Quadratic Equations: A Complete IGCSE Revision Guide | 二次方程：IGCSE 完整复习指南

📚 Quadratic Equations: A Complete IGCSE Revision Guide | 二次方程：IGCSE 完整复习指南

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. They appear in almost every paper, either as direct solving questions or as essential tools for tackling graphs, word problems, and simultaneous equations. Mastering this topic is a non-negotiable step towards securing a high grade.

二次方程是IGCSE数学中考查频率最高的主题之一。几乎所有试卷都会出现二次方程，无论是直接求解，还是作为解决图像、应用题和联立方程的重要工具。掌握这一主题是冲击高分不可或缺的一步。

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

Every quadratic equation can be written in the standard form ax² + bx + c = 0, where a, b and c are real constants and a ≠ 0. The term ‘quadratic’ comes from the Latin word quadratus, meaning ‘square’, because the highest power of the variable is 2. If a = 0, the equation degenerates into a linear equation, which is a different type of problem.

每个二次方程都可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 是实数常数，且 a ≠ 0。’二次’（quadratic）一词源自拉丁语 quadratus，意为’平方’，因为变量的最高次数为2。如果 a = 0，方程就退化为一次方程，属于完全不同类型的问题。

• The highest power of x is 2 | x 的最高次数为 2
• The graph is always a parabola | 图像始终是一条抛物线
• It has at most two real solutions | 至多有两个实数解
• a is the leading coefficient; c is the constant term | a 是首项系数；c 是常数项

For example, x² – 3x + 2 = 0 is a quadratic equation, but 2x – 5 = 0 is not, because its highest power is 1. Recognising the form correctly is the first step to choosing the right solution method.

例如，x² – 3x + 2 = 0 是一个二次方程，但 2x – 5 = 0 不是，因为它的最高次数是1。正确识别方程的形式是选择正确解法的第一步。

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2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer (or rational) roots. The method relies on the zero product property: if the product of two expressions equals zero, then at least one of the two expressions must be zero.

当二次方程具有简单的整数（或有理数）根时，因式分解是最快的方法。该方法依赖零积性质：若两个表达式的乘积为零，则其中至少有一个表达式为零。

Step-by-step procedure | 分步步骤：

• Step 1: Rearrange the equation so one side equals 0 | 步骤1：将方程整理为一侧等于0
• Step 2: Factorise the quadratic expression | 步骤2：对二次式进行因式分解
• Step 3: Set each factor equal to 0 | 步骤3：令每个因式等于0
• Step 4: Solve the resulting linear equations | 步骤4：解所得的一次方程

Worked example | 示例： Solve x² – 5x + 6 = 0

(x – 2)(x – 3) = 0

x – 2 = 0  or  x – 3 = 0

x = 2  or  x = 3

Check: substituting x = 2 gives 4 – 10 + 6 = 0 ✓; substituting x = 3 gives 9 – 15 + 6 = 0 ✓

检验：代入 x = 2 得 4 – 10 + 6 = 0 ✓；代入 x = 3 得 9 – 15 + 6 = 0 ✓

Also watch for the special case of the difference of two squares: x² – 9 = (x – 3)(x + 3) = 0, giving x = 3 or x = –3. If the coefficient of x² is not 1, such as 2x² + 5x + 2 = 0, look for a factor pair in the form (2x + 1)(x + 2) = 0.

还需要注意平方差特殊情况：x² – 9 = (x – 3)(x + 3) = 0，解得 x = 3 或 x = –3。如果 x² 的系数不是1，例如 2x² + 5x + 2 = 0，则要寻找形如 (2x + 1)(x + 2) = 0 的因式对。

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3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic as a perfect square plus a constant. This method is extremely useful when the quadratic does not factorise neatly, and it also directly reveals the vertex of the corresponding parabola.

配方法将二次式改写为一个完全平方加一个常数的形式。当二次式不易因式分解时，此方法非常有用，同时它还能直接揭示对应抛物线的顶点。

Worked example | 示例： Solve x² + 6x + 5 = 0

Step 1: Halve the coefficient of x: 6 ÷ 2 = 3, so use (x + 3)²

步骤1：将 x 的系数除以2：6 ÷ 2 = 3，因此构造 (x + 3)²

Step 2: Expand (x + 3)² = x² + 6x + 9; compare with x² + 6x + 5, we have 9 – 5 = 4 extra, so subtract 4:

步骤2：展开 (x + 3)² = x² + 6x + 9；与原式 x² + 6x + 5 比较，多出 9 – 5 = 4，因此减去4：

(x + 3)² – 4 = 0

Step 3: Solve | 步骤3：求解

(x + 3)² = 4

x + 3 = ±2

x = –3 ± 2 → x = –1 or x = –5

For the general case x² + bx + c = 0, the identity is:

对于一般情况 x² + bx + c = 0，恒等式为：

(x + b/2)² – (b/2)² + c = 0

When the leading coefficient is not 1, first divide the whole equation by a, then complete the square. This method always works and never requires guessing factor pairs.

当首项系数不为1时，先将整个方程除以 a，然后再配方。这个方法始终有效，且无需猜测因式对。

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4. The Quadratic Formula | 二次公式

The quadratic formula is the most universal tool for solving any quadratic equation. It is derived from completing the square and works for all cases, including those with irrational roots. You should memorise it exactly.

二次公式是解任何二次方程最通用的工具。它由配方法推导而来，适用于所有情况，包括无理数根的情况。你必须准确记住它。

x = (–b ± √(b² – 4ac)) / (2a)

Worked example | 示例： Solve 2x² + 3x – 5 = 0 using the formula.

Here a = 2, b = 3, c = –5. Substitute carefully:

这里 a = 2，b = 3，c = –5。仔细代入：

x = (–3 ± √(3² – 4 × 2 × (–5))) / (2 × 2)

x = (–3 ± √(9 + 40)) / 4

x = (–3 ± √49) / 4 = (–3 ± 7) / 4

x = 1  or  x = –5/2

Notice the ± symbol: it produces two values from one expression. Many students lose marks by forgetting this sign, so always write it explicitly. Also ensure you substitute negative values inside brackets to avoid sign errors — for example, writing –4 × 2 × (–5) rather than –4 × 2 × –5.

注意 ± 符号：它从一个表达式产生两个值。许多学生因忘记这个符号而丢分，所以一定要明确写出。同时在代入负值时务必加括号以避免符号错误，例如写成 –4 × 2 × (–5) 而不是 –4 × 2 × –5。

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5. The Discriminant | 判别式

The expression inside the square root of the quadratic formula, b² – 4ac, is called the discriminant and is denoted by Δ. It tells us the nature of the roots without actually solving the equation.

二次公式根号内的表达式 b² – 4ac 称为判别式，记作 Δ。它能在不解方程的情况下判断根的性质。

Δ = b² – 4ac

Value of Δ | Δ 的值	Nature of Roots | 根的性质	Graph Meaning | 图像含义
Δ > 0	Two distinct real roots | 两个不同实根	Parabola crosses the x-axis twice | 抛物线与x轴有两个交点
Δ = 0	One repeated real root | 一个重实根	Parabola touches the x-axis once | 抛物线与x轴相切于一点
Δ < 0	No real roots | 无实根	Parabola does not intersect the x-axis | 抛物线与x轴无交点

Worked example | 示例： Determine the nature of the roots of x² – 4x + 4 = 0. Here b² – 4ac = 16 – 16 = 0, so there is exactly one repeated root x = 2. Indeed, (x – 2)² = 0.

示例：判断 x² – 4x + 4 = 0 的根的性质。这里 b² – 4ac = 16 – 16 = 0，因此恰好有一个重根 x = 2。事实上，(x – 2)² = 0。

The discriminant is especially useful in problems that ask for a range of values of k, such as ‘find the value of k for which the equation x² + kx + 9 = 0 has two distinct real roots’. Simply set Δ > 0 and solve the resulting inequality.

判别式在求 k 的取值范围的问题中特别有用，例如’求使方程 x² + kx + 9 = 0 有两个不同实根的 k 值’。只需令 Δ > 0 并解所得的不等式即可。

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

The graph of y = ax² + bx + c is always a parabola. Understanding its key features helps you solve problems graphically and connect algebraic results to geometric meaning.

y = ax² + bx + c 的图像始终是一条抛物线。理解其关键特征有助于用图像法解决问题，并将代数结果与几何意义联系起来。

• If a > 0, the parabola opens upwards and has a minimum point | 若 a > 0，抛物线开口向上，有最小值点
• If a < 0, the parabola opens downwards and has a maximum point | 若 a < 0，抛物线开口向下，有最大值点
• The axis of symmetry is x = –b/(2a) | 对称轴为 x = –b/(2a)
• The vertex (turning point) lies on the axis of symmetry | 顶点（转折点）位于对称轴上
• The y-intercept is (0, c) | y 截距为 (0, c)
• The x-intercepts are the real roots of ax² + bx + c = 0 | x 截距是方程 ax² + bx + c = 0 的实根

Worked example | 示例： For y = x² – 4x + 3, the axis is x = 4/2 = 2. Substitute x = 2 to get y = 4 – 8 + 3 = –1, so the vertex is (2, –1). The roots are x = 1 and x = 3 (since x² – 4x + 3 = (x – 1)(x – 3)). The y-intercept is 3.

示例：对于 y = x² – 4x + 3，对称轴为 x = 4/2 = 2。代入 x = 2 得 y = 4 – 8 + 3 = –1，因此顶点为 (2, –1)。根为 x = 1 和 x = 3（因为 x² – 4x + 3 = (x – 1)(x – 3)）。y 截距为3。

Completing the square reveals the vertex directly: y = (x – 2)² – 1 shows the minimum at (2, –1). This is a key skill for sketching graphs quickly in the exam.

配方法可以直接揭示顶点：y = (x – 2)² – 1 显示最小值在 (2, –1)。这是在考试中快速作草图的关键技能。

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7. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0 with roots α and β, the following relationships hold. These are derived directly from expanding a(x – α)(x – β) = 0.

对于根为 α 和 β 的二次方程 ax² + bx + c = 0，以下关系成立。这些关系可以直接通过展开 a(x – α)(x – β) = 0 推导得出。

α + β = –b/a    and    αβ = c/a

α + β = –b/a    且    αβ = c/a

Worked example | 示例： Find a quadratic equation whose roots are 2 and –3.

示例：求一个根为 2 和 –3 的二次方程。

Sum of roots = 2 + (–3) = –1; product = 2 × (–3) = –6. Therefore the equation is:

根的和 = 2 + (–3) = –1；根的积 = 2 × (–3) = –6。因此方程为：

x² – (sum)x + (product) = 0 → x² + x – 6 = 0

x² – (和)x + (积) = 0 → x² + x – 6 = 0

This is extremely useful in reverse problems where you are given the roots and asked to reconstruct the equation, or when dealing with symmetric functions of the roots such as α² + β².

这在逆向问题中非常有用，比如已知根要求还原方程，或处理根的对称函数如 α² + β² 时。

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8. Quadratic Word Problems | 二次方程应用题

Many real-world problems naturally lead to quadratic equations. The key skills are translating the text into algebra, forming the correct equation, and rejecting any solution that does not make sense in the

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