📚 Quadratic Equations: Solving by Factorisation | 二次方程：因式分解法求解
In IGCSE Mathematics, quadratic equations appear in algebra, problem solving, and graph questions. This article explains how to solve them by factorising and using the zero product property. You will learn standard form, worked examples, special cases, and common mistakes to avoid.
在 IGCSE 数学中，二次方程出现在代数、应用题和图像问题中。本文讲解如何通过因式分解和零积性质求解二次方程。你将学习标准形式、典型例题、特殊情形以及需要避免的常见错误。
1. What is a Quadratic Equation? | 什么是二次方程？
0. The highest power of x is 2, which is why the graph of y = ax² + bx + c is a parabola.
二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。x 的最高次数是 2，因此 y = ax² + bx + c 的图像是一条抛物线。
In IGCSE exams, you may be asked to solve a quadratic by factorisation, by completing the square, by using the quadratic formula, or by reading roots from a graph.
在 IGCSE 考试中，你可能需要用因式分解法、配方法、求根公式或通过图像读取根来解二次方程。
ax² + bx + c = 0, a ≠ 0
2. Standard Form and the Zero Product Property | 标准形式与零积性质
d form. For example, x² + 5x = -6 must become x² + 5x + 6 = 0 before you try to factorise.
因式分解之前，一定要把方程改写为一边为零的形式，这称为标准形式。例如，x² + 5x = -6 必须先写成 x² + 5x + 6 = 0，然后再尝试因式分解。
of them must be zero. If (x + p)(x + q) = 0, then either x + p = 0 or x + q = 0, or both.
零积性质指出，如果两个式子的乘积为零，那么其中至少一个必须为零。如果 (x + p)(x + q) = 0，则 x + p = 0 或 x + q = 0，或两者都成立。
If p × q = 0, then p = 0 or q = 0
3. Factorising x² + bx + c | 因式分解 x² + bx + c
nd whose sum is b. Then write the quadratic as (x + m)(x + n), where mn = c and m + n = b.
当 x² 的系数为 1 时，寻找两个数，使它们的乘积等于 c，和等于 b。然后把二次式写成 (x + m)(x + n)，其中 mn = c，m + n = b。
x² + 5x + 6 = (x + 2)(x + 3) because 2 × 3 = 6 and 2 + 3 = 5.
x² - 5x + 6 = (x - 2)(x - 3) because -2 × -3 = 6 and -2 + -3 = -5.
x² + x - 6 = (x + 3)(x - 2) because 3 × -2 = -6 and 3 + -2 = 1.
中文对照：
x² + 5x + 6 = (x + 2)(x + 3)，因为 2 × 3 = 6 且 2 + 3 = 5。
x² - 5x + 6 = (x - 2)(x - 3)，因为 -2 × -3 = 6 且 -2 + -3 = -5。
x² + x - 6 = (x + 3)(x - 2)，因为 3 × -2 = -6 且 3 + -2 = 1。
4. Worked Example 1 | 例题 1
Solve x² + 7x + 12 = 0.
解方程 x² + 7x + 12 = 0。
Step 1: Find two numbers whose product is 12 and sum is 7. The numbers are 3 and 4.
步骤 1：找到两个数，乘积为 12，和为 7。这两个数是 3 和 4。
Step 2: Write the factorised form: (x + 3)(x + 4) = 0.
步骤 2：写出因式分解形式：(x + 3)(x + 4) = 0。
Step 3: Set each factor equal to zero: x + 3 = 0 or x + 4 = 0.
步骤 3：令每个因式为零：x + 3 = 0 或 x + 4 = 0。
Step 4: Solve each linear equation: x = -3 or x = -4.
步骤 4：解每个一元一次方程：x = -3 或 x = -4。
x² + 7x + 12 = 0 → x = -3 or x = -4
5. Worked Example 2 | 例题 2
Solve x² - 6x + 8 = 0.
解方程 x² - 6x + 8 = 0。
roduct is 8 and sum is -6. The numbers are -2 and -4 because -2 × -4 = 8 and -2 + -4 = -6.
找到两个数，乘积为 8，和为 -6。这两个数是 -2 和 -4，因为 -2 × -4 = 8 且 -2 + -4 = -6。
The factorised form is (x - 2)(x - 4) = 0.
因式分解形式为 (x - 2)(x - 4) = 0。
Using the zero product property: x - 2 = 0 or x - 4 = 0, so x = 2 or x = 4.
使用零积性质：x - 2 = 0 或 x - 4 = 0，所以 x = 2 或 x = 4。
x² - 6x + 8 = 0 → x = 2 or x = 4
6. Factorising when a ≠ 1 | 首项系数不为 1 时的因式分解
When the coefficient of x² is not 1, use the ac method. Multiply a and c, then find two numbers whose product is ac and whose sum is b. Split the middle term and factor by grouping.
当 x² 的系数不为 1 时，使用 ac 法。将 a 与 c 相乘，然后找到两个数，使它们的乘积为 ac，和为 b。拆分中间项并分组分解。
Example: Solve 2x² + 7x + 3 = 0.
例题：解方程 2x² + 7x + 3 = 0。
2, b = 7, c = 3, so ac = 2 × 3 = 6. We need two numbers with product 6 and sum 7: 6 and 1.
这里 a = 2，b = 7，c = 3，所以 ac = 2 × 3 = 6。我们需要两个数，乘积为 6，和为 7：6 和 1。
Rewrite 7x as 6x + 1x: 2x² + 6x + x + 3 = 0.
将 7x 改写为 6x + 1x：2x² + 6x + x + 3 = 0。
Group and factorise: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0.
分组并因式分解：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0。
Solve: 2x + 1 = 0 or x + 3 = 0, so x = -1/2 or x = -3.
求解：2x + 1 = 0 或 x + 3 = 0，所以 x = -1/2 或 x = -3。
2x² + 7x + 3 = 0 → x = -1/2 or x = -3
7. Difference of Two Squares | 平方差
of two squares: x² - a² = (x - a)(x + a). Recognising this form can save time in an exam.
因式分解的一个特殊情形是平方差：x² - a² = (x - a)(x + a)。识别这种形式可以在考试中节省时间。
For example, to solve x² - 25 = 0, write (x - 5)(x + 5) = 0. Then x = 5 or x = -5.
例如，解 x² - 25 = 0，写成 (x - 5)(x + 5) = 0。然后 x = 5 或 x = -5。
Another example: 4x² - 9 = (2x - 3)(2x + 3) = 0, giving x = 3/2 or x = -3/2.
另一个例子：4x² - 9 = (2x - 3)(2x + 3) = 0，得到 x = 3/2 或 x = -3/2。
x² - a² = (x - a)(x + a)
8. Removing a Common Factor First | 先提取公因式
Always check for a common factor before using another factorisation method. Taking out the common factor simplifies the quadratic.
在使用其他因式分解方法之前，始终检查是否有公因式。提取公因式可以简化二次式。
Example: Solve 3x² - 27 = 0.
例题：解方程 3x² - 27 = 0。
First factor out 3: 3(x² - 9) = 0. Divide both sides by 3: x² - 9 = 0.
先提取公因式 3：3(x² - 9) = 0。两边除以 3：x² - 9 = 0。
Now use the difference of two squares: (x - 3)(x + 3) = 0, so x = 3 or x = -3.
再使用平方差：(x - 3)(x + 3) = 0，所以 x = 3 或 x = -3。
9. Common Mistakes | 常见错误
One common mistake is to divide both sides of an equation by x. Since x could be zero, this may lose a root. Always use the zero product property instead.
一个常见错误是把方程两边同时除以 x。由于 x 可能为零，这样会丢失一个根。应使用零积性质，而不是除以含变量的因式。
² - 4x = 0, do not divide by x. Instead, factorise as x(x - 4) = 0, giving x = 0 or x = 4.
例如，在 x² - 4x = 0 中，不要除以 x。正确做法是因式分解为 x(x - 4) = 0，得到 x = 0 或 x = 4。
Another mistake is forgetting to set each factor equal to zero after factorising, or making sign errors when finding the two numbers.
另一个错误是分解后忘记令每个因式等于零，或者在找两个数时出现符号错误。
Always substitute your answers back into the original equation to check they work.
始终将答案代入原方程检验，确保它们成立。
10. Exam Tips and Summary | 考试技巧与总结
To solve a quadratic equation by factorisation, follow these steps: write the equation in standard form, factorise completely, set each factor equal to zero, solve the resulting linear equations, and check your answers by substitution.
用因式分解法解二次方程的步骤是：将方程写成标准形式，完全分解因式，令每个因式等于零，解得到的一元一次方程，并通过代入检验答案。
In the exam, show each step clearly. Many marks are awarded for method, even if a small arithmetic error occurs.
在考试中，清晰地展示每一步。即使出现小的计算错误，方法也能得到许多分数。
Remember the key forms:
记住关键形式：
x² + (m + n)x + mn = (x + m)(x + n)
x² - a² = (x - a)(x + a)
ax² + bx + c = (px + q)(rx + s), where pr = a, qs = c, and ps + qr = b
中文对照：
x² + (m + n)x + mn = (x + m)(x + n)
x² - a² = (x - a)(x + a)
ax² + bx + c = (px + q)(rx + s)，其中 pr = a，qs = c，ps + qr = b
ax² + bx + c = 0 → factorise → set each factor to 0 → solve
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