📚 Exercise 3D: Quadratic Inequalities | 练习3D：二次不等式
the form ax² + bx + c < 0, ax² + bx + c > 0, ax² + bx + c ≤ 0, or ax² + bx + c ≥ 0.
欢迎阅读本篇 AQA A-Level 数学练习3D 复习文章。本练习重点考查二次不等式，这是 AS 和 A-Level 纯数中经常出现的内容。我们将探索求解 ax²+bx+c<0、ax²+bx+c>0、ax²+bx+c≤0 或 ax²+bx+c≥0 型不等式的图像法、代数法和应用方法。
1. What is a Quadratic Inequality? | 什么是二次不等式？
≠ 0. Unlike a quadratic equation, which usually has two isolated roots, a quadratic inequality usually produces a range of x-values, often written as an interval or a union of intervals.
二次不等式是可以写成 ax²+bx+c<0、ax²+bx+c>0、ax²+bx+c≤0 或 ax²+bx+c≥0 形式的不等式，其中 a≠0。与二次方程通常有两个孤立根不同，二次不等式的解通常是一个 x 值范围，常写成一个区间或几个区间的并集。
2. The Graphical Approach | 图像法
iable starting point. If a > 0, the parabola opens upwards; if a < 0, it opens downwards. The x-coordinates where the curve crosses the x-axis are the roots or critical values. For a > 0 and y < 0, the solution is between the roots; for y > 0, the solution is outside the roots. For a < 0, these positions are reversed.
画出 y=ax²+bx+c 的草图是最可靠的入手方法。若 a>0，抛物线开口向上；若 a<0，开口向下。曲线与 x 轴交点的横坐标就是根或临界值。当 a>0 且 y<0 时，解在两个根之间；当 y>0 时，解在两个根之外。a<0 时则相反。
3. Critical Values from Factorisation | 因式分解求临界值
ues split the number line into three intervals: x < 2, 2 < x < 3, and x > 3. Test one value from each interval in the original inequality.
在许多 AQA 练习3D 的题目中，最快的方法是将二次式因式分解，并令其等于零求出临界值。例如，解 x²-5x+6<0，因式分解为 (x-2)(x-3)<0。临界值为 x=2 和 x=3。这两个值将数轴分成三个区间：x<2、2<x<3 和 x>3。从每个区间选一个值代入原不等式进行检验。
4. Using a Sign Table | 符号表法
A sign table organises the sign of each factor across the intervals. For (x - 2)(x - 3) < 0, the product is positive when both factors have the same sign, and negative when the factors have opposite signs. The table below shows the sign pattern.
符号表可以帮助整理每个因式在各区间上的符号。对于 (x-2)(x-3)<0，当两个因式同号时乘积为正，异号时乘积为负。下表给出了符号规律。
| Interval / 区间 | x < 2 | 2 < x < 3 | x > 3 |
|---|---|---|---|
| Sign of x - 2 / 符号 | negative / 负 | positive / 正 | positive / 正 |
| Sign of x - 3 / 符号 | negative / 负 | negative / 负 | positive / 正 |
| Product / 乘积 | positive / 正 | negative / 负 | positive / 正 |
nequalities < or >, the critical values are excluded; for ≤ or ≥, they are included.
因为题目要求乘积为负，所以解为 2<x<3。对于严格不等式 < 或 >，临界值不包括在内；对于 ≤ 或 ≥，临界值包括在内。
5. Completing the Square Method | 配方法
² - 9. The inequality becomes (x + 2)² - 9 ≤ 0, so (x + 2)² ≤ 9. Taking square roots gives |x + 2| ≤ 3, which expands to -3 ≤ x + 2 ≤ 3. Subtracting 2 from all three parts gives -5 ≤ x ≤ 1.
当二次式不易因式分解时，配方法是一种强有力的替代方法。例如，解 x²+4x-5≤0。配方：x²+4x-5=(x+2)²-9。不等式变为 (x+2)²-9≤0，因此
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