Solving Quadratic Equations by Factorising | 因式分解解一元二次方程

📚 Solving Quadratic Equations by Factorising | 因式分解解一元二次方程

Quadratic equations appear throughout IGCSE Mathematics, from simple problem-solving questions to graph sketching and word problems. One of the most direct and powerful methods for solving a quadratic equation is factorising, which uses the zero product property to find the roots of the equation.

一元二次方程在 IGCSE 数学中无处不在，从简单的解题到函数图像绘制和文字应用题都会涉及。因式分解法是求解一元二次方程最直接、最有效的方法之一，其核心依据是零乘积性质。

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 x is 2, which is why it is called ‘quadratic’.

一元二次方程是能写成以下形式的方程：

ax² + bx + c = 0

其中 a、b、c 为常数，且 a ≠ 0。方程中 x 的最高次数为 2，因此称为“二次”方程。

For example, x² + 5x + 6 = 0 is a quadratic equation, while 2x + 3 = 0 is linear because the highest power of x is 1.

例如，x² + 5x + 6 = 0 是一元二次方程，而 2x + 3 = 0 是一次方程，因为 x 的最高次数是 1。

2. The Zero Product Property | 零乘积性质

To solve a quadratic equation by factorising, we rely on a simple but crucial rule: if the product of two expressions is zero, then at least one of the expressions must be zero. In symbols:

用因式分解法解一元二次方程，关键依赖一个简单却极其重要的规则：如果两个表达式的乘积为零，那么其中至少有一个表达式必须为零。用符号表示：

If A × B = 0, then A = 0 or B = 0

若 A × B = 0，则 A = 0 或 B = 0

This property allows us to split a factorised quadratic into two simple linear equations, each of which can be solved easily.

这一性质允许我们将分解后的二次方程拆成两个简单的一次方程，分别求解即可。

3. Solving Quadratics with a = 1 (Monic Quadratics) | 求解 a = 1 的二次方程（首项系数为1）

When the coefficient of x² is 1, we look for two numbers that multiply to give c and add to give b. This is the most common type in IGCSE.

当 x² 的系数为 1 时，我们需要找到两个数，使它们的乘积等于 c，且它们的和等于 b。这是 IGCSE 中最常见的类型。

Example 1: Solve x² + 7x + 12 = 0

例 1：解方程 x² + 7x + 12 = 0

We need two numbers whose product is 12 and whose sum is 7. The numbers are 3 and 4, because 3 × 4 = 12 and 3 + 4 = 7. Therefore:

我们需要找两个数，乘积为 12，和为 7。这两个数是 3 和 4，因为 3 × 4 = 12 且 3 + 4 = 7。因此：

(x + 3)(x + 4) = 0

Using the zero product property:

根据零乘积性质：

x + 3 = 0 or x + 4 = 0

x = −3 or x = −4

So the solution set is x = −3 or x = −4.

所以解集为 x = −3 或 x = −4。

Example 2: Solve x² − 5x + 6 = 0

例 2：解方程 x² − 5x + 6 = 0

We need two numbers whose product is 6 and whose sum is −5. Since the sum is negative, both numbers must be negative. The numbers are −2 and −3:

我们需要两个数，乘积为 6，和为 −5。因为和为负数，所以这两个数必须都是负数。它们是 −2 和 −3：

(x − 2)(x − 3) = 0

Thus x = 2 or x = 3.

因此 x = 2 或 x = 3。

4. Solving Quadratics with a ≠ 1 (Non-Monic Quadratics) | 求解 a ≠ 1 的二次方程（首项系数不为1）

When the coefficient of x² is not 1, factorising requires more care. We look for factor pairs of a × c that sum to b, then split the middle term and factor by grouping.

当 x² 的系数不等于 1 时，因式分解需要更仔细。我们要找到 a × c 的因数对，使其和为 b，然后拆分中间项并分组因式分解。

Example: Solve 2x² + 7x + 3 = 0

例：解方程 2x² + 7x + 3 = 0

Here a = 2, b = 7, c = 3. We compute a × c = 2 × 3 = 6. We need two numbers whose product is 6 and sum is 7. The numbers are 6 and 1, because 6 + 1 = 7 and 6 × 1 = 6.

这里 a = 2，b = 7，c = 3。计算 a × c = 2 × 3 = 6。我们需要两个数，乘积为 6，和为 7。这两个数是 6 和 1，因为 6 + 1 = 7 且 6 × 1 = 6。

Split the middle term 7x into 6x + 1x:

把中间项 7x 拆分为 6x + 1x：

2x² + 6x + 1x + 3 = 0

Now factor by grouping:

现在分组因式分解：

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

Notice (x + 3) is common in both terms, so:

注意 (x + 3) 在两项中都有，因此：

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

Using the zero product property:

根据零乘积性质：

2x + 1 = 0 or x + 3 = 0

Therefore:

因此：

x = −½ or x = −3

5. The Difference of Two Squares | 平方差公式

A special quadratic form that factorises very neatly is the difference of two squares:

一种可以简洁因式分解的特殊二次形式是平方差公式：

a² − b² = (a + b)(a − b)

Example: Solve x² − 9 = 0

例：解方程 x² − 9 = 0

Rewrite 9 as 3², so x² − 9 = x² − 3². Using the difference of two squares:

将 9 写成 3²，所以 x² − 9 = x² − 3²。应用平方差公式：

(x + 3)(x − 3) = 0

Therefore x = −3 or x = 3.

因此 x = −3 或 x = 3。

This method is especially useful for equations such as 4x² − 25 = 0, since 4x² = (2x)² and 25 = 5².

这种方法对于 4x² − 25 = 0 这类方程尤其有用，因为 4x² = (2x)² 且 25 = 5²。

6. Perfect Square Quadratics | 完全平方二次方程

Sometimes a quadratic factorises into two identical brackets, called a perfect square trinomial. For example:

有时二次方程会分解成两个完全相同的括号，称为完全平方三项式。例如：

x² + 6x + 9 = (x + 3)²

Example: Solve x² − 10x + 25 = 0

例：解方程 x² − 10x + 25 = 0

Here x² − 10x + 25 = (x − 5)², because (−5) + (−5) = −10 and (−5) × (−5) = 25. Therefore:

这里 x² − 10x + 25 = (x − 5)²，因为 (−5) + (−5) = −10 且 (−5) × (−5) = 25。因此：

(x − 5)² = 0

So the only solution is x = 5. This is called a repeated root.

所以唯一解为 x = 5。这称为重根。

7. Equations Requiring Rearrangement | 需要整理的一元二次方程

Not every quadratic equation is given in the standard form ax² + bx + c = 0. You may need to expand brackets, collect like terms, or move all terms to one side before factorising.

并非所有一元二次方程都以标准形式 ax² + bx + c = 0 给出。可能需要先展开括号、合并同类项，或将所有项移到等号一侧，然后再进行因式分解。

Example: Solve x² = 5x

例：解方程 x² = 5x

First bring all terms to the left-hand side:

首先将所有项移到等号左侧：

x² − 5x = 0

Now factor out x:

然后提取公因式 x：

x(x − 5) = 0

Thus x = 0 or x = 5.

因此 x = 0 或 x = 5。

Important warning: Do not divide both sides by x, because you would lose the solution x = 0.

重要提醒：切勿两边同时除以 x，否则会丢失解 x = 0。

Example 2: Solve (x − 1)(x + 2) = 4

例 2：解方程 (x − 1)(x + 2) = 4

First expand the left-hand side:

先展开左侧：

x² + 2x − x − 2 = 4

x² + x − 2 = 4

Bring 4 to the left-hand side:

将 4 移到左侧：

x² + x − 6 = 0

Factorise:

分解因式：

(x + 3)(x − 2) = 0

Thus x = −3 or x = 2.

因此 x = −3 或 x = 2。

8. Solving Quadratic Equations from Word Problems | 从文字应用题中建立并求解一元二次方程

In exam questions, quadratic equations often arise from geometric or numerical contexts. You must translate the words into an equation, then solve by factorising.

在考试中，一元二次方程通常来自几何或数字背景。你必须将文字转化为方程，然后通过因式分解求解。

Example: The area of a rectangle is 40 cm². Its length is 3 cm longer than its width. Find the width.

例：一个矩形的面积为 40 cm²。它的长比宽长 3 cm。求宽。

Let the width be x cm. Then the length is (x + 3) cm. The area gives:

设宽为 x cm，则长为 (x + 3) cm。由面积可得：

x(x + 3) = 40

Expand and rearrange:

展开并整理：

x² + 3x − 40 = 0

We need two numbers whose product is −40 and sum is 3. These are 8 and −5:

我们需要两个数，乘积为 −40，和为 3。这两个数是 8 和 −5：

(x + 8)(x − 5) = 0

So x = −8 or x = 5. Since the width cannot be negative, the width is 5 cm.

因此 x = −8 或 x = 5。宽度不能为负数，所以宽度为 5 cm。

9. Common Mistakes and How to Avoid Them | 常见错误及其避免方法

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

很多学生在解一元二次方程时因一些细小但可以避免的错误而失分。以下是最常见的错误：

Forgetting to set the equation to zero before factorising. Always rearrange so that one side equals 0.
Dividing both sides by x or by a common factor containing x. This loses solutions. Instead, factorise fully.
Sign errors when expanding brackets. Check each term carefully.
Writing the wrong signs inside brackets. Verify your factors by expanding them back.
Ignoring a repeated root. If both brackets are identical, write the single root.

忘记在因式分解前将方程化为等于零的形式。务必先整理方程使一边为 0。
两边同时除以 x 或含有 x 的公因式。这会丢失解。应完整因式分解。
展开括号时符号错误。仔细检查每一项。
括号内符号写错。通过回代展开来验证因式是否正确。
忽略重根。如果两个括号完全相同，只需写出这一个根。

10. Practice Questions | 练习题目

Try these questions yourself before checking the answers.

请先自己尝试以下题目，再对照答案。

Solve x² + 9x + 20 = 0
Solve x² − 4x − 21 = 0
Solve 3x² − 10x + 3 = 0
Solve 2x² − 8 = 0
Solve x² = 6x
The sum of a number and its square is 30. Find the possible values of the number.

解方程 x² + 9x + 20 = 0
解方程 x² − 4x − 21 = 0
解方程 3x² − 10x + 3 = 0
解方程 2x² − 8 = 0
解方程 x² = 6x
一个数与它的平方之和为 30。求这个数可能的值。

Answers | 答案

x = −4 or x = −5
x = 7 or x = −3
x = 3 or x = ⅓
x = 2 or x = −2
x = 0 or x = 6
x = 5 or x = −6 (since x² + x = 30 → x² + x − 30 = 0 → (x + 6)(x − 5) = 0)

x = −4 或 x = −5
x = 7 或 x = −3
x = 3 或 x = ⅓
x = 2 或 x = −2
x = 0 或 x = 6
x = 5 或 x = −6（因为 x² + x = 30 → x² + x − 30 = 0 → (x + 6)(x − 5) = 0）

11. Quick Reference Summary | 快速参考总结

Type \| 类型	Method \| 方法	Example \| 示例
a = 1	Find two numbers with product c and sum b	x² + 5x + 6 = (x+2)(x+3)
a ≠ 1	Split middle term using factors of a×c	2x²+7x+3 = (2x+1)(x+3)
Difference of squares	a² − b² = (a+b)(a−b)	x²−9 = (x+3)(x−3)
Perfect square	(x ± k)²	x²−10x+25 = (x−5)²

Always check your solutions by substituting them back into the original equation. This simple step catches most arithmetic errors.

始终将解代回原方程进行验证。这个简单步骤能发现大多数计算错误。

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Solving Quadratic Equations by Factorising | 因式分解解一元二次方程