Example 2.1.1: Solving Quadratic Equations by Factorisation | 例题2.1.1：因式分解法求解二次方程

📚 Example 2.1.1: Solving Quadratic Equations by Factorisation | 例题2.1.1：因式分解法求解二次方程

Welcome to this revision note on Example 2.1.1 from the AQA A-Level Mathematics course. In this worked example, we solve the quadratic equation x² – 5x + 6 = 0 by factorisation. This example reinforces the fundamental techniques for solving quadratics, which are essential for many exam questions.

欢迎阅读 AQA A-Level 数学课程中例题 2.1.1 的复习笔记。在这个例题中，我们用因式分解法解二次方程 x² – 5x + 6 = 0。这个例子巩固了求解二次方程的基本技巧，这些技巧在许多考试题目中至关重要。

1. Introducing Example 2.1.1 | 例题2.1.1简介

Example 2.1.1 is a typical starter problem in the chapter on quadratic functions. It asks you to find the roots of a monic quadratic equation by factorisation. The solution proceeds by writing the quadratic as a product of two linear factors.

例题2.1.1是二次函数章节中典型的入门题。它要求你通过因式分解求一个首项系数为1的二次方程的根。解答过程是将二次式写成两个一次因式的乘积。

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2. Understanding Quadratic Equations | 理解二次方程

A quadratic equation is an equation of the form ax² + bx + c = 0, where a ≠ 0. In this example, a = 1, b = -5 and c = 6. The standard form allows us to apply systematic solution methods.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a ≠ 0。在本例中，a=1，b=-5，c=6。标准形式使我们能够应用系统性的求解方法。

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3. The Factorisation Method | 因式分解法

The factorisation method relies on finding two numbers whose product is ac and whose sum is b. For x² – 5x + 6, we need two numbers with product 6 and sum -5. These numbers are -2 and -3.

因式分解法依赖于寻找两个数，其乘积等于ac，和等于b。对于 x² – 5x + 6，我们需要找两个数，乘积为6，和为-5。这两个数是 -2 和 -3。

Hence the equation can be written as (x – 2)(x – 3) = 0. This factored form is the key to finding the roots.

因此原方程可写成 (x – 2)(x – 3) = 0。这个因式形式是求根的关键。

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4. Step-by-Step Solution | 分步解法

Now we solve each factor equal to zero. First, x – 2 = 0 gives x = 2. Second, x – 3 = 0 gives x = 3. Therefore the solution set is {2, 3}.

现在我们让每个因式等于零。首先，x – 2 = 0 得 x = 2。其次，x – 3 = 0 得 x = 3。因此解集为 {2, 3}。

(x – 2)(x – 3) = 0 ⇒ x = 2 or x = 3

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5. Checking the Answers | 验算答案

Substitute x = 2 into the original equation: 2² – 5(2) + 6 = 4 – 10 + 6 = 0. Similarly, for x = 3: 3² – 5(3) + 6 = 9 – 15 + 6 = 0. Both values satisfy the equation.

将 x=2 代入原方程：2² – 5×2 + 6 = 4 – 10 + 6 = 0。同样，x=3 时：3² – 5×3 + 6 = 9 – 15 + 6 = 0。两个值都满足方程。

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6. Alternative: Completing the Square | 其他方法：配方法

If factorisation is not obvious, completing the square is another reliable method. For x² – 5x + 6, we take half of -5 to get -2.5, so x² – 5x + 6 = (x – 2.5)² – 6.25 + 6 = (x – 2.5)² – 0.25. Setting this equal to zero gives (x – 2.5)² = 0.25, so x = 2.5 ± 0.5, which gives x = 2 or 3.

如果因式分解不明显，配方法是另一种可靠的方法。对于 x² – 5x + 6，取 -5 的一半得 -2.5，因此 x² – 5x + 6 = (x – 2.5)² – 6.25 + 6 = (x – 2.5)² – 0.25。令其等于0得 (x – 2.5)² = 0.25，所以 x = 2.5 ± 0.5，得到 x=2 或3。

x² – 5x + 6 = (x – 2.5)² – 0.25

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7. Alternative: Quadratic Formula | 其他方法：求根公式

The quadratic formula is always valid: for ax² + bx + c = 0, x = (-b ± √(b² – 4ac)) / (2a). Applying this with a = 1, b = -5, c = 6 gives x = (5 ± √(25 – 24)) / 2 = (5 ± 1) / 2, so x = 3 or 2.

求根公式始终成立：对于 ax² + bx + c = 0，x = (-b ± √(b² – 4ac)) / (2a)。代入 a=1, b=-5, c=6，得 x = (5 ± √(25 – 24)) / 2 = (5 ± 1) / 2，所以 x = 3 或 2。

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8. The Discriminant | 判别式

The discriminant is defined as Δ = b² – 4ac. For our equation, Δ = 25 – 24 = 1. Since Δ > 0 and a perfect square, the equation has two distinct rational roots, which matches our factorisation result.

判别式定义为 Δ = b² – 4ac。对于我们的方程，Δ = 25 – 24 = 1。因为 Δ > 0 且是完全平方数，所以方程有两个不相等的有理根，这符合因式分解的结果。

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9. Common Mistakes | 常见错误

A common mistake is to forget that the product equals zero, so each factor can be set to zero. Another error is misidentifying the signs when b is negative. Always check by expanding (x – 2)(x – 3) = x² – 3x – 2x + 6 = x² – 5x + 6.

一个常见错误是忘记乘积为零才能让每个因式等于零。另一个错误是在b为负时弄错符号。始终通过展开来检验：(x – 2)(x – 3) = x² – 3x – 2x + 6 = x² – 5x + 6。

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10. Graphical Interpretation | 图形意义

The roots x = 2 and x = 3 are the x-coordinates where the parabola y = x² – 5x + 6 crosses the x-axis. The axis of symmetry is at x = 2.5, the average of the roots.

根 x=2 和 x=3 是抛物线 y = x² – 5x + 6 与x轴交点的横坐标。对称轴位于 x = 2.5，即两个根的平均值。

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11. Practice Questions | 练习题

Try these practice problems to master the skill: (a) x² – 7x + 10 = 0, (b) x² + x – 12 = 0, (c) 2x² – 5x + 2 = 0. Answers: (a) 2 and 5, (b) -4 and 3, (c) 0.5 and 2.

试试这些练习题来掌握技巧：(a) x² – 7x + 10 = 0，(b) x² + x – 12 = 0，(c) 2x² – 5x + 2 = 0。答案：(a) 2 和 5，(b) -4 和 3，(c) 0.5 和 2。

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12. Summary | 总结

In summary, Example 2.1.1 demonstrates how to solve a quadratic equation by factorisation. This method is quick when the quadratic has simple integer roots. You should also be comfortable with completing the square and the quadratic formula as alternative strategies. Regular practice will help you recognise which method to use in the exam.

总之，例题2.1.1展示了如何用因式分解法求解二次方程。当二次方程有简单的整数根时，这种方法很快捷。你还应该熟练配方法和求根公式作为备选策略。定期练习将帮助你在考试中识别应使用哪种方法。

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