Example 4.1.1 | 示例 4.1.1

📚 Example 4.1.1 | 示例 4.1.1

In this revision note, we work through a typical AQA A-Level Mathematics Pure Core example: solving a quadratic equation by factorisation, then check the solution using the quadratic formula. This example corresponds to Section 4.1 of the AQA Pure Mathematics specification, where the focus is on algebraic methods for solving equations.

在本篇复习笔记中，我们将解析一道典型的 AQA A-Level 数学核心纯数例题：通过因式分解法解二次方程，并用二次公式检验解的正确性。该例题对应 AQA 纯数大纲第 4.1 节，重点是求解方程的代数方法。

1. The Problem | 问题

Solve the equation 2x² − 3x − 2 = 0.

解方程 2x² − 3x − 2 = 0。

2. Factorising the Quadratic | 因式分解二次式

We look for two numbers whose product is 2 × (−2) = −4 and whose sum is −3. The numbers −4 and 1 satisfy these conditions because −4 × 1 = −4 and −4 + 1 = −3.

我们寻找两个数，使它们的乘积为 2 × (−2) = −4，和为 −3。数字 −4 和 1 满足条件，因为 −4 × 1 = −4 且 −4 + 1 = −3。

Rewrite the middle term using these two numbers:

用这两个数改写中间项：

2x² − 4x + x − 2 = 0

Now factor by grouping:

现在分组因式分解：

2x(x − 2) + 1(x − 2) = 0

(2x + 1)(x − 2) = 0

3. Finding the Roots | 求根

Using the zero product property, we set each factor equal to zero:

利用零乘积性质，令每个因式等于零：

2x + 1 = 0 或 x − 2 = 0

Therefore:

因此：

x = −½ 或 x = 2

These are the two solutions of the equation.

这两个就是方程的解。

4. Checking with the Quadratic Formula | 用二次公式检验

The quadratic formula states that for an equation ax² + bx + c = 0,

二次公式指出，对于方程 ax² + bx + c = 0，

x = (−b ± √(b² − 4ac)) / 2a

Here, a = 2, b = −3, c = −2. Substitute:

这里 a = 2，b = −3，c = −2。代入：

x = (3 ± √((−3)² − 4 × 2 × (−2))) / (2 × 2)

= (3 ± √(9 + 16)) / 4 = (3 ± √25) / 4

= (3 ± 5) / 4

Thus:

因此：

When + : x = (3 + 5)/4 = 8/4 = 2

当取 + 时：x = (3 + 5)/4 = 8/4 = 2

When − : x = (3 − 5)/4 = −2/4 = −½

当取 − 时：x = (3 − 5)/4 = −2/4 = −½

Both methods give the same roots, confirming the solution.

两种方法得到相同的根，验证了解的正确性。

5. Alternative: Multiplying the Factors Back | 另一种方法：将因式乘回

It is often useful to expand the factored form to verify it matches the original quadratic:

将因式展开来验证与原二次式一致通常很有用：

(2x + 1)(x − 2) = 2x² − 4x + x − 2 = 2x² − 3x − 2

The expansion is correct, so the factors are valid.

展开正确，所以因式分解有效。

6. Graphical Interpretation | 图形解释

The equation 2x² − 3x − 2 = 0 represents a parabola. The solutions are the x-coordinates where the curve crosses the x-axis. Since the discriminant Δ = b² − 4ac = 25 > 0, there are two distinct real roots, which can be seen as the two intersection points at x = −½ and x = 2.

方程 2x² − 3x − 2 = 0 表示一条抛物线。解是曲线与 x 轴交点的 x 坐标。由于判别式 Δ = b² − 4ac = 25 > 0，存在两个不同的实数根，即 x = −½ 和 x = 2 处的两个交点。

7. Common Mistakes | 常见错误

When factorising, students often forget to include the factor of 2 in the first bracket, or incorrectly pair the middle terms. Always check that the product of the two bracket constants equals c (here −2) and the sum of outer and inner products equals b (here −3).

因式分解时，学生常常忘记第一个括号中的系数 2，或错误地配对中间项。务必检查两个括号中常数项的乘积等于 c（此处为 −2），且外项与内项乘积之和等于 b（此处为 −3）。

Another common error is dropping a negative sign when substituting into the quadratic formula. Write down every step and double-check the value of b² − 4ac.

另一个常见错误是在代入二次公式时丢失负号。请写下每一步并仔细检查 b² − 4ac 的值。

8. Summary | 总结

In this example, we solved 2x² − 3x − 2 = 0 by factorisation, obtaining x = −½ and x = 2. We then verified the solution using the quadratic formula. The same method applies to any quadratic equation with rational roots.

本例中，我们通过因式分解求解 2x² − 3x − 2 = 0，得到 x = −½ 和 x = 2。随后用二次公式进行了验证。该方法适用于任何具有有理根的二次方程。

For AQA A-Level Mathematics, always present your working clearly and state which method you are using. This ensures full marks for method and accuracy.

对于 AQA A-Level 数学，务必清晰展示解题过程并说明所用方法，这样可在方法和准确性上获得满分。

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Example 4.1.1 | 示例 4.1.1