Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are one of the most important topics in the Edexcel IGCSE Mathematics syllabus. They appear in algebra, coordinate geometry, and problem-solving questions. This article explains the key methods and common pitfalls to help you master them.

一元二次方程是 Edexcel IGCSE 数学大纲中最重要的内容之一。它不仅出现在代数题中，还涉及坐标几何和应用题。这篇文章将系统地讲解主要解法与常见易错点，帮助你掌握这一考点。

1. What Is a Quadratic Equation? | 什么是一元二次方程

A quadratic equation is an equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The highest power of the variable x is 2.

一元二次方程是能写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。未知数 x 的最高次数是 2。

For example, 2x² – 5x + 3 = 0 is quadratic, but x³ – 2x = 0 is not. When solving, you usually find two values of x that make the equation true.

例如，2x² – 5x + 3 = 0 是二次方程，而 x³ – 2x = 0 不是。解方程通常需要找到两个使等式成立的 x 值。

2. Solving by Factorisation | 因式分解法

Factorisation is the quickest method when the quadratic expression can be written as a product of two linear brackets. The key step is to use the zero-product property: if p × q = 0, then p = 0 or q = 0.

当二次式能写成两个一次括号相乘时，因式分解是最快的方法。关键步骤是使用零积性质：若 p × q = 0，则 p = 0 或 q = 0。

Example: Solve x² – 5x + 6 = 0. We look for two numbers that multiply to 6 and add to –5. These are –2 and –3, so the factorisation is:

例：解 x² – 5x + 6 = 0。我们要找两个数，它们相乘得 6，相加得 –5。这两个数是 –2 和 –3，因此因式分解为：

x² – 5x + 6 = (x – 2)(x – 3) = 0

Using the zero-product property, x – 2 = 0 or x – 3 = 0, so x = 2 or x = 3.

根据零积性质，x – 2 = 0 或 x – 3 = 0，所以 x = 2 或 x = 3。

3. Factorising When a ≠ 1 | 当 a ≠ 1 时的因式分解

When the coefficient of x² is not 1, you need to use a different method such as factorising by grouping. For example, solve 2x² + 7x + 3 = 0.

当 x² 的系数不是 1 时，需要使用分组分解等方法。例如，解 2x² + 7x + 3 = 0。

Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: these are 1 and 6. Then rewrite the middle term:

先计算 a × c：2 × 3 = 6。找两个数相乘得 6，相加得 7：这两个数是 1 和 6。然后改写中间项：

2x² + 7x + 3 = 2x² + x + 6x + 3

Group and factorise:

分组并因式分解：

x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)

So (2x + 1)(x + 3) = 0, which gives x = –½ or x = –3.

因此 (2x + 1)(x + 3) = 0，解得 x = –½ 或 x = –3。

4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This method is especially useful for solving equations that do not factorise easily.

配方法将二次式改写为 a(x + p)² + q 的形式。这种方法特别适用于不容易因式分解的方程。

Example: Solve x² + 6x + 2 = 0. Start by isolating the constant:

例：解 x² + 6x + 2 = 0。先移常数项：

x² + 6x = –2

Take half of 6, square it: (6/2)² = 9. Add 9 to both sides:

取 6 的一半并平方：(6/2)² = 9。两边加 9：

(x + 3)² = 7

Take square roots: x + 3 = ±√7, so x = –3 ± √7.

开平方根：x + 3 = ±√7，所以 x = –3 ± √7。

5. The Quadratic Formula | 二次公式法

For any quadratic equation ax² + bx + c = 0, the solutions are given by the quadratic formula:

对于任意一元二次方程 ax² + bx + c = 0，其解由二次公式给出：

x = [–b ± √(b² – 4ac)] / (2a)

This formula works for all quadratics, including those with irrational or complex roots, though complex roots are not required at IGCSE. You should memorise it carefully.

该公式适用于所有二次方程，包括含有无理根的情况（IGCSE 不要求复数根）。请务必牢记这个公式。

Example: Solve 3x² – 5x – 2 = 0. Here a = 3, b = –5, c = –2. Substitute:

例：解 3x² – 5x – 2 = 0。这里 a = 3，b = –5，c = –2。代入：

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

So x = 2 or x = –⅓.

因此 x = 2 或 x = –⅓。

6. The Discriminant | 判别式

The expression b² – 4ac is called the discriminant. It tells you the number and type of roots without solving the equation.

代数式 b² – 4ac 称为判别式。它能在不解方程的情况下告诉我们根的数量和类型。

If b² – 4ac > 0, there are two distinct real roots.
如果 b² – 4ac > 0，方程有两个不相等的实数根。
If b² – 4ac = 0, there is exactly one repeated real root.
如果 b² – 4ac = 0，方程有两个相等的实数根（即一个重根）。
If b² – 4ac < 0, there are no real roots.
如果 b² – 4ac < 0，方程没有实数根。

For example, the equation x² – 4x + 4 = 0 has discriminant 16 – 16 = 0, so it has one repeated root (x = 2).

例如，方程 x² – 4x + 4 = 0 的判别式为 16 – 16 = 0，因此它有一个重根（x = 2）。

7. Solving Word Problems | 解应用题

Quadratic equations often arise from geometric or numeric word problems. The key is to define a variable, form an equation, solve it, and check which answers make sense in the context.

一元二次方程经常出现在几何或数字应用题中。关键是设定变量、列出方程、解方程，并检查哪些答案在题目情境中合理。

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

例：一个长方形的长比宽多 3 cm，面积为 40 cm²。求宽。

Let the width be x. Then length is x + 3. Area gives:

设宽为 x，则长为 x + 3。面积方程为：

x(x + 3) = 40 ⇒ x² + 3x – 40 = 0

Factorise: (x + 8)(x – 5) = 0, so x = –8 or x = 5. Since width cannot be negative, the width is 5 cm.

因式分解得 (x + 8)(x – 5) = 0，所以 x = –8 或 x = 5。因为宽不能为负数，所以宽为 5 cm。

8. Sketching Quadratic Graphs | 二次函数图像

The graph of y = ax² + bx + c is a parabola. Its shape depends on the sign of a:

y = ax² + bx + c 的图像是一条抛物线。其开口方向取决于 a 的符号：

If a > 0, the parabola opens upwards (U-shaped).
若 a > 0，抛物线开口向上（U 形）。
If a < 0, it opens downwards (∩-shaped).
若 a < 0，抛物线开口向下（∩ 形）。

To sketch the graph, you need the roots (where y = 0), the y-intercept (c), and the turning point. The turning point’s x-coordinate is given by:

画草图时，需要知道根（y = 0 的点）、y 轴截距（c）以及顶点。顶点的 x 坐标为：

x = –b / (2a)

Substitute this value back into the equation to find the y-coordinate of the turning point.

将这个值代回原方程，即可得到顶点的 y 坐标。

9. Solving Quadratic Inequalities | 解一元二次不等式

Quadratic inequalities are solved by first finding the roots of the corresponding equation, then using a number line or the graph of the parabola to decide the solution interval.

解一元二次不等式时，先求对应方程的根，再借助数轴或抛物线图像确定解区间。

Example: Solve x² – x – 6 > 0. First factorise: (x – 3)(x + 2) > 0. The roots are x = 3 and x = –2.

例：解 x² – x – 6 > 0。先因式分解：(x – 3)(x + 2) > 0。根为 x = 3 和 x = –2。

Since the parabola opens upwards, the expression is positive outside the interval between the roots:

因为抛物线开口向上，所以表达式在两根之外为正：

x < –2 or x > 3

Remember to use strict inequalities if the original inequality is > or <; use closed intervals for ≥ or ≤.

注意：若原不等式为 > 或 <，则用严格不等号；若为 ≥ 或 ≤，则用闭区间。

10. Common Mistakes to Avoid | 常见错误提醒

Many students lose marks on quadratic equations due to avoidable errors. Here are the most common ones:

很多学生在解一元二次方程时因为可以避免的错误而失分。以下是最常见的几点：

Forgetting to rearrange the equation to the form ax² + bx + c = 0 before factorising or using the formula.
在因式分解或套用公式前，忘记将方程整理为 ax² + bx + c = 0 的形式。
Making sign errors when substituting into the quadratic formula, especially with negative b.
代入二次公式时出现符号错误，尤其当 b 为负数时。
Missing a solution when taking square roots – remember to include both ±.
开平方根时漏掉一个解——记得要取 ± 两个值。
Checking the discriminant too late; always verify that the quadratic has real roots before spending time solving.
太晚检查判别式；应在花时间求解前先确认方程是否有实数根。

11. Worked Exam-Style Question | 典型考试题精讲

Let us solve a typical Edexcel-style question step by step.

下面我们逐步解一道典型的 Edexcel 风格题目。

Question: Solve the equation 2x² – 3x – 5 = 0 by completing the square.

题目：用配方法解方程 2x² – 3x – 5 = 0。

First divide by 2: x² – (3/2)x – 5/2 = 0. Then add 5/2 to both sides:

首先两边除以 2：x² – (3/2)x – 5/2 = 0。然后把 5/2 移到右边：

x² – (3/2)x = 5/2

Half of 3/2 is 3/4; square it: 9/16. Add 9/16 to both sides:

3/2 的一半是 3/4；平方得 9/16。两边加 9/16：

(x – 3/4)² = 5/2 + 9/16 = 40/16 + 9/16 = 49/16

Take square roots: x – 3/4 = ±7/4. Thus x = 3/4 ± 7/4, giving x = 10/4 = 5/2 or x = –4/4 = –1.

开平方根：x – 3/4 = ±7/4。所以 x = 3/4 ± 7/4，得 x = 10/4 = 5/2 或 x = –4/4 = –1。

12. Summary and Revision Tips | 总结与复习建议

Make sure you can switch between all three solving methods fluently. Factorisation is fastest, the formula is reliable, and completing the square is useful for deriving the vertex form.

请确保你能熟练掌握三种解法之间的转换。因式分解最快，二次公式最可靠，配方法则有助于推导顶点式。

Practice with a variety of questions, including word problems and inequalities. The more you practise, the faster you will recognise which method to use.

多练习不同类型的题目，包括应用题和不等式。练习越多，你就能越快识别出该用哪种方法。

Finally, always check your answers by substituting them back into the original equation.

最后，记得将答案代回原方程进行检验。

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Solving Quadratic Equations | 解一元二次方程