Quadratic Equations | 二次方程

📚 Quadratic Equations | 二次方程

Quadratic equations are one of the most important topics in the Edexcel IGCSE Mathematics syllabus. Mastering them unlocks a wide range of exam questions, from factorisation to graph sketching and real-world problems. This guide covers every key skill you need, with worked examples and common pitfalls clearly explained.

二次方程是 Edexcel IGCSE 数学考纲中最重要的主题之一。掌握它，你就能应对从因式分解到函数图像、再到实际应用问题的大量考点。本指南覆盖你所需的每一个核心技能，并配有例题和常见陷阱的清晰讲解。

1. Standard Form | 标准形式

A quadratic equation is any equation that can be written in the form:

二次方程是指可以写成以下形式的任意方程：

ax² + bx + c = 0

where a, b and c are constants, and a ≠ 0. The term ax² is called the quadratic term, bx the linear term, and c the constant term. If a = 0, the equation becomes linear, not quadratic.

其中 a、b、c 为常数，且 a ≠ 0。ax² 称为二次项，bx 称为一次项，c 称为常数项。若 a = 0，方程就变成一次方程，而非二次方程。

For example, 2x² − 5x + 3 = 0 is a quadratic equation with a = 2, b = −5 and c = 3. The value x = 1 satisfies it because 2(1)² − 5(1) + 3 = 0.

例如，2x² − 5x + 3 = 0 是一个二次方程，其中 a = 2，b = −5，c = 3。x = 1 满足该方程，因为 2(1)² − 5(1) + 3 = 0。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the equation has simple integer roots. The idea is to rewrite ax² + bx + c as a product of two linear brackets, then use the fact that if AB = 0, then A = 0 or B = 0.

当方程具有简单的整数根时，因式分解通常是最快的方法。其思路是将 ax² + bx + c 改写为两个一次括号相乘，然后利用“若 AB = 0，则 A = 0 或 B = 0”的性质。

Example 1 | 例 1: Solve x² − 7x + 12 = 0.

例 1：解 x² − 7x + 12 = 0。

Find two numbers that multiply to 12 and add to −7. These are −3 and −4. Hence:

找两个数，乘积为 12，和为 −7。这两个数是 −3 和 −4。因此：

(x − 3)(x − 4) = 0

So x = 3 or x = 4.

所以 x = 3 或 x = 4。

Example 2 | 例 2: Solve 2x² + 5x − 3 = 0.

例 2：解 2x² + 5x − 3 = 0。

Here a ≠ 1, so we use factorisation by inspection or the AC method. Multiply a and c: 2 × (−3) = −6. Find two numbers that multiply to −6 and add to 5: these are 6 and −1. Rewrite:

这里 a ≠ 1，因此可以使用观察法或 AC 法分解。将 a 与 c 相乘：2 × (−3) = −6。找两个数，乘积为 −6，和为 5：它们是 6 和 −1。改写：

2x² + 6x − x − 3 = 0

Group and factor:

分组并提取公因式：

2x(x + 3) − 1(x + 3) = 0 → (2x − 1)(x + 3) = 0

Therefore x = ½ or x = −3.

因此 x = ½ 或 x = −3。

3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form p(x + q)² + r. This is especially useful for finding turning points and solving equations that do not factorise neatly.

配方法将二次式改写为 p(x + q)² + r 的形式。这在求顶点坐标以及解不能整洁分解的方程时特别有用。

Example | 例： Solve x² + 6x − 7 = 0 by completing the square.

例：用配方法解 x² + 6x − 7 = 0。

Halve the coefficient of x: 6 ÷ 2 = 3. Write:

将 x 的系数取半：6 ÷ 2 = 3。写成：

x² + 6x = (x + 3)² − 9

So the equation becomes:

于是方程变为：

(x + 3)² − 9 − 7 = 0 → (x + 3)² = 16

Taking square roots: x + 3 = ±4, so x = 1 or x = −7.

两边开平方：x + 3 = ±4，所以 x = 1 或 x = −7。

For a general quadratic ax² + bx + c, first factor out a from the first two terms, then complete the square inside the bracket.

对于一般二次式 ax² + bx + c，先从前两项中提出 a，再在括号内配方。

4. The Quadratic Formula | 求根公式

The quadratic formula solves any quadratic equation, even when factorisation is difficult or impossible. For ax² + bx + c = 0:

求根公式可以解任何二次方程，即使因式分解很困难或不可能时也适用。对于 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) / (2a)

Example | 例： Solve 3x² − 5x − 2 = 0.

例：解 3x² − 5x − 2 = 0。

Here a = 3, b = −5, c = −2. Substitute:

这里 a = 3，b = −5，c = −2。代入：

x = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6

So x = (5 + 7)/6 = 2, or x = (5 − 7)/6 = −⅓.

因此 x = (5 + 7)/6 = 2，或 x = (5 − 7)/6 = −⅓。

Always write the formula down first, substitute carefully, and check your signs. A common mistake is mis-handling the negative values of b or c.

先写出公式，再小心代入，并检查符号。常见错误是处理 b 或 c 的负值时出错。

5. The Discriminant | 判别式

The expression b² − 4ac inside the square root is called the discriminant, often denoted by Δ. Its value tells us the number and type of roots without solving the equation.

根号内的表达式 b² − 4ac 称为判别式，常用 Δ 表示。它的值可以告诉我们方程的根的数量和类型，而无需解方程。

Δ = b² − 4ac	Nature of roots \| 根的性质
Δ > 0	Two distinct real roots \| 两个不等实根
Δ = 0	One repeated real root \| 一个重根（两个相等实根）
Δ < 0	No real roots \| 无实根

Example | 例： For kx² − 4x + 2 = 0, find the value of k for which the equation has exactly one real root.

例：对于 kx² − 4x + 2 = 0，求使方程恰有一个实根的 k 值。

Set Δ = 0:

令 Δ = 0：

(−4)² − 4(k)(2) = 0 → 16 − 8k = 0 → k = 2

Always note the condition a ≠ 0. If k = 0 here, the equation is linear and has only one root automatically.

始终注意 a ≠ 0 的条件。若这里 k = 0，方程就是一次的，自然只有一个根。

6. Sketching Quadratics | 二次函数图像

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upward (a “U” shape); if a < 0, it opens downward (an "n" shape).

y = ax² + bx + c 的图像是一条抛物线。若 a > 0，开口向上（U 形）；若 a < 0，开口向下（n 形）。

To sketch a quadratic accurately, find three key features:

要准确地画出二次函数草图，需要找到三个关键特征：

The roots: solve ax² + bx + c = 0, or note if there are none.

根：解 ax² + bx + c = 0；若无实根则标注。
The y-intercept: when x = 0, y = c.

y 轴截距：当 x = 0 时，y = c。
The turning point: use completing the square or the formula x = −b/(2a).

顶点：使用配方法或公式 x = −b/(2a)。

Example | 例： Sketch y = x² − 4x + 3.

例：画出 y = x² − 4x + 3 的草图。

Factorise: y = (x − 1)(x − 3), so roots are x = 1 and x = 3. The y-intercept is 3. The turning point lies halfway between the roots: x = 2. Substituting gives y = 4 − 8 + 3 = −1. So the vertex is (2,−1).

因式分解：y = (x − 1)(x − 3)，所以根为 x = 1 和 x = 3。y 轴截距为 3。顶点位于两根中点：x = 2。代入得 y = 4 − 8 + 3 = −1。所以顶点为 (2,−1)。

7. Roots and Turning Points | 根与顶点

The x-coordinate of the turning point of y = ax² + bx + c is always:

y = ax² + bx + c 的顶点的 x 坐标始终为：

x = −b / (2a)

Substitute this x-value back into the equation to find the y-coordinate. This gives the minimum point when a > 0 and the maximum point when a < 0.

将该 x 值代回原方程，即可求出 y 坐标。当 a > 0 时这是最小值点；当 a < 0 时这是最大值点。

If the quadratic is written in completed-square form y = p(x + q)² + r, the turning point is simply (−q, r). This is often the quickest way to read it off directly.

若二次式写成配方法形式 y = p(x + q)² + r，顶点就是 (−q, r)。这通常是直接读取顶点坐标的最快方法。

The sum of the roots of ax² + bx + c = 0 is −b/a, and the product of the roots is c/a. These relationships are handy for quick checks or when one root is known.

二次方程 ax² + bx + c = 0 的两根之和为 −b/a，两根之积为 c/a。这些关系可用于快速检验或当已知一个根时求另一个根。

8. Solving Quadratic Inequalities | 二次不等式

To solve a quadratic inequality such as x² − x − 6 < 0, first solve the corresponding equation x² − x − 6 = 0. Factorise to get (x − 3)(x + 2) = 0, so x = 3 or x = −2.

解二次不等式（如 x² − x − 6 < 0）时，先解对应的方程 x² − x − 6 = 0。因式分解得 (x − 3)(x + 2) = 0，所以 x = 3 或 x = −2。

Because a > 0, the parabola opens upward and is negative between the roots. Therefore:

因为 a > 0，抛物线开口向上，在两个根之间取负值。因此：

−2 < x < 3

If the inequality were x² − x − 6 > 0, the solution would be x < −2 or x > 3. Always sketch a quick graph or test intervals to avoid sign errors.

若不等式为 x² − x − 6 > 0，则解为 x < −2 或 x > 3。建议随手画个草图或代入区间检验，以避免符号错误。

9. Real-World Applications | 实际应用

Quadratic equations appear in many contextual problems, such as projectile motion, area calculations, and economics. In IGCSE exams, you must translate a word problem into a quadratic equation and then solve it.

二次方程出现在许多情境题中，例如抛体运动、面积计算和经济学问题。在 IGCSE 考试中，你需要将文字题转化为二次方程并求解。

Example | 例： A rectangle has length 3 cm more than its width. Its area is 40 cm². Find its width.

例：一个长方形的长比宽多 3 cm，面积为 40 cm²。求其宽。

Let width = w. Then length = w + 3, and:

设宽为 w，则长为 w + 3，于是：

w(w + 3) = 40 → w² + 3w − 40 = 0

Factorise: (w + 8)(w − 5) = 0, so w = −8 or w = 5. Since width cannot be negative, w = 5 cm.

因式分解：(w + 8)(w − 5) = 0，所以 w = −8 或 w = 5。由于宽度不能为负，所以 w = 5 cm。

Always reject negative or impossible solutions in context, and always state units clearly.

在情境题中一定要舍去负数或不符合实际的解，并清楚写明单位。

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

Many marks are lost in quadratic questions through avoidable errors. Always check your solutions by substituting them back into the original equation.

在二次方程题目中，很多分数因可避免的错误而丢失。务必通过代回原方程来检验你的解。

Do not forget the ± sign when taking square roots.

开平方时不要忘记 ± 符号。
When using the quadratic formula, write down a, b and c first.

使用求根公式时，先写出 a、b、c 的值。
When factorising, check by expanding your brackets.

因式分解后，通过展开括号来检验。
When solving a problem in context, reject negative lengths or times.

解情境题时，舍去负的长度或时间。
Read whether the question asks for exact values or decimal approximations. Use π or surds where required.

看清题目要求精确值还是小数近似值。必要时保留 π 或根式。

11. Practice Questions | 练习

Try these questions before checking the answers below. They cover all the main skills from this guide.

先尝试以下题目，再核对下方答案。它们覆盖了本指南的所有核心技能。

Question 1 | 第 1 题: Solve x² − 5x − 14 = 0.

第 1 题：解 x² − 5x − 14 = 0。

Question 2 | 第 2 题: Solve 2x² + 3x − 2 = 0 using the quadratic formula.

第 2 题：用求根公式解 2x² + 3x − 2 = 0。

Question 3 | 第 3 题: Find the discriminant of x² − 4x + 8, and state the number of real roots.

第 3 题：求 x² − 4x + 8 的判别式，并说明实根的个数。

Question 4 | 第 4 题: A right-angled triangle has legs of lengths x and x + 2, and hypotenuse 10. Find x.

第 4 题：一个直角三角形的两条直角边长为 x 和 x + 2，斜边为 10。求 x。

Answers | 答案:

答案：

1. (x − 7)(x + 2) = 0 → x = 7 or x = −2.

1. (x − 7)(x + 2) = 0 → x = 7 或 x = −2。

2. x = (−3 ± √(9 + 16)) / 4 = (−3 ± 5) / 4 → x = ½ or x = −2.

2. x = (−3 ± √(9 + 16)) / 4 = (−3 ± 5) / 4 → x = ½ 或 x = −2。

3. Δ = (−4)² − 4(1)(8) = 16 − 32 = −16. Since Δ < 0, there are no real roots.

3. Δ = (−4)² − 4(1)(8) = 16 − 32 = −16。由于 Δ < 0，无实根。

4. x² + (x + 2)² = 100 → 2x² + 4x + 4 = 100 → x² + 2x − 48 = 0 → (x + 8)(x − 6) = 0 → x = 6 (since x > 0).

4. x² + (x + 2)² = 100 → 2x² + 4x + 4 = 100 → x² + 2x − 48 = 0 → (x + 8)(x − 6) = 0 → x = 6（因为 x > 0）。

With regular practice, quadratic equations become one of the most reliable topics for earning full marks in your Edexcel IGCSE Mathematics exam. Remember to master factorisation first, then the formula, and always check your answers.

通过定期练习，二次方程会成为你在 Edexcel IGCSE 数学考试中最容易拿满分的主题之一。记住先掌握因式分解，再掌握求根公式，并始终检验答案。

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Quadratic Equations | 二次方程