Category: cie-igcse-mathematics,cie-igcse-mathematics-cn

What Are Quadratic Equations? | 什么是二次方程？

二次方程是形式为 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a ≠ 0。这个名称来源于拉丁语 “quadratus”，意为”平方”，因为最高次项是 x²。在 IGCSE 数学课程中，二次方程是代数部分的核心内容，出现在多个考试局（包括 CIE、Edexcel 和 AQA）的课程大纲中，在 Paper 2 和 Paper 4 中都有考查。

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The name comes from the Latin word “quadratus,” meaning “square,” because the highest power of the variable is x². In the IGCSE Mathematics curriculum, quadratic equations form a core part of the algebra syllabus and appear across multiple exam boards (including CIE, Edexcel, and AQA) in both Paper 2 and Paper 4 assessments.

二次方程在现实世界中有广泛应用，从计算抛物运动轨迹到优化商业利润，从建筑设计到工程设计。理解二次方程不仅是应对考试的需要，更是培养数学思维和问题解决能力的关键一步。

Quadratic equations have extensive real-world applications, from calculating projectile trajectories to optimizing business profits, from architectural design to engineering problems. Understanding quadratic equations is not just about passing exams; it is a key step in developing mathematical thinking and problem-solving skills.

The Standard Form and Key Terminology | 标准形式和关键术语

二次方程的标准形式是 ax² + bx + c = 0。其中：a 是二次项系数（coefficient of x²），b 是一次项系数（coefficient of x），c 是常数项（constant term）。需要注意的是，二次方程的”根”（roots）或”解”（solutions）是使方程成立的 x 值，也就是二次函数图像与 x 轴交点的横坐标。

The standard form of a quadratic equation is ax² + bx + c = 0. Here: a is the coefficient of x² (the quadratic term), b is the coefficient of x (the linear term), and c is the constant term. It is important to note that the “roots” or “solutions” of a quadratic equation are the x-values that make the equation true, which correspond to the x-coordinates where the graph of the quadratic function intersects the x-axis.

在 IGCSE 考试中，你可能会看到二次方程以不同形式出现，例如：x² + 5x + 6 = 0（标准形式）、2x² = 8x + 10（需要整理）、或 (x + 2)(x – 3) = 0（因式分解形式）。理解如何在不同形式之间转换是解题的关键技能。

In IGCSE examinations, you may encounter quadratic equations in different forms, such as: x² + 5x + 6 = 0 (standard form), 2x² = 8x + 10 (needs rearranging), or (x + 2)(x – 3) = 0 (factorised form). Understanding how to convert between different forms is a critical skill for solving problems.

Method 1: Factorisation | 方法一：因式分解法

因式分解法是解决二次方程最基本且最优雅的方法之一。它的核心思想是将二次表达式分解为两个一次因式的乘积，然后利用”零乘积性质”（如果 ab = 0，则 a = 0 或 b = 0）来求解。这种方法特别适用于系数为整数且方程易于分解的情况。

Factorisation is one of the most fundamental and elegant methods for solving quadratic equations. The core idea is to break down the quadratic expression into the product of two linear factors, then use the “zero product property” (if ab = 0, then a = 0 or b = 0) to find the solutions. This method is particularly suitable when the coefficients are integers and the equation factorises easily.

让我们通过一个具体的 IGCSE 考试风格例题来说明：求解方程 x² + 7x + 12 = 0。第一步，我们需要找到两个数，它们的乘积等于常数项 12，它们的和等于一次项系数 7。经过尝试，我们发现 3 和 4 满足条件：3 × 4 = 12，3 + 4 = 7。因此，方程可以因式分解为 (x + 3)(x + 4) = 0。应用零乘积性质：x + 3 = 0 或 x + 4 = 0，从而得到 x = -3 或 x = -4。

Let us illustrate with a concrete IGCSE exam-style example: solve the equation x² + 7x + 12 = 0. Step one, we need to find two numbers whose product equals the constant term 12 and whose sum equals the linear coefficient 7. After trial, we find that 3 and 4 satisfy the conditions: 3 × 4 = 12, 3 + 4 = 7. Therefore, the equation factorises to (x + 3)(x + 4) = 0. Applying the zero product property: x + 3 = 0 or x + 4 = 0, giving x = -3 or x = -4.

当二次项系数 a ≠ 1 时的因式分解稍微复杂一些。例如，求解 2x² + 5x – 3 = 0。我们需要找到两个数，它们的乘积等于 a × c = 2 × (-3) = -6，它们的和等于 b = 5。这两个数是 6 和 -1。接下来，将中间项 5x 拆分为 6x – 1x：2x² + 6x – 1x – 3 = 0。然后分组分解：2x(x + 3) – 1(x + 3) = 0，提取公因式 (x + 3)：(2x – 1)(x + 3) = 0。因此 x = 1/2 或 x = -3。

Factorisation when the quadratic coefficient a ≠ 1 is slightly more involved. For example, solve 2x² + 5x – 3 = 0. We need to find two numbers whose product equals a × c = 2 × (-3) = -6 and whose sum equals b = 5. These numbers are 6 and -1. Next, split the middle term 5x into 6x – 1x: 2x² + 6x – 1x – 3 = 0. Then group and factor: 2x(x + 3) – 1(x + 3) = 0, extract the common factor (x + 3): (2x – 1)(x + 3) = 0. Hence x = 1/2 or x = -3.

因式分解法的优势在于快速直观，但局限性也很明显：并非所有二次方程都能方便地进行因式分解。当方程的根是无理数或分数时，因式分解法就变得困难甚至不可行。此时，我们就需要使用其他方法。

The advantage of factorisation is that it is quick and intuitive, but its limitation is also clear: not all quadratic equations can be factorised conveniently. When the roots are irrational or fractional, factorisation becomes difficult or even impossible. In such cases, we need to use other methods.

Method 2: The Quadratic Formula | 方法二：求根公式法

求根公式是解决任何二次方程 ax² + bx + c = 0 的通用方法。公式为：x = [-b ± √(b² – 4ac)] / 2a。这个公式是 IGCSE 数学课程中最重要的公式之一，学生需要在考试中熟练使用它。CIE IGCSE 试卷的公式表上通常印有这个公式，但学生仍然需要能够在不同情境下正确应用。

The quadratic formula is a universal method for solving any quadratic equation ax² + bx + c = 0. The formula is: x = [-b ± √(b² – 4ac)] / 2a. This is one of the most important formulas in the IGCSE Mathematics curriculum, and students need to use it proficiently in examinations. While the formula is usually printed on the formula sheet for CIE IGCSE papers, students still need to be able to apply it correctly in different contexts.

让我们通过例题来演示求根公式的使用。求解方程 3x² + 8x – 3 = 0。识别系数：a = 3, b = 8, c = -3。代入公式：x = [-8 ± √(8² – 4×3×(-3))] / (2×3) = [-8 ± √(64 + 36)] / 6 = [-8 ± √100] / 6 = [-8 ± 10] / 6。因此两个解为：x = (-8 + 10)/6 = 2/6 = 1/3，或 x = (-8 – 10)/6 = -18/6 = -3。

Let us demonstrate the use of the quadratic formula with an example. Solve the equation 3x² + 8x – 3 = 0. Identify the coefficients: a = 3, b = 8, c = -3. Substitute into the formula: x = [-8 ± √(8² – 4×3×(-3))] / (2×3) = [-8 ± √(64 + 36)] / 6 = [-8 ± √100] / 6 = [-8 ± 10] / 6. Therefore the two solutions are: x = (-8 + 10)/6 = 2/6 = 1/3, or x = (-8 – 10)/6 = -18/6 = -3.

使用求根公式时的常见错误包括：忘记分母是 2a 而写成 a、b² – 4ac 的计算中符号处理错误（尤其是当 c 为负数时）、以及未能正确简化最终答案。在 IGCSE 考试中，即使最终答案正确，计算过程中的符号错误也可能导致扣分。

Common mistakes when using the quadratic formula include: forgetting that the denominator is 2a rather than a, sign errors in the calculation of b² – 4ac (especially when c is negative), and failing to simplify the final answer correctly. In IGCSE examinations, sign errors in the working process can lead to mark deductions even if the final answer is correct.

Method 3: Completing the Square | 方法三：配方法

配方法是推导求根公式的基础，也是理解二次函数图像变换的关键方法。配方法的基本思想是将二次表达式 ax² + bx + c 转化为 a(x – h)² + k 的形式，其中 (h, k) 是二次函数图像的顶点坐标。这种形式不仅可以直接求解方程，还能帮助我们分析函数的最大值或最小值。

Completing the square is the foundation on which the quadratic formula is derived, and it is also a key method for understanding transformations of quadratic function graphs. The basic idea of completing the square is to transform the quadratic expression ax² + bx + c into the form a(x – h)² + k, where (h, k) are the coordinates of the vertex of the quadratic function graph. This form not only allows us to solve the equation directly, but also helps us analyse the maximum or minimum value of the function.

让我们以 x² + 6x + 2 = 0 为例来演示配方法。首先将常数项移到右边：x² + 6x = -2。然后，在两边加上 (b/2)² = (6/2)² = 9，使左边成为完全平方式：x² + 6x + 9 = -2 + 9，即 (x + 3)² = 7。接下来，两边开平方：x + 3 = ±√7。因此 x = -3 ± √7。这两个解为 -3 + √7 和 -3 – √7。在 IGCSE 考试中，通常要求答案保留根号形式或精确到三位有效数字。

Let us demonstrate completing the square with x² + 6x + 2 = 0. First, move the constant term to the right: x² + 6x = -2. Next, add (b/2)² = (6/2)² = 9 to both sides, making the left side a perfect square: x² + 6x + 9 = -2 + 9, giving (x + 3)² = 7. Then, take the square root of both sides: x + 3 = ±√7. Therefore x = -3 ± √7. The two solutions are -3 + √7 and -3 – √7. In IGCSE examinations, answers are typically required to be left in surd form or given correct to three significant figures.

配方法在需要找到函数顶点或确定函数值域时特别有用。例如，将 y = x² + 4x + 1 配方为 y = (x + 2)² – 3，我们可以直接看出顶点坐标为 (-2, -3)，最小值为 -3。这种能力在 IGCSE 的 Paper 4 中经常出现在涉及优化问题的长答题中。

Completing the square is particularly useful when you need to find the vertex of a function or determine the range of function values. For example, rewriting y = x² + 4x + 1 as y = (x + 2)² – 3 allows us to directly see that the vertex is (-2, -3) and the minimum value is -3. This ability frequently appears in IGCSE Paper 4 in long-answer questions involving optimisation problems.

The Discriminant and the Nature of Roots | 判别式和根的性质

判别式 Δ = b² – 4ac 是二次方程中一个极其强大的分析工具。它告诉我们方程根的数量和类型，而不需要我们实际求解方程。判别式在 IGCSE 考试大纲中是一个重要考点，尤其在 CIE 0607 和 Edexcel 4MA1 的试卷中经常出现。

The discriminant Δ = b² – 4ac is an extremely powerful analytical tool in quadratic equations. It tells us the number and type of roots without us needing to actually solve the equation. The discriminant is an important examination point in the IGCSE syllabus, appearing frequently in CIE 0607 and Edexcel 4MA1 papers.

判别式的三种情况：当 Δ > 0 时，方程有两个不同的实根（two distinct real roots）；当 Δ = 0 时，方程有两个相等的实根（two equal real roots），也称为一个重根（repeated root），此时二次函数图像与 x 轴相切；当 Δ < 0 时，方程没有实根（no real roots），二次函数图像完全位于 x 轴上方或下方。

The three cases of the discriminant: when Δ > 0, the equation has two distinct real roots; when Δ = 0, the equation has two equal real roots, also called a repeated root, at which point the quadratic graph touches the x-axis; when Δ < 0, the equation has no real roots, and the quadratic graph lies entirely above or below the x-axis.

IGCSE 考试中关于判别式的典型问题类型包括：确定方程根的性质、找出使方程具有特定根类型的参数取值范围、以及证明某个方程对于所有实数值始终具有实根（或始终没有实根）。例如：找出 k 的取值范围，使得方程 x² + kx + 9 = 0 有两个不同实根。解：Δ = k² – 4(1)(9) = k² – 36 > 0，因此 k² > 36，即 k < -6 或 k > 6。

Typical question types about the discriminant in IGCSE examinations include: determining the nature of the roots of an equation, finding the range of a parameter that makes the equation have a particular type of roots, and proving that a given equation always has real roots (or never has real roots) for all real values. Example: find the range of k such that the equation x² + kx + 9 = 0 has two distinct real roots. Solution: Δ = k² – 4(1)(9) = k² – 36 > 0, so k² > 36, giving k < -6 or k > 6.

Graphical Interpretation | 图形解释

二次函数 y = ax² + bx + c 的图像是一条抛物线（parabola）。理解二次方程的几何意义对于掌握整个主题至关重要。方程 ax² + bx + c = 0 的解正是抛物线 y = ax² + bx + c 与 x 轴（即直线 y = 0）交点的横坐标。

The graph of the quadratic function y = ax² + bx + c is a parabola. Understanding the geometric meaning of quadratic equations is essential for mastering the entire topic. The solutions of the equation ax² + bx + c = 0 are precisely the x-coordinates of the points where the parabola y = ax² + bx + c intersects the x-axis (that is, the line y = 0).

抛物线的开口方向由系数 a 决定：当 a > 0 时，抛物线向上开口（U 形），函数有最小值；当 a < 0 时，抛物线向下开口（倒 U 形），函数有最大值。对称轴（axis of symmetry）是直线 x = -b/(2a)，顶点（vertex）位于对称轴与抛物线的交点处。这些知识在 IGCSE 的图形绘制和函数分析题中至关重要。

The direction of opening of the parabola is determined by the coefficient a: when a > 0, the parabola opens upward (U-shape), and the function has a minimum value; when a < 0, the parabola opens downward (inverted U-shape), and the function has a maximum value. The axis of symmetry is the line x = -b/(2a), and the vertex is located at the intersection of the axis of symmetry and the parabola. This knowledge is crucial in IGCSE graph-sketching and function-analysis questions.

让我们通过一个具体例子来说明图形在解题中的作用。考虑方程 x² – 4x + 3 = 0。通过因式分解得到 (x – 1)(x – 3) = 0，解为 x = 1 和 x = 3。对应的函数图像 y = x² – 4x + 3 是一条开口向上的抛物线，与 x 轴交于 (1, 0) 和 (3, 0)，顶点位于 (2, -1)。图形直观地展示了这两个解是图像与 x 轴的交点。

Let us illustrate the role of graphs in problem-solving with a concrete example. Consider the equation x² – 4x + 3 = 0. By factorisation we obtain (x – 1)(x – 3) = 0, with solutions x = 1 and x = 3. The corresponding function graph y = x² – 4x + 3 is an upward-opening parabola, intersecting the x-axis at (1, 0) and (3, 0), with its vertex at (2, -1). The graph visually demonstrates that these two solutions are the points where the graph crosses the x-axis.

Common IGCSE Exam Question Types | IGCSE 常见考题类型

IGCSE 数学考试中关于二次方程的题目类型多种多样，具有清晰的模式。第一类是直接求解题：给出一个二次方程，要求使用指定方法（或因式分解，或求根公式，或配方法）求解。CIE 试卷经常明确要求使用特定方法，因此掌握所有三种方法非常重要。

IGCSE Mathematics examination questions on quadratic equations come in diverse but clear patterns. The first type is direct solving: given a quadratic equation, solve it using a specified method (factorisation, the quadratic formula, or completing the square). CIE papers frequently require the use of a specific method, so mastering all three methods is important.

第二类题目是将二次方程嵌入实际情境（word problem / real-world context）。常见情境包括：矩形的边长与面积问题（例如”一个矩形的长比宽大 3 厘米，面积为 40 平方厘米，求矩形的尺寸”）、抛射物运动的高度方程（例如”一个球从地面以初始速度 20 米/秒向上抛出，其高度 h = 20t – 5t²，问球何时回到地面”），以及产品利润的优化问题。

The second type embeds quadratic equations in real-world contexts (word problems). Common scenarios include: rectangle side-length and area problems (e.g., “The length of a rectangle is 3 cm more than its width and its area is 40 cm²; find the dimensions of the rectangle”), projectile motion height equations (e.g., “A ball is thrown upward from the ground with an initial velocity of 20 m/s; its height is h = 20t – 5t²; when does the ball return to the ground”), and product profit optimisation problems.

第三类是高阶思维题，通常出现在 Paper 4 的末尾部分。这类题目可能涉及：含参数的二次方程（找出使方程具有特定根的性质的参数范围）、二次方程与其他主题的交叉（如三角恒等式替换产生二次方程）、或二次不等式（利用二次函数图像的符号分布求解 ax² + bx + c > 0 型不等式）。

The third type involves higher-order thinking, typically appearing at the end of Paper 4. These may involve: quadratic equations with parameters (finding the parameter range that gives roots of a specific nature), intersection of quadratic equations with other topics (such as trigonometric identity substitution yielding quadratic equations), or quadratic inequalities (using the sign distribution of the quadratic graph to solve inequalities of the form ax² + bx + c > 0).

Exam Technique and Common Pitfalls | 考试技巧和常见陷阱

要在 IGCSE 数学考试中成功解决二次方程问题，除了掌握解题方法外，还需要养成良好的考试习惯和避免常见错误。以下是一些经过实践检验的考试策略。

To succeed in solving quadratic equation problems in IGCSE Mathematics examinations, beyond mastering the solution methods, you also need to develop good examination habits and avoid common mistakes. Here are some battle-tested examination strategies.

第一，在开始解题之前，始终将方程整理为标准形式 ax² + bx + c = 0。许多学生因为方程未整理就匆忙求解而犯下错误。第二，在代入求根公式之前，明确写出 a、b、c 的值 – 这三秒的额外工作可以帮助你在考试压力下避免符号错误。第三，始终检验你的答案，将求得的解代回原方程验证。如果方程左边不等于零，说明解有误。

First, always rearrange the equation into standard form ax² + bx + c = 0 before beginning to solve. Many students make mistakes by hurrying to solve without rearranging the equation first. Second, before substituting into the quadratic formula, explicitly write out the values of a, b, and c – this extra three seconds of work can help you avoid sign errors under examination pressure. Third, always check your answers by substituting the solutions back into the original equation. If the left-hand side does not equal zero, the solution is wrong.

第四，注意答案的精确度要求。IGCSE 题目通常会明确说明答案应该保留几位小数或几位有效数字，或者以分数或根号形式给出精确答案。不遵守精确度要求是常见的扣分原因。第五，在涉及判别式的问题中，不等式符号的方向（> 或 ≥、< 或 ≤）非常重要 - 仔细阅读题目中"两个不同实根"和"至少有一个实根"之间的区别。

Fourth, pay attention to accuracy requirements. IGCSE questions usually specify clearly whether answers should be given to a certain number of decimal places or significant figures, or as exact answers in fraction or surd form. Failure to follow accuracy requirements is a common cause of mark loss. Fifth, in discriminant problems, the direction of the inequality sign (> or ≥, < or ≤) matters greatly - read carefully to distinguish between "two distinct real roots" and "at least one real root."

Connections to Other IGCSE Topics | 与其他 IGCSE 知识点的联系

二次方程不是孤立的知识点。在 IGCSE 数学中，它与许多其他主题紧密相连。理解这些联系不仅能加深对二次方程的理解，还能在综合性题目中更灵活地应用。例如，二次函数与一次直线的交点问题转化为解二次方程；三角恒等式中 sin²x 的替换可以产生关于 sin x 的二次方程；代数分式方程的化简最终可能归结为二次方程的求解。

Quadratic equations are not an isolated topic. In IGCSE Mathematics, they connect closely with many other topics. Understanding these connections not only deepens your grasp of quadratic equations but also allows more flexible application in integrated questions. For example, finding the intersection of a quadratic function and a straight line reduces to solving a quadratic equation; substituting sin²x in trigonometric identities can yield a quadratic equation in sin x; simplifying algebraic fractional equations may ultimately reduce to solving a quadratic equation.

在微积分预备（pre-calculus）和进阶数学中，二次方程和二次函数也是更复杂概念的基础。在 IGCSE 附加数学（Additional Mathematics 0606）中，学生将学习用判别式法求曲线切线条件、使用配方法求二次函数的值域、以及解决涉及二次的联立方程组。打好 IGCSE 0580 或 0607 的二次方程基础，将为后续学习铺平道路。

In pre-calculus and further mathematics, quadratic equations and functions also serve as the foundation for more complex concepts. In IGCSE Additional Mathematics (0606), students will learn to use the discriminant to find tangent conditions for curves, apply completing the square to determine the range of quadratic functions, and solve simultaneous equations involving quadratics. Building a solid foundation in quadratic equations in IGCSE 0580 or 0607 paves the way for subsequent study.

二次方程还与统计学中的最小二乘回归有一定联系 – 回归线的参数估计过程涉及最小化一个二次函数形式的误差平方和。虽然 IGCSE 阶段不要求理解这种深层联系，但知道不同数学分支之间存在内在统一性是培养数学素养的重要一环。

Quadratic equations also connect to least squares regression in statistics – the parameter estimation process for a regression line involves minimising a sum of squared errors that takes the form of a quadratic function. While IGCSE level does not require understanding this deep connection, knowing that different branches of mathematics share an underlying unity is an important part of developing mathematical literacy.

Practice Problems | 练习题

以下是一组精选的 IGCSE 风格练习题，涵盖不同的解法和难度级别。建议独立完成所有题目，然后对照答案，找出自己的薄弱环节。

Below is a curated set of IGCSE-style practice problems covering different methods and difficulty levels. It is recommended that you complete all problems independently, then check against the answers to identify your weak areas.

基础题（Basic Level）：

1. 求解 x² + 8x + 15 = 0（因式分解法）
2. 求解 x² – 5x – 6 = 0（因式分解法）
3. 求解 2x² + 3x – 2 = 0（求根公式法）

Basic Level:

1. Solve x² + 8x + 15 = 0 (by factorisation)
2. Solve x² – 5x – 6 = 0 (by factorisation)
3. Solve 2x² + 3x – 2 = 0 (by quadratic formula)

中等题（Intermediate Level）：

4. 将 x² + 10x + 3 配方为 (x + p)² + q 的形式
5. 确定方程 x² + 2x + 5 = 0 的根的性质
6. 求解 3x² – 7x – 6 = 0，答案保留三位有效数字

Intermediate Level:

4. Express x² + 10x + 3 in the form (x + p)² + q
5. Determine the nature of the roots of x² + 2x + 5 = 0
6. Solve 3x² – 7x – 6 = 0, giving answers correct to three significant figures

难题（Advanced Level）：

7. 找出 k 的取值范围，使得方程 x² + (k + 2)x + (k + 1) = 0 有两个不同实根
8. 一个矩形的对角线长度为 13 厘米，长为 6 厘米，求宽（保留两位小数）

Advanced Level:

7. Find the range of k such that x² + (k + 2)x + (k + 1) = 0 has two distinct real roots
8. The diagonal of a rectangle is 13 cm and its length is 6 cm; find the width (correct to two decimal places)

Summary | 总结

二次方程是 IGCSE 数学代数部分的核心内容，掌握三种主要解法（因式分解法、求根公式法、配方法）是取得好成绩的关键。判别式 Δ = b² – 4ac 提供了一种无需实际求解就能判断方程根的性质的强大工具。图形理解为代数解法提供了直观的几何验证。通过系统的练习和对常见错误的警觉，学生可以在这个重要主题上建立扎实的能力，为后续的数学学习打下坚实基础。

Quadratic equations form a core component of the IGCSE Mathematics algebra syllabus, and mastering the three main solution methods (factorisation, the quadratic formula, and completing the square) is key to achieving good results. The discriminant Δ = b² – 4ac provides a powerful tool for determining the nature of a quadratic equation’s roots without actually solving the equation. Graphical understanding offers intuitive geometric verification for algebraic solutions. Through systematic practice and vigilance against common mistakes, students can build solid competence in this important topic, laying a strong foundation for subsequent mathematical study.

Introduction to Quadratic Equations — 二次方程简介

Quadratic equations are one of the most important topics in IGCSE Mathematics (CIE 0580/0980). 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. Understanding quadratic equations is essential not only for the IGCSE exam but also for A-Level Mathematics, physics, engineering, and many real-world applications ranging from projectile motion to economic modeling.

二次方程是IGCSE数学（CIE 0580/0980）中最重要的主题之一。二次方程是任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数，且 a ≠ 0。理解二次方程不仅对IGCSE考试至关重要，对A-Level数学、物理、工程以及从抛射运动到经济建模的许多实际应用也同样关键。

The Standard Form and Key Terminology — 标准形式与关键术语

Every quadratic equation follows the standard form ax² + bx + c = 0. Here, ‘a’ is the coefficient of the quadratic term (x²), ‘b’ is the coefficient of the linear term (x), and ‘c’ is the constant term. The value of ‘a’ determines whether the parabola opens upward (a > 0) or downward (a < 0). The quadratic term x² is what distinguishes quadratic equations from linear equations, and it is this squared term that gives the graph its characteristic U-shaped curve called a parabola.

每个二次方程都遵循标准形式 ax² + bx + c = 0。其中，’a’ 是二次项（x²）的系数，’b’ 是一次项（x）的系数，’c’ 是常数项。’a’ 的值决定了抛物线是向上开口（a > 0）还是向下开口（a < 0）。二次项 x² 是二次方程与一次方程的区别所在，正是这个平方项使图像呈现出特有的U形曲线，称为抛物线。

Key vocabulary you must master for IGCSE includes: roots (also called solutions or x-intercepts), vertex (the turning point of the parabola), axis of symmetry, discriminant, and completing the square. Knowing these terms precisely will help you read exam questions correctly and structure your answers effectively.

IGCSE必须掌握的关键词汇包括：根（也称为解或x轴截距）、顶点（抛物线的转折点）、对称轴、判别式、以及配方法。准确理解这些术语将帮助你正确阅读考题并有效组织答案。

Solving Quadratic Equations by Factorisation — 因式分解法解二次方程

Factorisation is the simplest method for solving quadratic equations and is often the first approach you should try in an IGCSE exam. The idea is to rewrite ax² + bx + c = 0 as (px + q)(rx + s) = 0, then use the zero product property: if the product of two expressions equals zero, then at least one of them must be zero. This gives two linear equations to solve.

因式分解法是解二次方程最简单的方法，通常是IGCSE考试中你应该首先尝试的方法。其思路是将 ax² + bx + c = 0 改写为 (px + q)(rx + s) = 0，然后利用零乘积性质：如果两个表达式的乘积为零，那么其中至少有一个必须为零。这样就得到了两个一元一次方程来求解。

For example, to solve x² + 5x + 6 = 0, we find two numbers that multiply to 6 and add to 5. These numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3) = 0. Therefore x + 2 = 0 or x + 3 = 0, giving solutions x = -2 and x = -3. Always check your answers by substituting them back into the original equation. For x = -2: (-2)² + 5(-2) + 6 = 4 – 10 + 6 = 0. For x = -3: (-3)² + 5(-3) + 6 = 9 – 15 + 6 = 0. Both solutions are correct.

例如，解方程 x² + 5x + 6 = 0，我们找出两个数，它们相乘得6，相加得5。这两个数是2和3，所以 x² + 5x + 6 = (x + 2)(x + 3) = 0。因此 x + 2 = 0 或 x + 3 = 0，得到解 x = -2 和 x = -3。务必通过将答案代回原方程来检验。对于 x = -2：(-2)² + 5(-2) + 6 = 4 – 10 + 6 = 0。对于 x = -3：(-3)² + 5(-3) + 6 = 9 – 15 + 6 = 0。两个解均正确。

Common pitfalls include forgetting to set the equation equal to zero before factorising, misidentifying the signs of factors, and overlooking the case where a ≠ 1. When the leading coefficient is not 1 (e.g., 2x² + 7x + 3 = 0), you need to consider factor pairs of a × c = 6 that add to b = 7, giving 6 and 1, then split the middle term: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. Solutions: x = -1/2 and x = -3.

常见错误包括：因式分解前忘记将方程设为零、误判因式的符号、以及忽略 a ≠ 1 的情况。当首项系数不为1时（例如 2x² + 7x + 3 = 0），你需要考虑 a × c = 6 的因数对中相加等于 b = 7 的组合，即6和1，然后拆分中间项：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。解为：x = -1/2 和 x = -3。

Completing the Square — 配方法

Completing the square is a powerful technique that works for any quadratic equation, including those that cannot be factorised easily. The method transforms ax² + bx + c into the form a(x + p)² + q, from which the solution can be obtained directly. The key step is to add and subtract (b/2a)² to create a perfect square trinomial.

配方法是一种强大的技巧，适用于任何二次方程，包括那些不容易因式分解的方程。该方法将 ax² + bx + c 转化为 a(x + p)² + q 的形式，从中可以直接求解。关键步骤是加减 (b/2a)² 来创建完全平方三项式。

Let us work through a complete example: solve x² + 6x + 2 = 0 by completing the square. First, move the constant: x² + 6x = -2. Next, take half the coefficient of x (half of 6 is 3) and square it (3² = 9). Add this to both sides: x² + 6x + 9 = -2 + 9, which gives (x + 3)² = 7. Taking the square root of both sides: x + 3 = ±√7. Therefore x = -3 ± √7. The two solutions are x = -3 + √7 and x = -3 – √7. These are exact (surd) answers, which IGCSE examiners prefer to decimal approximations unless the question specifies otherwise.

让我们通过一个完整例子来演示：用配方法解 x² + 6x + 2 = 0。首先，移项常数：x² + 6x = -2。然后，取x系数的一半（6的一半是3）并平方（3² = 9）。将这个值加到等式两边：x² + 6x + 9 = -2 + 9，得到 (x + 3)² = 7。对两边开平方根：x + 3 = ±√7。因此 x = -3 ± √7。两个解为 x = -3 + √7 和 x = -3 – √7。这些是精确（根式）答案，除非题目另有说明，IGCSE考官更倾向于精确答案而非小数近似。

For equations where a ≠ 1, first divide through by a to make the leading coefficient 1 before applying the completing-the-square procedure. For instance, to solve 2x² + 8x + 3 = 0, divide by 2: x² + 4x + 3/2 = 0, then proceed as above.

对于 a ≠ 1 的方程，先整体除以 a 使首项系数为1，再应用配方法。例如，解 2x² + 8x + 3 = 0，先除以2：x² + 4x + 3/2 = 0，然后按照上述步骤进行。

The Quadratic Formula — 二次公式

The quadratic formula is the most general method for solving quadratic equations, derived directly from completing the square on the standard form ax² + bx + c = 0. The formula states that x = [-b ± √(b² – 4ac)] / 2a. This formula works for every quadratic equation without exception, making it the safest choice when you are unsure about factorisation or when coefficients are large or involve fractions.

二次公式是解二次方程最通用的方法，直接由标准形式 ax² + bx + c = 0 配方法推导而来。公式为 x = [-b ± √(b² – 4ac)] / 2a。该公式适用于所有二次方程，没有例外，因此当你对因式分解不确定或系数较大或涉及分数时，它是最安全的选择。

Let us apply the formula to solve 3x² – 5x – 2 = 0. Identify a = 3, b = -5, c = -2. Substitute into the formula: x = [5 ± √((-5)² – 4(3)(-2))] / 2(3) = [5 ± √(25 + 24)] / 6 = [5 ± √49] / 6 = [5 ± 7] / 6. So x = (5 + 7)/6 = 12/6 = 2, and x = (5 – 7)/6 = -2/6 = -1/3. The solutions are x = 2 and x = -1/3. Verify: for x = 2, 3(4) – 5(2) – 2 = 12 – 10 – 2 = 0; for x = -1/3, 3(1/9) – 5(-1/3) – 2 = 1/3 + 5/3 – 2 = 2 – 2 = 0.

让我们用公式解 3x² – 5x – 2 = 0。确定 a = 3, b = -5, c = -2。代入公式：x = [5 ± √((-5)² – 4(3)(-2))] / 2(3) = [5 ± √(25 + 24)] / 6 = [5 ± √49] / 6 = [5 ± 7] / 6。所以 x = (5 + 7)/6 = 12/6 = 2，以及 x = (5 – 7)/6 = -2/6 = -1/3。解为 x = 2 和 x = -1/3。验证：对于 x = 2，3(4) – 5(2) – 2 = 12 – 10 – 2 = 0；对于 x = -1/3，3(1/9) – 5(-1/3) – 2 = 1/3 + 5/3 – 2 = 2 – 2 = 0。

Memorise the quadratic formula and practise using it with both positive and negative coefficients. IGCSE questions often require you to give answers to a specified degree of accuracy (e.g., 2 decimal places) or to leave them in surd form. Read the question carefully for guidance on the required format.

熟记二次公式，并练习在正负系数下使用它。IGCSE题目通常要求你将答案保留到指定精度（如2位小数）或保留根式形式。仔细阅读题目，以确定所需格式。

The Discriminant — 判别式

The discriminant, represented by the Greek letter Delta (Δ) and defined as Δ = b² – 4ac, tells us about the nature of the roots of a quadratic equation without actually solving it. This is a frequently tested concept in IGCSE Mathematics, particularly in papers 2 and 4 (extended). The discriminant can produce three possible outcomes, each corresponding to a different type of solution.

判别式用希腊字母 Δ 表示，定义为 Δ = b² – 4ac，它告诉我们二次方程根的性质，而无需实际求解。这是IGCSE数学中经常考查的概念，特别是在试卷2和试卷4（扩展）中。判别式可以产生三种可能的结果，每种对应不同类型的解。

When Δ > 0, the quadratic equation has two distinct real roots. Geometrically, the parabola crosses the x-axis at two different points. When Δ = 0, there is exactly one real root (sometimes called a repeated or double root), meaning the parabola touches the x-axis at exactly one point (the vertex lies on the x-axis). When Δ < 0, there are no real roots; the parabola does not intersect the x-axis at all. In this case, the equation has two complex conjugate roots, but these are beyond the IGCSE syllabus and would simply be stated as "no real solutions".

当 Δ > 0 时，二次方程有两个不同的实根。几何上，抛物线与x轴在两个不同点相交。当 Δ = 0 时，恰好有一个实根（有时称为重根或二重根），意味着抛物线恰好在一个点接触x轴（顶点在x轴上）。当 Δ < 0 时，没有实根；抛物线完全不与x轴相交。这种情况下，方程有两个共轭复根，但这超出了IGCSE大纲范围，通常只需回答"无实数解"。

Exam questions often ask: “Find the value of k for which the equation has equal roots” or “Determine the range of values of k for which the equation has no real roots.” For equal roots, set Δ = 0 and solve for the unknown parameter. For no real roots, set Δ < 0 and solve the inequality. For instance, to find k such that x² + kx + 9 = 0 has equal roots, set Δ = k² - 4(1)(9) = 0, giving k² = 36, so k = ±6.

考题常问：”求使方程有等根的k值”或”确定使方程无实根的k的取值范围。” 对于等根，设 Δ = 0 并求解未知参数。对于无实根，设 Δ < 0 并解不等式。例如，求使 x² + kx + 9 = 0 有等根的k值，设 Δ = k² - 4(1)(9) = 0，得出 k² = 36，所以 k = ±6。

Graphing Quadratic Functions — 二次函数图像

The graph of a quadratic function y = ax² + bx + c is a parabola, and being able to sketch or interpret these graphs is a core IGCSE skill. The shape and position of the parabola are determined entirely by the values of a, b, and c. The sign of ‘a’ determines whether the parabola opens upward (a > 0, giving a minimum point) or downward (a < 0, giving a maximum point).

二次函数 y = ax² + bx + c 的图像是一条抛物线，能够绘制或解读这些图像是IGCSE的核心技能。抛物线的形状和位置完全由 a、b 和 c 的值决定。’a’ 的符号决定抛物线是向上开口（a > 0，有最小值点）还是向下开口（a < 0，有最大值点）。

The y-intercept of the parabola is simply the constant term c, because when x = 0, y = a(0)² + b(0) + c = c. The x-intercepts (roots) are found by solving ax² + bx + c = 0. The axis of symmetry is the vertical line x = -b/(2a), and the vertex (turning point) has coordinates (-b/(2a), f(-b/(2a))). To sketch a parabola accurately, you should plot at least five points: the vertex, the y-intercept, and two or three additional points on each side of the axis of symmetry.

抛物线的y轴截距就是常数项c，因为当 x = 0 时，y = a(0)² + b(0) + c = c。x轴截距（根）通过解 ax² + bx + c = 0 求得。对称轴是垂直线 x = -b/(2a)，顶点（转折点）坐标为 (-b/(2a), f(-b/(2a)))。要准确绘制抛物线，你至少需要绘制五个点：顶点、y轴截距，以及对称轴两侧各两到三个附加点。

Worked example: sketch the graph of y = x² – 4x + 3. Here a = 1 > 0, so the parabola opens upward (U-shaped, minimum). The y-intercept is c = 3. Factorise to find roots: x² – 4x + 3 = (x – 1)(x – 3) = 0, so roots are x = 1 and x = 3. The axis of symmetry is x = -(-4)/(2×1) = 2. The vertex has x-coordinate 2, and y = 2² – 4(2) + 3 = 4 – 8 + 3 = -1. So the vertex is at (2, -1), which is the minimum point. Plot these five key points and draw a smooth curve through them.

示例：绘制 y = x² – 4x + 3 的图像。这里 a = 1 > 0，所以抛物线向上开口（U形，最小值）。y轴截距为 c = 3。因式分解求根：x² – 4x + 3 = (x – 1)(x – 3) = 0，所以根为 x = 1 和 x = 3。对称轴为 x = -(-4)/(2×1) = 2。顶点x坐标为2，y = 2² – 4(2) + 3 = 4 – 8 + 3 = -1。所以顶点坐标为 (2, -1)，这是最小值点。绘制这五个关键点并通过它们画一条平滑曲线。

Real-World Applications — 实际应用

Quadratic equations appear in many real-world contexts, and IGCSE exam papers frequently include word problems that require setting up and solving a quadratic equation. Projectile motion is one of the most common applications: the height h of an object thrown upward with initial velocity u from height h₀ is given by h = -4.9t² + ut + h₀ (using g = 9.8 m/s²). This quadratic function lets us calculate maximum height, time of flight, and when the object hits the ground.

二次方程出现在许多实际情境中，IGCSE试卷经常包含需要建立和求解二次方程的应用题。抛射运动是最常见的应用之一：以初速度u从高度h₀向上抛出的物体的高度h由 h = -4.9t² + ut + h₀ 给出（使用 g = 9.8 m/s²）。这个二次函数使我们能够计算最大高度、飞行时间以及物体何时落地。

Other applications include: maximising the area of a rectangular enclosure given a fixed perimeter (a classic optimisation problem), calculating profit in business scenarios where revenue and cost are modelled by quadratic functions, finding the dimensions of a right-angled triangle using the Pythagorean theorem, and determining the braking distance of a vehicle which varies with the square of its speed. Recognising when a problem can be modelled by a quadratic equation is a valuable problem-solving skill.

其他应用包括：在给定固定周长的情况下最大化矩形围栏面积（经典的优化问题），在商业场景中计算利润（其中收入和成本由二次函数建模），使用勾股定理计算直角三角形边长，以及确定随速度平方变化的车辆制动距离。识别何时可以用二次方程对问题进行建模是一项宝贵的解题技巧。

Example problem: A rectangular garden has a perimeter of 40 m. If its area is 96 m², find its dimensions. Let the length be x and the width be (20 – x), since 2(x + width) = 40 gives width = 20 – x. Area = x(20 – x) = 96, so 20x – x² = 96, or x² – 20x + 96 = 0. Factorising: (x – 8)(x – 12) = 0, so x = 8 or x = 12. The dimensions are 8 m by 12 m (or 12 m by 8 m).

例题：一个矩形花园的周长为40米。如果其面积为96平方米，求其尺寸。设长度为x，宽度为 (20 – x)，因为 2(x + 宽) = 40 得出宽度 = 20 – x。面积 = x(20 – x) = 96，所以 20x – x² = 96，即 x² – 20x + 96 = 0。因式分解：(x – 8)(x – 12) = 0，所以 x = 8 或 x = 12。尺寸为 8米 × 12米（或 12米 × 8米）。

Solving Quadratic Inequalities — 解二次不等式

Quadratic inequalities extend the concept of solving quadratic equations to finding ranges of x values that satisfy an inequality such as x² + 3x – 10 > 0 or 2x² – 5x – 3 <= 0. The standard approach involves three steps: solve the corresponding quadratic equation to find the critical values, sketch the parabola (or use a sign table) to determine where the expression is positive or negative, and then write the solution as intervals.

二次不等式将解二次方程的概念扩展到寻找满足不等式的x值范围，例如 x² + 3x – 10 > 0 或 2x² – 5x – 3 <= 0。标准方法包括三个步骤：解相应的二次方程找出临界值，绘制抛物线（或使用符号表）确定表达式何处为正或负，然后将解写为区间。

Let us solve x² + 3x – 10 > 0. First, factorise: (x + 5)(x – 2) = 0, so the critical values are x = -5 and x = 2. Since a = 1 > 0, the parabola opens upward. For x < -5, the expression is positive. For -5 < x < 2, the expression is negative. For x > 2, the expression is positive. Therefore, the solution to x² + 3x – 10 > 0 is x < -5 or x > 2. In interval notation: (-∞, -5) ∪ (2, ∞). Always check a test point: for x = 0 (between -5 and 2), 0² + 3(0) – 10 = -10, which is negative, confirming our analysis.

我们来解 x² + 3x – 10 > 0。首先，因式分解：(x + 5)(x – 2) = 0，所以临界值为 x = -5 和 x = 2。由于 a = 1 > 0，抛物线向上开口。当 x < -5 时，表达式为正。当 -5 < x < 2 时，表达式为负。当 x > 2 时，表达式为正。因此，x² + 3x – 10 > 0 的解为 x < -5 或 x > 2。用区间表示：(-∞, -5) ∪ (2, ∞)。务必检查测试点：对于 x = 0（在-5和2之间），0² + 3(0) – 10 = -10，为负，验证了我们的分析。

Simultaneous Equations Involving Quadratics — 涉及二次方程的联立方程组

IGCSE Extended Mathematics frequently tests the ability to solve simultaneous equations where one equation is linear and the other is quadratic. The method is substitution: express one variable in terms of the other from the linear equation, substitute this expression into the quadratic equation, and solve the resulting quadratic in a single variable. This topic bridges algebra and coordinate geometry, as the solutions correspond to the intersection points of a line and a parabola.

IGCSE扩展数学经常考查解一个方程为线性、另一个方程为二次的联立方程组的能力。方法是代入法：从线性方程中用一个变量表示另一个变量，将这个表达式代入二次方程，然后求解得到的单变量二次方程。这个主题连接了代数和坐标几何，因为解对应于直线和抛物线的交点。

Example: solve the system y = 2x + 1 and y = x² – 2x + 3. Substitute y from the first equation into the second: 2x + 1 = x² – 2x + 3. Rearrange: 0 = x² – 4x + 2, or x² – 4x + 2 = 0. Using the quadratic formula: x = [4 ± √(16 – 8)] / 2 = [4 ± √8] / 2 = [4 ± 2√2] / 2 = 2 ± √2. When x = 2 + √2, y = 2(2 + √2) + 1 = 5 + 2√2. When x = 2 – √2, y = 2(2 – √2) + 1 = 5 – 2√2. The two intersection points are (2 + √2, 5 + 2√2) and (2 – √2, 5 – 2√2).

例题：解方程组 y = 2x + 1 和 y = x² – 2x + 3。将第一个方程的y代入第二个方程：2x + 1 = x² – 2x + 3。整理：0 = x² – 4x + 2，即 x² – 4x + 2 = 0。使用二次公式：x = [4 ± √(16 – 8)] / 2 = [4 ± √8] / 2 = [4 ± 2√2] / 2 = 2 ± √2。当 x = 2 + √2 时，y = 2(2 + √2) + 1 = 5 + 2√2。当 x = 2 – √2 时，y = 2(2 – √2) + 1 = 5 – 2√2。两个交点为 (2 + √2, 5 + 2√2) 和 (2 – √2, 5 – 2√2)。

Transformations of Quadratic Graphs — 二次函数图像的变换

Understanding how changes to the equation affect the graph is an essential IGCSE skill. The basic parabola y = x² can be transformed through translations, reflections, and stretches. A vertical translation of k units is given by y = x² + k (shift up if k > 0, down if k < 0). A horizontal translation of h units is given by y = (x - h)² (shift right if h > 0, left if h < 0).

理解方程变化如何影响图像是IGCSE的一项基本技能。基本抛物线 y = x² 可以通过平移、反射和伸缩来变换。垂直平移k个单位由 y = x² + k 给出（若 k > 0 则向上平移，若 k < 0 则向下平移）。水平平移h个单位由 y = (x - h)² 给出（若 h > 0 则向右平移，若 h < 0 则向左平移）。

A reflection in the x-axis is given by y = -x², which flips the parabola upside down. A vertical stretch by factor p is given by y = px². If p > 1, the parabola becomes narrower; if 0 < p < 1, it becomes wider. The general vertex form y = a(x - h)² + k combines all these transformations: h gives the horizontal shift, k the vertical shift, and a controls the direction and steepness. The vertex is at (h, k).

关于x轴的反射由 y = -x² 给出，它将抛物线上下翻转。垂直伸缩因子p由 y = px² 给出。若 p > 1，抛物线变窄；若 0 < p < 1，抛物线变宽。通用的顶点形式 y = a(x - h)² + k 结合了所有这些变换：h 表示水平位移，k 表示垂直位移，a 控制方向和陡峭程度。顶点坐标为 (h, k)。

Exam Tips and Common Mistakes — 考试技巧与常见错误

IGCSE examiners consistently report the same mistakes year after year. The most common error is failing to set the equation equal to zero before applying any solving method. Students often try to factorise an expression like x² + 5x + 6 on its own without setting it equal to zero, which yields an expression but not a solution. Always write “= 0” explicitly before beginning to solve.

IGCSE考官年复一年地报告相同的错误。最常见的错误是在应用任何解法之前未能将方程设为零。学生常常试图单独因式分解 x² + 5x + 6 这样的表达式而不将其设为零，这只能得到一个表达式而非一个解。在开始解题前，务必明确写出”= 0″。

Another frequent mistake concerns signs when using the quadratic formula, particularly when b is negative. Remember that -b means the opposite sign of b. If b = -5, then -b = +5. Students also frequently lose marks by not giving answers in the requested form: if the question asks for answers to 2 decimal places, do not leave them as surds or fractions. Finally, always check that your answers are reasonable. A projectile’s height should never be negative in a physical context, and a length or area should always be positive.

另一个常见错误涉及使用二次公式时的符号问题，特别是当b为负时。记住 -b 表示b的相反符号。若 b = -5，则 -b = +5。学生还经常因未按要求的格式给出答案而丢分：如果题目要求保留2位小数，不要留下根式或分数形式。最后，务必检查答案是否合理。在物理情境中，抛射物的高度绝不应为负，长度或面积应始终为正。

For the extended (papers 2 and 4) examination, you should be comfortable with all four methods of solving quadratics: factorisation, completing the square, the quadratic formula, and graphical methods. The extended paper often includes questions on quadratic inequalities, simultaneous equations (one linear, one quadratic), and problems requiring the discriminant, all of which are beyond the core (papers 1 and 3) syllabus.

对于扩展（试卷2和4）考试，你应该熟练掌握所有四种解二次方程的方法：因式分解法、配方法、二次公式法和图像法。扩展试卷常包括二次不等式、联立方程组（一个线性一个二次）以及需要判别式的题目，这些都超出了核心（试卷1和3）大纲的范围。

Practice Questions — 练习题

Try these questions to test your understanding. Full worked solutions are provided below.

尝试以下题目来检验你的理解。完整解答过程见下方。

Question 1: Solve x² – 7x + 12 = 0 by factorisation.

题目1：用因式分解法解 x² – 7x + 12 = 0。

Question 2: Solve 2x² + 3x – 5 = 0 using the quadratic formula, giving your answers correct to 2 decimal places.

题目2：用二次公式解 2x² + 3x – 5 = 0，答案保留2位小数。

Question 3: Find the value(s) of k for which the equation x² + kx + 16 = 0 has equal roots.

题目3：求使方程 x² + kx + 16 = 0 有等根的k值。

Question 4: The height h metres of a ball thrown from the ground is given by h = -5t² + 20t, where t is the time in seconds. Find the maximum height reached and the time when the ball returns to the ground.

题目4：从地面抛出的球的高度h（米）由 h = -5t² + 20t 给出，其中t是以秒为单位的时间。求达到的最大高度和球回到地面的时间。

Solutions — 解答

Solution 1: x² – 7x + 12 = 0 factorises as (x – 3)(x – 4) = 0, giving x = 3 or x = 4.

解答1：x² – 7x + 12 = 0 因式分解为 (x – 3)(x – 4) = 0，得到 x = 3 或 x = 4。

Solution 2: For 2x² + 3x – 5 = 0, a = 2, b = 3, c = -5. x = [-3 ± √(9 + 40)] / 4 = [-3 ± √49] / 4 = [-3 ± 7] / 4. x = 1.00 or x = -2.50 (to 2 d.p.).

解答2：对于 2x² + 3x – 5 = 0，a = 2, b = 3, c = -5。x = [-3 ± √(9 + 40)] / 4 = [-3 ± √49] / 4 = [-3 ± 7] / 4。x = 1.00 或 x = -2.50（保留2位小数）。

Solution 3: For equal roots, b² – 4ac = 0, so k² – 4(1)(16) = 0, giving k² = 64, so k = ±8.

解答3：等根需要 b² – 4ac = 0，所以 k² – 4(1)(16) = 0，得出 k² = 64，所以 k = ±8。

Solution 4: The function h = -5t² + 20t is a downward-opening parabola. The maximum occurs at the vertex: t = -b/(2a) = -20/(2(-5)) = 2 seconds. Maximum height h = -5(2)² + 20(2) = -20 + 40 = 20 metres. The ball returns to the ground when h = 0: -5t² + 20t = 0, so -5t(t – 4) = 0, giving t = 0 (launch) or t = 4 seconds (return).

解答4：函数 h = -5t² + 20t 是一条向下开口的抛物线。最大值出现在顶点：t = -b/(2a) = -20/(2(-5)) = 2秒。最大高度 h = -5(2)² + 20(2) = -20 + 40 = 20米。球回到地面时 h = 0：-5t² + 20t = 0，所以 -5t(t – 4) = 0，得到 t = 0（发射）或 t = 4秒（返回）。

The Relationship between Roots and Coefficients — 根与系数的关系

For a quadratic equation ax² + bx + c = 0 with roots α and β, there are two fundamental relationships known as Vieta’s formulas (or Vieta’s theorem). The sum of the roots α + β = -b/a, and the product of the roots αβ = c/a. These formulas provide a powerful shortcut for checking your answers and for solving problems where you need to find a quadratic equation given its roots, without having to solve the equation from scratch.

对于有根 α 和 β 的二次方程 ax² + bx + c = 0，有两个基本关系，称为韦达定理（Vieta’s formulas）。根之和 α + β = -b/a，根之积 αβ = c/a。这些公式为检验答案和解决给定根求二次方程的问题提供了强大的捷径，无需从头解方程。

Let us verify Vieta’s formulas with a concrete example. Take the equation x² – 5x + 6 = 0. The roots are x = 2 and x = 3. The sum of roots: 2 + 3 = 5, and -b/a = -(-5)/1 = 5. The product of roots: 2 × 3 = 6, and c/a = 6/1 = 6. Both formulas are confirmed. If you are ever unsure whether your factorisation is correct, quickly check the sum and product against Vieta’s formulas – this takes seconds and can save you marks in an exam.

让我们用一个具体例子验证韦达定理。取方程 x² – 5x + 6 = 0。根为 x = 2 和 x = 3。根之和：2 + 3 = 5，而 -b/a = -(-5)/1 = 5。根之积：2 × 3 = 6，而 c/a = 6/1 = 6。两个公式都得到了验证。如果你不确定因式分解是否正确，用韦达定理快速检验和与积 – 这只需几秒钟，却能在考试中为你保住分数。

A common IGCSE question type asks: “The roots of the equation 3x² + kx + 12 = 0 are equal. Find the value of k.” Using the discriminant, k² – 4(3)(12) = 0 gives k² = 144, so k = ±12. But another approach uses Vieta: if the roots are equal, then each root = α, and the sum 2α = -k/3 and the product α² = 12/3 = 4, so α = ±2. Then k = -3(2α) = -6α, giving k = -12 or k = 12. Both methods are valid; the discriminant is usually faster.

常见的IGCSE题型：”方程 3x² + kx + 12 = 0 的根相等，求k的值。” 使用判别式，k² – 4(3)(12) = 0 得出 k² = 144，所以 k = ±12。但另一种方法使用韦达定理：若根相等，则每根 = α，和 2α = -k/3，积 α² = 12/3 = 4，所以 α = ±2。然后 k = -3(2α) = -6α，得出 k = -12 或 k = 12。两种方法都有效；判别式通常更快。

Quadratic Equations with Fractional Coefficients — 带分数系数的二次方程

When a quadratic equation contains fractions, the best strategy is to multiply every term by the lowest common denominator (LCD) to clear all fractions before attempting to solve. This converts the equation into the standard integer-coefficient form that is much easier to work with. For example, consider the equation (x/2)² + (3x/4) – 1 = 0. First, expand: x²/4 + 3x/4 – 1 = 0. Multiply every term by 4: x² + 3x – 4 = 0. Now factorise: (x – 1)(x + 4) = 0, giving x = 1 or x = -4.

当二次方程包含分数时，最佳策略是在尝试求解前将每一项乘以最小公分母（LCD）以清除所有分数。这将方程转化为更容易处理的标准整数系数形式。例如，考虑方程 (x/2)² + (3x/4) – 1 = 0。首先展开：x²/4 + 3x/4 – 1 = 0。每项乘以4：x² + 3x – 4 = 0。现在因式分解：(x – 1)(x + 4) = 0，得到 x = 1 或 x = -4。

A more challenging example: solve (x + 1)/(x – 2) = 3x/(x + 1). This is a rational equation that reduces to a quadratic. Cross-multiply: (x + 1)² = 3x(x – 2). Expand: x² + 2x + 1 = 3x² – 6x. Rearrange: 0 = 2x² – 8x – 1, or 2x² – 8x – 1 = 0. Using the quadratic formula with a = 2, b = -8, c = -1: x = [8 ± √(64 + 8)] / 4 = [8 ± √72] / 4 = [8 ± 6√2] / 4 = 2 ± (3√2)/2. Always check that these solutions do not make any denominator zero. Here, the denominators are x – 2 and x + 1. For x = 2 ± (3√2)/2, neither denominator is zero, so both solutions are valid.

一个更具挑战性的例子：解 (x + 1)/(x – 2) = 3x/(x + 1)。这是一个可化为二次方程的有理方程。交叉相乘：(x + 1)² = 3x(x – 2)。展开：x² + 2x + 1 = 3x² – 6x。整理：0 = 2x² – 8x – 1，即 2x² – 8x – 1 = 0。使用二次公式，a = 2, b = -8, c = -1：x = [8 ± √(64 + 8)] / 4 = [8 ± √72] / 4 = [8 ± 6√2] / 4 = 2 ± (3√2)/2。务必检查这些解不会使任何分母为零。这里分母为 x – 2 和 x + 1。对于 x = 2 ± (3√2)/2，两个分母都不为零，所以两个解均有效。

Using Technology: Graphical Calculators and Software — 使用科技：图形计算器和软件

The IGCSE syllabus expects students to be familiar with using graphing technology to explore quadratic functions. While you must be able to sketch graphs by hand, a graphical calculator or graphing software (such as Desmos or GeoGebra) is an invaluable tool for checking your work and for tackling complex problems where exact algebraic solutions are impractical. You should know how to use the trace, zoom, and root-finding features of your calculator.

IGCSE大纲期望学生熟悉使用图形技术来探索二次函数。虽然你必须能够手绘图像，但图形计算器或图形软件（如Desmos或GeoGebra）是检验你的作业和处理代数精确解不实用的复杂问题的宝贵工具。你应该了解如何使用计算器的追踪、缩放和求根功能。

To find the roots of y = x² – 3x – 10 using a graphical calculator, plot the function, then use the “zero” or “root” function (often accessed via the CALC menu) to find the x-intercepts. The calculator will return x = -2 and x = 5. To find the minimum point, use the “minimum” function: select a left bound, a right bound, and a guess near the vertex. The calculator returns the vertex coordinates. You can verify algebraically: x² – 3x – 10 = (x + 2)(x – 5), giving roots -2 and 5; vertex at x = -(-3)/(2×1) = 1.5, y = (1.5)² – 3(1.5) – 10 = 2.25 – 4.5 – 10 = -12.25.

使用图形计算器求 y = x² – 3x – 10 的根，先绘制函数，然后使用”零点”或”求根”功能（常通过CALC菜单访问）来找到x轴截距。计算器将返回 x = -2 和 x = 5。要找最小值点，使用”最小值”功能：选择左边界、右边界和靠近顶点的猜测值。计算器返回顶点坐标。你可以代数验证：x² – 3x – 10 = (x + 2)(x – 5)，根为-2和5；顶点在 x = -(-3)/(2×1) = 1.5，y = (1.5)² – 3(1.5) – 10 = 2.25 – 4.5 – 10 = -12.25。

Linking Quadratics to Other IGCSE Topics — 二次方程与IGCSE其他主题的联系

Quadratic equations do not exist in isolation within the IGCSE syllabus. They connect deeply with several other major topics, and examiners often design questions that span multiple topic areas. Understanding these connections will help you tackle the more challenging, multi-step problems that appear in the extended papers. The most important links are with coordinate geometry, functions, trigonometry, and sequences.

二次方程在IGCSE大纲中并非孤立存在。它们与其他几个主要主题有着深刻的联系，考官经常设计跨越多个主题领域的题目。理解这些联系将帮助你应对扩展试卷中更具挑战性的多步骤问题。最重要的联系是与坐标几何、函数、三角学和数列。

In coordinate geometry, the intersection of a line y = mx + c with a curve y = f(x) leads to a quadratic equation. Finding where a line is tangent to a parabola (i.e., touches at exactly one point) requires setting the discriminant to zero. In functions, the domain and range of quadratic functions, along with inverse functions (which require restricting the domain to make the quadratic one-to-one), are standard extended-paper topics. In trigonometry, equations like 2sin²θ + sinθ – 1 = 0 are quadratic in sinθ and are solved by substitution: let y = sinθ, solve the quadratic for y, then solve for θ. In sequences, the nth term of a quadratic sequence takes the form an² + bn + c, and finding the formula requires solving simultaneous equations.

在坐标几何中，直线 y = mx + c 与曲线 y = f(x) 的交点导致一个二次方程。求直线与抛物线相切的位置（即恰好接触于一点）需要将判别式设为零。在函数中，二次函数的定义域和值域，以及反函数（需要限制定义域使二次函数成为一一对应），是标准扩展试卷主题。在三角学中，像 2sin²θ + sinθ – 1 = 0 这样的方程是sinθ的二次方程，通过代换求解：令 y = sinθ，解关于y的二次方程，然后解出θ。在数列中，二次数列的第n项形式为 an² + bn + c，求公式需要解联立方程组。

Summary — 总结

Quadratic equations and their graphs form a substantial part of the IGCSE Mathematics syllabus, appearing in both core and extended papers. The key skills to master are: solving quadratics by factorisation, completing the square, and the quadratic formula; understanding the discriminant and its role in determining the nature of roots; graphing quadratic functions and identifying key features including the vertex, axis of symmetry, and intercepts; solving quadratic inequalities and simultaneous equations involving quadratics; and applying quadratic models to real-world problems. Systematic practice and careful attention to common pitfalls such as sign errors, missing the zero-product step, and inappropriate decimal rounding will build the confidence needed to excel on this topic in the IGCSE examination.

二次方程及其图像是IGCSE数学大纲的重要组成部分，出现在核心和扩展试卷中。需要掌握的关键技能包括：通过因式分解法、配方法和二次公式解二次方程；理解判别式及其在确定根的性质中的作用；绘制二次函数图像并识别关键特征，包括顶点、对称轴和截距；解二次不等式和涉及二次方程的联立方程组；以及将二次模型应用于实际问题。系统性的练习和对常见错误（如符号错误、遗漏零乘积步骤、不恰当的小数舍入）的仔细注意，将建立起在IGCSE考试中出色应对该主题所需的信心。