Example 4.4.2: Solving Quadratic Inequalities | 例4.4.2：求解二次不等式

📚 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。

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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。

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2. Rewrite in Standard Form | 化为标准形式

The inequality is already given in the standard form ax² + bx + c > 0, with a = 1, b = −5, c = 6. Since a > 0, the parabola 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。

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3. Find Critical Values | 求临界值

To solve the inequality, first solve the corresponding quadratic equation x² − 5x + 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

These critical values divide the real number line into three intervals: (−∞, 2), (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。

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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。

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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。

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6. Graphical Interpretation | 图形解释

Consider the graph of y = x² − 5x + 6. It is a parabola intersecting the x-axis at x = 2 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。

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7. Verification | 验证

We check a value inside the solution set, say x = 5. Substituting: 5² − 25 + 6 = 6 > 0, which is true. We check a value outside the solution 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。

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8. Summary and Key Points | 总结与要点

To solve 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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