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

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📚 Quadratic Equations: 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. This comprehensive revision guide covers every method you need to solve quadratic equations, understand their graphs, and apply them to real-world problems. Whether you are aiming for a Grade 9 or striving to pass, mastering quadratics is essential for success.

二次方程是 IGCSE 数学中最高频考查的主题之一。本完整复习指南涵盖了解二次方程所需的所有方法、理解其图像以及将其应用于实际问题。无论你是冲刺 9 分还是力求及格，掌握二次方程都是成功的关键。

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

A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The highest power of the variable is 2, which is why it is called “quadratic” (from the Latin word quadratus, meaning square).

二次方程是指任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数是 2，因此被称为”二次”（源自拉丁语 quadratus，意为”平方”）。

Key characteristics you must recognise:

  • The coefficient a must never be zero — if a = 0, the equation becomes linear (bx + c = 0).

  • A quadratic equation can have at most two solutions (roots), which may be real and distinct, real and equal, or not real (complex).

  • Every quadratic equation corresponds to a parabola when graphed.

  • 系数 a 绝不能为零——如果 a = 0，方程就变成一次方程（bx + c = 0）。

  • 二次方程最多有两个解（根），它们可能是不相等的实数、相等的实数，或非实数（复数）。

  • 每个二次方程在图像上都对应一条抛物线。

2. Solving by Factorisation | 因式分解法求解

Factorisation is the fastest method when the quadratic has simple integer factors. The principle is based on the zero product property: if p × q = 0, then either p = 0 or q = 0 (or both).

因式分解法是当二次方程具有简单整数因子时最快的解法。其原理基于零积性质：如果 p × q = 0，那么 p = 0 或 q = 0（或两者都为零）。

Worked Example | 例题： Solve x² + 5x + 6 = 0

x² + 5x + 6 = (x + 2)(x + 3) = 0

Therefore: x + 2 = 0 → x = −2, or x + 3 = 0 → x = −3. The solutions are x = −2 and x = −3.

因此：x + 2 = 0 → x = −2，或 x + 3 = 0 → x = −3。解为 x = −2 和 x = −3。

When the coefficient of x² is not 1, the process is slightly more involved:

当 x² 的系数不为 1 时，过程稍微复杂一些：

Worked Example | 例题： Solve 2x² + 7x + 3 = 0

Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7 — those are 6 and 1. Rewrite the middle term: 2x² + 6x + x + 3. Factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. Hence x = −½ or x = −3.

将 a 和 c 相乘：2 × 3 = 6。找到两个乘积为 6 且和为 7 的数——即 6 和 1。重写中间项：2x² + 6x + x + 3。分组因式分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。

Common mistakes to avoid | 常见错误提醒：

  • Forgetting to rearrange the equation to equal zero before factorising.

  • Dropping the negative sign when moving terms across the equals sign.

  • 因式分解前忘记将方程整理为等于零的形式。

  • 移项时漏掉负号。

3. Solving by Completing the Square | 配方法求解

Completing the square rewrites a quadratic in the form (x + p)² + q. This method works for every quadratic equation and is especially useful when factorisation is not possible. It also directly reveals the vertex (turning point) of the corresponding parabola.

配方法将二次方程改写为 (x + p)² + q 的形式。这种方法适用于所有二次方程，在无法因式分解时尤其有用。它还能直接揭示对应抛物线的顶点（转折点）。

Worked Example | 例题： Solve x² − 6x + 5 = 0 by completing the square.

Step 1: Move the constant term to the right: x² − 6x = −5.
Step 2: Take half of the coefficient of x (−6 ÷ 2 = −3), square it ((−3)² = 9), and add to both sides:
x² − 6x + 9 = −5 + 9 → (x − 3)² = 4.
Step 3: Square root both sides: x − 3 = ±2.
Step 4: Solve: x = 3 + 2 = 5 or x = 3 − 2 = 1.

步骤 1：将常数项移到右边：x² − 6x = −5。
步骤 2：取 x 系数的一半（−6 ÷ 2 = −3），平方（(−3)² = 9），加到两边：
x² − 6x + 9 = −5 + 9 → (x − 3)² = 4。
步骤 3：两边开平方根：x − 3 = ±2。
步骤 4：求解：x = 3 + 2 = 5 或 x = 3 − 2 = 1。

When a ≠ 1, first factor out a from the x² and x terms, then complete the square inside the brackets.

当 a ≠ 1 时，先从 x² 和 x 项中提出 a，然后在括号内配方。

Vertex form | 顶点形式： The expression (x + p)² + q tells us the vertex is at (−p, q). If a > 0, this vertex is a minimum; if a < 0, it is a maximum.

顶点形式： 表达式 (x + p)² + q 告诉我们顶点在 (−p, q)。如果 a > 0，该顶点是最小值点；如果 a < 0，则是最大值点。

4. The Quadratic Formula | 二次公式

The quadratic formula is the most universal method — it can solve any quadratic equation, even when factorisation fails. For ax² + bx + c = 0, the solutions are given by:

二次公式是最通用的方法——它可以解决任何二次方程，即使因式分解失效也能适用。对于 ax² + bx + c = 0，解由下式给出：

x = (−b ± √(b² − 4ac)) ÷ (2a)

Worked Example | 例题： Solve 2x² − 3x − 2 = 0 using the quadratic formula.

Here a = 2, b = −3, c = −2.
Substitute into the formula:

这里 a = 2，b = −3，c = −2。
代入公式：

x = (−(−3) ± √((−3)² − 4 × 2 × (−2))) ÷ (2 × 2)

x = (3 ± √(9 + 16)) ÷ 4 = (3 ± √25) ÷ 4 = (3 ± 5) ÷ 4

So x = (3 + 5) ÷ 4 = 2, or x = (3 − 5) ÷ 4 = −½. The solutions are x = 2 and x = −½.

因此 x = (3 + 5) ÷ 4 = 2，或 x = (3 − 5) ÷ 4 = −½。解为 x = 2 和 x = −½。

Memory tip | 记忆技巧： “Negative b, plus or minus the square root, of b squared minus four a c, all over two a” — sing it to the tune of “Pop Goes the Weasel” to remember it easily.

记忆技巧： “负 b，加减根号，b 方减 4ac，除以 2a”——可以配上”Pop Goes the Weasel”的旋律来轻松记忆。

5. The Discriminant: What b² − 4ac Tells Us | 判别式：b² − 4ac 告诉我们什么

The expression inside the square root, b² − 4ac, is called the discriminant. It determines the nature and number of roots without actually solving the equation.

根号内的表达式 b² − 4ac 称为判别式。它无需真正解方程即可确定根的性质和数量。

Discriminant Value | 判别式值 Nature of Roots | 根的性质 Graph Meaning | 图像意义
b² − 4ac > 0 Two distinct real roots | 两个不相等的实数根 Parabola crosses x-axis twice | 抛物线与 x 轴有两个交点
b² − 4ac = 0 One repeated real root | 一个重根（两个相等的实数根） Parabola touches x-axis once (tangent) | 抛物线与 x 轴相切（一个交点）
b² − 4ac < 0 No real roots | 无实数根 Parabola does not intersect x-axis | 抛物线与 x 轴无交点

Worked Example | 例题： Determine the nature of the roots of 3x² + 2x + 5 = 0.

Here a = 3, b = 2, c = 5.
b² − 4ac = 2² − 4 × 3 × 5 = 4 − 60 = −56 < 0.

Since the discriminant is negative, the equation has no real roots — the parabola lies entirely above the x-axis (since a > 0) without touching it.

这里 a = 3，b = 2，c = 5。
b² − 4ac = 2² − 4 × 3 × 5 = 4 − 60 = −56 < 0。

由于判别式为负，方程无实数根——抛物线完全位于 x 轴上方（因为 a > 0），与 x 轴无交点。

6. Sum and Product of Roots | 根的和与积

For any quadratic equation ax² + bx + c = 0 with roots α (alpha) and β (beta), the following relationships hold — these are derived from the factorised form and are powerful shortcuts in exams:

对于任何根为 α（阿尔法）和 β（贝塔）的二次方程 ax² + bx + c = 0，以下关系成立——这些关系由因式分解形式推导而来，是考试中强大的捷径：

Sum of roots: α + β = −b ÷ a

Product of roots: α × β = c ÷ a

Worked Example | 例题： The equation 2x² + 5x − 3 = 0 has roots α and β. Find α + β and α × β.

Using the formulas: α + β = −5 ÷ 2 = −2.5, and α × β = −3 ÷ 2 = −1.5.

These relationships also allow us to construct a quadratic equation from given roots. If the roots are α and β, the equation is:

利用公式：α + β = −5 ÷ 2 = −2.5，且 α × β = −3 ÷ 2 = −1.5。

这些关系还允许我们从已知根构造二次方程。如果根是 α 和 β，则方程为：

x² − (α + β)x + (α × β) = 0

For example, to find a quadratic with roots 4 and −3: x² − (4 + (−3))x + (4 × (−3)) = x² − x − 12 = 0.

例如，要找到根为 4 和 −3 的二次方程：x² − (4 + (−3))x + (4 × (−3)) = x² − x − 12 = 0。

7. 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 is crucial for both sketching and interpreting questions:

y = ax² + bx + c 的图像总是一条抛物线。理解其关键特征是绘制草图和理解题目的关键：

  • Shape (开口方向): If a > 0, the parabola opens upwards (U-shaped); if a < 0, it opens downwards (n-shaped).

  • Axis of symmetry (对称轴): The vertical line x = −b ÷ (2a) divides the parabola into two mirror images.

  • Vertex (顶点): The turning point lies on the axis of symmetry. Its y-coordinate is found by substituting x = −b ÷ (2a) into the equation.

  • y-intercept (y 截距): The point (0, c) where the curve crosses the y-axis.

  • x-intercepts (x 截距): The roots of the equation, found by solving ax² + bx + c = 0.

  • 开口方向： 如果 a > 0，抛物线开口向上（U 形）；如果 a < 0，开口向下（∩ 形）。

  • 对称轴： 垂直线 x = −b ÷ (2a) 将抛物线分成两个镜像部分。

  • 顶点： 转折点位于对称轴上。其 y 坐标通过将 x = −b ÷ (2a) 代入方程求得。

  • y 截距： 曲线与 y 轴的交点 (0, c)。

  • x 截距： 即方程的根，通过解 ax² + bx + c = 0 得到。

Sketching guide | 草图绘制步骤：

Step 1: Determine the shape from the sign of a.
Step 2: Mark the y-intercept (0, c).
Step 3: Solve the quadratic to find x-intercepts (if real roots exist).
Step 4: Find the vertex using x = −b ÷ (2a).
Step 5: Draw a smooth curve through these points.

步骤 1：根据 a 的符号确定开口方向。
步骤 2：标出 y 截距 (0, c)。
步骤 3：解二次方程找到 x 截距（如果存在实数根）。
步骤 4：用 x = −b ÷ (2a) 求出顶点。
步骤 5：过这些点绘制平滑曲线。

8. Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities, such as x² − 5x + 6 < 0 or 2x² + 3x − 2 ≥ 0, are solved using the roots and the graph's shape. The key rule: between the roots, the sign is opposite to a; outside the roots, the sign matches a (when a > 0).

二次不等式，如 x² − 5x + 6 < 0 或 2x² + 3x − 2 ≥ 0，通过根和图像形状来解。关键规则：在两根之间，符号与 a 相反；在两根之外，符号与 a 相同（当 a > 0 时）。

Worked Example | 例题： Solve x² − 5x + 6 < 0.

First factorise: (x − 2)(x − 3) < 0. The roots are x = 2 and x = 3.

Since a = 1 > 0, the parabola opens upward. The expression is negative between the roots. Therefore the solution is 2 < x < 3.

先因式分解：(x − 2)(x − 3) < 0。根为 x = 2 和 x = 3。

因为 a = 1 > 0，抛物线开口向上。在两根之间表达式为负。因此解为 2 < x < 3。

For “greater than” inequalities (e.g., x² − 5x + 6 > 0), the solution would be the union of the two outer regions: x < 2 or x > 3.

对于”大于”型不等式（如 x² − 5x + 6 > 0），解为两个外侧区域的并集：x < 2 或 x > 3。

Critical warning | 重要警示： If you multiply or divide an inequality by a negative number, you must reverse the inequality sign!

重要警示： 如果将不等式乘以或除以负数，必须反转不等号！

9. Solving Simultaneous Equations with a Quadratic | 含二次方程的联立方程组

IGCSE often pairs a linear equation with a quadratic equation. The standard method is substitution: solve the linear equation for one variable, then substitute into the quadratic equation.

IGCSE 经常将线性方程与二次方程配对考查。标准方法是代入法：从线性方程解出一个变量，然后代入二次方程。

Worked Example | 例题： Solve the simultaneous equations:
y = x + 1 and y = x² − 3x + 5.

Step 1: Since both equal y, set them equal: x + 1 = x² − 3x + 5.
Step 2: Rearrange: x² − 4x + 4 = 0.
Step 3: Factorise: (x − 2)² = 0, so x = 2 (repeated root).
Step 4: Substitute back: y = 2 + 1 = 3.

The solution is x = 2, y = 3. Geometrically, this means the line is tangent to the parabola (touches it at exactly one point).

步骤 1：因为两者都等于 y，令其相等：x + 1 = x² − 3x + 5。
步骤 2：整理：x² − 4x + 4 = 0。
步骤 3：因式分解：(x − 2)² = 0，所以 x = 2（重根）。
步骤 4：代回：y = 2 + 1 = 3。

解为 x = 2，y = 3。几何上，这意味着直线与抛物线相切（恰好接触于一点）。

If the resulting quadratic has two roots, the line cuts the parabola at two points. If it has no real roots, the line never meets the parabola.

如果所得二次方程有两个根，则直线与抛物线相交于两点。如果无实数根，则直线与抛物线永不相交。

10. Word Problems: Applying Quadratics | 应用题：二次方程的实际应用

Examiners love testing quadratics in real-world contexts. The most common applications include area problems, projectile motion, and number puzzles. The key is to translate the words into an equation carefully.

考官喜欢在实际情境中考查二次方程。最常见的应用包括面积问题、抛体运动和数字谜题。关键是要仔细将文字转化为方程。

Worked Example | 例题（Area | 面积）： A rectangle has a length that is 3 cm longer than its width. Its area is 40 cm². Find its dimensions.

Let the width be x cm. Then the length is (x + 3) cm.
Area = x(x + 3) = 40 → x² + 3x − 40 = 0.
Factorise: (x + 8)(x − 5) = 0 → x = −8 or x = 5.

Since length cannot be negative, discard x = −8. The width is 5 cm and the length is 8 cm.

设宽为 x 厘米，则长为 (x + 3) 厘米。
面积 = x(x + 3) = 40 → x² + 3x − 40 = 0。
因式分解：(x + 8)(x − 5) = 0 → x = −8 或 x = 5。

由于长度不能为负，舍去 x = −8。宽为 5 厘米，长为 8 厘米。

Worked Example | 例题（Projectile | 抛体）： A ball is thrown upward with height h metres given by h = 20t − 5t², where t is time in seconds. When does the ball hit the ground?

The ball hits the ground when h = 0: 20t − 5t² = 0 → 5t(4 − t) = 0 → t = 0 or t = 4.

The ball is on the ground at t = 0 (thrown) and again at t = 4 seconds. Therefore the ball hits the ground after 4 seconds.

球落地时 h = 0：20t − 5t² = 0 → 5t(4 − t) = 0 → t = 0 或 t = 4。

球在 t = 0（抛出时）和 t = 4 秒时在地面上。因此球在4 秒后落地。

11. Typical Exam Pitfalls and How to Avoid Them | 典型考试陷阱及应对策略

Even top students lose marks on quadratics due to avoidable errors. Here are the most frequent pitfalls and the strategies to overcome them:

即使是优秀学生也会因为在二次方程上犯下可避免的错误而丢分。以下是最常见的陷阱及应对策略：

Pitfall | 陷阱 Solution | 解决方案
Dividing both sides by x (losing a root) | 两边同除以 x（丢失一个根） Never divide by a variable; always factorise instead | 绝不用变量去除，始终使用因式分解
Forgetting to set equation to zero | 忘记将方程整理为零 Always rearrange to ax² + bx + c = 0 first | 始终先整理为 ax² + bx + c = 0
Sign errors in the quadratic formula | 二次公式中符号错误 Write out a, b, c explicitly before substituting | 代入前明确写出 a、b、c 的值
Forgetting negative roots in word problems | 在应用题中忘记舍去负根 Check whether the context allows negative values | 检查题目情境是否允许负值
Misdrawing the parabola shape | 抛物线开口方向画错 Always check sign of coefficient a before sketching | 画图前务必检查系数 a 的符号

Exam strategy | 考试策略： When a quadratic appears in a non-calculator paper, first try factorisation. If factorisation is not evident within 30 seconds, switch to completing the square or the quadratic formula. On a calculator paper, the quadratic formula is always reliable — but ensure you input negative numbers correctly using brackets.

考试策略： 当二次方程出现在非计算器试卷中时，先尝试因式分解。如果在 30 秒内看不出因式分解，切换到配方法或二次公式。在允许使用计算器的试卷中，二次公式始终可靠——但务必使用括号正确输入负数。

12. Rapid-Fire Practice Questions | 快速练习题目

Test yourself with these questions. Attempt each one before checking the answer.

用这些问题自我测试。先尝试作答再看答案。

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