📚 Example 3.5.1: Solving Quadratic Inequalities | 示例 3.5.1：解二次不等式
Quadratic inequalities are a core part of the AQA A Level Mathematics specification, appearing regularly in both AS and A2 papers. In this worked example, we explore a typical question: solve the inequality x² - 5x + 6 > 0. Understanding how to factorise, locate critical values, and test intervals is essential for accurate solutions.
二次不等式是 AQA A Level 数学考纲中的核心内容，在 AS 和 A2 试卷中都经常出现。在这个例题中，我们探讨一个典型问题：解不等式 x² - 5x + 6 > 0。理解如何因式分解、确定临界值并检验区间，对于准确求解至关重要。
1. What Is a Quadratic Inequality? | 什么是二次不等式？
sually a range of x-values rather than a single number. For Example 3.5.1, the inequality x² - 5x + 6 > 0 asks us to find all real x for which the quadratic is strictly positive.
二次不等式包含一个二次表达式以及 <、>、≤ 或 ≥ 中的一个符号。解通常是一个 x 值的范围，而不是单个数值。对于示例 3.5.1，不等式 x² - 5x + 6 > 0 要求我们找出所有使该二次式严格为正的实数 x。
Unlike linear inequalities, the solution set often consists of two separate intervals. This happens because the graph of a quadratic is a parabola, which can cross the x-axis twice and change sign at each crossing.
与线性不等式不同，二次不等式的解集通常由两个独立的区间组成。这是因为二次函数的图像是一条抛物线，它可以两次穿过 x 轴，并在每个交点处改变符号。
2. The Standard Form of Example 3.5.1 | 示例 3.5.1 的标准形式
In our example, the inequality is already in standard form with a = 1, b = -5, and c = 6.
在解任何二次不等式之前，最好将其写成标准形式：ax² + bx + c > 0、ax² + bx + c < 0、ax² + bx + c ≥ 0 或 ax² + bx + c ≤ 0。在我们的示例中，不等式已经是标准形式，其中 a = 1，b = -5，c = 6。
x² - 5x + 6 > 0
Since the coefficient of x² is positive, the parabola opens upwards. This means the quadratic will be positive outside the interval between its two roots and negative inside that interval.
由于 x² 的系数为正，抛物线开口向上。这意味着该二次式在其两个根之间的区间外为正，在该区间内为负。
3. Factorising the Quadratic Expression | 因式分解二次表达式
The first algebraic step is to factorise the quadratic expression. We look for two numbers that multiply to 6 and add to -5. These numbers are -2 and -3, so the factorised form is:
第一步代数运算是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，因此因式分解形式为：
x² - 5x + 6 = (x - 2)(x - 3)
The roots of the related quadratic equation x² - 5x + 6 = 0 are therefore x = 2 and x = 3.
这种因式分解是有效的，因为 (x - 2)(x - 3) 展开后为 x² - 3x - 2x + 6 = x² - 5x + 6。因此，相关二次方程 x² - 5x + 6 = 0 的根为 x = 2 和 x = 3。
4. Locating Critical Values on the Number Line | 在数轴上定位临界值
cal values because they are the only points where the quadratic expression can change sign. We place these values on a number line, dividing the real line into three intervals:
根 x = 2 和 x = 3 被称为临界值，因为它们是该二次表达式唯一可能改变符号的点。我们将这些值放在数轴上，将实数轴分成三个区间：
- (-∞, 2)
- (2, 3)
- (3, ∞)
he solution for a strict inequality (> or <), but they would be included for ≤ or ≥.
对于严格不等式（> 或 <），临界值本身不包含在解集中，但对于 ≤ 或 ≥，它们会被包含。
5. Using a Sign Table to Determine the Solution | 使用符号表确定解集
We now test the sign of the product (x - 2)(x - 3) in each interval. Choose a convenient test value in each interval and substitute it into the factorised expression.
现在我们检验乘积 (x - 2)(x - 3) 在每个区间内的符号。在每个区间中选择一个方便的测试值，并将其代入因式分解后的表达式。
| Interval | Test value | Sign of (x - 2)(x - 3) |
| x < 2 | x = 0 | (-)(-) = positive |
| 2 < x < 3 | x = 2.5 | (+)(-) = negative |
| x > 3 | x = 4 | (+)(+) = positive |
Since the inequality is x² - 5x + 6 > 0, we need the intervals where the sign is positive. From the table, these are x < 2 and x > 3.
由于不等式是 x² - 5x + 6 > 0，我们需要符号为正的区间。根据表格，这些区间是 x < 2 和 x > 3。
6. Graphical Interpretation of the Solution | 解的图形解释
pening upwards with x-intercepts at 2 and 3. The part of the graph above the x-axis corresponds to y > 0, which occurs when x < 2 or x > 3.
y = x² - 5x + 6 的图像是一条开口向上的抛物线，与 x 轴的交点为 2 和 3。图像位于 x 轴上方的部分对应 y > 0，这发生在 x < 2 或 x > 3 时。
Sketching the graph is a useful check for the sign table. If the parabola opens upwards, the graph is above the x-axis outside the roots and below the x-axis between the roots.
绘制草图是检验符号表的一种有用方法。如果抛物线开口向上，则在根的外部图像位于 x 轴上方，在根之间图像位于 x 轴下方。
7. Writing the Final Answer in Set Notation | 用集合符号写出最终答案
The solution set can be written in several acceptable forms. In set notation, we write:
解集可以用几种可接受的形式表示。用集合符号，我们写成：
{ x ∈ ℝ : x < 2 or x > 3 }
In interval notation, the same solution is:
用区间符号，同样的解为：
(-∞, 2) ∪ (3, ∞)
mple 3.5.1, x = 2 and x = 3 are not part of the solution because the inequality is strict.
对于 > 或 <，始终使用严格不等式；对于 ≥ 或 ≤，则包含端点。对于示例 3.5.1，x = 2 和 x = 3 不属于解集，因为该不等式是严格的。
8. Worked Variation: Reversing the Inequality Sign | 变式练习：不等号反向
itical values are the same, but this time we need the intervals where the quadratic is negative or zero.
现在考虑相关的不等式 x² - 5x + 6 ≤ 0。因式分解和临界值相同，但这次我们需要该二次式为负或为零的区间。
gative when 2 < x < 3. It equals zero at x = 2 and x = 3. Therefore the solution is:
根据符号表，该二次式在 2 < x < 3 时为负。它在 x = 2 和 x = 3 处等于零。因此解为：
2 ≤ x ≤ 3
In interval notation, this is [2, 3]. The inclusion of endpoints is a frequent source of marks in AQA exams, so always check the inequality sign carefully.
用区间符号表示为 [2, 3]。端点的包含是 AQA 考试中常见的得分点，因此务必仔细检查不等号。
9. Common Errors to Avoid in AQA Exams | AQA 考试中要避免的常见错误
x = 3. A quadratic inequality almost always has a range of solutions, not discrete values.
一个常见错误是将不等式当作方程来处理，只写出 x = 2 或 x = 3。二次不等式几乎总是有一个解的范围，而不是离散的数值。
- Always factorise or use the quadratic formula to find critical values first.
- Do not divide both sides of an inequality by x or an expression that could be negative.
- When using a sign table, test one value from each interval, not the critical values themselves.
- Remember to reverse the inequality sign if you multiply or divide by a negative number.
中文要点：
- 务必先因式分解或使用求根公式找到临界值。
- 不要将不等式两边同时除以 x 或可能为负的表达式。
- 使用符号表时，从每个区间取一个测试值，而不是临界值本身。
- 如果乘以或除以一个负数，记得反转不等号。
10. Summary and Exam Tips | 总结与应试技巧
To solve a quadratic inequality like x² - 5x + 6 > 0, follow these steps: write the inequality in standard form, factorise the quadratic, find the critical values, test each interval using a sign table or graph, and then write the solution using set or interval notation.
要解像 x² - 5x + 6 > 0 这样的二次不等式，请遵循以下步骤：将不等式写成标准形式，对二次式进行因式分解，找到临界值，使用符号表或图像检验每个区间，然后用集合或区间符号写出解。
In AQA A Level Mathematics, questions on quadratic inequalities often appear as part of larger problem-solving tasks, such as finding the domain of a function or solving optimisation problems. Practising the sign table method will help you gain confidence and avoid careless mistakes.
在 AQA A Level 数学中，二次不等式的题目通常作为更大的问题解决任务的一部分出现，例如求函数的定义域或解决优化问题。练习符号表方法将帮助你建立信心并避免粗心错误。
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