Example 5.4.1: Solving a Quadratic Inequality | 示例5.4.1：解二次不等式

📚 Example 5.4.1: Solving a Quadratic Inequality | 示例5.4.1：解二次不等式

Welcome to Example 5.4.1 from the AQA A-Level Mathematics syllabus. In this worked example, we explore a classic quadratic inequality and develop a clear, step-by-step method for finding its solution set. This example will strengthen your algebraic manipulation and graphical interpretation skills, both essential for the Pure Mathematics examinations.

欢迎来到AQA A-Level数学课程中的示例5.4.1。在本示例中，我们将探讨一个经典的二次不等式，并逐步建立求解其解集的方法。该示例将增强你的代数运算和图像解读能力，这两者对纯数学考试至关重要。

1. Introduction to the Example | 例题介绍

Consider the quadratic inequality: x² − 5x + 6 > 0. This is a simple yet powerful example because it shows how algebraic factorisation and a graphical understanding combine to produce the full solution set. We are asked to find all real values of x that satisfy this inequality.

考虑二次不等式：x² − 5x + 6 > 0。这是一个简单但极具代表性的例子，因为它展示了代数因式分解与图像理解相结合，从而得到完整解集的过程。我们需要找出满足该不等式的所有实数 x。

The inequality is strict (greater than 0), so the roots where the expression equals 0 will not be included in the solution. We must be careful to use open intervals in our final answer.

该不等式是严格不等式（大于0），因此表达式等于0的根不包含在解集中。在最终答案中，我们必须小心使用开区间。

2. The Standard Form | 标准形式

A quadratic inequality is typically written in the form ax² + bx + c > 0 (or <, ≥, ≤ 0). Our example already has the standard form with a = 1, b = −5, and c = 6. Since a = 1 > 0, the parabola opens upward, which tells us the graph is U-shaped.

二次不等式通常写成 ax² + bx + c > 0（或 <、≥、≤ 0）的形式。我们的示例已经是标准形式，其中 a = 1，b = −5，c = 6。由于 a = 1 > 0，抛物线开口向上，这说明图像呈U形。

Before solving, always check whether the coefficient of x² is positive. If it is negative, the direction of the inequality will need careful handling when interpreting the graph. In this example, the positive leading coefficient simplifies our analysis.

在求解之前，务必检查 x² 的系数是否为正。如果为负，在解读图像时需要小心处理不等号的方向。在本示例中，正的首项系数简化了我们的分析。

3. Factorisation | 因式分解

To solve x² − 5x + 6 > 0, we first factorise the quadratic expression. We look for two numbers that multiply to give 6 and add to give −5. These numbers are −2 and −3 because (−2) × (−3) = 6 and (−2) + (−3) = −5.

为了求解 x² − 5x + 6 > 0，我们首先对二次表达式进行因式分解。我们寻找两个数，它们相乘得6，相加得−5。这两个数是−2和−3，因为(−2) × (−3) = 6，且(−2) + (−3) = −5。

Therefore, the expression can be written as (x − 2)(x − 3). The inequality becomes:

因此，该表达式可以写成 (x − 2)(x − 3)。不等式变为：

(x − 2)(x − 3) > 0

Factorisation is the key step because it reveals the roots of the quadratic equation, which are also the boundary points of the inequality solution.

因式分解是关键的一步，因为它揭示了二次方程的根，这些根也是不等式解集的边界点。

4. Critical Values | 临界值

Setting each factor equal to zero gives the critical values. From x − 2 = 0 we get x = 2, and from x − 3 = 0 we get x = 3. These are the x-coordinates where the function f(x) = x² − 5x + 6 crosses the x-axis.

令每个因子等于零，可得到临界值。由 x − 2 = 0 得 x = 2，由 x − 3 = 0 得 x = 3。这些是函数 f(x) = x² − 5x + 6 与 x 轴交点的 x 坐标。

The critical values divide the real number line into three distinct intervals: (−∞, 2), (2, 3), and (3, ∞). In each of these intervals, the product (x − 2)(x − 3) has a constant sign — either positive or negative.

临界值将实数轴划分为三个不同的区间：(−∞, 2)、(2, 3) 和 (3, ∞)。在每个区间内，乘积 (x − 2)(x − 3) 的符号保持不变——要么为正，要么为负。

We do not include x = 2 and x = 3 in the solution because the inequality is strict ( > 0, not ≥ 0). These points are where the expression equals 0.

因为不等式是严格的（> 0，而不是 ≥ 0），所以解集中不包含 x = 2 和 x = 3。这些点是表达式等于0的地方。

5. Graph Sketch | 图像草图

A quick sketch of the graph helps us visualise the solution. The function f(x) = x² − 5x + 6 is a parabola opening upward with x-intercepts at 2 and 3. The vertex lies midway between the intercepts at x = 2.5; substituting gives f(2.5) = 6.25 − 12.5 + 6 = −0.25, so the vertex is (2.5, −0.25).

快速画出图像有助于我们直观理解解集。函数 f(x) = x² − 5x + 6 是一条开口向上的抛物线，x 截距为2和3。顶点位于截距中间 x = 2.5 处；代入得 f(2.5) = 6.25 − 12.5 + 6 = −0.25，因此顶点为 (2.5, −0.25)。

Since the parabola opens upward, the graph is above the x-axis to the left of x = 2 and to the right of x = 3. It is below the x-axis between x = 2 and x = 3. We are interested in where the graph is above the x-axis, i.e., where f(x) > 0.

由于抛物线开口向上，图像在 x = 2 的左侧和 x = 3 的右侧位于 x 轴上方。在 x = 2 和 x = 3 之间，图像位于 x 轴下方。我们关注图像在 x 轴上方的部分，即 f(x) > 0 的区域。

From the graph, it is immediately clear that the solution consists of two separate intervals: x < 2 or x > 3. This visual approach confirms the algebraic method.

从图像中我们可以立即看出，解集由两个独立的区间组成：x < 2 或 x > 3。这种视觉方法验证了代数方法的结果。

6. Solving by Interval Testing | 区间测试法

If you prefer not to sketch a graph, you can test the sign of the product in each interval. Choose a test value from each of the three intervals and substitute it into (x − 2)(x − 3).

如果你不喜欢画图，也可以在每个区间内测试乘积的符号。从三个区间中各取一个测试值，代入 (x − 2)(x − 3)。

Test x = 0 in (−∞, 2): (0 − 2)(0 − 3) = (−2)(−3) = 6 > 0

在 (−∞, 2) 中取 x = 0： (0 − 2)(0 − 3) = (−2)(−3) = 6 > 0
Test x = 2.5 in (2, 3): (2.5 − 2)(2.5 − 3) = (0.5)(−0.5) = −0.25 < 0

在 (2, 3) 中取 x = 2.5： (2.5 − 2)(2.5 − 3) = (0.5)(−0.5) = −0.25 < 0
Test x = 4 in (3, ∞): (4 − 2)(4 − 3) = (2)(1) = 2 > 0

在 (3, ∞) 中取 x = 4： (4 − 2)(4 − 3) = (2)(1) = 2 > 0

The sign table below summarises the results:

下面的符号表总结了结果：

Interval \| 区间	(−∞, 2)	(2, 3)	(3, ∞)
Sign \| 符号	+	−	+

7. Writing the Final Solution | 写出最终解集

The product (x − 2)(x − 3) is positive in the intervals (−∞, 2) and (3, ∞). Therefore, the solution to x² − 5x + 6 > 0 is:

乘积 (x − 2)(x − 3) 在区间 (−∞, 2) 和 (3, ∞) 内为正。因此，x² − 5x + 6 > 0 的解集为：

x < 2 或 x > 3

In set notation, we write the solution as {x : x < 2} ∪ {x : x > 3}, or simply as the union of open intervals:

用集合符号表示，我们写作 {x : x < 2} ∪ {x : x > 3}，或者简写为开区间的并集：

(−∞, 2) ∪ (3, ∞)

Notice that we use parentheses (open brackets) because the endpoints 2 and 3 are not included. If the inequality had been ≥ 0, we would use square brackets.

请注意，我们使用圆括号（开括号），因为端点2和3不包含在内。如果不等式是 ≥ 0，则应使用方括号。

8. Alternative: Sign Table | 备选：符号表

A sign table is a systematic way to determine the sign of each factor individually. We draw a table with rows for each factor and the product, and columns for the intervals.

符号表是一种系统性的方法，可以分别确定每个因子的符号。我们画一张表，每行代表一个因子和乘积，每列代表一个区间。

For x < 2, both x − 2 and x − 3 are negative, so their product is positive. For 2 < x < 3, x − 2 is positive but x − 3 is negative, so the product is negative. For x > 3, both factors are positive, so the product is positive.

当 x < 2 时，x − 2 和 x − 3 均为负，因此乘积为正。当 2 < x < 3 时，x − 2 为正但 x − 3 为负，因此乘积为负。当 x > 3 时，两个因子均为正，因此乘积为正。

Factor \| 因子	x < 2	2 < x < 3	x > 3
x − 2	−	+	+
x − 3	−	−	+
(x − 2)(x − 3)	+	−	+

The sign table method is particularly useful for inequalities with more than two factors, where sketching may be difficult.

符号表法对于因子超过两个的不等式特别有用，因为此时画图可能比较困难。

9. Checking Your Answer | 检查答案

It is always good practice to verify your solution by substituting a value from each interval back into the original inequality. For example, choose x = 1 (which is less than 2): 1² − 5(1) + 6 = 1 − 5 + 6 = 2 > 0, which is true.

始终建议通过将每个区间内的值代入原不等式来验证答案。例如，取 x = 1（小于2）：1² − 5(1) + 6 = 1 − 5 + 6 = 2 > 0，成立。

Choose x = 2.5 (between 2 and 3): 2.5² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25, which is not greater than 0, so the inequality is false as expected.

取 x = 2.5（在2和3之间）：2.5² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25，不大于0，因此不等式不成立，符合预期。

Choose x = 4 (greater than 3): 4² − 5(4) + 6 = 16 − 20 + 6 = 2 > 0, which is true. These checks confirm that the solution set is correct.

取 x = 4（大于3）：4² − 5(4) + 6 = 16 − 20 + 6 = 2 > 0，成立。这些检验确认了解集是正确的。

10. Common Pitfalls | 常见陷阱

One common mistake is to write the solution as a single interval, such as (2, ∞) or (−∞, 3). This is incorrect because the inequality is not satisfied between 2 and 3. The correct solution must be written as two separate intervals.

一个常见错误是将解集写成单一区间，例如 (2, ∞) 或 (−∞, 3)。这是不正确的，因为不等式在2和3之间不成立。正确的解集必须写成两个独立的区间。

Another pitfall is forgetting to reverse the inequality sign when multiplying or dividing by a negative number. In this example we did not need to do that, but if the coefficient of x² had been negative, we might have chosen to multiply by −1 and reverse the inequality.

另一个陷阱是当乘以或除以负数时忘记反向改变不等号。在本示例中我们不需要这样做，但如果 x² 的系数为负，我们可能会选择乘以 −1 并反转不等号。

Finally, do not confuse the solution of the inequality with the solution of the equation x² − 5x + 6 = 0, which is x = 2 or x = 3. The inequality asks for intervals, not points.

最后，不要将不等式的解与方程 x² − 5x + 6 = 0 的解混淆，方程的解是 x = 2 或 x = 3。不等式要求的是区间，而不是点。

11. Extension: Strict vs Non-strict Inequalities | 扩展：严格与非严格不等式

If the original inequality had been x² − 5x + 6 ≥ 0, the solution would include the endpoints 2 and 3. In that case, the solution set would be (−∞, 2] ∪ [3, ∞). The square brackets indicate that the endpoints are included.

如果原不等式是 x² − 5x + 6 ≥ 0，则解集将包含端点2和3。在这种情况下，解集为 (−∞, 2] ∪ [3, ∞)。方括号表示包含端点。

Similarly, the inequality x² − 5x + 6 < 0 has the solution 2 < x < 3, which is the open interval (2, 3). Understanding the difference between strict and non-strict inequalities is crucial for correct notation.

类似地，不等式 x² − 5x + 6 < 0 的解为 2 < x < 3，即开区间 (2, 3)。理解严格不等式与非严格不等式之间的区别对于正确书写符号至关重要。

Always read the inequality sign carefully in the exam. A small change from ‘>’ to ‘≥’ changes the entire meaning of the solution.

考试时务必仔细阅读不等号。’ > ‘ 到 ‘ ≥ ‘ 的微小变化会改变整个解集的意义。

12. Summary | 总结

In Example 5.4.1, we solved the quadratic inequality x² − 5x + 6 > 0 by factorising, finding critical values, testing intervals, and using a graph sketch. The solution is x < 2 or x > 3, written as (−∞, 2) ∪ (3, ∞).

在示例5.4.1中，我们通过因式分解、找到临界值、测试区间以及绘制图像草图，求解了二次不等式 x² − 5x + 6 > 0。解为 x < 2 或 x > 3，写作 (−∞, 2) ∪ (3, ∞)。

Remember the three-step process: factorise, find the roots, then determine the sign on each side of the roots. This method applies to all quadratic inequalities and is a reliable strategy in the A-Level Mathematics examination.

请记住三步流程：因式分解、求根，然后确定根两侧的符号。这种方法适用于所有二次不等式，是A-Level数学考试中可靠的策略。

We hope this worked example has clarified the technique. Practice with similar inequalities to build confidence and speed.

希望这个示例已经阐明了相关技巧。通过练习类似的不等式，可以增强信心并提高速度。

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Example 5.4.1: Solving a Quadratic Inequality | 示例5.4.1：解二次不等式