IGCSE Mathematics: Quadratic Equations and Graphs — IGCSE 数学：二次方程与图像

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。