📚 Example 3.5.2: Solving Quadratic Inequalities | 二次不等式求解
In this AQA A-Level Mathematics worked example, we tackle a core Pure topic: solving a quadratic inequality. Example 3.5.2 asks us to find all real values of x for which a quadratic expression is greater than zero. This skill is examined across AS and A-Level, often as part of coordinate geometry, curve sketching, and optimisation problems.
在这道 AQA A-Level 数学例题中，我们处理一个纯数学核心主题：解二次不等式。例题 3.5.2 要求我们找出使一个二次式大于零的所有实数 x。这项技能在 AS 和 A-Level 考试中都会出现，常见于坐标几何、曲线作图以及优化问题中。
1. The Example Statement | 例题题干
Example 3.5.2 asks: Solve the inequality x² - 5x + 6 > 0.
例题 3.5.2 要求：解不等式 x² - 5x + 6 > 0。
The expression on the left is a quadratic in standard form. We need to determine where its graph lies above the x-axis, because the inequality is strictly greater than zero.
左边是一个标准形式的二次式。我们需要确定它的图像在 x 轴上方的位置，因为该不等式是严格大于零。
2. Standard Form and Leading Coefficient | 标准形式与二次项系数
the inequality is in the standard form ax² + bx + c > 0. Here a = 1, b = -5, and c = 6.
首先，检查不等式是否为标准形式 ax² + bx + c > 0。这里 a = 1，b = -5，c = 6。
Because a is positive, the parabola opens upwards. This means the graph has a minimum point and rises to positive values on both sides far from the vertex.
因为 a 为正，抛物线开口向上。这意味着图像有一个最低点，并且在远离顶点两侧趋向正无穷。
For upward-opening quadratics, the expression is positive outside the roots and negative between the roots.
对于开口向上的二次函数，二次式在两根之外为正，在两根之间为负。
3. Factorising the Quadratic | 分解二次式
We factorise the left-hand side by looking for two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3.
我们通过寻找两个乘积为 6、和为 -5 的数来分解左边。这两个数是 -2 和 -3。
x² - 5x + 6 = (x - 2)(x - 3)
The inequality can therefore be rewritten as (x - 2)(x - 3) > 0.
因此原不等式可以改写为 (x - 2)(x - 3) > 0。
4. Identifying Critical Values | 确定临界值
o find the critical values. The equation x - 2 = 0 gives x = 2, and x - 3 = 0 gives x = 3.
令每个因式等于零，求出临界值。方程 x - 2 = 0 得 x = 2，x - 3 = 0 得 x = 3。
values split the real number line into three open intervals: (-∞, 2), (2, 3), and (3, ∞).
这些临界值将实数轴分成三个开区间：(-∞, 2)、(2, 3) 和 (3, ∞)。
cause the inequality is strict, neither x = 2 nor x = 3 is included in the final solution.
因为该不等式是严格的，x = 2 和 x = 3 都不包含在最终解中。
5. Sketching the Parabola | 绘制抛物线草图
e graph of y = (x - 2)(x - 3). It crosses the x-axis at x = 2 and x = 3 and opens upwards.
画出 y = (x - 2)(x - 3) 的图像。它在 x = 2 和 x = 3 处与 x 轴相交，并开口向上。
The portion above the x-axis corresponds to y > 0, while the portion below corresponds to y < 0.
在 x 轴上方的部分对应 y > 0，而在 x 轴下方的部分对应 y < 0。
From the sketch, the graph is above the x-axis when x is less than 2 or greater than 3.
从图像可以看出，当 x 小于 2 或大于 3 时，图像位于 x 轴上方。
6. Testing Intervals | 区间测试
To confirm the sign of (x - 2)(x - 3) in each interval, substitute a test value from each region.
为了确认 (x - 2)(x - 3) 在每个区间的符号，从每个区域代入一个测试值。
| Interval | Test value | Sign of (x - 2)(x - 3) |
|---|---|---|
| (-∞, 2) | 0 | (-)(-) = positive |
| (2, 3) | 2.5 | (+)(-) = negative |
| (3, ∞) | 4 | (+)(+) = positive |
d the other is negative, so the product is negative. For x = 4, both factors are positive.
当 x = 0 时，两个因式都为负，乘积为正。当 x = 2.5 时，一个因式为正而另一个为负，乘积为负。当 x = 4 时，两个因式都为正。
7. Writing the Solution Set | 写出解集
We require the product to be greater than zero, so we choose the intervals where the sign is positive.
我们需要乘积大于零，因此选择符号为正的区间。
The solution is therefore x < 2 or x > 3.
因此解为 x < 2 或 x > 3。
In interval notation, this is written as (-∞, 2) ∪ (3, ∞).
用区间符号，可以写作 (-∞, 2) ∪ (3, ∞)。
8. Boundary Point Check | 边界点检查
x = 2, the left-hand side is (2 - 2)(2 - 3) = 0 × (-1) = 0, which does not satisfy > 0.
在 x = 2 处，左边为 (2 - 2)(2 - 3) = 0 × (-1) = 0，不满足 > 0。
At x = 3, the product is (3 - 2)(3 - 3) = 1 × 0 = 0, which is also not greater than zero.
在 x = 3 处，乘积为 (3 - 2)(3 - 3) = 1 × 0 = 0，也不大于零。
This confirms that both critical values are excluded and the open intervals are correct.
这确认了两个临界值都被排除，并且开区间是正确的。
9. Alternative Method: Sign Table | 替代方法：符号表
A sign table organises the same information in a compact way, especially for more complex products or quotients.
符号表可以以紧凑的方式整理相同信息，特别适用于更复杂的乘积或商式。
| Factor | x < 2 | 2 < x < 3 | x > 3 |
|---|---|---|---|
| x - 2 | - | + | + |
| x - 3 | - | - | + |
| (x - 2)(x - 3) | + | - | + |
For x < 2, both factors are negative, so the product is positive. For 2 < x < 3, one factor is positive and one is negative, so the product is negative. For x > 3, both factors are positive.
当 x < 2 时，两个因式为负，乘积为正。当 2 < x < 3 时，一个因式为正一个为负，乘积为负。当 x > 3 时，两个因式都为正。
This method gives the same conclusion but can be faster once you are confident with sign rules.
这种方法得出相同结论，而且一旦你熟悉符号规则，它可能更快。
10. Common Mistakes | 常见错误
he solution. Since the original symbol is > and not ≥, the boundaries must be excluded.
一个常见错误是将临界值包含在解中。因为原符号是 > 而不是 ≥，边界必须被排除。
Another error is deciding that the answer is always between the roots. This is true for a negative-leading-coefficient quadratic with > 0, but here a is positive.
另一个错误是认为答案总是在两根之间。对于开口向下且 > 0 的二次式确实如此，但这里 a 为正。
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