📚 A-Level Mathematics Example 1.7.1: Solving Quadratic Inequalities | A-Level 数学例 1.7.1：解二次不等式
Quadratic inequalities are a core topic in A-Level Mathematics, especially for the AQA syllabus. In this worked example, we will solve a typical quadratic inequality step by step, using both algebraic manipulation and a sign table. This approach is widely tested in examinations and forms the foundation for more advanced topics such as curve sketching and optimisation.
二次不等式是 A-Level 数学的核心主题，尤其是在 AQA 考纲中。在本题例中，我们将逐步求解一个典型的二次不等式，运用代数变换和符号表两种方法。这类方法在考试中广为考查，也是学习曲线草图与最优化等进阶主题的基础。
1. Understanding the Problem | 理解问题
Suppose Example 1.7.1 asks us to solve the inequality:
假设例 1.7.1 要求我们解不等式：
x² − 5x + 6 < 0
This inequality means that we need to find every real number x for which the quadratic expression x² − 5x + 6 takes a negative value. In other words, we are looking for the interval(s) on the x-axis where the corresponding parabola lies strictly below the horizontal axis.
这个不等式意味着我们需要找出所有使二次表达式 x² − 5x + 6 取负值的实数 x。换句话说，我们要在 x 轴上找到对应的抛物线严格位于水平轴下方的区间。
quality is usually a range of values. The roots help us split the real number line into distinct intervals, which we then test individually.
一个常见错误是将不等式视为方程而只写出 x = 2 或 x = 3。实际上，不等式的解通常是一个取值范围。根帮助我们分割实数轴为若干区间，然后逐一检验各区间的符号。
2. Step 1: Factorise the Quadratic | 第一步：因式分解二次式
The first step is to factorise the quadratic expression, if possible. We look for two numbers that multiply to give +6 (the constant term) and add to give −5 (the coefficient of x). These numbers are −2 and −3.
第一步是尽可能对二次表达式进行因式分解。我们需要寻找两个数，它们相乘为 +6（常数项），相加为 −5（x 的系数）。这两个数是 −2 和 −3。
x² − 5x + 6 = (x − 2)(x − 3)
(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6. This confirms that the factorisation is correct.
你应该总是通过展开括号来验证因式分解是否正确：(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6。这证实分解结果是正确的。
If the quadratic does not factorise easily, you may need to use the quadratic formula or complete the square. However, for this example, factorisation is straightforward.
如果二次式不容易因式分解，你可能需要使用二次公式或配方法。然而，在本例中因式分解非常直接。
3. Step 2: Find the Roots | 第二步：求根
he roots (also called zeros or critical values) of the quadratic equation x² − 5x + 6 = 0.
现在我们令每个因子等于零，以求出二次方程 x² − 5x + 6 = 0 的根（也称为零点或临界值）。
- x − 2 = 0 ⟹ x = 2
x − 2 = 0 ⟹ x = 2 - x − 3 = 0 ⟹ x = 3
x − 3 = 0 ⟹ x = 3
These two roots divide the x-axis into three separate intervals: x < 2, 2 < x < 3, and x > 3. On each interval, the product (x − 2)(x − 3) will have a consistent sign — either positive or negative.
这两个根将 x 轴分为三个独立的区间：x < 2、2 < x < 3 和 x > 3。在每个区间内，乘积 (x − 2)(x − 3) 的符号是一致的——要么为正，要么为负。
Note that for a strict inequality '<', the roots themselves are not included in the solution set. Therefore, we will use open intervals when writing the final answer.
请注意，对于严格不等式 '<'，根本身不包括在解集中。因此，在写最终答案时我们将使用开区间。
4. Step 3: Build a Sign Table | 第三步：构建符号表
A sign table is a practical way to determine the sign of each factor and their product across the intervals. We choose a test value in each interval and evaluate the signs of (x − 2) and (x − 3).
符号表是一个实用的工具，用于确定每个因子及其乘积在各区间内的符号。我们在每个区间内选取一个测试值，并计算 (x − 2) 和 (x − 3) 的符号。
| Interval 区间 |
Test x 测试 x |
Sign of (x − 2) (x − 2) 的符号 |
Sign of (x − 3) (x − 3) 的符号 |
Sign of product 乘积的符号 |
| x < 2 | 1 | − | − | + |
| 2 < x < 3 | 2.5 | + | − | − |
| x > 3 | 4 | + | + | + |
e). The product of two negative numbers is positive, so the quadratic is positive on the interval x < 2.
例如，当 x = 1 时，(1 − 2) = −1（负），(1 − 3) = −2（负）。两个负数相乘为正，因此二次式在区间 x < 2 上为正。
The sign table clearly displays that the product is negative only in the middle interval, 2 < x < 3. This is exactly the region we are looking for because our inequality requires the expression to be less than zero.
符号表清楚地显示乘积仅在中间的区间 2 < x < 3 上为负。这正是我们要找的区域，因为我们的不等式要求表达式小于零。
5. Step 4: Interpret the Signs | 第四步：解读符号
isely when 2 < x < 3. No other interval gives a negative product. What about the endpoints? x = 2 and x = 3, the expression equals zero, which does not satisfy the strict inequality '<'. Therefore, these points are excluded.
从符号表可以看出，(x − 2)(x − 3) < 0 恰好发生在 2 < x < 3 时。其他区间不产生负乘积。那么端点如何呢？在 x = 2 和 x = 3 处，表达式等于零，这不满足严格不等式 '<'。因此，这些点被排除在外。
This interpretation is consistent with the behaviour of a positive coefficient quadratic: it is positive outside the roots and negative between the roots. Since the coefficient of x² is 1 (which is positive), the parabola opens upwards, confirming our findings.
这一解读与正系数二次式的行为一致：在根之外为正，在根之间为负。由于 x² 的系数为 1（正数），抛物线开口向上，这证实了我们的发现。
3) = 0.5 × (−0.5) = −0.25, which is indeed negative. This supports the sign table result.
快速心算检验：在 2 和 3 之间任选一个数，比如 x = 2.5。则 (2.5 − 2)(2.5 − 3) = 0.5 × (−0.5) = −0.25，确实是负数。这支持符号表的结果。
6. Step 5: Write the Solution Set | 第五步：写出解集
The solution to the inequality x² − 5x + 6 < 0 is the set of all x that lie strictly between 2 and 3. There are several acceptable ways to express this answer.
不等式 x² − 5x + 6 < 0 的解是所有严格位于 2 和 3 之间的 x 的集合。有几种可接受的表达方式。
2 < x < 3
In interval notation, this is written as:
用区间记号表示，即：
x ∈ (2, 3)
, which would include the endpoints and therefore be incorrect for this strict inequality.
在 AQA 考试中，两种形式均可接受，只要你使用圆括号或严格不等式来表明端点不被包含。常见错误是写成 [2, 3] 或 2 ≤ x ≤ 3，那将包括端点，因此对于这个严格不等式是错误的。
the inequality were ≤, then the solution would be 2 ≤ x ≤ 3, including the roots. But for '<', we exclude them.
务必仔细阅读题目：如果不等式是 ≤，则解为 2 ≤ x ≤ 3，包括根在内。但对于 '<'，我们排除它们。
7. Alternative Method: Sketching the Graph | 替代方法：画出函数图像
and x = 3. Since the coefficient of x² is positive, the parabola opens upwards, forming a 'U' shape.
另一个有力的方法是画出 y = x² − 5x + 6 的图像。抛物线在 x = 2 和 x = 3 处穿过 x 轴。由于 x² 的系数为正，抛物线开口向上，形成 'U' 形。
The inequality x² − 5x + 6 < 0 asks for the portions of the curve below the x-axis. Looking at the sketch, the curve is below the x-axis exactly between the two roots. Hence, the solution is 2 < x < 3.
不等式 x² − 5x + 6 < 0 要求曲线在 x 轴下方的部分。从图像上看，曲线恰好位于两个根之间的 x 轴下方。因此，解为 2 < x < 3。
Sketching is particularly useful when dealing with quadratic inequalities that have irrational roots or when you need to verify your algebraic solution. However, in a written exam, you must show the algebraic steps to earn full marks; a graph alone may not be sufficient.
当处理具有无理根的二次不等式或需要验证代数解时，草图法尤其有用。然而，在笔试中，你必须展示代数步骤才能获得满分；仅画图可能不够。
.5. The vertex lies below the x-axis, at (2.5, −0.25), which is consistent with the negative values between the roots.
要精确作图，你应确定根、y 截距（即 6）以及对称轴 x = 2.5。顶点位于 x 轴下方，坐标为 (2.5, −0.25)，这与根之间为负值一致。
8. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Even strong students often lose marks on quadratic inequalities. Here are the most frequent errors, along with advice to avoid them.
即使是优秀的学生也常在二次不等式上失分。以下是最常见的错误及避免建议。
- Treating the inequality as an equation: Always remember that the solution is an interval, not a single value. Factorise, find the roots, then test intervals.
将不等式视为方程：始终记住解是一个区间，而不是单一值。因式分解、求根、然后测试区间。 - Wrong interval notation: : Use round brackets for strict inequalities and square brackets for non-strict ones. For example, x² − 5x + 6 ≤ 0 gives [2, 3], while the '<' version gives (2, 3).
错误的区间记号：严格不等式用圆括号，非严格不等式用方括号。例如，x² − 5x + 6 ≤ 0 给出 [2, 3]，而 '<' 版本给出 (2, 3)。 - Incorrect sign table: When constructing a sign table, always choose test values that are clearly inside each interval and avoid using the roots themselves as tests.
符号表错误：构建符号表时，务必选择明确位于各区间内的测试值，避免使用根本身作为测试。 - Misinterpreting the leading coefficient: If the coefficient of x² is negative, the parabola opens downwards, and the signs are reversed. For example, −x² + 5x − 6 > 0 is equivalent to x² − 5x + 6 < 0 (multiply by −1 and flip the inequality).
误解首项系数：如果 x² 的系数为负，抛物线开口向下，符号会反向。例如，−x² + 5x − 6 > 0 等价于 x² − 5x + 6 < 0（乘以 −1 并翻转不等式）。 - Skipping simplification: Before factorising, always expand and collect like terms. If the inequality is given in a non-standard form, rearrange it to the form ax² + bx + c < 0 first.
跳步化简：在因式分解前，务必展开并合并同类项。如果不等式以非标准形式给出，先整理为 ax² + bx + c < 0 的形式。
To check your final answer, substitute a value from the solution interval into the original inequality. Also substitute a value outside the interval to ensure it does not satisfy the inequality.
为检验最终答案，将解区间内的一个值代入原不等式。同时代入区间外的值，确保它不满足该不等式。
9. Exam Tips for AQA A-Level Mathematics | AQA A-Level 数学考试提示
For AQA A-Level Mathematics, quadratic inequalities typically appear in the Pure Mathematics papers. They may be standalone questions or part of a larger problem involving functions, graphs, or modelling.
对于 AQA A-Level 数学，二次不等式通常出现在纯数学试卷中。它们可能是独立题目，也可能作为涉及函数、图像或建模的大型问题的一部分。
- Show all working: AQA mark schemes award method marks. Even if your final answer is wrong, correct intermediate steps can earn partial credit.
展示所有步骤：AQA 评分方案会授予方法分。即使最终答案错误，正确的中间步骤也能获得部分分数。 - Know the standard form: Learn to factorise quadratics quickly, including the difference of two squares and perfect squares. This speeds up your solution process.
掌握标准形式：学会快速分解二次式，包括平方差和完全平方。这能加快你的解题过程。 - Use interval notation correctly: l with infinity signs: e.g., x > 2 is written as (2, ∞), not (2, ∞) with a square bracket.
正确使用区间记号：练习用不等式和区间两种形式书写答案。注意无穷符号：例如，x > 2 写为 (2, ∞)，而不是带方括号。 - Combine with other topics: Quadratic inequalities often appear alongside discriminant analysis. For instance, you might be asked to find the range of values of k for which a quadratic has no real roots. This requires setting the discriminant Δ < 0 and solving a quadratic inequality.
与其他主题结合：二次不等式常与判别式分析一起出现。例如，你可能会被要求求出使二次式无实根时 k 的取值范围。这需要令判别式 Δ < 0 并解二次不等式。 - Remember the graph: Visualising the parabola can help you quickly state the solution. A quick sketch is a valuable tool during the exam, even if not officially required.
记住图像：想象抛物线有助于你快速写出解。即使在考试中不是必须，快速草图也是一个有价值的工具。
Practice with past papers and focus on the precision of your notation. The difference between (2, 3) and [2, 3] can be a whole mark in a question worth only a couple of marks.
通过历年真题进行练习，并注重符号的精确性。在满分仅几分的题目中，(2, 3) 与 [2, 3] 的差别可能价值整整一分。
10. Practice Questions | 练习题目
To consolidate your understanding, attempt the following questions. Use the same step-by-step method: factorise, find roots, test intervals, and write the solution in interval notation.
为巩固理解，请尝试以下问题。使用同样的逐步方法：因式分解、求根、测试区间、并用区间记号写出解。
- Question 1: Solve x² + x − 6 > 0. Hint: factorises to (x + 3)(x − 2). The solution is x < −3 or x > 2.
题目 1：解 x² + x − 6 > 0。提示：可分解为 (x + 3)(x − 2)。解为 x < −3 或 x > 2。 - Question 2: (2x + 1)(x − 3). The roots are −0.5 and 3. Since ≤ is used, the solution is −0.5 ≤ x ≤ 3.
题目 2：解 2x² − 5x − 3 ≤ 0。分解为 (2Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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