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Solving Quadratic Equations | 解二次方程

2026-08-30 09:56 · 本文部分内容由AI辅助创作

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper. Mastering this topic requires a clear understanding of algebraic methods, the discriminant, and graphical interpretation. This revision guide will walk you through every essential skill step by step.

二次方程几乎是每一份 IGCSE 数学试卷中都会出现的考点。掌握这一主题,需要清楚理解代数方法、判别式以及图像意义。本复习指南将一步步带你掌握每一个核心技巧。


1. Introduction to Quadratic Equations | 二次方程简介

A quadratic equation is an equation in which the highest power of the unknown is 2. Its general form is shown below.

二次方程是未知数最高次数为 2 的方程,其一般形式如下。

ax² + bx + c = 0, where a ≠ 0

Here, a is the coefficient of x², b is the coefficient of x, and c is a constant term. Every quadratic equation has up to two solutions, which are called roots.

其中 a 是 x² 的系数,b 是 x 的系数,c 是常数项。每个二次方程最多有两个解,称为根。


2. Standard Form and Coefficients | 标准形式与系数

ax² + bx + c = 0. This makes it easy to identify coefficients and choose the best method.

在解任何二次方程之前,一定要先将它整理成标准形式 ax² + bx + c = 0。这样可以方便地确定各项系数,并选择最适合的方法。

EquationRearranged formabc
2x² + 3x = 52x² + 3x − 5 = 023−5
x² = 4 − 2xx² + 2x − 4 = 012−4

Notice that the sign of each term is included. A negative constant is written with the minus sign.

注意每一项的正负号都要保留。当常数项为负数时,要写上负号。


3. Solving by Factorisation | 因式分解法

Factorisation is the quickest method when the quadratic has simple integer factors. The key principle is the zero product property: if the product of two factors is 0, then at least one factor must be 0.

当二次式具有简单的整数因式时,因式分解是最快捷的方法。其核心原理是零乘积性质:如果两个因式的乘积为 0,那么至少有一个因式必须为 0。

Example: solve x² − 5x + 6 = 0.

例:解方程 x² − 5x + 6 = 0。

(x − 2)(x − 3) = 0

Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

因此 x − 2 = 0 或 x − 3 = 0,所以 x = 2 或 x = 3。

Always expand your answer to check that the factors produce the original equation.

求解后一定要展开因式来检查是否与原方程一致。


4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This method is especially useful when the quadratic cannot be factorised easily.

配方法将二次式改写为 (x + p)² + q 的形式。当二次式不易因式分解时,这种方法特别有用。

Example: solve x² + 6x + 2 = 0.

例:解方程 x² + 6x + 2 = 0。

First, halve the coefficient of x: 6 ÷ 2 = 3. Then write:

首先,将 x 的系数取半:6 ÷ 2 = 3。然后写出:

(x + 3)² − 9 + 2 = 0

(x + 3)² = 7

Taking square roots gives x + 3 = ±√7, so x = −3 ± √7.

两边开平方得 x + 3 = ±√7,因此 x = −3 ± √7。

Remember that the square root of a positive number has two values: one positive and one negative.

记住,一个正数的平方根有两个值:一个正,一个负。


5. The Quadratic Formula | 求根公式

The quadratic formula works for every quadratic equation, including those with irrational or no real roots. It is derived from completing the square.

求根公式适用于所有二次方程,包括具有无理根或无实根的方程。它由配方法推导而来。

x = (−b ± √b² − 4ac) / (2a)

For example, solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3.

例如,解方程 2x² − 4x − 3 = 0。这里 a = 2,b = −4,c = −3。

Substitute carefully:

代入时要小心:

x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (2 ± √10) / 2

Always simplify the square root and the fraction completely in your final answer.

最终答案中一定要化简平方根和分数。


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6. The Discriminant | 判别式

The expression b² − 4ac inside the quadratic formula is called the discriminant. It tells us the number and type of roots without solving the equation.

求根公式中的 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。

This is a favourite exam topic, so practise identifying the discriminant quickly.

这是考试中非常常见的考点,所以要多练习快速计算判别式。


7. Roots and Coefficients | 根与系数的关系

with roots α and β, the sum and product of the roots relate directly to the coefficients.

对于方程 ax² + bx + c = 0,若其根为 α 和 β,则根的和与积与系数有直接关系。

Sum of roots: α + β = −b/a

Product of roots: αβ = c/a

These relationships let you find the roots when one root is known, or form a new quadratic with given roots.

这些关系可以帮助你在已知一个根时求出另一个根,或者根据给定的根构造一个新的二次方程。


8. Solving Word Problems | 应用题

Quadratic equations often model areas, projectile motion, and number problems. The first step is to translate the words into an equation.

二次方程常用来建立面积、抛体运动以及数字问题的数学模型。第一步是将文字转化为方程。

Example: A rectangle has length 3 cm longer than its width. Its area is 28 cm². Find the width.

例:一个矩形的长比宽多 3 cm,面积为 28 cm²。求宽。

Let width = x. Then length = x + 3, so x(x + 3) = 28.

设宽为 x,则长为 x + 3,所以 x(x + 3) = 28。

x² + 3x − 28 = 0 → (x + 7)(x − 4) = 0 → x = 4

not be negative. Always check whether every algebraic solution makes sense in the context.

我们舍去 x = −7,因为长度不能为负。一定要检查每个代数解是否符合实际意义。


9. Graphical Interpretation | 图像意义

. The roots of the equation are the x-coordinates of the points where the parabola crosses the x-axis.

二次函数 y = ax² + bx + c 的图像是抛物线。方程 ax² + bx + c = 0 的根就是抛物线与 x 轴交点的 x 坐标。

Sketch the graph whenever possible to check that your roots are reasonable.

只要可能就画出草图,以检查你的根是否合理。


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

Many students lose marks on quadratic equations because of small but avoidable errors.

许多学生在二次方程上失分,是因为一些细小但可以避免的错误。


11. Practice Questions | 练习

Test yourself on these typical IGCSE questions.

请完成下面这些典型的 IGCSE 练习题。

  1. Solve x² − 3x − 10 = 0 by factorisation.

    用因式分解法解 x² − 3x − 10 = 0。

  2. + 5x − 1 = 0 using the quadratic formula, giving your answers correct to 2 decimal places.

    用求根公式解 2x² + 5x − 1 = 0,答案精确到两位小数。

  3. Find the values of k such that the equation kx² + 4x + 1 = 0 has two distinct real roots.

    求使得方程 kx² + 4x + 1 = 0 有两个不同实数根时的 k 值范围。

Write your working out step by step, then check your answers using substitution.

写出逐步计算过程,然后用代入法检查你的答案。


Published by TutorHao | Mathematics Revision Series | aleveler.com

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