −b ÷ (2a). This formula helps you find the vertex quickly.

对称轴是一条经过顶点的竖直直线。它的方程是 x = −b ÷ (2a)。这个公式可以帮助你快速找到顶点。

Axis of symmetry: x = −b ÷ (2a)

对称轴：x = −b ÷ (2a)

e when x = 0, y = c. The x-intercepts are the real roots of the equation ax² + bx + c = 0.

y 轴截距是常数项 c，因为当 x = 0 时，y = c。x 轴截距是方程 ax² + bx + c = 0 的实数根。

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8. Roots, Intercepts and Turning Points | 根、截距与顶点

4x + 3 = 0. Factorising gives (x − 1)(x − 3) = 0, so the x-intercepts are x = 1 and x = 3.

对于 y = x² − 4x + 3，其根可通过解 x² − 4x + 3 = 0 得到。因式分解得 (x − 1)(x − 3) = 0，所以 x 轴截距为 x = 1 和 x = 3。

-axis at (0, 3). The axis of symmetry is x = −(−4) ÷ (2×1) = 2, halfway between the roots.

y 轴截距为 c = 3，所以图像在 (0, 3) 处穿过 y 轴。对称轴为 x = −(−4) ÷ (2×1) = 2，正好位于两个根的中点。

= −1, so the turning point is (2, −1). This point is a minimum because a = 1 is positive.

将 x = 2 代入方程得 y = 2² − 4(2) + 3 = −1，所以顶点为 (2, −1)。由于 a = 1 为正，因此该点为最低点。

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9. Common Mistakes and Exam Tips | 常见错误与考试技巧

x² − 3x = 4 by simply writing x(x − 3) = 4; you must first rewrite it as x² − 3x − 4 = 0.

一个常见错误是在因式分解前忘记把方程化为零。你不能把 x² − 3x = 4 直接写成 x(x − 3) = 4；必须先把方程改写为 x² − 3x − 4 = 0。

t x = 0. For example, x² − 5x = 0 should be solved as x(x − 5) = 0, giving x = 0 or x = 5.

另一个错误是两边同时除以 x，从而丢失根 x = 0。例如，x² − 5x = 0 应先写成 x(x − 5) = 0，得到 x = 0 或 x = 5。

use brackets: if b = −4, then −b = 4. Simplify the square root completely where possible.

在使用求根公式时，要特别注意负号，尤其是当 b 为负数时。始终使用括号：如果 b = −4，那么 −b = 4。在可能的情况下，要把根号化简到最简形式。

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10. Worked Example | 例题精讲

2x² + 3x − 2 = 0 using two different methods, and explain which method is more efficient.

用两种不同方法解 2x² + 3x − 2 = 0，并说明哪一种方法更高效。

4x − x − 2 = 0, then factor by grouping: 2x(x + 2) − 1(x + 2) = 0, so (2x − 1)(x + 2) = 0.

方法一：因式分解法。寻找两个数，使它们相乘得 2 × (−2) = −4，相加得 3。这两个数是 4 和 −1。将中间项拆开：2x² + 4x − x − 2 = 0，然后分组分解：2x(x + 2) − 1(x + 2) = 0，所以 (2x − 1)(x + 2) = 0。

Setting each factor to zero gives 2x − 1 = 0 or x + 2 = 0, so x = 1/2 or x = −2.

令每个因式等于零，得 2x − 1 = 0 或 x + 2 = 0，所以 x = 1/2 或 x = −2。

a. Here a = 2, b = 3 and c = −2. The formula gives x = (−3 ± √(3² − 4(2)(−2))) ÷ 4 = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4, so x = 1/2 or x = −2.

方法二：求根公式。这里 a = 2，b = 3，c = −2。公式给出 x = (−3 ± √(3² − 4(2)(−2))) ÷ 4 = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4，所以 x = 1/2 或 x = −2。

Both methods give the same roots, but factorisation is faster here because the expression factorises neatly. When the coefficients are large or messy, the quadratic formula is more reliable.

两种方法得到相同的根，但这里因式分解更快，因为该二次式能很整齐地分解。当系数较大或比较杂乱时，求根公式更加可靠。

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