📚 Example 4.4.2: Solving Quadratic Inequalities | 例4.4.2：求解二次不等式
In this worked example, we solve a quadratic inequality step by step. The method shown here is essential for AQA A-Level Mathematics, particularly when dealing with domain and range questions, optimisation problems, and curve sketching.
在本例题中，我们将逐步求解一个二次不等式。这里展示的方法对 AQA A-Level 数学至关重要，尤其在处理定义域与值域问题、优化问题以及曲线作图时经常用到。
1. Problem Statement | 题目陈述
Solve the inequality x² − 5x + 6 > 0 for all real values of x.
解不等式 x² − 5x + 6 > 0，其中 x 为全体实数。
We are asked to find every real number x that makes the quadratic expression strictly greater than zero. This is a classic example of a quadratic inequality in standard form.
我们需要找出所有使该二次表达式严格大于零的实数 x。这是一个标准形式的二次不等式经典例子。
2. Rewrite in Standard Form | 化为标准形式
rabola opens upwards. This fact will be useful when interpreting the solution graphically.
所给不等式已经是标准形式 ax² + bx + c > 0，其中 a = 1，b = −5，c = 6。由于 a > 0，抛物线开口向上。这个事实在图形解释时非常有用。
x² − 5x + 6 > 0
标准形式已就绪
No rearrangement is required. We now move directly to solving the associated quadratic equation.
无需重新整理。我们直接求解对应的二次方程。
3. Find Critical Values | 求临界值
+ 6 = 0. Factorising gives (x − 2)(x − 3) = 0, so the critical values are x = 2 and x = 3.
为了解不等式，首先解对应的二次方程 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0，因此临界值为 x = 2 和 x = 3。
(x − 2)(x − 3) = 0 → x = 2, 3
因式分解得根
), (2, 3), and (3, ∞). On each interval, the sign of the quadratic expression is constant.
这些临界值将实数轴分成三个区间：(−∞, 2)、(2, 3) 和 (3, ∞)。在每个区间上，二次表达式的符号保持不变。
4. Sign Diagram | 符号图
We test one point from each interval to determine the sign of x² − 5x + 6.
我们在每个区间中选取一个测试点，以确定 x² − 5x + 6 的符号。
- Interval (−∞, 2), test x = 0: 0² − 0 + 6 = 6 > 0, so positive.
- 区间 (−∞, 2)，取 x = 0：0² − 0 + 6 = 6 > 0，故为正。
- Interval (2, 3), test x = 2.5: 2.5² − 12.5 + 6 = 0.25 > 0? Actually 6.25 − 12.5 + 6 = −0.25 < 0, so negative.
- 区间 (2, 3)，取 x = 2.5：2.5² − 12.5 + 6 = −0.25 < 0，故为负。
- Interval (3, ∞), test x = 4: 4² − 20 + 6 = 2 > 0, so positive.
- 区间 (3, ∞)，取 x = 4：4² − 20 + 6 = 2 > 0，故为正。
Thus the expression is positive for x < 2 and for x > 3, and negative for 2 < x < 3.
因此表达式在 x < 2 和 x > 3 时为正，在 2 < x < 3 时为负。
5. Write the Solution Set | 写出解集
Because the inequality asks for x² − 5x + 6 > 0, we take the intervals where the expression is strictly positive. The endpoints are not included since the inequality is strict.
因为不等式要求 x² − 5x + 6 > 0，我们取表达式严格为正的区间。端点为开区间，因为不等式是严格的。
x ∈ (−∞, 2) ∪ (3, ∞)
解集为 (−∞, 2) ∪ (3, ∞)
This is the final answer in set-builder notation. In interval notation, the solution is the union of two open intervals.
这是用集合描述法给出的最终答案。在区间表示法中，解集是两个开区间的并集。
6. Graphical Interpretation | 图形解释
and x = 3. Above the x-axis, the y-values are positive, which corresponds to our solution.
考虑 y = x² − 5x + 6 的图象。它是与 x 轴相交于 x = 2 和 x = 3 的抛物线。在 x 轴上方的部分，y 值为正，这正是我们的解。
y > 0 ⇔ x < 2 or x > 3
y > 0 等价于 x < 2 或 x > 3
The graph lies above the x-axis exactly on the two outer intervals. The vertex of the parabola lies below the x-axis between the roots.
图象恰好在两个外侧区间位于 x 轴上方。抛物线的顶点位于两根之间的 x 轴下方。
7. Verification | 验证
n set, say x = 2.5: 6.25 − 12.5 + 6 = −0.25, which is not > 0. The solution is consistent.
我们验证解集内的一个值，例如 x = 5。代入得 5² − 25 + 6 = 6 > 0，成立。再验证解集外的一个值，例如 x = 2.5：6.25 − 12.5 + 6 = −0.25，不大于 0，不成立。解集一致。
This verification step helps to catch sign errors and reinforces the logic of the sign-diagram method.
验证步骤有助于发现符号错误并强化符号图法的逻辑。
8. Summary and Key Points | 总结与要点
olve a quadratic inequality of the form ax² + bx + c > 0 (or ≥, <, ≤), follow these steps:
解形如 ax² + bx + c > 0（或 ≥、<、≤）的二次不等式时，遵循以下步骤：
- Ensure the quadratic is in standard form.
- 确保二次式化为标准形式。
- Solve the corresponding quadratic equation to find critical values.
- 解对应的二次方程求临界值。
- Draw a sign diagram or test points in the intervals.
- 绘制符号图或测试区间内的点。
- Select the intervals satisfying the inequality sign.
- 选取满足不等式符号的区间。
- Write the solution using interval notation, remembering strict vs non-strict inequalities.
- 使用区间表示法写出解集，注意严格不等式与不严格不等式的区别。
This method appears frequently in AQA exam questions, especially when combined with functions, domains, and optimisation. Mastery of quadratic inequalities is therefore essential for success.
此方法在 AQA 考试中频繁出现，尤其是在与函数、定义域和优化问题结合时。因此，熟练掌握二次不等式对取得好成绩至关重要。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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