Mastering Quadratic Equations and Functions | IGCSE 二次方程与二次函数

📚 Mastering Quadratic Equations and Functions | IGCSE 二次方程与二次函数

Quadratic equations and functions form one of the most important topics in the IGCSE Mathematics syllabus. Whether you are aiming for a grade A* or simply trying to build a solid algebraic foundation, understanding quadratics is essential. In this article, we will explore every key aspect of quadratic equations, from factorisation to the quadratic formula, from completing the square to sketching graphs, and finally to real-life applications.

二次方程与二次函数是 IGCSE 数学大纲中最重要的内容之一。无论你是为了冲击 A* 高分，还是希望打下扎实的代数基础，掌握二次函数都至关重要。本文将系统讲解二次方程的每一个关键知识点，从因式分解到求根公式，从配方法到图像画法，再到实际应用。

1. Expanding and Factorising Quadratics | 展开与因式分解二次式

A quadratic expression in x is written in the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. The word ‘quadratic’ comes from the Latin word ‘quadratus’, meaning square, because the highest power of x is 2.

二次表达式的一般形式写作 ax² + bx + c，其中 a、b、c 为常数，且 a ≠ 0。”二次”一词源自拉丁语 “quadratus”，意为”平方”，因为表达式中 x 的最高次数是 2。

Expanding two linear factors is the first skill. For example:

展开两个一次因式是第一个核心技能。例如：

(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15

This uses the distributive law: each term in the first bracket multiplies every term in the second bracket. A useful shortcut is FOIL: First, Outer, Inner, Last.

这里用到了分配律：第一个括号中的每一项分别乘以第二个括号中的每一项。一个实用的口诀是 FOIL：First（首项）、Outer（外项）、Inner（内项）、Last（末项）。

Factorising is the reverse process. To factorise x² + 8x + 15, we look for two numbers that multiply to give 15 and add to give 8. The numbers are 3 and 5, so:

因式分解是展开的逆过程。要将 x² + 8x + 15 因式分解，我们需要找到两个数，它们的乘积为 15，和为 8。这两个数就是 3 和 5，因此：

x² + 8x + 15 = (x + 3)(x + 5)

When the coefficient of x² is not 1, the process is slightly more involved. Consider 2x² + 7x + 3. We multiply the coefficient of x² (which is 2) by the constant term (which is 3), giving 6. We then find two numbers that multiply to 6 and add to 7 — they are 1 and 6. Rewrite the middle term:

当 x² 的系数不为 1 时，过程会稍微复杂一些。以 2x² + 7x + 3 为例，我们将 x² 的系数（即 2）乘以常数项（即 3），得到 6。然后寻找两个数，它们的乘积为 6，和为 7——这两个数就是 1 和 6。改写中间项：

2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

2. Solving by Factorisation | 因式分解法解方程

Once a quadratic equation is factorised, we apply the zero product property: if the product of two factors is zero, then at least one of the factors must be zero. For example, to solve x² − 5x + 6 = 0:

当一个二次方程完成因式分解后，我们使用零积性质：如果两个因式的乘积为零，那么至少有一个因式必须为零。例如，解方程 x² − 5x + 6 = 0：

x² − 5x + 6 = (x − 2)(x − 3) = 0

Therefore, either x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3. Always check your answers by substituting each value back into the original equation.

因此，要么 x − 2 = 0，要么 x − 3 = 0，解得 x = 2 或 x = 3。务必把每个解代回原方程验证。

For a quadratic with two identical roots, such as x² − 6x + 9 = 0, we have (x − 3)² = 0, so x = 3 is a repeated root. The graph of y = x² − 6x + 9 just touches the x-axis at one point.

对于具有两个相等根的二次方程，如 x² − 6x + 9 = 0，我们有 (x − 3)² = 0，因此 x = 3 是重根。函数 y = x² − 6x + 9 的图像与 x 轴仅相切于一点。

Some quadratics cannot be factorised using integers; for these, we need the quadratic formula or completing the square, which we will cover next.

有些二次式无法用整数进行因式分解；对于这些情况，我们需要使用求根公式或配方法，这将在接下来讨论。

3. The Quadratic Formula | 二次求根公式

The quadratic formula provides a universal method for solving any quadratic equation of the form ax² + bx + c = 0. It is derived by completing the square on the general form:

求根公式为解任意形如 ax² + bx + c = 0 的二次方程提供了一种通用方法。它是通过对一般形式进行配方推导得到的：

x = (−b ± √(b² − 4ac)) ÷ (2a)

Here the symbol ± means that we calculate two values: one with a plus sign and one with a minus sign. Let us apply the formula to solve 2x² − 4x − 3 = 0. Here a = 2, b = −4 and c = −3.

这里的符号 ± 表示我们需要计算两个值：一个取加号，一个取减号。让我们用该公式求解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。

x = (4 ± √((−4)² − 4 × 2 × (−3))) ÷ (2 × 2) = (4 ± √(16 + 24)) ÷ 4 = (4 ± √40) ÷ 4

Since √40 = 2√10, the two solutions can be written as x = 1 + (√10)/2 and x = 1 − (√10)/2. In your exam, you may be asked to give your answer to a specified number of decimal places.

由于 √40 = 2√10，两个解可以写作 x = 1 + (√10)/2 和 x = 1 − (√10)/2。在考试中，题目可能要求你将答案保留到指定的小数位数。

When using the quadratic formula, be extremely careful with negative signs. If b is negative, then −b becomes positive, as shown in the example above.

使用求根公式时，要特别小心负号。如果 b 为负数，那么 −b 就变为正数，如上例所示。

4. Completing the Square | 配方法

Completing the square is another powerful technique that rewrites a quadratic in the form a(x + p)² + q. This form reveals the turning point of the graph immediately and is also used to derive the quadratic formula.

配方法是另一种强力技巧，它将二次式改写为 a(x + p)² + q 的形式。这种形式能直接显示图像的顶点坐标，也用于推导求根公式。

Consider x² + 6x + 2. We take half of the coefficient of x, which is 3, and write:

以 x² + 6x + 2 为例。我们取 x 系数的一半，即 3，然后写出：

x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7

The key idea is that (x + 3)² = x² + 6x + 9, so we subtract 9 to compensate for the extra term. The vertex of the curve y = (x + 3)² − 7 is at (−3, −7).

关键思想是 (x + 3)² = x² + 6x + 9，因此我们减去 9 来补偿多余的部分。曲线 y = (x + 3)² − 7 的顶点坐标为 (−3, −7)。

When the coefficient of x² is not 1, factor it out first. For example, for 2x² − 8x + 5:

当 x² 的系数不为 1 时，应先将该系数提取出来。例如，对于 2x² − 8x + 5：

2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3

The vertex here is at (2, −3), and because a = 2 is positive, the curve opens upwards and has a minimum point.

此处的顶点为 (2, −3)，由于 a = 2 为正数，曲线开口向上，具有最小值点。

5. The Discriminant | 判别式

The discriminant is the expression b² − 4ac inside the square root of the quadratic formula. It tells us how many real roots a quadratic equation has, without actually solving the equation.

判别式是求根公式根号内的表达式 b² − 4ac。它告诉我们二次方程有多少个实数根，而无需实际解方程。

Discriminant 判别式	Number of Real Roots 实数根的个数	Graph Interpretation 图像意义
b² − 4ac > 0	Two distinct real roots 两个不等实根	The curve crosses the x-axis at two points 曲线与 x 轴相交于两点
b² − 4ac = 0	One repeated real root 一个重根	The curve touches the x-axis at one point 曲线与 x 轴相切于一点
b² − 4ac < 0	No real roots 无实数根	The curve does not intersect the x-axis 曲线不与 x 轴相交

For example, for the equation x² − 4x + 5 = 0, the discriminant is (−4)² − 4 × 1 × 5 = 16 − 20 = −4, which is negative. Hence the equation has no real roots, and the graph lies entirely above the x-axis.

例如，对方程 x² − 4x + 5 = 0，判别式为 (−4)² − 4 × 1 × 5 = 16 − 20 = −4，为负数。因此该方程无实数根，图像完全位于 x 轴上方。

Many exam questions ask you to ‘find the range of values of k for which the equation has two distinct real roots’. In such cases you set the discriminant to be greater than zero and solve the resulting inequality.

许多考试题目会要求你”求 k 的取值范围，使得该方程有两个不同的实数根”。此时你需要令判别式大于零，然后解所得的不等式。

6. Sum and Product of Roots | 根与系数的关系

For a quadratic equation ax² + bx + c = 0 with roots α and β, there is a beautiful relationship between the roots and the coefficients:

对于具有根 α 和 β 的二次方程 ax² + bx + c = 0，根与系数之间存在优美的关系：

α + β = −b ÷ a

α × β = c ÷ a

This is known as Vieta’s formulas and is extremely useful. For example, if we know that the sum of the roots of a quadratic equation is 7 and the product is 12, we can immediately write the equation as x² − 7x + 12 = 0 (assuming a = 1), which factorises to (x − 3)(x − 4) = 0.

这就是韦达定理，非常实用。例如，如果我们知道某二次方程的两根之和为 7，两根之积为 12，就可以直接写出方程 x² − 7x + 12 = 0（假设 a = 1），其因式分解为 (x − 3)(x − 4) = 0。

These relationships are particularly helpful when you are given roots and asked to form an equation, or when you need to find expressions involving both roots without solving the original equation.

这些关系在给定根求方程，或需要计算涉及两个根的表达式而无需解原方程时特别有用。

7. Sketching Quadratic Graphs | 二次函数图像画法

The graph of a quadratic function is a parabola. The sign of the leading coefficient a determines the direction of the opening: if a > 0, the parabola opens upwards and has a minimum point; if a < 0, it opens downwards and has a maximum point.

二次函数的图像是一条抛物线。首项系数 a 的符号决定开口方向：当 a > 0 时，抛物线开口向上，存在最小值点；当 a < 0 时，开口向下，存在最大值点。

To sketch a quadratic graph accurately, you need to identify four key features:

要准确画出二次函数图像，你需要确定四个关键特征：

The y-intercept: set x = 0, giving y = c. y 轴截距：令 x = 0，得 y = c。
The x-intercepts: set y = 0 and solve the quadratic equation. x 轴截距：令 y = 0，解二次方程。
The vertex: found by completing the square or using x = −b ÷ (2a). 顶点：通过配方法或使用 x = −b ÷ (2a) 求得。
The axis of symmetry: the vertical line x = −b ÷ (2a). 对称轴：垂直直线 x = −b ÷ (2a)。

For instance, to sketch y = x² − 2x − 3, we note that the y-intercept is −3. Factorising gives (x − 3)(x + 1), so the x-intercepts are x = 3 and x = −1. The axis of symmetry lies halfway between the roots, at x = 1. Substituting x = 1 gives y = 1 − 2 − 3 = −4, so the vertex is at (1, −4).

例如，要画 y = x² − 2x − 3 的图像，我们注意 y 轴截距为 −3。因式分解得 (x − 3)(x + 1)，因此 x 轴截距为 x = 3 和 x = −1。对称轴位于两个根的正中间，即 x = 1。将 x = 1 代入得 y = 1 − 2 − 3 = −4，因此顶点为 (1, −4)。

Always label your axes, mark the scale, and clearly indicate the vertex and intercepts on your sketch. A well-labelled diagram earns valuable method marks in the exam.

画图时务必标注坐标轴、标明刻度，并清楚标示顶点和截距。一份标注完善的图像能在考试中为你赢得宝贵的方法分。

8. Applications and Word Problems | 实际应用与文字题

Quadratic equations appear frequently in geometry and real-world problems. A classic example involves area and perimeter. Suppose a rectangular garden has an area of 50 square metres and its length is 5 metres longer than its width. Let the width be w metres. Then the length is w + 5 metres, and:

二次方程在几何和实际问题中频繁出现。一个经典的例子涉及面积和周长。假设一个矩形花园的面积为 50 平方米，其长比宽长 5 米。设宽为 w 米，则长为 w + 5 米，于是：

w(w + 5) = 50 ⇒ w² + 5w − 50 = 0

Factorising gives (w + 10)(w − 5) = 0, so w = 5 or w = −10. Since a width cannot be negative, the answer is w = 5 metres. The length is therefore 10 metres.

因式分解得 (w + 10)(w − 5) = 0，因此 w = 5 或 w = −10。由于宽度不能为负数，答案为 w = 5 米。因此长为 10 米。

Another important application is projectile motion. If a ball is thrown upwards with an initial speed such that its height h metres after t seconds is given by h = 20t − 5t², then to find when the ball hits the ground we solve:

另一个重要的应用是抛体运动。如果一个小球以某个初速度被向上抛出，其高度 h（米）与时间 t（秒）的关系为 h = 20t − 5t²，那么要找到球何时落地，我们解：

20t − 5t² = 0 ⇒ −5t(t − 4) = 0 ⇒ t = 0 or t = 4

The ball hits the ground after 4 seconds. Exam questions often combine quadratics with Pythagoras’ theorem as well; for example, finding a side length when the other two sides are given as algebraic expressions.

小球在 4 秒后落地。考试题目还常将二次方程与勾股定理结合；例如，已知另外两条边为代数表达式，求某一侧边的长度。

9. Common Mistakes and Tips | 常见错误与提分技巧

Many students lose marks on quadratic questions due to small but avoidable errors. Here are the most common pitfalls and how to avoid them.

许多学生在二次函数题目上失分，往往是因为一些细小但可以避免的错误。以下是最常见的陷阱以及如何避免它们。

Forgetting to set the equation to zero. Before factorising or applying the quadratic formula, always rearrange the equation so that one side equals zero.
忘记将方程化为零。在因式分解或使用求根公式之前，务必整理方程，使一边等于零。
Missing negative signs. Always check the sign of each coefficient carefully, particularly when substituting into the quadratic formula.
漏掉负号。务必仔细检查每个系数的符号，尤其是在将系数代入求根公式时。
Incorrect cancelling. Remember that you cannot cancel a term from one side of the equation unless you do the same on the other side.
约分错误。记住，除非在方程两边进行相同的操作，否则不能单独消去某一项。
Drawing graphs without labels. An unlabelled sketch may not receive full marks. Always mark the vertex, intercepts and axis of symmetry.
画图不标注。没有标注的图像可能无法获得满分。务必标出顶点、截距和对称轴。

Practice is the key to mastering quadratics. Try to solve every problem using two different methods — for example, factorisation and the quadratic formula — and check that you obtain the same answers. This deepens your understanding and exposes any mistakes.

练习是掌握二次函数的关键。尝试用两种不同的方法解答每一道题——例如因式分解法和求根公式——并检查是否得到相同的答案。这样可以加深你的理解，并暴露可能存在的错误。

In summary, quadratic equations are a central pillar of IGCSE mathematics. By mastering expansion, factorisation, the quadratic formula, completing the square, the discriminant and graph sketching, you equip yourself with tools that appear in countless examinations and future mathematical studies. Keep a clear checklist of the steps for each method, and do not forget to verify your solutions.

总而言之，二次方程是 IGCSE 数学的核心支柱。通过掌握展开、因式分解、求根公式、配方法、判别式和图像画法，你将拥有在无数考试和未来数学学习中都会用到的工具。为每种方法建立清晰的步骤清单，并且别忘了验证你的答案。

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Mastering Quadratic Equations and Functions | IGCSE 二次方程与二次函数