📚 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。
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
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）
or 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
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。
od 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
u 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。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图