Example 3.3.1: Solving Quadratic Equations by Factorisation | 例3.3.1：用因式分解法解一元二次方程

📚 Example 3.3.1: Solving Quadratic Equations by Factorisation | 例3.3.1：用因式分解法解一元二次方程

This worked example demonstrates the standard factorisation method for solving quadratic equations of the form ax² + bx + c = 0, where a = 1. It is a core skill in the AQA A-Level Mathematics specification and appears in many pure mathematics and applied contexts. We will break the solution into clear steps, discuss common mistakes, and provide verification techniques.

本例题演示如何用标准因式分解法求解形如 ax² + bx + c = 0（其中 a = 1）的一元二次方程。这是 AQA A-Level 数学考纲中的核心技能，在纯数学和应用数学的许多场景中都会出现。我们将把解法拆分为清晰的步骤，讨论常见错误，并提供检验方法。

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1. The Problem | 题目

Solve the quadratic equation:

x² − 5x + 6 = 0

The equation is already in standard form with a = 1, b = −5, and c = 6. Our goal is to find all real values of x that satisfy the equation.

该方程已经是一般形式，其中 a = 1，b = −5，c = 6。我们的目标是求出满足方程的所有实数 x 值。

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2. The Factorisation Method | 因式分解法

We look for two numbers that multiply to give c = 6 and add to give b = −5. Since the product is positive and the sum is negative, both numbers must be negative.

我们寻找两个数，使它们的乘积等于 c = 6，且它们的和等于 b = −5。由于乘积为正而和为负，这两个数必同为负数。

• Possible factor pairs of 6: (1, 6), (2, 3), (−1, −6), (−2, −3).

  6 的可能因数对： (1, 6)、(2, 3)、(−1, −6)、(−2, −3)。

• Among these, the pair (−2, −3) has sum −5.

  其中，对 (−2, −3) 的和为 −5。

Therefore, the factorised form is (x − 2)(x − 3) = 0.

因此，因式分解的形式为 (x − 2)(x − 3) = 0。

(x − 2)(x − 3) = 0

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3. Setting Each Factor to Zero | 令每个因式为零

If the product of two expressions is zero, then at least one of the expressions must be zero. This is known as the zero product property.

如果两个表达式的乘积为零，则至少其中一个表达式必须为零。这称为零积性质。

• x − 2 = 0 ⇒ x = 2

  x − 2 = 0 ⇒ x = 2

• x − 3 = 0 ⇒ x = 3

  x − 3 = 0 ⇒ x = 3

Thus the solution set is x = 2 or x = 3.

因此解集为 x = 2 或 x = 3。

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4. Verification | 验证

Substitute each root back into the original equation to check:

将每个根代回原方程进行检验：

x value x 值	Substitution 代入计算	Result 结果
x = 2	(2)² − 5(2) + 6 = 4 − 10 + 6 = 0	✓
x = 3	(3)² − 5(3) + 6 = 9 − 15 + 6 = 0	✓

Both values satisfy the equation, confirming correct factorisation.

两个值均满足方程，确认因式分解正确。

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5. Graphical Interpretation | 图像解释

The quadratic function y = x² − 5x + 6 is a parabola opening upwards. The solutions we found are the x-coordinates of the points where the curve crosses the x-axis.

二次函数 y = x² − 5x + 6 是一条开口向上的抛物线。我们求出的解正是曲线与 x 轴交点的 x 坐标。

• Axis of symmetry: x = (2 + 3)/2 = 2.5

  对称轴： x = (2 + 3)/2 = 2.5

• Vertex y-coordinate: (2.5)² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25

  顶点纵坐标： (2.5)² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25

The minimum point is (2.5, −0.25), and the roots are equidistant from the axis of symmetry.

最低点为 (2.5, −0.25)，两根关于对称轴等距。

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6. Common Mistakes | 常见错误

Students often make the following errors when solving this type of equation:

学生在解此类方程时常犯以下错误：

• Wrong sign selection: For x² − 5x + 6, choosing numbers 2 and 3 gives (x + 2)(x + 3), which expands to x² + 5x + 6 — incorrect because the middle term must be −5x.

  符号选择错误： 对于 x² − 5x + 6，选择 2 和 3 会得到 (x + 2)(x + 3)，展开为 x² + 5x + 6 —— 不正确，因为一次项必须是 −5x。

• Forgetting the zero product property: Some students divide both sides by (x − 2), losing the root x = 2. This is invalid because division by zero is undefined.

  忘记零积性质： 有些学生两边同除以 (x − 2)，从而丢掉了根 x = 2。这是不可行的，因为除以零无定义。

• Sign errors when expanding: Always expand (x − 2)(x − 3) to verify: x² − 3x − 2x + 6 = x² − 5x + 6.

  展开时符号错误： 务必展开 (x − 2)(x − 3) 验证： x² − 3x − 2x + 6 = x² − 5x + 6。

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7. When Factorisation Works | 因式分解法的适用条件

The method works perfectly when the quadratic has rational roots, i.e., when the discriminant Δ = b² − 4ac is a perfect square. Here Δ = (−5)² − 4(1)(6) = 25 − 24 = 1, which is a perfect square, confirming that factorisation is possible.

当二次方程有有理根时，即判别式 Δ = b² − 4ac 为完全平方数时，因式分解法完美适用。本例中 Δ = (−5)² − 4(1)(6) = 25 − 24 = 1，是完全平方数，确认可以因式分解。

• If Δ > 0 and a perfect square → two distinct rational roots.

  若 Δ > 0 且为完全平方数 → 两个不同的有理根。

• If Δ > 0 but not a perfect square → two distinct irrational roots (use quadratic formula).

  若 Δ > 0 但不是完全平方数 → 两个不同的无理根（使用求根公式）。

• If Δ = 0 → one repeated root.

  若 Δ = 0 → 一个重根。

• If Δ < 0 → no real roots.

  若 Δ < 0 → 无实数根。

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8. Extension: General Form ax² + bx + c = 0 | 拓展：一般形式 ax² + bx + c = 0

When a ≠ 1, we use a slightly different approach. For example, solve 2x² + 5x − 3 = 0.

当 a ≠ 1 时，我们采用略有不同的方法。例如，求解 2x² + 5x − 3 = 0。

We need two numbers that multiply to a × c = 2 × (−3) = −6 and add to b = 5. The pair (6, −1) satisfies these conditions.

我们需要两个数，使它们的乘积等于 a × c = 2 × (−3) = −6，且和为 b = 5。数对 (6, −1) 满足这些条件。

2x² + 6x − x − 3 = 0

Now factor by grouping:

现在使用分组分解：

2x(x + 3) − 1(x + 3) = 0

(2x − 1)(x + 3) = 0

Thus x = 1/2 or x = −3.

因此 x = 1/2 或 x = −3。

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9. Practice Problems | 练习题目

Try solving these quadratics by factorisation:

尝试用因式分解法求解下列二次方程：

1. x² − 7x + 12 = 0

   答案： x = 3, x = 4

2. x² + 3x − 10 = 0

   答案： x = 2, x = −5

3. x² − 9 = 0

   答案： x = 3, x = −3（平方差公式）

4. x² + 6x + 9 = 0

   答案： x = −3（重根）

Remember to expand your factorised answer to check each one.

记得展开因式分解后的结果检验每一题。

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Summary | 总结

Example 3.3.1 shows how to solve x² − 5x + 6 = 0. The process is: identify factor pairs of c whose sum is b, write the factorised form, apply the zero product property, and solve each linear factor. This method is efficient, reliable, and fundamental to A-Level Mathematics.

例 3.3.1 展示了如何求解 x² − 5x + 6 = 0。过程为：找出 c 的因数对，使其和为 b，写出因式分解形式，应用零积性质，然后解每个一次因式。这种方法高效、可靠，是 A-Level 数学的基础。

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