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

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

Quadratic equations and functions form one of the most important topics in IGCSE Mathematics. From solving equations to sketching parabolic graphs, a solid understanding of this topic is essential for both Paper 2 and Paper 4. In this article, we will explore every key concept step by step, with worked examples and common exam pitfalls.

二次方程与二次函数是 IGCSE 数学中最核心的专题之一。无论是解方程还是绘制抛物线图像，扎实掌握这一章节对 Paper 2 和 Paper 4 都至关重要。在本文中，我们将逐步剖析每一个核心概念，配套典型例题与常见失分点分析。

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

A quadratic equation is a polynomial equation of degree 2, written in the general form:

ax² + bx + c = 0，其中 a ≠ 0

Here, a is the coefficient of x², b is the coefficient of x, and c is the constant term. The condition a ≠ 0 is essential — if a = 0, the equation becomes linear, not quadratic. Examples of quadratic equations include x² − 5x + 6 = 0 and 2x² + 3x − 1 = 0.

二次方程是最高次数为 2 的多项式方程，其一般形式为：

ax² + bx + c = 0，其中 a ≠ 0

其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。条件 a ≠ 0 至关重要——若 a = 0，方程退化为一次方程而非二次方程。例如 x² − 5x + 6 = 0 和 2x² + 3x − 1 = 0 都是二次方程。

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

Before solving quadratic equations, you must be fluent in two algebraic skills: expanding brackets and factorising. Expanding means removing brackets, while factorising is the reverse process — writing the expression as a product of two brackets.

在解二次方程之前，你必须熟练掌握两项代数技能：去括号展开与因式分解。展开是去掉括号，而因式分解是相反的过程——将表达式写成两个括号相乘的形式。

To expand (x + p)(x + q), use the distributive law:

(x + p)(x + q) = x² + (p + q)x + pq

For example, (x + 3)(x − 5) = x² − 2x − 15, since 3 + (−5) = −2 and 3 × (−5) = −15.

使用分配律展开 (x + p)(x + q)：

(x + p)(x + q) = x² + (p + q)x + pq

例如 (x + 3)(x − 5) = x² − 2x − 15，因为 3 + (−5) = −2，3 × (−5) = −15。

To factorise x² + bx + c, look for two numbers whose product is c and whose sum is b. For x² + 7x + 12, the numbers 3 and 4 give product 12 and sum 7, so x² + 7x + 12 = (x + 3)(x + 4).

因式分解 x² + bx + c 时，需找到两个数，使其乘积为 c、和为 b。对于 x² + 7x + 12，数 3 和 4 的乘积为 12、和为 7，因此 x² + 7x + 12 = (x + 3)(x + 4)。

When the coefficient of x² is not 1, such as 2x² + 7x + 3, factorising requires extra care. Look for two brackets (2x + m)(x + n) such that the product of the outer and inner terms sums to 7x. Here, 2x² + 7x + 3 = (2x + 1)(x + 3).

当 x² 的系数不为 1 时，例如 2x² + 7x + 3，因式分解需要格外小心。寻找两个括号 (2x + m)(x + n)，使得外项与内项乘积之和为 7x。这里 2x² + 7x + 3 = (2x + 1)(x + 3)。

3. Solving by Factorisation | 用因式分解法求解

Once a quadratic equation is factorised, solving becomes straightforward using the zero product property: if pq = 0, then p = 0 or q = 0.

一旦二次方程完成因式分解，利用零乘积性质求解就变得直接：如果 pq = 0，则 p = 0 或 q = 0。

Solve x² − 5x + 6 = 0 by factorisation. First, factorise: x² − 5x + 6 = (x − 2)(x − 3). Then set each factor to zero:

用因式分解法解 x² − 5x + 6 = 0。首先因式分解：x² − 5x + 6 = (x − 2)(x − 3)。然后令每个因子为零：

(x − 2)(x − 3) = 0 ⇒ x − 2 = 0 或 x − 3 = 0 ⇒ x = 2 或 x = 3

Always remember to rearrange the equation into the standard form ax² + bx + c = 0 before factorising. For example, to solve x² = 4x + 5, first rewrite it as x² − 4x − 5 = 0, then factorise to get (x − 5)(x + 1) = 0, giving x = 5 or x = −1.

在因式分解前，务必先将方程整理成标准形式 ax² + bx + c = 0。例如解 x² = 4x + 5 时，先改写为 x² − 4x − 5 = 0，再分解为 (x − 5)(x + 1) = 0，得 x = 5 或 x = −1。

4. The Quadratic Formula | 二次公式

Some quadratic equations cannot be factorised easily using integers. In such cases, the quadratic formula provides a reliable universal method:

有些二次方程难以用整数因式分解。此时，二次公式提供了一种通用的可靠解法：

x = (−b ± √(b² − 4ac)) / 2a

This formula gives the solutions of any quadratic equation ax² + bx + c = 0. The symbol ± means that we compute two values: one with the plus sign and one with the minus sign.

该公式给出任意二次方程 ax² + bx + c = 0 的解。符号 ± 表示我们计算两个值：一个取加号，另一个取减号。

Solve 2x² + 3x − 2 = 0 using the quadratic formula. Here a = 2, b = 3, c = −2. Substituting into the formula gives:

用二次公式解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入公式得：

x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √(9 + 16)) / 4 = (−3 ± 5) / 4

Therefore x = (−3 + 5)/4 = 1/2 or x = (−3 − 5)/4 = −2. Always check your solutions by substituting back into the original equation.

因此 x = (−3 + 5)/4 = 1/2 或 x = (−3 − 5)/4 = −2。务必通过代回原方程来验证解。

5. Completing the Square | 配方法

Completing the square is a technique used to rewrite a quadratic expression in the form a(x − h)² + k. This form reveals the turning point of the parabola directly and is also used to solve equations.

配方法是将二次表达式改写为 a(x − h)² + k 形式的技术。这种形式直接揭示抛物线的顶点，同时也可用于解方程。

For a simple quadratic x² + bx, add and subtract (b/2)² to complete the square. For example:

对于简单的二次式 x² + bx，通过加减 (b/2)² 来完成配方。例如：

x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8

To solve (x + 3)² − 8 = 0, rearrange to (x + 3)² = 8, then x + 3 = ±√8, so x = −3 ± 2√2. When the coefficient of x² is not 1, factor it out first before completing the square.

解 (x + 3)² − 8 = 0 时，移项得 (x + 3)² = 8，则 x + 3 = ±√8，所以 x = −3 ± 2√2。当 x² 的系数不为 1 时，先提取该系数再配方。

6. Graphs of Quadratic Functions | 二次函数图像

The graph of a quadratic function y = ax² + bx + c is a parabola. The sign of a determines the direction of the curve:

二次函数 y = ax² + bx + c 的图像是一条抛物线。a 的符号决定曲线的开口方向：

If a > 0, the parabola opens upwards and the turning point is a minimum.

若 a > 0，抛物线开口向上，顶点为最小值点。
If a < 0, the parabola opens downwards and the turning point is a maximum.

若 a < 0，抛物线开口向下，顶点为最大值点。

Key features of the graph include the y-intercept at (0, c), the x-intercepts (roots of the equation ax² + bx + c = 0), and the axis of symmetry, which is the vertical line passing through the turning point.

图像的关键特征包括 y 轴截距 (0, c)、x 轴截距（即方程 ax² + bx + c = 0 的根）以及对称轴——通过顶点的竖直直线。

To sketch a quadratic graph, find the y-intercept, the roots (if they exist), and the turning point, then plot these key points and draw a smooth curve.

绘制二次函数草图时，先求出 y 轴截距、根（若存在）和顶点，标出关键点后画平滑曲线。

7. The Discriminant | 判别式

The expression b² − 4ac inside the quadratic formula is called the discriminant, denoted by Δ. It determines the nature of the roots without solving the equation fully.

二次公式中的 b² − 4ac 称为判别式，记作 Δ。它无需完全解方程即可判断根的性质。

There are three cases to memorise:

以下三种情形必须牢记：

判别式 Δ	根的性质	图像特征
Δ = b² − 4ac > 0	两个不同的实根	抛物线与 x 轴有两个交点
Δ = b² − 4ac = 0	一个重根（相等实根）	抛物线与 x 轴相切
Δ = b² − 4ac < 0	无实根	抛物线与 x 轴不相交

For example, for 2x² − 4x + 2 = 0, the discriminant is (−4)² − 4 × 2 × 2 = 16 − 16 = 0, so the equation has exactly one repeated root, x = 1.

例如，对于 2x² − 4x + 2 = 0，判别式为 (−4)² − 4 × 2 × 2 = 16 − 16 = 0，因此方程恰有一个重根 x = 1。

8. Finding the Turning Point | 求顶点坐标

The turning point of a quadratic function can be found by completing the square. In the form y = a(x − h)² + k, the turning point is (h, k).

二次函数的顶点可通过配方法求得。在 y = a(x − h)² + k 形式中，顶点为 (h, k)。

Find the turning point of y = x² − 4x + 5. Complete the square: x² − 4x + 5 = (x − 2)² − 4 + 5 = (x − 2)² + 1. Therefore the turning point is (2, 1), and since a = 1 > 0, it is a minimum.

求 y = x² − 4x + 5 的顶点。配方：x² − 4x + 5 = (x − 2)² − 4 + 5 = (x − 2)² + 1。因此顶点为 (2, 1)，且因为 a = 1 > 0，该点为最小值点。

Alternatively, the x-coordinate of the turning point can be found using the formula x = −b/(2a). Substituting this x-value into the original equation gives the y-coordinate.

另一种方法：顶点的 x 坐标可用公式 x = −b/(2a) 求得，将其代回原方程即可得到 y 坐标。

9. Real-World Applications | 实际应用

Quadratic functions model many real-world situations, such as projectile motion, area optimisation, and profit maximisation. Recognising how to translate a word problem into a quadratic equation is a key exam skill.

二次函数可以建模许多实际情境，如抛射运动、面积最优化和利润最大化。将文字题转化为二次方程是关键的考试技能。

A rectangle has a length 4 cm greater than its width, and its area is 60 cm². Let the width be x cm. Then the length is x + 4 cm, and x(x + 4) = 60, which simplifies to x² + 4x − 60 = 0. Factorising gives (x + 10)(x − 6) = 0, so x = 6 or x = −10. Since a width cannot be negative, the width is 6 cm and the length is 10 cm.

一个矩形的长比宽多 4 cm，面积为 60 cm²。设宽为 x cm，则长为 x + 4 cm，于是 x(x + 4) = 60，化简得 x² + 4x − 60 = 0。分解得 (x + 10)(x − 6) = 0，所以 x = 6 或 x = −10。由于宽度不能为负，宽为 6 cm，长为 10 cm。

In projectile motion, the height h of an object thrown upwards can be modelled by h = −5t² + 20t + 1. The object hits the ground when h = 0, which requires solving the quadratic equation −5t² + 20t + 1 = 0. Using the quadratic formula gives the positive time value.

在抛射运动中，物体高度 h 可以用 h = −5t² + 20t + 1 建模。物体落地时 h = 0，即需要解方程 −5t² + 20t + 1 = 0。使用二次公式即可求得正的时间值。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Avoiding common pitfalls can save you valuable marks in the exam. Below are the most frequent errors students make with quadratics, along with practical advice.

避免常见失分点可以在考试中为你保住宝贵的分数。以下是学生在二次方程章节中最常犯的错误及实用建议。

Forgetting to rearrange the equation to the form ax² + bx + c = 0 before applying any method. Always move all terms to one side first.

忘记先将方程整理为 ax² + bx + c = 0 形式再使用任何解法。务必先将所有项移到同一边。
Losing solutions by dividing by a common factor that contains x, such as dividing x² = 3x by x. This loses the root x = 0. Instead, rearrange to x² − 3x = 0 and factorise.

在解 x² = 3x 时直接除以 x 而丢失解 x = 0。正确做法是移项得 x² − 3x = 0 再因式分解。
When using the quadratic formula, check that the value of c is substituted with its correct sign. In 2x² + 3x − 2 = 0, c = −2, not 2.

使用二次公式时，注意 c 的符号。在 2x² + 3x − 2 = 0 中，c = −2 而非 2。
When completing the square, remember to balance the equation by subtracting the same value that you add.

配方时，记住加上多少就必须减去多少以保持等式平衡。
Always check whether your final answers satisfy the original equation — this catches most arithmetic errors.

始终验证最终答案是否满足原方程——这能检查出绝大多数计算错误。

With consistent practice across all five techniques — factorisation, the quadratic formula, completing the square, the discriminant, and graphing — you will build the confidence needed to tackle any quadratic question in the IGCSE exam.

通过持续练习全部五种技巧——因式分解、二次公式、配方法、判别式和图像绘制——你将建立解决 IGCSE 考试中任何二次函数问题的信心。

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