IGCSE Mathematics: Quadratic Equations Complete Guide — IGCSE 数学：二次方程完全指南

一、什么是二次方程？二次方程的标准形式 | What is a Quadratic Equation? The Standard Form

二次方程是代数学中最基础也最重要的内容之一，在IGCSE数学课程中占据核心地位。一个二次方程的标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。这里的 x² 项是二次项，bx 是一次项，c 是常数项。如果 a = 0，方程就退化为一次方程，不再是二次方程。理解标准形式是解决二次方程所有问题的基础。

A quadratic equation is one of the most fundamental and important topics in algebra, occupying a central position in the IGCSE Mathematics curriculum. The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The x² term is the quadratic term, bx is the linear term, and c is the constant term. If a = 0, the equation degenerates into a linear equation and is no longer quadratic. Understanding the standard form is the foundation for solving all problems involving quadratic equations.

在IGCSE考试中，二次方程可以以多种形式出现。有时题目直接给出标准形式的方程要求求解，有时则需要你先通过代数变换将方程整理成标准形式。例如，将 3x² = 5x + 2 整理为标准形式：移项得到 3x² – 5x – 2 = 0。能够熟练地识别和整理二次方程，是解题的第一步。

In IGCSE exams, quadratic equations can appear in various forms. Sometimes the question directly provides an equation in standard form and asks you to solve it; other times, you need to rearrange the equation into standard form through algebraic manipulation first. For example, rearrange 3x² = 5x + 2 into standard form: move all terms to one side to get 3x² – 5x – 2 = 0. Being able to identify and rearrange quadratic equations fluently is the first step to solving them.

二、因式分解法：将二次三项式分解为两个一次因式 | Factorisation: Breaking the Quadratic into Two Linear Factors

因式分解法（Factorisation）是解二次方程最基本的方法，也是IGCSE考试中最常用的方法之一。其核心思想是将二次表达式 ax² + bx + c 写成两个一次因式的乘积形式，即 (px + q)(rx + s) = 0，然后利用”零乘积性质”得出 px + q = 0 或 rx + s = 0，最后解这两个一次方程即可得到原二次方程的解。

Factorisation is the most basic method for solving quadratic equations and one of the most commonly used approaches in IGCSE exams. The core idea is to express the quadratic expression ax² + bx + c as the product of two linear factors, i.e. (px + q)(rx + s) = 0, then apply the “zero product property” to obtain px + q = 0 or rx + s = 0. Solving these two linear equations gives the solutions to the original quadratic equation.

举例说明：解方程 x² + 5x + 6 = 0。我们需要找到两个数，使得它们的和为 5（b 的值），乘积为 6（c 的值）。这两个数是 2 和 3。因此 x² + 5x + 6 可以分解为 (x + 2)(x + 3)。令每个因式等于零：x + 2 = 0 得 x = -2；x + 3 = 0 得 x = -3。答案为 x = -2 或 x = -3。

Let us illustrate with an example: solve x² + 5x + 6 = 0. We need to find two numbers whose sum is 5 (the value of b) and whose product is 6 (the value of c). These two numbers are 2 and 3. Therefore, x² + 5x + 6 factorises as (x + 2)(x + 3). Setting each factor equal to zero: x + 2 = 0 gives x = -2; x + 3 = 0 gives x = -3. The answer is x = -2 or x = -3.

当 a ≠ 1 时，因式分解会变得更复杂。例如解 2x² + 7x + 3 = 0：先找两个数使其和为 7（即 b），乘积为 2 × 3 = 6（即 a × c）。这两个数是 6 和 1。然后将一次项 7x 拆分为 6x + x：2x² + 6x + x + 3 = 0。分组提取公因式：2x(x + 3) + 1(x + 3) = 0，提取 (x + 3) 得 (x + 3)(2x + 1) = 0。所以 x = -3 或 x = -½。

When a ≠ 1, factorisation becomes more involved. For example, solve 2x² + 7x + 3 = 0: first find two numbers whose sum is 7 (b) and whose product is 2 × 3 = 6 (a × c). These numbers are 6 and 1. Then split the linear term 7x into 6x + x: 2x² + 6x + x + 3 = 0. Group and factorise by grouping: 2x(x + 3) + 1(x + 3) = 0, then extract (x + 3) to get (x + 3)(2x + 1) = 0. Therefore, x = -3 or x = -½.

三、配方法：将一个二次项系数为1的表达式配成完全平方 | Completing the Square: Turning the Expression into a Perfect Square

配方法（Completing the Square）是解二次方程的第二种标准方法。虽然在某些考试中因式分解更快，但配方法具有普适性 – 即使方程无法因式分解，配方法仍然有效。更重要的是，配方法是推导二次公式（Quadratic Formula）的基础，也是理解二次函数图像顶点坐标的关键工具。

Completing the Square is the second standard method for solving quadratic equations. While factorisation may be faster in some exam questions, completing the square has universal applicability – even when an equation cannot be factorised, completing the square still works. More importantly, completing the square is the foundation for deriving the Quadratic Formula and a key tool for understanding the vertex coordinates of quadratic function graphs.

配方法的基本步骤如下：对于形如 x² + bx + c = 0 的方程（a = 1），将常数项 c 移到等式右边得到 x² + bx = -c。然后在等式两边同时加上 (b/2)²，使左边成为一个完全平方 trinomial：(x + b/2)²。最后两边开平方根求解。

The basic steps for completing the square are as follows: for an equation of the form x² + bx + c = 0 (where a = 1), move the constant term c to the right side to get x² + bx = -c. Then add (b/2)² to both sides, making the left side a perfect square trinomial: (x + b/2)². Finally, take the square root of both sides to solve for x.

举例：用配方法解 x² + 6x + 4 = 0。首先移项得 x² + 6x = -4。(6/2)² = 9，两边同时加 9：x² + 6x + 9 = -4 + 9，即 (x + 3)² = 5。开平方得 x + 3 = ±√5，所以 x = -3 ± √5。这是精确解，考试中通常保留根号形式。

Example: solve x² + 6x + 4 = 0 by completing the square. First move the constant: x² + 6x = -4. (6/2)² = 9, add 9 to both sides: x² + 6x + 9 = -4 + 9, i.e. (x + 3)² = 5. Take the square root: x + 3 = ±√5, so x = -3 ± √5. These are exact solutions; in exams, you should usually leave them in surd form.

当 a ≠ 1 时，需要先将方程两边同时除以 a，使二次项系数变为 1，然后再进行配方法操作。例如 2x² + 8x + 5 = 0，先除以 2 得 x² + 4x + 2.5 = 0，移项 x² + 4x = -2.5，加 4 得 (x + 2)² = 1.5，x = -2 ± √1.5。

When a ≠ 1, you must first divide both sides of the equation by a to make the coefficient of the quadratic term 1, then proceed with completing the square. For example, 2x² + 8x + 5 = 0: divide by 2 to get x² + 4x + 2.5 = 0, rearrange to x² + 4x = -2.5, add 4 to get (x + 2)² = 1.5, giving x = -2 ± √1.5.

四、二次公式的推导与直接使用 | Deriving and Using the Quadratic Formula

二次公式（Quadratic Formula）是解决一切二次方程的通用工具。它的推导过程直接来自配方法：从标准形式 ax² + bx + c = 0（a ≠ 0）出发，通过配方法得到 x = [-b ± √(b² – 4ac)] / (2a)。这个公式是IGCSE数学中最著名的公式之一，考试中既可能直接给出让你代入使用，也可能要求你通过配方法自行推导。

The Quadratic Formula is a universal tool for solving any quadratic equation. Its derivation comes directly from completing the square: starting from the standard form ax² + bx + c = 0 (a ≠ 0), completing the square yields x = [-b ± √(b² – 4ac)] / (2a). This formula is one of the most famous in IGCSE Mathematics. In exams, it may be given to you for direct substitution, or you may be asked to derive it yourself via completing the square.

使用二次公式时，关键是正确识别 a、b、c 的值。以 3x² – 7x + 2 = 0 为例：a = 3，b = -7，c = 2。代入公式：x = [7 ± √(49 – 24)] / 6 = [7 ± √25] / 6 = [7 ± 5] / 6。x = (7 + 5)/6 = 2，或 x = (7 – 5)/6 = 1/3。注意 b = -7 时，-b = 7，许多学生在符号上犯错。

When using the quadratic formula, the key is correctly identifying the values of a, b, and c. Take 3x² – 7x + 2 = 0 as an example: a = 3, b = -7, c = 2. Substitute into the formula: x = [7 ± √(49 – 24)] / 6 = [7 ± √25] / 6 = [7 ± 5] / 6. Thus x = (7 + 5)/6 = 2, or x = (7 – 5)/6 = 1/3. Note that when b = -7, -b = 7 – many students make sign errors at this step.

二次公式的一个重要优势在于，它能够处理因式分解无效的情形。例如 x² + x + 1 = 0，判别式 b² – 4ac = 1 – 4 = -3 < 0，说明该方程没有实数解。二次公式在这种情况下会给出包含虚数单位 i 的复数解，但在IGCSE阶段，你只需要判断"无实数解"即可。

One important advantage of the quadratic formula is that it handles cases where factorisation fails. For example, x² + x + 1 = 0: the discriminant b² – 4ac = 1 – 4 = -3 < 0, indicating that the equation has no real solutions. The quadratic formula would give complex solutions involving the imaginary unit i, but at the IGCSE level, you only need to conclude "no real solutions."

五、判别式 Δ = b² – 4ac 与根的性质 | The Discriminant: Determining the Nature of Roots

判别式（Discriminant）Δ = b² – 4ac 是二次公式中根号下的部分。它决定了二次方程根的数量和性质，是IGCSE考试中经常单独考查的知识点。具体规则如下：当 Δ > 0 时，方程有两个不同的实数根；当 Δ = 0 时，方程有一个实数根（重根，或说两个相等的实数根）；当 Δ < 0 时，方程没有实数根。

The discriminant, Δ = b² – 4ac, is the expression under the square root sign in the quadratic formula. It determines the number and nature of the roots of a quadratic equation and is a frequently examined topic in IGCSE. The specific rules are: when Δ > 0, the equation has two distinct real roots; when Δ = 0, the equation has one real root (a repeated root, or two equal real roots); when Δ < 0, the equation has no real roots.

判别式在”参数范围”类题目中尤为重要。例如：已知方程 x² + 2kx + 9 = 0 有两个相等的实数根，求 k 的值。由 Δ = 0 得 (2k)² – 4 × 1 × 9 = 0，即 4k² – 36 = 0，k² = 9，所以 k = ±3。这类题目在IGCSE扩展卷（Extended Paper）中经常出现。

The discriminant is especially important in “parameter range” questions. For example: given that the equation x² + 2kx + 9 = 0 has two equal real roots, find the value of k. Setting Δ = 0 gives (2k)² – 4 × 1 × 9 = 0, i.e. 4k² – 36 = 0, k² = 9, so k = ±3. These types of questions frequently appear in IGCSE Extended Papers.

此外，判别式还可以结合图像分析来出题。当 Δ > 0 时，二次函数的图像与 x 轴相交于两个不同的点；Δ = 0 时图像与 x 轴相切（顶点在 x 轴上）；Δ < 0 时图像完全在 x 轴上方或下方，不与 x 轴相交。理解这种对应关系对解答图像变换题非常有帮助。

Moreover, the discriminant can be combined with graph analysis in exam questions. When Δ > 0, the quadratic function’s graph intersects the x-axis at two distinct points; when Δ = 0, the graph touches the x-axis (the vertex lies on the x-axis); when Δ < 0, the graph is entirely above or below the x-axis and does not intersect it. Understanding this correspondence is very helpful for solving graph transformation problems.

六、二次函数的图像：抛物线、顶点与对称轴 | Graphs of Quadratic Functions: Parabolas, Vertices, and Axes of Symmetry

二次函数 y = ax² + bx + c 的图像是一条抛物线（Parabola）。a 的正负决定了抛物线的开口方向：a > 0 时开口向上（U 形，有最小值），a < 0 时开口向下（倒 U 形，有最大值）。对称轴方程始终为 x = -b/(2a)，这也是顶点（Vertex）的 x 坐标。

The graph of a quadratic function y = ax² + bx + c is a parabola. The sign of a determines the direction of opening: a > 0 opens upward (U-shaped, with a minimum point), a < 0 opens downward (inverted U-shaped, with a maximum point). The axis of symmetry is always x = -b/(2a), which is also the x-coordinate of the vertex.

顶点坐标可以通过公式 (-b/(2a), f(-b/(2a))) 直接求得，也可以利用配方法将一般式转化为顶点式 y = a(x – h)² + k，其中 (h, k) 即为顶点坐标。例如 y = x² – 4x + 3：配方得 y = (x – 2)² – 1，顶点为 (2, -1)，对称轴为 x = 2。图像与 y 轴的交点为 (0, 3)，与 x 轴的交点即方程 x² – 4x + 3 = 0 的解 x = 1 和 x = 3。

The vertex coordinates can be found directly using the formula (-b/(2a), f(-b/(2a))), or by completing the square to convert the general form into vertex form y = a(x – h)² + k, where (h, k) are the vertex coordinates. For example, y = x² – 4x + 3: completing the square gives y = (x – 2)² – 1, with vertex at (2, -1) and axis of symmetry at x = 2. The graph intercepts the y-axis at (0, 3), and the x-intercepts are at x = 1 and x = 3 – the solutions to x² – 4x + 3 = 0.

在IGCSE考试中，常出现”sketching”（草图绘制）题目，要求你画出抛物线的大致形状并标注关键特征：顶点、截距和对称轴。你不需要画出精确到像素的图像，但形状、截距位置和对称性必须正确体现。

In IGCSE exams, “sketching” questions often appear, requiring you to draw the approximate shape of a parabola and label key features: the vertex, intercepts, and axis of symmetry. You do not need a pixel-perfect graph, but the shape, intercept positions, and symmetry must be correctly represented.

七、二次方程与不等式的结合 | Quadratic Equations and Inequalities

二次不等式（Quadratic Inequality）是IGCSE扩展卷的常见题型。解决思路是：先将不等式化为与零比较的形式（如 ax² + bx + c > 0），然后解对应的二次方程 ax² + bx + c = 0 得到临界值，最后通过数轴测试各区间符号来确定解集。

Quadratic inequalities are a common question type in IGCSE Extended Papers. The solution approach is: first rewrite the inequality to compare with zero (e.g. ax² + bx + c > 0), then solve the corresponding quadratic equation ax² + bx + c = 0 to find the critical values, and finally test the sign in each interval on a number line to determine the solution set.

举例：解不等式 x² – 5x + 6 > 0。先解方程 x² – 5x + 6 = 0，因式分解得 (x – 2)(x – 3) = 0，x = 2 或 x = 3。这两个临界值将数轴分为三个区间：(-∞, 2), (2, 3), (3, +∞)。测试每个区间：当 x = 0（在 (-∞, 2) 内）时，0² – 0 + 6 = 6 > 0 ✓；当 x = 2.5（在 (2, 3) 内）时，6.25 – 12.5 + 6 = -0.25 < 0 ✗；当 x = 4（在 (3, +∞) 内）时，16 - 20 + 6 = 2 > 0 ✓。因此解集为 x < 2 或 x > 3。

Example: solve the inequality x² – 5x + 6 > 0. First solve the equation x² – 5x + 6 = 0: factorising gives (x – 2)(x – 3) = 0, so x = 2 or x = 3. These two critical values divide the number line into three intervals: (-∞, 2), (2, 3), (3, +∞). Test each interval: when x = 0 (in (-∞, 2)), 0² – 0 + 6 = 6 > 0 ✓; when x = 2.5 (in (2, 3)), 6.25 – 12.5 + 6 = -0.25 < 0 ✗; when x = 4 (in (3, +∞)), 16 - 20 + 6 = 2 > 0 ✓. Therefore, the solution set is x < 2 or x > 3.

当二次不等式包含等号时（如 ax² + bx + c ≥ 0），解集应包含等号对应的点（临界值）。x² – 5x + 6 ≥ 0 的解集为 x ≤ 2 或 x ≥ 3。理解”大于取两边，小于取中间”的口诀有助于快速判断 – 但这只适用于 a > 0 且开口向上的情形。

When the quadratic inequality includes an equality sign (e.g. ax² + bx + c ≥ 0), the solution set should include the points where equality holds (the critical values). The solution set for x² – 5x + 6 ≥ 0 is x ≤ 2 or x ≥ 3. Understanding the mnemonic “greater than: take the outside intervals; less than: take the middle interval” helps with quick judgment – but this only applies when a > 0 and the parabola opens upward.

八、二次方程的实际应用题 | Real-World Applications of Quadratic Equations

二次方程在实际生活中有广泛的应用。IGCSE考试中常见的应用题类型包括：面积问题（如矩形花园的面积与周长约束）、抛体运动问题（如将球抛向空中的高度函数 h = -5t² + 20t + 1）、优化问题（如最大利润或最小成本）等。

Quadratic equations have wide real-world applications. Common application question types in IGCSE exams include: area problems (e.g. the area and perimeter constraints of a rectangular garden), projectile motion problems (e.g. the height function of a ball thrown into the air h = -5t² + 20t + 1), and optimisation problems (e.g. maximum profit or minimum cost).

典型例题：一个矩形花园的长比宽多 4 米，面积为 60 平方米，求花园的长和宽。设宽为 x 米，则长为 (x + 4) 米。面积方程：x(x + 4) = 60，即 x² + 4x – 60 = 0。因式分解得 (x + 10)(x – 6) = 0，x = -10（舍去，长度不能为负）或 x = 6。所以宽为 6 米，长为 10 米。

Typical example: a rectangular garden’s length is 4 metres more than its width, and its area is 60 square metres. Find the length and width. Let the width be x metres, then the length is (x + 4) metres. The area equation is x(x + 4) = 60, i.e. x² + 4x – 60 = 0. Factorising gives (x + 10)(x – 6) = 0, so x = -10 (reject, length cannot be negative) or x = 6. Therefore, the width is 6 m and the length is 10 m.

处理应用题时，务必检查解的合理性。二次方程通常会给出两个数学上的解，但实际场景中通常只有一个符合物理意义。检查项目包括：长度是否为正、时间是否在合理范围内、数值是否满足题目条件。在答题时，建议用一句话明确指出你拒绝了哪个解以及拒绝的原因。

When handling application problems, always check the reasonableness of your solutions. Quadratic equations typically yield two mathematical solutions, but in real-world scenarios, usually only one makes physical sense. Check items include: whether lengths are positive, whether times fall within reasonable ranges, and whether the values satisfy the given conditions. In your answer, it is recommended to state explicitly in one sentence which solution you rejected and why.

九、二次方程解题技巧与常见易错点 | Exam Techniques and Common Pitfalls in Quadratic Equations

在IGCSE考试中，二次方程题目的常见失分原因包括：符号错误（特别是处理负的 b 值时 -b 的符号）、忘记 a ≠ 0 的条件、混淆判别式公式（将 b² – 4ac 写成 b² + 4ac 或 b – 4ac）、以及在因式分解后忘记分情况讨论（只写出一个解）。

In IGCSE exams, common reasons for losing marks on quadratic equation questions include: sign errors (especially the sign of -b when b is negative), forgetting the condition a ≠ 0, mixing up the discriminant formula (writing b² + 4ac or b – 4ac instead of b² – 4ac), and forgetting to consider separate cases after factorisation (only writing one solution).

解题建议：第一，拿到题目后先识别方程是否已经是标准形式，如果不是先整理；第二，判断哪种解法最高效：如果系数简单且可以快速因式分解就用因式分解法，否则使用二次公式；第三，完成后务必代入原方程检验，这是性价比最高的防错手段；第四，对于涉及参数的题目，区分”两个相等实数根”（Δ = 0）、”两个不同实数根”（Δ > 0）和”没有实数根”（Δ < 0）这三种情况。

Exam tips: first, upon seeing the question, check whether the equation is already in standard form; if not, rearrange it first. Second, determine which method is most efficient: use factorisation if the coefficients are simple and the expression factorises quickly, otherwise use the quadratic formula. Third, always substitute your solutions back into the original equation to verify – this is the most cost-effective error-prevention technique. Fourth, for parameter-based questions, distinguish between “two equal real roots” (Δ = 0), “two distinct real roots” (Δ > 0), and “no real roots” (Δ < 0).

另一个关键技巧是：在处理二次不等式时，画出二次函数的草图非常有助于确定解集。即使只是一个粗略的草图，也能帮助你判断抛物线开口方向以及哪些区间满足不等式条件。这个过程只需 30 秒，却能大幅降低符号错误率。

Another key technique: when handling quadratic inequalities, sketching a rough graph of the quadratic function is immensely helpful for determining the solution set. Even a rough sketch helps you judge the direction of the parabola’s opening and which intervals satisfy the inequality. This process takes only 30 seconds but can dramatically reduce sign errors.

十、二次方程组：一个一次方程加一个二次方程 | Simultaneous Equations: One Linear and One Quadratic

在IGCSE扩展卷中，经常出现二次方程组（Simultaneous Equations with Quadratics）的题目。最常见的类型是一个一次方程和一个二次方程的组合，例如 y = 2x + 1 和 y = x² + x – 3。这类题目的解法是代入法（Substitution）：将一次方程中的 y 表达式代入二次方程，化为只含 x 的一元二次方程，求解 x 后再回代求得 y。

In IGCSE Extended Papers, simultaneous equations with quadratics appear frequently. The most common type combines one linear equation and one quadratic equation, for example, y = 2x + 1 and y = x² + x – 3. The solution method is substitution: substitute the expression for y from the linear equation into the quadratic equation, reducing it to a quadratic in x alone. Solve for x, then substitute back to find y.

完整例题演示：解方程组 y = 2x + 1 和 y = x² + x – 3。代入：2x + 1 = x² + x – 3。整理为标准二次方程：0 = x² – x – 4，即 x² – x – 4 = 0。使用二次公式：a = 1，b = -1，c = -4，x = [1 ± √(1 + 16)] / 2 = [1 ± √17] / 2。求得 x₁ ≈ 2.56，x₂ ≈ -1.56。分别回代 y = 2x + 1 得对应的 y 值：(2.56, 6.12) 和 (-1.56, -2.12)。注意每组解必须用括号成对给出。

Full worked example: solve the simultaneous equations y = 2x + 1 and y = x² + x – 3. Substitute: 2x + 1 = x² + x – 3. Rearrange to standard quadratic form: 0 = x² – x – 4, i.e. x² – x – 4 = 0. Apply the quadratic formula: a = 1, b = -1, c = -4, giving x = [1 ± √(1 + 16)] / 2 = [1 ± √17] / 2. We obtain x₁ ≈ 2.56 and x₂ ≈ -1.56. Substitute back into y = 2x + 1 to get the corresponding y values: (2.56, 6.12) and (-1.56, -2.12). Note that each pair of solutions must be given as an ordered pair in brackets.

考试中常见的另一种变体是：两个方程都需要进行变形。例如 x² + y² = 25 和 x + y = 7。这类题目通常先将一次方程变形为 y = 7 – x，然后代入圆的方程 x² + (7 – x)² = 25，展开整理后解二次方程。最终得到两组解，对应于直线与圆的两个交点。

Another common variant in exams is where both equations require manipulation. For example, x² + y² = 25 and x + y = 7. In such cases, first rearrange the linear equation to y = 7 – x, then substitute into the circle equation: x² + (7 – x)² = 25. Expand and simplify to obtain a quadratic equation. The final answer yields two solution pairs, corresponding to the two intersection points of the line and the circle.

十一、根与系数的关系：韦达定理 | Relationship Between Roots and Coefficients: Vieta’s Formulas

韦达定理（Vieta’s Formulas）描述了二次方程 ax² + bx + c = 0 的两个根 α 和 β 与系数 a、b、c 之间的关系。具体来说，两根之和 α + β = -b/a，两根之积 αβ = c/a。这一定理在不需要直接解方程的情况下就能得到根的相关信息，在IGCSE扩展卷中是一个重要的进阶考点。

Vieta’s Formulas describe the relationships between the two roots α and β of the quadratic equation ax² + bx + c = 0 and the coefficients a, b, c. Specifically, the sum of the roots α + β = -b/a, and the product of the roots αβ = c/a. This theorem allows you to obtain information about the roots without solving the equation directly, making it an important advanced topic in IGCSE Extended Papers.

应用举例：已知方程 2x² – 8x + k = 0 的两根之差为 4，求 k 的值。由韦达定理得 α + β = 8/2 = 4，αβ = k/2。已知 |α – β| = 4，结合 (α – β)² = (α + β)² – 4αβ 可得 16 = 16 – 2k，所以 2k = 0，k = 0。验证：方程变为 2x² – 8x = 0，即 2x(x – 4) = 0，两根为 0 和 4，差为 4，符合条件。

Application example: given that the difference between the two roots of 2x² – 8x + k = 0 is 4, find the value of k. By Vieta’s Formulas, α + β = 8/2 = 4, and αβ = k/2. Given that |α – β| = 4, combine with (α – β)² = (α + β)² – 4αβ to get 16 = 16 – 2k, so 2k = 0, k = 0. Verification: the equation becomes 2x² – 8x = 0, i.e. 2x(x – 4) = 0, with roots 0 and 4, difference 4 – satisfying the condition.

韦达定理的另一常见应用是构造方程：已知两个根的值，求对应的二次方程。例如，已知 α = 3 和 β = -2，求以 α 和 β 为根的二次方程。两根之和为 1，两根之积为 -6，因此方程为 x² – x – 6 = 0（注意：二次项系数取 1 时，方程可写为 x² – (α + β)x + αβ = 0）。

Another common application of Vieta’s Formulas is constructing an equation from its roots: given the values of the two roots, find the corresponding quadratic equation. For example, given α = 3 and β = -2, find the quadratic equation whose roots are α and β. The sum of the roots is 1, and the product is -6, so the equation is x² – x – 6 = 0 (note: when the leading coefficient is 1, the equation can be written as x² – (α + β)x + αβ = 0).

Summary | 总结

二次方程是IGCSE数学中最核心的代数主题之一，掌握好它对于后续学习函数、微积分和高等数学至关重要。我们从标准形式出发，学习了因式分解法、配方法和二次公式三种基本解法，理解了判别式如何揭示根的性质，探究了二次函数图像的关键特征，并练习了二次不等式和实际应用题。熟练掌握这些工具和概念，你将在IGCSE数学考试中自信地面对所有类型的二次方程题目。

Quadratic equations are one of the most central algebraic topics in IGCSE Mathematics, and mastering them is crucial for subsequent study of functions, calculus, and advanced mathematics. We have covered the standard form, the three fundamental solution methods (factorisation, completing the square, and the quadratic formula), understood how the discriminant reveals the nature of roots, explored key features of quadratic function graphs, and practised quadratic inequalities and real-world application problems. With proficiency in these tools and concepts, you will tackle all types of quadratic equation questions with confidence in the IGCSE Mathematics exam.

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