📚 Example 2.5.1: Solving Quadratic Inequalities | 例题2.5.1：解二次不等式
Example 2.5.1 is a classic AQA A-Level Mathematics question on quadratic inequalities. In this tutorial, we solve x² − 5x + 6 > 0 step by step and explore the reasoning behind each step.
例题2.5.1是一道经典的 AQA A-Level 数学二次不等式题。本教程将逐步解 x² − 5x + 6 > 0，并深入探讨每一步背后的逻辑。
1. The Problem Statement | 题目陈述
We are asked to solve the inequality x² − 5x + 6 > 0. This means we must find all real values of x that make the quadratic expression positive.
题目要求解不等式 x² − 5x + 6 > 0。也就是说，我们要找出所有能使这个二次表达式为正的实数 x。
At first glance, we cannot simply take square roots or divide by x, because the unknown appears in both the squared and linear terms. We need an algebraic or graphical strategy.
乍一看，我们不能直接开平方或除以 x，因为未知数同时出现在二次项和一次项中。我们需要代数方法或图像方法。
2. Factorising the Quadratic | 因式分解二次表达式
The first step is to factorise. We look for two numbers whose product is 6 and whose sum is −5. These numbers are −2 and −3.
第一步是因式分解。我们要找两个数，乘积为 6，和为 −5。这两个数就是 −2 和 −3。
x² − 5x + 6 = (x − 2)(x − 3)
We verify by expanding: (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6, which matches the original expression.
我们通过展开验证：(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6，与原式一致。
3. Finding Critical Values | 求临界值
Critical values occur where the quadratic equals zero. We set (x − 2)(x − 3) = 0 and solve:
临界值出现在二次函数等于零的地方。令 (x − 2)(x − 3) = 0 并求解：
x − 2 = 0 ⇒ x = 2, or x − 3 = 0 ⇒ x = 3
These two numbers, 2 and 3, partition the real number line into three intervals: x < 2, 2 < x < 3, and x > 3.
这两个数 2 和 3 把实数轴分成三个区间：x < 2、2 < x < 3 和 x > 3。
4. Testing the Intervals | 检验区间符号
On each interval, the sign of the product (x − 2)(x − 3) is constant, so we choose one test value per interval.
在每个区间内，乘积 (x − 2)(x − 3) 的符号保持恒定，因此我们在每个区间各取一个测试值。
| Interval | Test value | Sign of (x − 2)(x − 3) |
| x < 2 | x = 0 | (−)(−) = + |
| 2 < x < 3 | x = 2.5 | (+)(−) = − |
| x > 3 | x = 4 | (+)(+) = + |
The product is positive when x < 2 and when x > 3. It is negative in the middle interval.
乘积在 x < 2 和 x > 3 时为正，在中间区间为负。
5. Writing the Solution Set | 写出解集
Because the inequality is strict, > 0, the critical values x = 2 and x = 3 are not included. Only the positive intervals belong to the solution.
由于不等式是严格大于零，临界值 x = 2 和 x = 3 不包含在解集中。只有正区间属于解集。
x < 2 or x > 3
In set notation, we write x ∈ (−∞, 2) ∪ (3, ∞). This is the exact answer.
用集合记号写作 x ∈ (−∞, 2) ∪ (3, ∞)。这就是最终答案。
6. Graphical Interpretation | 图像解释
The graph of y = x² − 5x + 6 is an upward-opening parabola with x-intercepts at 2 and 3. Its vertex lies at x = 2.5, and the value there is −0.25.
函数 y = x² − 5x + 6 的图像是一条开口向上的抛物线，与 x 轴交于 2 和 3。其顶点在 x = 2.5 处，对应的函数值为 −0.25。
The parabola is above the x-axis for x < 2 and x > 3, and below the x-axis for 2 < x < 3. This exactly matches our algebraic solution.
抛物线在 x < 2 和 x > 3 时位于 x 轴上方，在 2 < x < 3 时位于 x 轴下方。这与我们的代数解完全吻合。
7. Verification | 验证答案
We can verify the answer by substituting values from each region into the original inequality.
我们可以从每个区域取值代入原不等式进行验证。
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For x = 0: 0² − 5(0) + 6 = 6 > 0, so x < 2 works.
取 x = 0：0² − 5(0) + 6 = 6 > 0，说明 x < 2 可行。
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For x = 2.5: 2.5² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25 < 0, so the middle interval does not work.
取 x = 2.5：2.5² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25 < 0，说明中间区间不行。
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For x = 4: 4² − 5(4) + 6 = 16 −
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