Solving Quadratic Inequalities | 解二次不等式

📚 Solving Quadratic Inequalities | 解二次不等式

Welcome to this revision guide on solving quadratic inequalities — a core skill in AQA A-Level Mathematics. In this article, we will work through Example 1.7.3 in detail, explore two reliable solution methods, and highlight common errors so you can avoid them in your exams.

欢迎阅读本篇关于解二次不等式的复习指南——这是 AQA A-Level 数学中的核心技能。在本文中，我们将详细解析 Example 1.7.3，探讨两种可靠的解题方法，并指出常见错误以帮助你在考试中避免失分。

1. What Is a Quadratic Inequality? | 什么是二次不等式？

A quadratic inequality involves a quadratic expression and one of the inequality signs: <, >, ≤, or ≥. The general form is ax² + bx + c < 0 (or > 0, ≤ 0, ≥ 0), where a ≠ 0. Solving such an inequality means finding the range of x-values for which the statement is true.

二次不等式涉及一个二次表达式和以下不等式符号之一：<（小于）、>（大于）、≤（小于或等于）或 ≥（大于或等于）。其一般形式为 ax² + bx + c < 0（或 > 0、≤ 0、≥ 0），其中 a ≠ 0。解这类不等式就是找出使该不等式成立的 x 的取值范围。

Unlike a quadratic equation, which usually gives discrete roots, a quadratic inequality typically yields an interval (or union of intervals) of x-values. For instance, solving x² − 5x + 6 < 0 does not give “x equals something” but rather “2 < x < 3”.

与通常给出离散根的二次方程不同，二次不等式通常给出的是一个区间（或区间的并集）的 x 值。例如，解 x² − 5x + 6 < 0 得到的不是”x 等于某值”，而是”2 < x < 3″。

2. Why Example 1.7.3 Matters | 为什么 Example 1.7.3 很重要

In the AQA Pure Mathematics specification, Example 1.7.3 is a benchmark problem designed to test your ability to combine factorisation, sign analysis, and interval notation. It typically asks you to solve an inequality such as x² − 5x + 6 < 0 or 2x² + 3x − 2 ≥ 0. Mastering this example builds a foundation for later topics like curve sketching, optimisation, and even calculus-based reasoning.

在 AQA 纯数学考纲中，Example 1.7.3 是一个标杆题目，旨在考查你综合运用因式分解、符号分析和区间记法的能力。它通常要求你解一个类似 x² − 5x + 6 < 0 或 2x² + 3x − 2 ≥ 0 的不等式。掌握这个例题能为后续课题奠定基础，如曲线草图、最优化问题，乃至基于微积分的推理。

3. The Standard Form and Factorisation | 标准形式与因式分解

The first step in solving any quadratic inequality is to rearrange it into standard form. This means bringing all terms to one side so that the other side is zero. For example, if you are given x² + 1 < 4x, you must rewrite it as x² − 4x + 1 < 0 before proceeding.

解任何二次不等式的第一步都是将其整理为标准形式。这意味着将所有项移到一边，使另一边为零。例如，若题目给出 x² + 1 < 4x，你必须先将其改写为 x² − 4x + 1 < 0 再进行后续步骤。

Once in standard form, try to factorise the quadratic expression. If the expression factorises cleanly, the roots are immediately visible. For instance, x² − 5x + 6 factorises as (x − 2)(x − 3), giving roots x = 2 and x = 3. These roots are the critical values that divide the x-axis into intervals.

一旦整理成标准形式，尝试对二次表达式进行因式分解。如果表达式能干净地分解，根就一目了然。例如，x² − 5x + 6 可分解为 (x − 2)(x − 3)，得到根 x = 2 和 x = 3。这些根就是划分 x 轴区间的临界值。

4. Method One: The Sign Diagram Approach | 方法一：符号图表法

The sign diagram method is systematic and works for all factorisable quadratic inequalities. Begin by drawing a horizontal line representing the x-axis. Mark the critical roots on this line. In the case of x² − 5x + 6 < 0, mark x = 2 and x = 3.

符号图表法系统性强，适用于所有可因式分解的二次不等式。首先画一条水平线代表 x 轴，在线上标出临界根。对于 x² − 5x + 6 < 0，标出 x = 2 和 x = 3。

Next, choose a test value from each of the three intervals: (−∞, 2), (2, 3), and (3, ∞). For the interval (−∞, 2), pick x = 0. Substitute into (x − 2)(x − 3): (−2)(−3) = 6, which is positive. For the interval (2, 3), pick x = 2.5: (0.5)(−0.5) = −0.25, which is negative. For the interval (3, ∞), pick x = 4: (2)(1) = 2, which is positive.

接下来，在三个区间 (−∞, 2)、(2, 3) 和 (3, ∞) 中各取一个测试值。对于区间 (−∞, 2)，取 x = 0。代入 (x − 2)(x − 3) 得 (−2)(−3) = 6，为正。对于区间 (2, 3)，取 x = 2.5：得 (0.5)(−0.5) = −0.25，为负。对于区间 (3, ∞)，取 x = 4：得 (2)(1) = 2，为正。

Since we need the expression to be less than zero, we select the interval where the product is negative, namely 2 < x < 3. If the inequality were ≤ 0, we would include the endpoints, giving 2 ≤ x ≤ 3.

因为我们需要表达式小于零，所以选择乘积为负的区间，即 2 < x < 3。若不等式是 ≤ 0，我们应包含端点，得到 2 ≤ x ≤ 3。

Sign pattern for (x − 2)(x − 3): positive for x < 2, negative for 2 < x < 3, positive for x > 3.

(x − 2)(x − 3) 的符号规律：x < 2 为正，2 < x < 3 为负，x > 3 为正。

5. Method Two: The Graphical Approach | 方法二：图像法

The graphical method relies on visualising the parabola y = ax² + bx + c. For a > 0, the parabola opens upward like a U-shape. The portions of the curve below the x-axis correspond to y < 0, while the portions above correspond to y > 0.

图像法依赖于视化抛物线 y = ax² + bx + c。当 a > 0 时，抛物线开口向上呈 U 形。曲线在 x 轴下方的部分对应 y < 0，而在上方的部分对应 y > 0。

For y = x² − 5x + 6, the parabola crosses the x-axis at x = 2 and x = 3. Since a = 1 > 0, the curve is below the x-axis precisely between the two roots. Thus, the solution to x² − 5x + 6 < 0 is visibly 2 < x < 3.

对于 y = x² − 5x + 6，抛物线在 x = 2 和 x = 3 处穿过 x 轴。由于 a = 1 > 0，曲线恰好位于两个根之间的 x 轴下方。因此，x² − 5x + 6 < 0 的解直观可见为 2 < x < 3。

For the case a < 0 (a downward-opening parabola), the inequality reverses: y < 0 occurs outside the interval between the roots. For example, for −x² + 4x − 3 < 0, the roots are x = 1 and x = 3, and the solution is x < 1 or x > 3.

对于 a < 0（开口向下的抛物线）的情况，不等号方向相反：y < 0 出现在两根之外。例如，对于 −x² + 4x − 3 < 0，根为 x = 1 和 x = 3，解为 x < 1 或 x > 3。

6. Worked Example: Example 1.7.3 in Full | 完整解析 Example 1.7.3

Let us now work through the full solution of a typical Example 1.7.3 problem: Solve the inequality 2x² + 3x − 2 ≥ 0.

现在我们来完整解一道典型的 Example 1.7.3 题目：解不等式 2x² + 3x − 2 ≥ 0。

Step 1: Factorise — Find two numbers whose product is 2 × (−2) = −4 and whose sum is 3. The numbers are 4 and −1. Splitting the middle term: 2x² + 4x − x − 2 = 2x(x + 2) − 1(x + 2) = (2x − 1)(x + 2).

步骤 1：因式分解 —— 找到两个数，其乘积为 2 × (−2) = −4，其和为 3。这两个数是 4 和 −1。拆分中间项：2x² + 4x − x − 2 = 2x(x + 2) − 1(x + 2) = (2x − 1)(x + 2)。

Step 2: Find the critical roots — Set (2x − 1)(x + 2) = 0, giving x = ½ and x = −2.

步骤 2：求临界根 —— 令 (2x − 1)(x + 2) = 0，得 x = ½ 和 x = −2。

Step 3: Analyse the sign — The coefficient a = 2 > 0, so the parabola opens upward. The product is positive when x < −2 or x > ½, and negative when −2 < x < ½. Verify with test values: choose x = −3: (2(−3) − 1)(−3 + 2) = (−7)(−1) = 7 > 0 ✓. Choose x = 0: (−1)(2) = −2 < 0 ✓. Choose x = 1: (1)(3) = 3 > 0 ✓.

步骤 3：符号分析 —— 系数 a = 2 > 0，所以抛物线开口向上。当 x < −2 或 x > ½ 时乘积为正，当 −2 < x < ½ 时乘积为负。用测试值验证：取 x = −3：(2(−3) − 1)(−3 + 2) = (−7)(−1) = 7 > 0 ✓。取 x = 0：(−1)(2) = −2 < 0 ✓。取 x = 1：(1)(3) = 3 > 0 ✓。

Step 4: Write the answer — Since the inequality is ≥ 0, we include the roots. The solution is x ≤ −2 or x ≥ ½. In interval notation: (−∞, −2] ∪ [½, ∞).

步骤 4：写出答案 —— 因为不等式是 ≥ 0，我们包含根。解为 x ≤ −2 或 x ≥ ½。用区间记法表示为：(−∞, −2] ∪ [½, ∞)。

Final answer: x ∈ (−∞, −2] ∪ [½, ∞)

最终答案：x ∈ (−∞, −2] ∪ [½, ∞)

7. Strict vs. Non-Strict Inequalities | 严格不等式与非严格不等式

One of the most frequent mistakes students make is confusing strict (< and >) with non-strict (≤ and ≥) inequalities. Strict inequalities exclude the critical roots from the solution set, while non-strict inequalities include them.

学生最常犯的错误之一是混淆严格不等式（< 和 >）与非严格不等式（≤ 和 ≥）。严格不等式将临界根排除在解集之外，而非严格不等式则将临界根包含在内。

Inequality Type \| 不等式类型	Example \| 示例	Solution \| 解
Strict \| 严格	x² − 5x + 6 < 0	2 < x < 3
Non-strict \| 非严格	x² − 5x + 6 ≤ 0	2 ≤ x ≤ 3
Strict \| 严格	x² − 5x + 6 > 0	x < 2 or x > 3
Non-strict \| 非严格	x² − 5x + 6 ≥ 0	x ≤ 2 or x ≥ 3

In graphical terms, a strict inequality uses open circles on the number line (indicating the roots are not included), while a non-strict inequality uses closed circles (indicating the roots are included).

在图像表示中，严格不等式在数轴上使用空心圆（表示不包含根），而非严格不等式使用实心圆（表示包含根）。

8. Dealing with Unfactorisable Quadratics | 处理不可因式分解的二次式

When the quadratic cannot be factorised over the integers, you can still solve the inequality using the quadratic formula. For ax² + bx + c = 0, the roots are given by:

当二次式无法在整数范围内因式分解时，你仍然可以通过求根公式来解不等式。对于 ax² + bx + c = 0，根由下式给出：

x = (−b ± √(b² − 4ac)) / 2a

If the discriminant (b² − 4ac) is positive, there are two distinct real roots, and the sign analysis proceeds exactly as before. If the discriminant is zero, the parabola touches the x-axis at a single point; the inequality x² + 2x + 1 > 0 then holds for all x except x = −1. If the discriminant is negative, the parabola never touches the x-axis; the inequality is either always true or never true, depending on the sign of a.

如果判别式 (b² − 4ac) 为正，则有两个不同的实根，符号分析与之前完全相同。如果判别式为零，抛物线在 x 轴上仅触及一点；不等式 x² + 2x + 1 > 0 对所有 x 都成立，除了 x = −1。如果判别式为负，抛物线永远不会触及 x 轴；不等式要么恒真，要么恒假，取决于 a 的符号。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Let us examine four typical errors that students make when solving quadratic inequalities, along with strategies to avoid them.

让我们来审视学生在解二次不等式时常见的四类错误，以及相应的避免策略。

Error 1: Multiplying or dividing by a variable. Unlike linear inequalities, you should never multiply or divide both sides by an expression containing x, because you do not know whether it is positive or negative. Always rearrange to one side first.
错误 1：乘以或除以一个变量。与线性不等式不同，切勿将两边同时乘以或除以含有 x 的表达式，因为你不知道其正负。始终先整理到一边。

Error 2: Writing the answer as a single interval. For a > 0, the solution to (x − 2)(x − 3) > 0 is x < 2 or x > 3 — not 3 < x < 2. These are two disjoint intervals and must be written separately.
错误 2：将答案写成一个单一区间。对于 a > 0，(x − 2)(x − 3) > 0 的解是 x < 2 或 x > 3——而不是 3 < x < 2。这是两个不相交的区间，必须分开书写。

Error 3: Forgetting to reverse the inequality when a < 0. If you multiply both sides by −1 to make a positive, you must reverse the inequality sign. For example, −x² + 4 < 0 becomes x² − 4 > 0.
错误 3：当 a < 0 时忘记反转不等号。如果两边同时乘以 −1 以使其为正，必须反转不等号。例如，−x² + 4 < 0 变为 x² − 4 > 0。

Error 4: Ignoring the endpoints. Always check whether the original inequality uses <, >, ≤, or ≥, and adjust your answer accordingly. Missing the equality condition loses marks in AQA exams.
错误 4：忽略端点。始终检查原不等式使用的是 <、>、≤ 还是 ≥，并相应调整答案。遗漏等号条件会在 AQA 考试中丢分。

10. Practice Problems | 练习题

Test your understanding with these five problems. Work through each one fully before checking your answers.

通过以下五道题来测试你的理解。每道题先完整作答，再核对答案。

1. Solve x² − 9 > 0.

1. 解 x² − 9 > 0。

2. Solve x² + 4x + 3 ≤ 0.

2. 解 x² + 4x + 3 ≤ 0。

3. Solve 3x² − 10x + 3 < 0.

3. 解 3x² − 10x + 3 < 0。

4. Solve 4x² − 4x + 1 ≥ 0 (hint: think about the discriminant).

4. 解 4x² − 4x + 1 ≥ 0（提示：思考判别式）。

5. Solve −2x² + 8x − 6 < 0.

5. 解 −2x² + 8x − 6 < 0。

Answers: 1) x < −3 or x > 3 2) −3 ≤ x ≤ −1 3) 1/3 < x < 3 4) all real x 5) x < 1 or x > 3

答案：1) x < −3 或 x > 3 2) −3 ≤ x ≤ −1 3) 1/3 < x < 3 4) 所有实数 x 5) x < 1 或 x > 3

11. Summary and Exam Tips | 总结与考试技巧

To summarise, solving a quadratic inequality always follows the same four-step process: rearrange to standard form, factorise (or use the quadratic formula), sketch the sign pattern or parabola, and write the final answer using the correct inequality sign and interval notation.

总而言之，解二次不等式始终遵循相同的四步流程：整理为标准形式，因式分解（或使用求根公式），画出符号图或抛物线，以及用正确的不等号和区间记法写出最终答案。

In the AQA exam, marks are typically awarded for each stage of the process. Therefore, even if you cannot factorise, you can still earn method marks by using the quadratic formula and sign analysis. Always state your critical values explicitly and show your test values when using the sign diagram method.

在 AQA 考试中，通常每个阶段都会给予过程分。因此，即使你无法因式分解，你仍然可以通过使用求根公式和符号分析获得方法分。使用符号图法时，务必明确写出你的临界值，并展示测试值。

Finally, remember that practice is the key to mastery. Work through as many examples as you can, including both factorisable and unfactorisable quadratics, and always verify your answer by substituting a value from your solution interval back into the original inequality.

最后，请记住熟能生巧。尽可能多地练习各种例题，包括可因式分解和不可因式分解的二次不等式，并且始终通过将解区间中的某个值代回原不等式来验证你的答案。

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Solving Quadratic Inequalities | 解二次不等式