Example 2.4.1: Solving Quadratic Inequalities | 示例2.4.1：解二次不等式

📚 Example 2.4.1: Solving Quadratic Inequalities | 示例2.4.1：解二次不等式

In AQA A-Level Mathematics, Example 2.4.1 typically introduces the systematic method for solving quadratic inequalities. This article walks through a detailed worked example, explains each step with clear reasoning, and highlights common pitfalls to avoid in the exam.

在AQA A-Level数学中，示例2.4.1通常介绍解二次不等式的系统方法。本文将逐步剖析一个完整的例题，解释每一步的推理过程，并强调考试中应避免的常见错误。

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

A quadratic inequality is a mathematical statement that relates a quadratic expression to zero (or another expression) using one of the symbols , ≤ or ≥. The general form is \(ax² + bx + c < 0\), where a ≠ 0.

二次不等式是用符号、≤ 或 ≥ 将二次表达式与零（或其他表达式）联系起来的一种数学陈述。其一般形式为 ax² + bx + c < 0，其中 a ≠ 0。

Unlike a quadratic equation, which has specific solutions, a quadratic inequality usually has a range of solutions that can be expressed as an interval or a union of intervals.

与具有特定解的二次方程不同，二次不等式通常有一系列解，这些解可用区间或区间的并集来表示。

2. The Standard Form and Our Example | 标准形式与示例

Before solving, we must ensure the quadratic expression is written in standard form with all terms on one side of the inequality sign. Our Example 2.4.1 asks us to solve:

在求解前，我们必须确保二次表达式以标准形式书写，所有项都在不等号的一侧。示例2.4.1要求我们解：

x² − 5x + 6 < 0

Here the coefficient of x² is positive, which is important for sketching the graph later. We need to find all real values of x that make this expression negative.

此处 x² 的系数为正，这对后续绘制图形很重要。我们需要找出所有使该表达式为负的实数 x 值。

3. Step 1: Factorise the Quadratic | 第一步：因式分解二次式

We look for two numbers that multiply to give +6 and add to give −5. Those numbers are −2 and −3. Therefore:

我们要寻找两个数，它们相乘得 +6，相加得 −5。这两个数是 −2 和 −3。因此：

x² − 5x + 6 = (x − 2)(x − 3)

Factorisation is the key to locating the critical values, also called roots or zeros of the quadratic expression.

因式分解是找到关键值（也称为二次表达式的根或零点）的关键。

4. Step 2: Find the Critical Values | 第二步：求临界值

Set the factorised expression equal to zero to find the boundaries of the intervals:

令因式分解后的表达式等于零，以找到区间的边界：

(x − 2)(x − 3) = 0 ⇒ x = 2 or x = 3

These two values split the real number line into three intervals: x < 2, 2 < x < 3, and x > 3.

这两个值将实数轴分成三个区间：x < 2、2 < x < 3 和 x > 3。

5. Step 3: Test Each Interval | 第三步：测试每个区间

Pick a convenient test point from each interval and substitute it into the factorised expression. The sign of the result tells us whether the inequality holds.

从每个区间选取一个方便的测试点，代入因式分解后的表达式。结果的符号告诉我们该不等式是否成立。

For x < 2, choose x = 0: (0 − 2)(0 − 3) = (−2)(−3) = +6 > 0. Not part of the solution.

对于 x < 2，取 x = 0：(0 − 2)(0 − 3) = (−2)(−3) = +6 > 0。不属于解。
For 2 < x < 3, choose x = 2.5: (2.5 − 2)(2.5 − 3) = (0.5)(−0.5) = −0.25 < 0. This interval works.

对于 2 < x < 3，取 x = 2.5：(2.5 − 2)(2.5 − 3) = (0.5)(−0.5) = −0.25 < 0。该区间满足。
For x > 3, choose x = 4: (4 − 2)(4 − 3) = (2)(1) = +2 > 0. Not part of the solution.

对于 x > 3，取 x = 4：(4 − 2)(4 − 3) = (2)(1) = +2 > 0。不属于解。

If the product is negative, the inequality x² − 5x + 6 < 0 is satisfied.

如果乘积为负，则不等式 x² − 5x + 6 < 0 成立。

6. Step 4: Write the Solution Set | 第四步：写出解集

From the interval test, the inequality holds only when x lies strictly between 2 and 3. Since the original inequality uses

由区间测试可知，只有当 x 严格位于 2 和 3 之间时，不等式才成立。由于原不等式使用

2 < x < 3

In interval notation, the solution is (2, 3). On a number line, we use open circles at 2 and 3 and shade the segment between them.

用区间记号表示为 (2, 3)。在数轴上，我们在 2 和 3 处画空心圆，并涂阴影连接这两点之间的线段。

7. Graphical Verification | 图形验证

The graph of y = x² − 5x + 6 is a parabola opening upwards because the coefficient of x² is positive. Its x-intercepts are 2 and 3.

y = x² − 5x + 6 的图像是一条开口向上的抛物线，因为 x² 的系数为正。其与 x 轴的交点为 2 和 3。

The curve lies below the x-axis exactly between the intercepts, confirming that the expression is negative for 2 < x < 3.

曲线恰好位于两个交点之间的x轴下方，这证实了当 2 < x < 3 时表达式为负。

Sketching the graph is a quick and reliable way to check your algebraic solution in the exam.

在考试中，绘制草图是检查代数解的快速而可靠的方法。

8. Handling Other Inequality Signs | 其他不等号的处理

If the inequality were ≤ or ≥, the endpoints would be included in the solution set. For example, x² − 5x + 6 ≤ 0 would have the solution 2 ≤ x ≤ 3, written [2, 3] in interval notation.

如果不等式是 ≤ 或 ≥，端点将包含在解集中。例如，x² − 5x + 6 ≤ 0 的解为 2 ≤ x ≤ 3，用区间记号写为 [2, 3]。

When the inequality sign is reversed, such as x² − 5x + 6 > 0, the solution becomes x < 2 or x > 3. In interval notation, this is (−∞, 2) ∪ (3, ∞).

当不等号反向时，例如 x² − 5x + 6 > 0，解变为 x < 2 或 x > 3。用区间记号表示为 (−∞, 2) ∪ (3, ∞)。

9. Common Mistakes | 常见错误

Dividing both sides of an inequality by a negative number without reversing the sign. This is not usually needed when solving quadratics, but careful sign handling is essential.

将不等式两边除以负数时没有反转不等号。这在解二次不等式时通常不需要，但谨慎处理符号至关重要。
Forgetting to check whether the quadratic has no real roots. If the discriminant is negative, the solution is either all real numbers or none, depending on the sign of a.

忘记检查二次式是否有实数根。如果判别式为负，则解要么是全体实数，要么是无解，这取决于 a 的符号。
Including the endpoints when the inequality is strict (< or >). Remember that open intervals exclude endpoints; closed intervals include them.

当不等式是严格不等式（< 或 >）时仍包含端点。请记住，开区间排除端点；闭区间包含端点。

10. Summary and Exam Tips | 总结与考试技巧

To solve a quadratic inequality, always start by rearranging into standard form, then factorise, find the critical values, test the intervals, and write the final answer clearly. A quick sketch of the parabola can prevent many errors.

解二次不等式时，务必先整理成标准形式，然后因式分解，求临界值，测试区间，并清晰写出最终答案。快速画出抛物线草图可以避免许多错误。

In AQA examinations, marks are often awarded for method, so show all working including the factorisation and the interval test, even if you can spot the answer by inspection.

在AQA考试中，方法往往得分，因此请展示所有过程，包括因式分解和区间测试，即使你能直接看出答案。

Practice with different coefficient signs and unequal roots to build confidence. The technique remains the same as long as you always check the original inequality.

练习不同系数符号和不等根的情况以增强信心。只要始终检查原不等式，技巧始终相同。

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Example 2.4.1: Solving Quadratic Inequalities | 示例2.4.1：解二次不等式