Solving Quadratic Equations by Factorising | 用因式分解解二次方程

📚 Solving Quadratic Equations by Factorising | 用因式分解解二次方程

A quadratic equation is one of the most important topics in IGCSE Mathematics. Solving quadratic equations by factorising is a core skill that appears in almost every exam paper. This method is often the fastest and most elegant way to find the roots of a quadratic equation when the expression factorises nicely.

二次方程是 IGCSE 数学中最重要的考点之一。用因式分解解二次方程是几乎每份试卷都会出现的核心技能。当表达式可以顺利分解时，这种方法是求二次方程根最快、最简洁的方式。

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1. What is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is an equation of 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’ (from the Latin ‘quadratus’, meaning square).

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。未知数 x 的最高次数是 2，因此称为”二次”方程（源自拉丁语 ‘quadratus’，意为”平方”）。

For example, x² – 5x + 6 = 0 is a quadratic equation. Here a = 1, b = -5 and c = 6.

例如，x² – 5x + 6 = 0 就是一个二次方程，其中 a = 1，b = -5，c = 6。

A quadratic equation can have two distinct real roots, one repeated root, or no real roots. In this article, we focus on equations that can be solved by factorising, which usually gives two distinct real roots.

二次方程可以有两个不同的实根、一个重根，或者没有实根。本文将重点讨论可以用因式分解求解的方程，这类方程通常会有两个不同的实根。

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2. The Zero Product Property | 零乘积性质

The key idea behind solving quadratic equations by factorising is the zero product property: if the product of two expressions is zero, then at least one of the expressions must be zero. In symbols: if A × B = 0, then A = 0 or B = 0.

用因式分解解二次方程的关键思想是零乘积性质：如果两个表达式的乘积为零，那么至少有一个表达式必须为零。用符号表示：如果 A × B = 0，则 A = 0 或 B = 0。

This property is simple but extremely powerful. Once a quadratic expression is factorised into two brackets, we can set each bracket equal to zero and solve the resulting linear equations.

这个性质看似简单，却极其强大。一旦二次表达式被分解为两个括号，我们就可以令每个括号等于零，然后解出所得的线性方程。

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3. Factorising Quadratics of the Form x² + bx + c | 因式分解 x² + bx + c 形式的二次式

When the coefficient of x² is 1 (that is, a = 1), we look for two numbers that multiply to give c and add to give b.

当 x² 的系数为 1（即 a = 1）时，我们要找两个数，它们的乘积等于 c，和等于 b。

For example, to factorise x² + 5x + 6, we need two numbers that multiply to 6 and add to 5. The numbers are 2 and 3, so:

例如，要因式分解 x² + 5x + 6，我们需要找到两个数，乘积为 6，和为 5。这两个数是 2 和 3，因此：

x² + 5x + 6 = (x + 2)(x + 3)

Always check by expanding: (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6 ✓

务必通过展开来验证：(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6 ✓

When b or c is negative, the signs become important. For x² – x – 12, we need two numbers that multiply to -12 and add to -1. The numbers are -4 and 3:

当 b 或 c 为负数时，符号就变得很重要。对于 x² – x – 12，我们需要两个数，乘积为 -12，和为 -1。这两个数是 -4 和 3：

x² – x – 12 = (x – 4)(x + 3)

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4. Solving x² + bx + c = 0 | 解 x² + bx + c = 0

Now let us solve the equation x² – 5x + 6 = 0 step by step.

现在我们逐步求解方程 x² – 5x + 6 = 0。

Step 1: Factorise the quadratic expression. We need two numbers that multiply to 6 and add to -5. These numbers are -2 and -3.

第一步：因式分解二次表达式。我们需要两个数，乘积为 6，和为 -5。这两个数是 -2 和 -3。

x² – 5x + 6 = (x – 2)(x – 3) = 0

Step 2: Apply the zero product property.

第二步：应用零乘积性质。

x – 2 = 0 or x – 3 = 0

Step 3: Solve each linear equation.

第三步：分别解线性方程。

x = 2 or x = 3

Therefore, the solutions of x² – 5x + 6 = 0 are x = 2 and x = 3.

因此，方程 x² – 5x + 6 = 0 的解为 x = 2 和 x = 3。

Always substitute your answers back into the original equation to check. For x = 2: 4 – 10 + 6 = 0 ✓ For x = 3: 9 – 15 + 6 = 0 ✓

务必将答案代回原方程进行验证。当 x = 2 时：4 – 10 + 6 = 0 ✓；当 x = 3 时：9 – 15 + 6 = 0 ✓

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5. Factorising Quadratics of the Form ax² + bx + c | 因式分解 ax² + bx + c 形式的二次式

When a ≠ 1, the factorisation is slightly harder. There are several methods, but the most common for IGCSE is the ‘split the middle term’ method, also called the product-sum method.

当 a ≠ 1 时，因式分解会稍微复杂一些。常用的方法有”拆分中间项”法，也称为乘积-和法，这是 IGCSE 中最常见的方法。

Consider 2x² + 7x + 3. We first multiply a and c: 2 × 3 = 6. We then look for two numbers that multiply to 6 and add to b = 7. These numbers are 6 and 1.

考虑 2x² + 7x + 3。我们先将 a 和 c 相乘：2 × 3 = 6。然后找两个数，乘积为 6，和为 b = 7。这两个数是 6 和 1。

Next, we split the middle term and factorise by grouping:

接下来，将中间项拆分，并用分组法进行因式分解：

2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3)

2x² + 7x + 3 = (2x + 1)(x + 3)

Check by expanding: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓

展开验证：(2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓

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6. Solving ax² + bx + c = 0 | 解 ax² + bx + c = 0

Let us solve the equation 2x² – 5x – 3 = 0.

让我们来解方程 2x² – 5x – 3 = 0。

Multiply a and c: 2 × (-3) = -6. We need two numbers that multiply to -6 and add to b = -5. These numbers are -6 and 1.

将 a 和 c 相乘：2 × (-3) = -6。我们需要两个数，乘积为 -6，和为 b = -5。这两个数是 -6 和 1。

Split the middle term and factorise by grouping:

拆分中间项并用分组法因式分解：

2x² – 5x – 3 = 2x² – 6x + x – 3 = 2x(x – 3) + 1(x – 3) = (2x + 1)(x – 3)

So the equation becomes (2x + 1)(x – 3) = 0.

因此方程为 (2x + 1)(x – 3) = 0。

By the zero product property:

根据零乘积性质：

2x + 1 = 0 or x – 3 = 0

Solving each gives:

分别解得：

x = -½ or x = 3

So the solutions are x = -½ and x = 3.

因此解为 x = -½ 和 x = 3。

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7. Difference of Two Squares | 平方差

A special case worth knowing is the difference of two squares: a² – b² = (a + b)(a – b). This is extremely useful when solving equations of the form x² – k² = 0.

有一个值得记住的特殊情形：平方差公式 a² – b² = (a + b)(a – b)。这在解形如 x² – k² = 0 的方程时非常有用。

For example, x² – 9 = 0 can be written as (x + 3)(x – 3) = 0, giving x = -3 or x = 3.

例如，x² – 9 = 0 可以写成 (x + 3)(x – 3) = 0，得到 x = -3 或 x = 3。

Similarly, 4x² – 25 = 0 factors as (2x + 5)(2x – 5) = 0, giving x = -5/2 or x = 5/2.

同样，4x² – 25 = 0 可分解为 (2x + 5)(2x – 5) = 0，得到 x = -5/2 或 x = 5/2。

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8. Common Factor First | 先提公因式

Before attempting any factorisation, always check whether there is a common factor that can be taken out first. This simplifies the expression and makes the subsequent factorisation easier.

在进行任何因式分解之前，一定要先检查是否存在公因式可以提出。这样可以简化表达式，使后续分解更容易。

For example, 2x² – 8x = 0 can be factorised as 2x(x – 4) = 0. Then x = 0 or x = 4.

例如，2x² – 8x = 0 可以分解为 2x(x – 4) = 0，于是 x = 0 或 x = 4。

Similarly, 3x² – 12 = 0 becomes 3(x² – 4) = 3(x + 2)(x – 2) = 0, giving x = -2 or x = 2.

同样，3x² – 12 = 0 可以化为 3(x² – 4) = 3(x + 2)(x – 2) = 0，得到 x = -2 或 x = 2。

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9. Common Mistakes and Tips | 常见错误与技巧

Here are some common mistakes students make, along with tips to avoid them:

以下是学生常犯的一些错误以及避免建议：

• Mistake 1: Forgetting to rearrange the equation into the form ax² + bx + c = 0 before factorising. Always move all terms to one side first.
• 错误一：在因式分解前忘记将方程整理为 ax² + bx + c = 0 的形式。务必先将所有项移到等号一边。
• Mistake 2: Getting the signs wrong when factorising. Practice with negative values and always expand to check.
• 错误二：因式分解时弄错符号。多练习含负值的题目，并且始终展开验证。
• Mistake 3: Forgetting that x = 0 is a valid solution when a common factor of x is taken out.
• 错误三：当提出公因式 x 时，忘记 x = 0 也是一个有效解。
• Mistake 4: Thinking that x² = 9 gives only x = 3. Remember that x = -3 is also a solution.
• 错误四：认为 x² = 9 只有 x = 3 这一个解。记住 x = -3 也是解。

Tips for success: Always write out each step clearly. After factorising, expand to check your brackets are correct. After solving, substitute back into the original equation to verify your answers.

成功技巧：每一步都要写清楚。因式分解后展开以检查括号是否正确。解出答案后代回原方程验证。

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10. Practice Problems | 练习题

Try the following problems to consolidate your understanding. Answers are provided in the table.

请尝试以下练习来巩固理解，答案已在表格中给出。

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