📚 Example 2.7.1: Solving Quadratic Inequalities | 示例 2.7.1：求解二次不等式
In AQA A-Level Mathematics, quadratic inequalities appear in the Pure core and are frequently tested as part of algebraic techniques. Example 2.7.1 focuses on solving the inequality x² - 5x + 6 > 0. The method relies on factorising the quadratic, identifying the critical values where the expression equals zero, and then using the shape of the parabola or a sign table to decide which intervals of x satisfy the inequality.
在 AQA A-Level 数学中，二次不等式属于纯数学核心内容，并且经常作为代数技巧的一部分被考查。示例 2.7.1 重点求解不等式 x² - 5x + 6 > 0。解题方法依赖于因式分解二次式、确定使表达式等于零的临界值，然后利用抛物线的形状或符号表来判断哪些 x 区间满足该不等式。
1. The Problem Statement | 题目表述
Consider the inequality x² - 5x + 6 > 0. We want to find all real values of x for which the quadratic expression is strictly greater than zero. Because the inequality sign is strict, the boundary values themselves will not be included in the final answer. A complete solution must therefore state two things: the numerical boundary values, and whether they are included or excluded.
考虑不等式 x² - 5x + 6 > 0。我们要找出使该二次表达式严格大于零的所有实数 x。由于不等号为严格大于，边界值本身不会包含在最终答案中。因此一个完整的解答必须说明两件事：数值边界值，以及这些边界值是被包含还是被排除。
la opens upwards, which immediately tells us something useful about where the expression is positive and where it is negative.
该表达式是标准形式的二次式 ax² + bx + c，其中 a = 1、b = -5、c = 6。由于 a 为正，抛物线开口向上，这立刻告诉我们关于该表达式在何处为正、在何处为负的有用信息。
2. Step 1: Solve the Related Quadratic Equation | 第一步：求解相关的二次方程
. So we begin by solving x² - 5x + 6 = 0. The solutions of this equation are exactly the values of x where the quadratic expression changes sign, so they mark the boundaries of the solution intervals.
几乎所有二次不等式题目的第一步，都是把不等号换成等号，并求解所得的方程。因此我们首先求解 x² - 5x + 6 = 0。这个方程的解正是二次表达式改变符号的 x 值，因此它们标记了解集的边界。
These boundary values are often called the critical values or the roots of the quadratic. If the quadratic does not factorise neatly, the quadratic formula can be used instead:
这些边界值通常称为临界值或二次方程的根。如果二次式不能整齐地因式分解，也可以改用求根公式：
x = (−b ± √b² − 4ac) ÷ 2a
For this example, however, factorisation is quick and exact, so the quadratic formula is unnecessary. Recognising a factorable quadratic saves time in the exam and reduces the risk of arithmetic slips.
不过对于本例来说，因式分解既快又精确，因此不需要用求根公式。在考试中识别出可因式分解的二次式可以节省时间，并降低算术失误的风险。
3. Factorising x² - 5x + 6 | 因式分解 x² - 5x + 6
since (−2) × (−3) = 6 and (−2) + (−3) = −5. Therefore the quadratic factorises as follows:
我们需要两个数，它们的乘积为 +6，和为 −5。−2 和 −3 这一对符合条件，因为 (−2) × (−3) = 6，且 (−2) + (−3) = −5。因此二次式可以这样分解：
x² - 5x + 6 = (x - 2)(x - 3)
g: (x - 2)(x - 3) = x² - 3x - 2x + 6 = x² - 5x + 6, which matches the original expression.
令每个括号等于零，得到临界值 x = 2 和 x = 3。你随时可以通过展开来检验因式分解：(x - 2)(x - 3) = x² - 3x - 2x + 6 = x² - 5x + 6，与原表达式一致。
- Product check: (−2) × (−3) = +6, matching the constant term. | 乘积检验：(−2) × (−3) = +6，与常数项一致。
- Sum check: (−2) + (−3) = −5, matching the coefficient of x. | 和检验：(−2) + (−3) = −5，与 x 的系数一致。
- Expansion check: the expanded form returns x² - 5x + 6. | 展开检验：展开后重新得到 x² - 5x + 6。
4. Critical Values and the Number Line | 临界值与数轴
e the real number line into three separate regions. Because the expression is a continuous function, its sign can only change at these critical values, so within each region the sign is constant. This is the key idea that makes the number line method work.
临界值 x = 2 和 x = 3 把实数轴分成三个互不重叠的区域。由于该表达式是连续函数，它的符号只可能在这些临界值处改变，因此每个区域内部的符号是恒定的。这正是数轴法能够成立的关键思想。
| Region | 区域 | Test value | 检验值 | Sign of (x - 2)(x - 3) | 符号 |
| x < 2 | x = 0 | (−)(−) = positive | 正 |
| 2 < x < 3 | x = 2.5 | (+)(−) = negative | 负 |
| x > 3 | x = 4 | (+)(+) = positive | 正 |
Notice that the sign pattern for a quadratic with a positive leading coefficient and two distinct roots is always positive, then negative, then positive as you move from left to right along the number line. This alternating pattern is a direct consequence of the upward-opening parabola crossing the x-axis twice.
请注意，对于首项系数为正且有两个不同根的二次式，沿数轴从左到右的符号模式总是正、负、正。这种交替模式是开口向上的抛物线两次穿过 x 轴的直接结果。
5. The Parabola Sketch Method | 抛物线草图法
e x-axis at x = 2 and x = 3. The turning point lies halfway between the roots, at x = 2.5, where y = (2.5)² - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25.
另一种通常更快的方法是画出抛物线 y = x² - 5x + 6 的草图。由于 a = 1 > 0，曲线是开口向上的 U 形，并在 x = 2 和 x = 3 处穿过 x 轴。顶点位于两根的中点，即 x = 2.5 处，此时 y = (2.5)² - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25。
x-axis between them. Since the inequality asks where the expression is greater than zero, the required parts of the curve are the two outer branches.
观察草图可知，曲线在 x = 2 左侧和 x = 3 右侧位于 x 轴上方，在两者之间下落到 x 轴下方。由于不等式要求表达式大于零，所以需要的部分是曲线的外侧两支。
Sketch: U-shape crossing at x = 2 and x = 3, minimum at (2.5, −0.25)
A quick sketch is usually worth the thirty seconds it takes, because it makes the direction of the answer visually obvious and protects you against the classic mistake of writing the solution the wrong way round.
花三十秒画一个快速草图通常是值得的，因为它让答案的方向在视觉上一目了然，并能保护你不犯下把解集写反的经典错误。
6. The Sign Table Method | 符号表法
The sign table method formalises the reasoning above. Write the factorised form (x - 2)(x - 3) at the top, list the two factors on separate rows, and mark the critical values as columns where each factor is zero. Then determine the sign of each factor in each interval.
符号表法把上述推理形式化。把因式分解后的形式 (x - 2)(x - 3) 写在顶部，把两个因式分别列在不同行，并把临界值标记为各因式为零的列。然后判断每个因式在每个区间内的符号。
| Factor | 因式 | x < 2 | 2 < x < 3 | x > 3 |
| x - 2 | − | + | + |
| x - 3 | − | − | + |
| Product | 乘积 | + | − | + |
Multiplying the signs row by row gives the sign of the whole expression. The product is positive on the first and third intervals, so the inequality x² - 5x + 6 > 0 is satisfied there. The sign table is particularly reliable when the factors are more complicated, for example when one factor is itself a linear expression with a negative coefficient.
逐行相乘即可得到整个表达式的符号。乘积在第一和第三个区间为正，因此不等式 x² - 5x + 6 > 0 在这些区间成立。当因式较为复杂时——例如某个因式本身是系数为负的一次式——符号表就特别可靠。
7. Writing the Solution Set | 写出解集
ation, the solution is x < 2 or x > 3. Using set notation, the solution set is {x : x < 2} ∪ {x : x > 3}, and in interval notation it is (−∞, 2) ∪ (3, ∞).
把两个为正的区域合并就得到最终答案。用不等式记号表示，解为 x < 2 或 x > 3。用集合记号表示，解集为 {x : x < 2} ∪ {x : x > 3}；用区间记号表示，则为 (−∞, 2) ∪ (3, ∞)。
Solution: x < 2 or x > 3
The word "or" is essential here, because the solution set is the union of two disjoint intervals. Writing "and" would be wrong, since no single value of x can be simultaneously less than 2 and greater than 3. The union symbol ∪ performs the same job as the word "or" in set notation.
这里 "or"（或）这个词是必不可少的，因为解集是两个不相交区间的并集。写成 "and"（且）则是错误的，因为没有任何一个 x 值能同时小于 2 又大于 3。集合记号中的并集符号 ∪ 起到的正是 "or" 的作用。
Edexcel and AQA mark schemes usually accept either the inequality form or the interval form, but the inequality form with the word "or" is the safest choice if you are unsure. Whichever you choose, make the two branches clearly separate.
Edexcel 和 AQA 的评分标准通常同时接受不等式形式和区间形式，但如果你不确定，带 "or" 的不等式形式是最稳妥的选择。无论选择哪种，都要把两支清楚地区分开。
8. Strict versus Non-Strict Inequalities | 严格不等式与非严格不等式
ld be x ≤ 2 or x ≥ 3, with the endpoints included. On a number line diagram, strict inequalities are shown with open circles and non-strict inequalities with filled circles.
本例中的不等式使用严格符号 >，它排除了临界值。如果题目要求解 x² - 5x + 6 ≥ 0，答案就会是 x ≤ 2 或 x ≥ 3，端点被包含在内。在数轴图上，严格不等式用空心圆表示，非严格不等式用实心圆表示。
| Inequality | 不等式 | Solution set | 解集 | Endpoints | 端点 |
| x² - 5x + 6 > 0 | x < 2 or x > 3 | excluded | 排除 |
| x² - 5x + 6 ≥ 0 | x ≤ 2 or x ≥ 3 | included | 包含 |
| x² - 5x + 6 < 0 | 2 < x < 3 | excluded | 排除 |
| x² - 5x + 6 ≤ 0 | 2 ≤ x ≤ 3 | included | 包含 |
Notice how the direction of the inequality completely changes the shape of the answer: greater than zero gives two outer rays, while less than zero gives a single bounded interval. Getting the direction right is the single most important step in the whole method.
注意不等号的方向如何彻底改变答案的形状：大于零给出两条外侧射线，而小于零给出一个单一的有界区间。判断正确的方向是整个方法中最重要的一步。
9. Common Errors and How to Avoid Them | 常见错误与避免方法
Examiners report the same handful of mistakes year after year on quadratic inequality questions. The most common is writing 2 < x < 3 as the answer to a "greater than zero" inequality, which is exactly the opposite of the correct solution. A quick sketch of the parabola prevents this error almost every time.
考官年复一年地在二次不等式题目上看到同样几种错误。最常见的是把 "大于零" 的不等式答案写成 2 < x < 3，而这恰好与正确解相反。快速画一张抛物线草图几乎每次都能避免这个错误。
- Sign flip error: writing the inner interval when the outer intervals are required, or vice versa. | 方向翻转错误：需要外侧区间时写了内侧区间，或反之。
- ividing both sides by x loses the solution x = 0 and produces a linear inequality instead. | 除以 x 的错误：两边同除以 x 会丢失解 x = 0，并产生一个一次不等式。
- Forgetting to factorise fully: a partially factorised quadratic gives wrong critical values. | 未完全分解：未完全因式分解的二次式会给出错误的临界值。
- Ignoring a negative leading coefficient: if a < 0 the parabola opens downwards and the sign pattern reverses. | 忽略负的首项系数：若 a < 0，抛物线开口向下，符号模式会反转。
- Mishandling the endpoints when the inequality is non-strict. | 在非严格不等式时错误处理端点。
x = 2.5 gives −0.25 > 0 (false) and x = 4 gives 2 > 0 (true), confirming the solution x < 2 or x > 3.
检验答案的一个简便方法是从每个区域取一个检验值代回原表达式。例如 x = 0 得到 6 > 0（真），x = 2.5 得到 −0.25 > 0（假），x = 4 得到 2 > 0（真），从而确认解为 x < 2 或 x > 3。
10. How AQA Exam Questions Present This | AQA 考题如何呈现此类问题
In AQA A-Level Mathematics, quadratic inequalities rarely appear in isolation. They are normally embedded in a longer question, for example after a student has found the coordinates of the points where a curve crosses the x-axis, or as part of a problem asking for the set of values for which a function is increasing or positive.
在 AQA A-Level 数学中，二次不等式很少单独出现。它们通常嵌入在较长的题目中，例如在学生求出曲线与 x 轴交点坐标之后，或作为求函数递增区间或取正值的取值范围问题的一部分。
Typical instruction words include "find the set of values of x for which", "solve the inequality" and "hence determine the range of values". The word "hence" is a strong hint that the factorisation or critical values found earlier in the question must be reused rather than recomputed.
典型的指令词包括 "find the set of values of x for which"（求 x 的取值范围）、"solve the inequality"（解不等式）和 "hence determine the range of values"（由此确定取值范围）。"hence"（由此）一词是强烈的提示，说明题目前面求出的因式分解或临界值必须被重复使用，而不是重新计算。
Marks are usually awarded as follows: one mark for correct factorisation, one mark for both critical values, and one mark for the correct final intervals with the correct inequality signs. Dropping the "or" or using the wrong strictness typically costs the final mark.
分数通常这样分配：因式分解正确得 1 分，两个临界值都正确得 1 分，最终区间与不等号方向都正确得 1 分。漏写 "or" 或严格性判断错误通常会丢掉最后 1 分。
11. Generalising the Method | 方法的推广
The same three-step routine works for any quadratic inequality: rearrange to make one side zero, factorise and find the critical values, then use the shape of the parabola to select the correct intervals. For ax² + bx + c > 0 with a > 0 and two distinct roots r₁ < r₂, the solution is always x < r₁ or x > r₂.
同样的三步流程适用于任何二次不等式：整理使一边为零，因式分解并求出临界值，然后利用抛物线形状选出正确区间。对于 ax² + bx + c > 0，若 a > 0 且有两个不同根 r₁ < r₂，解永远是 x < r₁ 或 x > r₂。
If a < 0, the simplest approach is to multiply the whole inequality by −1 and reverse the inequality sign, turning the problem back into the standard case with a positive leading coefficient. Alternatively, note that a downward parabola is positive between its roots rather than outside them.
若 a < 0，最简单的方法是把整个不等式乘以 −1 并反转不等号，把问题重新化为首项系数为正的标准情形。或者也可以记住：开口向下的抛物线是在两根之间为正，而不是在两根之外为正。
a > 0: positive outside the roots | a < 0: positive between the roots
When the discriminant b² − 4ac is negative, the parabola never crosses the x-axis. If a > 0 the expression is positive for all real x, so x² + x + 1 > 0 has every real number as a solution; if a < 0 the expression is negative for all x and the inequality has no solution.
当判别式 b² − 4ac 为负时，抛物线永不穿过 x 轴。若 a > 0，表达式对所有实数 x 都为正，因此 x² + x + 1 > 0 的解是全体实数；若 a < 0，表达式对所有 x 都为负，该不等式无解。
12. Practice Questions and Answers | 练习题与答案
Work through the following questions using the method developed above. Attempt each one before reading the answer, and always sketch the parabola if you are unsure about the direction of the inequality.
请用上面建立的方法完成下列题目。先自己尝试再阅读答案，若对不等号方向不确定，一定要画出抛物线草图。
- Solve x² - 7x + 12 > 0. Answer: x < 3 or x > 4. | 解 x² - 7x + 12 > 0。答案：x < 3 或 x > 4。
- Solve x² - 7x + 12 ≤ 0. Answer: 3 ≤ x ≤ 4. | 解 x² - 7x + 12 ≤ 0。答案：3 ≤ x ≤ 4。
- Solve x² + 3x - 10 ≥ 0. Answer: x ≤ −5 or x ≥ 2. | 解 x² + 3x - 10 ≥ 0。答案：x ≤ −5 或 x ≥ 2。
- Solve 2x² - 5x + 2 < 0. Answer: ½ < x < 2. | 解 2x² - 5x + 2 < 0。答案：½ < x < 2。
- Solve −x² + 4x - 3 > 0. Answer: 1 < x < 3. | 解 −x² + 4x - 3 > 0。答案：1 < x < 3。
These five questions cover the main variations you will meet: two positive roots, a negative leading coefficient, a non-integer root from a factor pair such as (2x - 1)(x - 2), and a mixture of strict and non-strict signs. Mastering all five gives you a reliable toolkit for any quadratic inequality in the AQA Pure papers.
这五道题涵盖了你会遇到的主要变式：两个正根、负的首项系数、来自 (2x - 1)(x - 2) 这类因式对的非整数根，以及严格与非严格符号的混合。掌握全部五道题，你就拥有了应对 AQA 纯数学试卷中任何二次不等式的可靠工具箱。
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