📚 Example 1.4.2: Solving Quadratic Inequalities | 示例1.4.2：解二次不等式
Quadratic inequalities are a fundamental topic in A-Level mathematics. They appear in many contexts, including curve sketching, optimisation, and modelling. In this article, we will work through a specific example, labelled Example 1.4.2 from our revision series, to illustrate a robust and systematic method for solving quadratic inequalities. We will explain every step in detail and highlight the reasoning behind each operation.
二次不等式是A-Level数学中的一个基础主题。它们出现在许多情境中，包括曲线作图、优化和建模。在本文中，我们将通过一个具体示例（本复习系列中的示例1.4.2）来演示解二次不等式的一种稳健且系统的方法。我们将详细解释每一步，并强调每个操作背后的推理。
1. The Standard Form | 标准形式
A quadratic inequality is an expression of the form ax² + bx + c < 0, where the inequality sign can also be ≤, >, or ≥. Here, a, b and c are real numbers, and a must not be zero; otherwise the expression becomes linear. It is essential to rearrange any given quadratic inequality so that one side is zero before applying the solution method. For example, x² – x – 6 < 0 is already in standard form, with a = 1, b = -1 and c = -6.
二次不等式是形如 ax² + bx + c < 0 的表达式，其中不等式符号也可以是 ≤、> 或 ≥。这里，a、b、c 是实数，且 a 不能为零；否则表达式退化为线性。在应用解法之前，必须将任何给定的二次不等式重新排列，使其一边为零。例如，x² – x – 6 < 0 已经是标准形式，其中 a = 1，b = -1，c = -6。
2. The General Method | 一般方法
The general method for solving a quadratic inequality can be summarised in five steps. First, rearrange the inequality so that all terms are on one side, leaving zero on the other side. Second, factorise the quadratic expression, if possible, into a product of two linear factors. Third, find the critical values by setting each linear factor equal to zero. Fourth, determine the sign of the product on each interval created by the critical values. Fifth, select the intervals that satisfy the original inequality and write the solution using interval notation or set notation.
解二次不等式的一般方法可以概括为五个步骤。首先，重新排列不等式，使所有项都在一边，另一边为零。其次，如果可能，将二次表达式因式分解为两个一次因式的乘积。第三，令每个一次因式等于零，求出临界值。第四，确定临界值划分出的每个区间上乘积的符号。第五，选择满足原不等式的区间，并用区间表示法或集合表示法写出解。
3. Example 1.4.2 — The Problem | 示例1.4.2 — 问题
Let us consider a concrete problem from our revision series. The inequality we need to solve is x² – x – 6 < 0. This is a typical example in the AQA A-Level mathematics specification, often appearing in the first year of study. Our goal is to find all real values of x for which the quadratic expression is negative.
让我们考虑复习系列中的一个具体问题。我们需要解的不等式是 x² – x – 6 < 0。这是AQA A-Level数学大纲中的一个典型示例，通常出现在第一年的学习中。我们的目标是找到使该二次表达式为负的所有实数 x 值。
4. Factorization | 因式分解
To factorise x² – x – 6, we look for two numbers that multiply to the constant term -6 and add to the coefficient of x, which is -1. The pair of numbers that works is -3 and 2, because (-3) × 2 = -6 and (-3) + 2 = -1. Therefore, we can write the quadratic as (x – 3)(x + 2). This factorisation is a crucial step because it allows us to identify the points where the expression equals zero.
要对 x² – x – 6 进行因式分解，我们需要找到两个数，它们相乘等于常数项 -6，相加等于 x 的系数 -1。满足条件的一对数是 -3 和 2，因为 (-3) × 2 = -6 且 (-3) + 2 = -1。因此，我们可以将二次式写成 (x – 3)(x + 2)。这种因式分解是关键步骤，因为它使我们能够确定表达式等于零的点。
5. Critical Values | 临界值
Once the quadratic is factorised, we set each factor equal to zero to find the critical values. For x – 3 = 0, we get x = 3. For x + 2 = 0, we get x = -2. These values are called critical values, roots, or zeros of the quadratic. They split the real number line into three disjoint intervals: (-∞, -2), (-2, 3), and (3, ∞). These intervals are the only regions where the sign of the product could change, so we test a point from each interval.
一旦二次式被因式分解，我们令每个因式等于零来求临界值。由 x – 3 = 0，得 x = 3。由 x + 2 = 0，得 x = -2。这些值被称为临界值、根或二次式的零点。它们将实数轴分成三个不相交的区间：(-∞, -2)、(-2, 3) 和 (3, ∞)。这些区间是乘积符号可能发生变化的唯一区域，因此我们从每个区间中选取一个测试点来检验。
6. Sign Diagram | 符号图
To determine the sign of (x – 3)(x + 2) on each interval, we choose a test value from each interval. For the interval (-∞, -2), we choose x = -3. Substituting, we get (-3 – 3)(-3 + 2) = (-6)(-1) = 6, which is positive, so the product is > 0 on this interval. For the interval (-2, 3), we choose x = 0. Substituting gives (0 – 3)(0 + 2) = (-3)(2) = -6, which is negative. For the interval (3, ∞), we choose x = 4. Substituting gives (4 – 3)(4 + 2) = (1)(6) = 6, which is positive. This information is often displayed in a sign diagram, with plus and minus signs placed on a number line.
为了确定 (x – 3)(x + 2) 在每段区间上的符号，我们从每个区间中选取一个测试值。对于区间 (-∞, -2)，我们选取 x = -3。代入得 (-3 – 3)(-3 + 2) = (-6)(-1) = 6，为正，因此乘积在该区间上 > 0。对于区间 (-2, 3)，我们选取 x = 0。代入得 (0 – 3)(0 + 2) = (-3)(2) = -6，为负。对于区间 (3, ∞)，我们选取 x = 4。代入得 (4 – 3)(4 + 2) = (1)(6) = 6，为正。这些信息通常用符号图来展示，在数轴上标注正号和负号。
7. Solution Interval | 解区间
The original inequality is (x – 3)(x + 2) < 0, which asks for the product to be negative. From our sign testing, the product is negative on the interval (-2, 3). Therefore, the solution to the inequality is the open interval -2 < x < 3. In interval notation, we write x ∈ (-2, 3). It is important to use round brackets because the inequality is strict (<), meaning the critical values themselves are not included in the solution set.
原不等式为 (x – 3)(x + 2) < 0，即要求乘积为负。从我们的符号测试中，乘积在区间 (-2, 3) 上为负。因此，不等式的解是开区间 -2 < x < 3。用区间表示法写作 x ∈ (-2, 3)。注意必须使用圆括号，因为不等式是严格的 (<)，意味着临界值本身不包含在解集中。
8. Graphical Interpretation | 图形解释
A graphical perspective can help solidify the algebraic result. The function y = x² – x – 6 is a quadratic with a positive leading coefficient, so its graph is a parabola that opens upwards. The roots of the function are x = -2 and x = 3, which are precisely the critical values we found. On the interval between the roots, the parabola lies below the x-axis, meaning the output is negative. For x < -2 or x > 3, the parabola lies above the x-axis. Thus, the solution to x² – x – 6 < 0 is exactly the set of x-values for which the curve is below the x-axis.
图形的角度有助于巩固代数的结果。函数 y = x² – x – 6 是一个二次函数，其首项系数为正，因此图像是开口向上的抛物线。该函数的根是 x = -2 和 x = 3，这正是我们求出的临界值。在两个根之间，抛物线位于 x 轴下方，即函数值为负。当 x < -2 或 x > 3 时，抛物线位于 x 轴上方。因此，x² – x – 6 < 0 的解正是曲线位于 x 轴下方所对应的 x 值集合。
9. Common Mistakes | 常见错误
Many students make predictable errors when solving quadratic inequalities. One common mistake is to divide or multiply both sides by a negative number without reversing the inequality sign. In this example, we did not need to multiply by a negative number, but if we had, the sign would need to be flipped. Another mistake is to treat the inequality as if it were an equation and write x = -2 or x = 3 as the final answer. Remember that the solution is an interval, not a set of discrete points. A third mistake is to include the critical values when the inequality is strict. Since we have <, the endpoints are excluded. If the inequality were ≤, then we would use closed brackets and include the roots.
许多学生在解二次不等式时会犯可预见的错误。一个常见的错误是两边乘以或除以负数时没有反转不等式符号。在本例中，我们不需要乘以负数，但如果需要，就必须翻转不等式符号。另一个错误是将不等式当作方程来解，最终写成 x = -2 或 x = 3。请记住，解是一个区间，而不是一组离散的点。第三个错误是在不等式为严格的时候包含了临界值。由于我们处理的是 <，端点应被排除。如果不等式是 ≤，那么我们就使用闭区间并包含根。
10. Practice Exercise | 练习
Now that we have worked through Example 1.4.2, let us practise a similar problem. Solve the inequality x² + 2x – 8 ≥ 0. Factorising gives (x + 4)(x – 2) ≥ 0. The critical values are x = -4 and x = 2. Testing the intervals, we find the product is positive on (-∞, -4) and (2, ∞), and negative on (-4, 2). Since we need the product to be greater than or equal to zero, the solution is x ≤ -4 or x ≥ 2, written as (-∞, -4] ∪ [2, ∞). Remember to include the critical values because the inequality is non-strict.
在完成示例1.4.2之后，让我们练习一个类似的问题。解不等式 x² + 2x – 8 ≥ 0。因式分解得 (x + 4)(x – 2) ≥ 0。临界值为 x = -4 和 x = 2。测试区间，我们发现乘积在 (-∞, -4) 和 (2, ∞) 上为正，在 (-4, 2) 上为负。由于我们需要乘积大于或等于零，所以解为 x ≤ -4 或 x ≥ 2，写作 (-∞, -4] ∪ [2, ∞)。记住要包含临界值，因为不等式是非严格的。
11. Summary | 总结
In this revision article, we have used Example 1.4.2 to demonstrate the complete process of solving a quadratic inequality. The key steps are: rearrange to standard form, factorise, find critical values, test signs, and write the solution interval. We have also discussed the graphical interpretation and common pitfalls. Mastering this technique will help you solve more complex inequalities, such as those involving rational functions or absolute values, which appear later in the A-Level course.
在这篇复习文章中，我们通过示例1.4.2演示了解决二次不等式的完整过程。关键步骤是：整理成标准形式、因式分解、求临界值、测试符号、写出解区间。我们还讨论了图形解释和常见陷阱。掌握这一技巧将帮助您解决更复杂的不等式，例如涉及有理函数或绝对值的不等式，这些会在A-Level课程的后续部分出现。
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