Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in almost every IGCSE Mathematics exam. They test your algebraic manipulation, logical thinking, and ability to apply abstract methods to real-world problems. In this article, you will learn the essential techniques for solving quadratic equations, with worked examples and common pitfalls to avoid.

二次方程几乎出现在每一场 IGCSE 数学考试中。它考查你的代数运算能力、逻辑思维，以及将抽象方法应用于实际问题的能力。本文将帮助你掌握解二次方程的核心技巧，并提供经典例题与常见易错点分析。

1. Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a ≠ 0, and a, b, c are real numbers. The highest power of x is 2, which is why it is called “quadratic”.

二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a ≠ 0，a、b、c 为实数。未知数 x 的最高次数为 2，因此称为“二次”方程。

If a = 0, the equation becomes linear. | 如果 a = 0，方程就变成了一次方程。
Every quadratic equation has at most two real solutions. | 每个二次方程至多有两个实数解。

ax² + bx + c = 0

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic has simple integer factors. You rewrite the equation as a product of two linear expressions, then set each factor equal to zero.

当二次方程有简单的整数因式时，因式分解法通常是最快捷的方法。你将方程改写为两个一次表达式的乘积，然后令每个因式等于零。

Example: Solve x² – 5x + 6 = 0.

示例：解 x² – 5x + 6 = 0。

We look for two numbers that multiply to 6 and add to -5. These are -2 and -3. Thus (x – 2)(x – 3) = 0, so x = 2 or x = 3.

我们需要找到两个数，它们相乘得 6，相加得 -5。这两个数是 -2 和 -3。因此 (x – 2)(x – 3) = 0，所以 x = 2 或 x = 3。

(x – 2)(x – 3) = 0 ⇒ x = 2 or x = 3

3. Completing the Square | 配方法

Completing the square rewrites the equation in the form (x + p)² = q. This method is particularly useful when the equation cannot be factorised easily, and it also helps in finding the vertex of a parabola.

配方法将方程改写为 (x + p)² = q 的形式。当方程不易因式分解时，此方法非常有用，同时也有助于求抛物线的顶点。

Example: Solve x² + 6x – 7 = 0 by completing the square.

示例：用配方法解 x² + 6x – 7 = 0。

Move constant: x² + 6x = 7. | 移常数项：x² + 6x = 7。
Add (6/2)² = 9 to both sides: x² + 6x + 9 = 16. | 两边加上 (6/2)² = 9：x² + 6x + 9 = 16。
Write as square: (x + 3)² = 16. | 写成完全平方：(x + 3)² = 16。
Solve: x + 3 = ±4 ⇒ x = 1 or x = -7. | 解得：x + 3 = ±4 ⇒ x = 1 或 x = -7。

(x + 3)² = 16 ⇒ x = 1 or x = -7

4. The Quadratic Formula | 求根公式

The quadratic formula works for every quadratic equation. It is derived from completing the square and is expressed as:

求根公式适用于所有二次方程。它由配方法推导而来，表达为：

x = (-b ± √(b² – 4ac)) / 2a

Given ax² + bx + c = 0, you substitute the values of a, b and c directly into the formula.

对于 ax² + bx + c = 0，你只需将 a、b、c 的值直接代入公式。

Example: Solve 2x² + 3x – 5 = 0 using the formula.

示例：用公式解 2x² + 3x – 5 = 0。

Here a = 2, b = 3, c = -5. Then b² – 4ac = 9 + 40 = 49, √49 = 7. So x = (-3 ± 7) / 4, giving x = 1 or x = -2.5.

此时 a = 2，b = 3，c = -5。则 b² – 4ac = 9 + 40 = 49，√49 = 7。所以 x = (-3 ± 7) / 4，得到 x = 1 或 x = -2.5。

5. The Discriminant | 判别式

The expression b² – 4ac is called the discriminant. It tells you how many real solutions a quadratic equation has without solving it.

表达式 b² – 4ac 被称为判别式。它无需解方程即可判断二次方程有多少个实数解。

Discriminant Δ = b² – 4ac	Number of real solutions
Δ > 0	Two distinct real solutions
Δ = 0	One repeated real solution
Δ < 0	No real solutions

中文对照：Δ > 0 有两个不同实数解；Δ = 0 有一个重根；Δ < 0 没有实数解。

6. Solving by Graphing | 图像法

To solve a quadratic equation graphically, you draw the curve y = ax² + bx + c and find the x-coordinates where the curve crosses the x-axis. These are the roots.

用图像法解二次方程时，先画出曲线 y = ax² + bx + c，然后找出曲线与 x 轴交点的横坐标。这些横坐标就是方程的根。

If the curve touches the x-axis at one point, there is one repeated root. If it does not touch the x-axis, there are no real roots.

如果曲线与 x 轴相切于一点，则有一个重根；如果曲线不与 x 轴相交，则没有实数根。

This method is useful for estimating solutions, but in exams you usually need exact answers obtained by algebraic methods.

这种方法适用于估算解，但在考试中通常需要用代数方法求出精确答案。

7. Word Problems Involving Quadratics | 二次方程应用题

Many exam problems describe a real-life situation that leads to a quadratic equation. You must define the unknown, set up the equation, solve it, and then interpret both solutions in the context of the problem.

许多考试题目描述一个实际情境，最终会得到一个二次方程。你需要设未知数、建立方程、求解，并在问题情境中解释两个解的含义。

Example: A rectangle has length (x + 3) cm and width x cm. Its area is 70 cm². Find x.

示例：一个长方形的长为 (x + 3) 厘米，宽为 x 厘米。面积为 70 平方厘米。求 x。

Setup: x(x + 3) = 70 ⇒ x² + 3x – 70 = 0. Factorise: (x + 10)(x – 7) = 0, so x = -10 or x = 7. Since a length cannot be negative, x = 7 cm.

建立方程：x(x + 3) = 70 ⇒ x² + 3x – 70 = 0。因式分解：(x + 10)(x – 7) = 0，所以 x = -10 或 x = 7。由于长度不能为负，因此 x = 7 厘米。

8. Quadratic Functions and Their Graphs | 二次函数及其图像

A quadratic function y = ax² + bx + c always produces a parabola. The sign of a determines whether it opens upward (a > 0) or downward (a < 0).

二次函数 y = ax² + bx + c 的图像总是抛物线。a 的正负决定开口方向：a > 0 时开口向上，a < 0 时开口向下。

The vertex lies on the axis of symmetry x = -b/(2a). | 顶点位于对称轴 x = -b/(2a) 上。
The y-intercept is c. | y 轴截距为 c。
The x-intercepts are the real roots of the equation. | x 轴截距是方程的实数根。

Axis of symmetry: x = -b / (2a)

9. Finding the Vertex by Completing the Square | 用配方法求顶点

When a quadratic function is written as y = a(x – h)² + k, the vertex is simply (h, k). This form is called the vertex form.

当二次函数写成 y = a(x – h)² + k 时，顶点就是 (h, k)。这种形式称为顶点式。

Example: Express y = x² – 4x + 1 in vertex form.

示例：将 y = x² – 4x + 1 化为顶点式。

Complete the square: y = (x² – 4x + 4) – 4 + 1 = (x – 2)² – 3. So the vertex is (2, -3).

配方：y = (x² – 4x + 4) – 4 + 1 = (x – 2)² – 3。因此顶点为 (2, -3)。

This method is especially important for questions about maximum or minimum values of a quadratic function.

这种方法在求二次函数最大值或最小值的问题中尤其重要。

10. Solving Quadratic Inequalities | 二次不等式

Quadratic inequalities may also appear, such as x² – 5x + 6 > 0. The approach is to first solve the corresponding equation, then test intervals on a number line.

考试中也可能出现二次不等式，例如 x² – 5x + 6 > 0。解法是先求对应方程的根，然后在数轴上检验区间。

For x² – 5x + 6 = 0, roots are x = 2 and x = 3. The parabola opens upward, so the inequality is positive when x < 2 or x > 3.

对于 x² – 5x + 6 = 0，根为 x = 2 和 x = 3。抛物线开口向上，因此不等式大于零在 x < 2 或 x > 3 时成立。

x² – 5x + 6 > 0 ⇒ x < 2 or x > 3

11. Common Mistakes | 常见易错点

Students often lose marks on quadratic equation questions because of small but critical errors. Here are the most frequent ones you should avoid.

学生在解二次方程时常常因为一些微小但关键的失误而丢分。以下是最常见的错误，请务必避免。

Forgetting to rearrange into standard form before factorising or using the formula. | 在因式分解或使用公式前忘记将方程化为标准形式。
Miscopying signs: always check whether b or c is negative. | 抄错符号：务必确认 b 或 c 是否为负。
Dividing both sides by x, which loses the solution x = 0. | 两边同时除以 x，导致丢失解 x = 0。
Using the discriminant incorrectly: remember Δ = b² – 4ac, not b² – 4a. | 错误使用判别式：请记住 Δ = b² – 4ac，而不是 b² – 4a。
In word problems, accepting a negative solution without checking whether it makes sense. | 在应用题中，不检查负解是否合理就直接接受。

12. Exam Strategies and Practice | 考试策略与练习

In the Edexcel IGCSE exam, quadratic questions appear in both Paper 1 and Paper 2. Always show your working clearly, because method marks are awarded even if your final answer is wrong.

在 Edexcel IGCSE 考试中，二次方程题目在 Paper 1 和 Paper 2 中都会出现。请务必清晰写出解题过程，因为即使最终答案错误，步骤分仍然可得。

Recommended approach:

推荐方法：

Write the equation in standard form. | 将方程写成标准形式。
Try factorisation first. If it is easy, solve immediately. | 先尝试因式分解；如果容易，直接求解。
If factorisation is not obvious, use completing the square or the quadratic formula. | 如果因式分解不明显，使用配方法或求根公式。
Use the discriminant to check the number of solutions before solving. | 求解前用判别式检查解的个数。
For word problems, interpret both algebraic answers and reject any impossible values. | 对于应用题，解释两个代数解并排除不可能的值。

Try the following practice question: Solve 3x² – 2x – 1 = 0 by any method. Answer: x = 1 or x = -1/3.

试试下面的练习题：用任何方法解 3x² – 2x – 1 = 0。答案：x = 1 或 x = -1/3。

Mastering quadratics takes regular practice. Use past papers and check every step carefully. You will soon find these questions both predictable and profitable.

掌握二次方程需要定期练习。使用历年真题并仔细检查每一步。你很快会发现这类题目既有规律又容易得分。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Quadratic Equations | 解二次方程