Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are a central topic in Edexcel IGCSE Mathematics. You will meet them in algebra, graphs, geometry and word problems. This revision guide explains the main methods for solving quadratic equations, with clear worked examples and exam-style advice.

一元二次方程是 Edexcel IGCSE 数学的核心内容。你会在代数、图像、几何和应用题中遇到它们。本复习指南详细讲解解一元二次方程的主要方法，并配有清晰的例题和考试建议。

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

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

一元二次方程可以写成 ax² + bx + c = 0 的形式，其中 a、b、c 是常数，并且 a ≠ 0。x 的最高次数是 2，因此称为“二次”。

For example, 2x² − 5x − 3 = 0 is a quadratic equation because it has an x² term, an x term and a constant term.

例如，2x² − 5x − 3 = 0 是一元二次方程，因为它同时含有 x² 项、x 项和常数项。

In the Edexcel IGCSE course, you are expected to solve quadratic equations using factorisation, completing the square and the quadratic formula. You should also understand the discriminant and how to apply these methods to real-world problems.

在 Edexcel IGCSE 课程中，你需要掌握用因式分解、配方法和求根公式来解一元二次方程，同时理解判别式，并能够将这些方法用于解决实际问题。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic expression can be written as a product of two brackets. The key idea is that if two factors multiply to give zero, then at least one of them must be zero.

当二次表达式可以写成两个括号相乘时，因式分解往往是最快的方法。核心思想是：如果两个因式相乘等于零，那么至少其中一个因式必为零。

Follow these steps:

按照以下步骤操作：

Rearrange the equation so that one side is zero.

重新整理方程，使一边等于 0。
Factorise the quadratic expression completely.

将二次表达式完全因式分解。
Set each factor equal to zero.

令每个因式分别等于 0。
Solve the resulting linear equations.

解所得的一次方程。

Example: Solve x² − 7x + 10 = 0.

例：解方程 x² − 7x + 10 = 0。

x² − 7x + 10 = (x − 2)(x − 5) = 0

Therefore x − 2 = 0 or x − 5 = 0, so x = 2 or x = 5.

因此 x − 2 = 0 或 x − 5 = 0，所以 x = 2 或 x = 5。

Example: Solve 2x² + 7x + 3 = 0.

例：解方程 2x² + 7x + 3 = 0。

2x² + 7x + 3 = (2x + 1)(x + 3) = 0

So 2x + 1 = 0 or x + 3 = 0, giving x = −1/2 or x = −3.

因此 2x + 1 = 0 或 x + 3 = 0，得到 x = −1/2 或 x = −3。

3. Special Case: Difference of Two Squares | 特殊情形：平方差

When a quadratic has the form x² − a², it can be factorised using the difference of two squares:

当二次方程具有 x² − a² 的形式时，可以利用平方差公式因式分解：

x² − a² = (x − a)(x + a)

Example: Solve x² − 16 = 0.

例：解方程 x² − 16 = 0。

x² − 16 = (x − 4)(x + 4) = 0

So x = 4 or x = −4.

所以 x = 4 或 x = −4。

This also works when the x² term has a coefficient. For example, 4x² − 25 = (2x − 5)(2x + 5).

当 x² 项有系数时也同样适用。例如，4x² − 25 = (2x − 5)(2x + 5)。

Learn to spot this pattern because it saves time in exams.

在考试中要学会快速识别这种形式，因为它可以节省时间。

4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form (x + p)² + q. This is useful when factorisation is difficult and when you need to find maximum or minimum points.

配方法将二次表达式改写为 (x + p)² + q 的形式。当因式分解困难，或者需要求最大值、最小值时，配方法非常有用。

For x² + bx + c = 0, use the identity:

对于 x² + bx + c = 0，使用恒等式：

x² + bx + c = (x + b/2)² − (b/2)² + c

Example: Solve x² + 6x + 1 = 0 by completing the square.

例：用配方法解方程 x² + 6x + 1 = 0。

(x + 3)² − 9 + 1 = 0

(x + 3)² − 8 = 0

So (x + 3)² = 8, which gives x + 3 = ±√8, so x = −3 ± √8. Since √8 = 2√2, we can write x = −3 ± 2√2.

因此 (x + 3)² = 8，于是 x + 3 = ±√8，所以 x = −3 ± √8。由于 √8 = 2√2，可以写成 x = −3 ± 2√2。

If the coefficient of x² is not 1, divide the whole equation by that coefficient before completing the square.

如果 x² 的系数不是 1，需要先将方程两边同时除以该系数，再进行配方。

5. The Quadratic Formula | 二次求根公式

The quadratic formula can solve any quadratic equation, even when factorisation is not easy. For a quadratic equation ax² + bx + c = 0, the solutions are:

求根公式可以解任意一元二次方程，即使因式分解不方便时也能使用。对于一元二次方程 ax² + bx + c = 0，解为：

x = (−b ± √(b² − 4ac)) / (2a)

Example: Solve 2x² + 3x − 2 = 0 using the quadratic formula.

例：用求根公式解方程 2x² + 3x − 2 = 0。

Here a = 2, b = 3 and c = −2. Substitute into the formula:

这里 a = 2，b = 3，c = −2。代入公式：

x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2)

x = (−3 ± √(9 + 16)) / 4 = (−3 ± √25) / 4

So x = (−3 + 5) / 4 = 1/2 or x = (−3 − 5) / 4 = −2.

所以 x = (−3 + 5) / 4 = 1/2 或 x = (−3 − 5) / 4 = −2。

Always check whether your answer can be simplified. If the value under the square root is a perfect square, the quadratic might have factorised easily.

检查答案能否化简。如果根号内的值是一个完全平方数，那么这个二次方程通常也可以用因式分解法求解。

6. The Discriminant | 判别式

The expression b² − 4ac inside the quadratic formula is called the discriminant. It tells us how many real roots a quadratic equation has.

求根公式中的 b² − 4ac 部分称为判别式。它告诉我们一元二次方程有多少个实数根。

Δ = b² − 4ac

The table below summarises the three cases:

下表总结了三种情况：

Discriminant 判别式	Number of Real Roots 实数根个数
Δ > 0	Two distinct real roots 两个不相等的实数根
Δ = 0	One repeated real root 一个相等的实数根
Δ < 0	No real roots 没有实数根

Example: For x² − 4x + 4 = 0, a = 1, b = −4 and c = 4.

例：对于 x² − 4x + 4 = 0，a = 1，b = −4，c = 4。

Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0

Since Δ = 0, the equation has exactly one repeated root. Indeed, x² − 4x + 4 = (x − 2)², so x = 2.

因为 Δ = 0，所以方程有一个相等的实数根。事实上，x² − 4x + 4 = (x − 2)²，所以 x = 2。

Example: For x² + 2x + 5 = 0, Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16, so there are no real roots.

例：对于 x² + 2x + 5 = 0，Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16，因此没有实数根。

7. Solving Word Problems | 应用题

Many exam questions describe a situation that can be modelled with a quadratic equation. The process is:

许多考试题目会描述一个可以用二次方程建模的实际情境。解题过程是：

Define the unknown variable clearly.

明确设出未知数。
Write an equation using the information given.

根据已知信息写出方程。
Solve the equation using a suitable method.

选择合适的方法解方程。
Check whether both answers make sense in the context.

结合实际问题判断两个答案是否都有意义。

Example: A rectangle is 3 cm longer than it is wide. Its area is 28 cm². Find its width.

例：一个长方形的长比宽多 3 cm，面积为 28 cm²。求它的宽。

Let the width be x cm. Then the length is (x + 3) cm, so:

设宽为 x cm，则长为 (x + 3) cm，因此：

x(x + 3) = 28

x² + 3x − 28 = 0

(x + 7)(x − 4) = 0

So x = −7 or x = 4. A width cannot be negative, so the width is 4 cm and the length is 7 cm.

所以 x = −7 或 x = 4。宽度不能为负数，因此宽为 4 cm，长为 7 cm。

Always reject negative answers when the variable represents a length, time or quantity that cannot be negative.

当未知数表示长度、时间或不能为负的数量时，一定要舍去负值。

8. Choosing the Best Method | 方法选择

Different methods are useful in different situations. The table below can help you decide quickly in an exam.

不同方法适用于不同情况。下表可以帮助你在考试中快速做出选择。

Method 方法	When to Use 使用时机
Factorisation 因式分解	When the expression factorises quickly, especially if a = 1. 当表达式很容易因式分解，尤其当 a = 1 时。
Difference of two squares 平方差	When the quadratic has only x² and a constant, with no x term. 当二次方程只含 x² 项和常数项，没有 x 项时。
Completing the square 配方法	When you need the turning point, or when a = 1 and b is even. 当需要求顶点，或当 a = 1 且 b 为偶数时。
Quadratic formula 求根公式	When the expression is difficult to factorise, or when the question asks for answers to 2 decimal places. 当表达式难以因式分解，或题目要求答案保留两位小数时。

The quadratic formula always works, but it takes longer and gives more opportunity for arithmetic mistakes. Factorisation is quicker when the expression is simple.

求根公式总是有效，但计算更慢，也更容易出现算术错误。当表达式简单时，因式分解更快。

9. Common Mistakes and Tips | 常见错误与提示

Many students lose marks for avoidable errors. Here are the most common mistakes and how to avoid them.

许多学生因为可避免的错误而失分。以下是最常见的错误以及避免方法。

Forgetting to rearrange the equation so that one side is 0 before factorising.

在因式分解之前，忘记将方程整理成一侧为 0。
Getting signs wrong when factorising, for example writing (x − 3)(x + 4) instead of (x + 3)(x − 4).

因式分解时符号出错，例如将 (x + 3)(x − 4) 错写成 (x − 3)(x + 4)。
Forgetting that a quadratic equation can have two solutions, so always include ± when using the formula or completing the square.

忘记二次方程可能有两个解，因此在使用求根公式或配方法时，不要漏写 ±。
Dividing both sides by x when x is unknown, which loses one of the two roots.

当 x 是未知数时，两边同时除以 x，这样会丢失其中一个根。
Using the quadratic formula but forgetting that −b is also negative when b is negative.

使用求根公式时，忘记当 b 为负数时 −b 也是负数。

Tip: Always check your solutions by substituting them back into the original equation.

提示：将解代回原方程进行检验是一个好习惯。

10. Practice Questions | 练习

Try these questions on your own before looking at the answers.

请先独立完成以下练习，再查看答案。

Solve x² − 9x + 20 = 0.

解方程 x² − 9x + 20 = 0。
Solve 2x² − 7x + 3 = 0.

解方程 2x² − 7x + 3 = 0。
Solve x² − 8x + 1 = 0 by completing the square.

用配方法解方程 x² − 8x + 1 = 0。
Solve 9x² − 16 = 0.

解方程 9x² − 16 = 0。
A right-angled triangle has hypotenuse 10 cm. One leg is x cm and the other leg is (x + 2) cm. Find x.

一个直角三角形斜边为 10 cm，一条直角边为 x cm，另一条直角边为 (x + 2) cm。求 x。

Answers:

答案：

x = 4 or x = 5.

x = 4 或 x = 5。
x = 1/2 or x = 3.

x = 1/2 或 x = 3。
(x −

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Solving Quadratic Equations | 解一元二次方程