📚 Solving Quadratic Inequalities: A Complete Guide | 二次不等式求解全攻略：从入门到精通
Quadratic inequalities look intimidating on paper, but once you understand what the parabola is actually doing, they collapse into a short and reliable routine: rearrange, solve the matching equation, sketch, read off the region, and write the answer cleanly. This guide covers the full method used by GCSE Higher Tier and A Level students, together with the special cases and examiner traps that quietly decide grades every exam season.
二次不等式在纸面上看着吓人，但只要你真正明白抛物线在做什么，它就会变成一个短小可靠的固定流程：整理、解对应的方程、画草图、读出满足条件的区域、规范写出答案。本文完整梳理 GCSE Higher Tier 与 A Level 学生需要掌握的方法，并补充那些每逢考季都在悄悄决定分数的特殊情形与考官陷阱。
1. What Are Quadratic Inequalities? | 什么是二次不等式
A quadratic inequality is an inequality that contains a term in x² and no higher power of x. It always has a non-zero coefficient a, which means the graph involved is a parabola rather than a straight line. The four standard shapes you will meet are shown below.
二次不等式指的是含有 x² 项、且不含更高次项的不等式。它的二次项系数 a 一定不为零，因此涉及的图像是抛物线而不是直线。考试中你会遇到的四种标准形式如下。
+ c ≥ 0 ax² + bx + c < 0 ax² + bx + c ≤ 0 (a ≠ 0)
e interval, such as 2 ≤ x ≤ 4, or a union of two intervals, such as x < −3 or x > ½.
它与二次方程最本质的区别在于：答案不是一个或两个数，而是一整段取值。这个解集通常是一个区间，例如 2 ≤ x ≤ 4；也可能是两个区间的并集，例如 x < −3 或 x > ½。
2. Why This Topic Matters for GCSE and A Level | 为什么它是考试核心考点
Quadratic inequalities appear on the GCSE Higher Tier papers of Edexcel, AQA and OCR, and they reappear in A Level Pure Mathematics Year 1 under the algebra and functions topic. They typically carry three to five marks, and those marks are among the easiest on the paper once the method is automatic.
二次不等式出现在 Edexcel、AQA 与 OCR 的 GCSE Higher Tier 试卷上，也会在 A Level 纯数学第一年的代数与函数章节中再次出现。它通常占 3 到 5 分，而一旦方法变成自动化反应，这几乎是整份卷子里最容易拿到的分数。
The topic is also a gateway skill. The same reasoning is tested through the discriminant, graph transformations, and hidden quadratic inequalities later in the course, so weakness here tends to spread into other chapters.
同时它还是一个"前置技能"。后续课程中的判别式、图像变换以及"伪二次"不等式，考的都是同一套推理，所以这里的薄弱往往会扩散到其他章节。
3. Step 1: Rearrange to Standard Form | 第一步：整理成标准形式
Always begin by rearranging so that all terms sit on the left and the right-hand side is exactly zero. This makes the left-hand side a single quadratic expression that you can factorise, and it lets you compare it directly with the shape of the graph.
永远先从整理开始：把所有项移到左边，让右边恰好等于零。这样左边就是一个可以因式分解的二次表达式，并且可以直接与图像形状对照。
2x² + 5x > 3 → 2x² + 5x − 3 > 0
Two rules keep you safe at this stage. First, never divide both sides by x, because x may be zero or negative and you would lose solutions or flip the inequality by accident. Second, if the coefficient of x² is negative, you may multiply both sides by −1 to make it positive, but you must reverse the inequality sign when you do so.
这一步有两条安全规则。第一，千万不要两边同时除以 x，因为 x 可能为零或负数，你会丢解或者不小心改变不等号方向。第二，若 x² 的系数为负，可以两边同乘 −1 使其变正，但必须同时把不等号反向。
4. Step 2: Solve the Equation to Find Critical Values | 第二步：解方程求出临界值
d the critical values, also known as roots, zeros, or boundary points, and they are exactly the x-intercepts of the parabola.
此时暂时忽略不等号，解与它对应的二次方程 ax² + bx + c = 0。得到的解称为临界值，也叫根、零点或边界点，它们正是抛物线与 x 轴的交点横坐标。
x² + 2x − 8 = 0 → (x + 4)(x − 2) = 0 → x = −4, x = 2
= −3. Always write the two critical values in ascending order before you move on, since every later step depends on knowing which one is smaller.
当二次项系数不是 1 时，因式分解的逻辑完全相同。例如 2x² + 5x − 3 = 0 可分解为 (2x − 1)(x + 3) = 0，得到 x = ½ 与 x = −3。请务必在继续之前把两个临界值按从小到大排列，因为后面每一步都依赖"谁更小"。
5. When Factorising Fails: Formula and Completing the Square | 无法因式分解时：公式法与配方
Not every quadratic factorises with integers. When it does not, use the quadratic formula, which works for every case and gives exact answers in surd form.
并不是每个二次式都能用整数因式分解。做不到时就用求根公式，它对所有情形都有效，并给出根式形式的精确答案。
x = (−b ± √b² − 4ac) / (2a)
The expression under the square root, b² − 4ac, is the discriminant, written as Δ. It tells you how many critical values exist before you even finish the calculation, and it also signals whether the parabola crosses, touches, or misses the x-axis.
根号下的表达式 b² − 4ac 就是判别式，记作 Δ。它在你算完之前就告诉你会有几个临界值，同时也预示了抛物线与 x 轴是相交、相切还是不相交。
e exact. For instance, x² + 2x − 8 becomes (x + 1)² − 9, and x² − 4x + 1 = 0 gives x = 2 ± √3, which is approximately 0.268 or 3.732.
配方法是第三条路径，在答案需要精确形式时尤其好用。例如 x² + 2x − 8 可写成 (x + 1)² − 9；而 x² − 4x + 1 = 0 得到 x = 2 ± √3，约等于 0.268 或 3.732。
6. Step 3: Sketch the Parabola | 第三步：画出抛物线草图
A quick sketch removes almost all of the guesswork. You only need four pieces of information: the direction of opening, the two x-intercepts, the y-intercept, and the position of the vertex. The sign of a decides the direction: a > 0 opens upward like a ∪, while a < 0 opens downward like a ∩.
一张快速草图几乎能消除所有猜测。你只需要四项信息：开口方向、两个 x 轴交点、y 轴交点，以及顶点位置。开口方向由 a 的符号决定：a > 0 开口向上（形如 ∪），a < 0 开口向下（形如 ∩）。
x lies on the axis of symmetry, which is always the vertical line halfway between the two roots.
y 轴截距就是把 x = 0 代入后得到的常数项 c。顶点位于对称轴上，而对称轴永远是两条根正中间的那条竖直线。
x = (r₁ + r₂) / 2 or equivalently x = −b / (2a)
The sketch does not need to be beautiful or to scale. It needs to be honest: the right number of intercepts, in the right order, with the correct opening direction.
草图不需要画得漂亮或按比例，但必须"诚实"：交点数量正确、位置顺序正确、开口方向正确。
7. Step 4: Read Off the Satisfying Region | 第四步：读出满足条件的区域
With the sketch in front of you, the inequality becomes a question about position: is the curve above or below the x-axis? A condition such as > 0 means the part of the curve above the axis, while < 0 means the part below it.
有了草图，不等式就变成一道关于位置的问题：曲线在 x 轴上方还是在下方？形如 > 0 的条件对应曲线位于 x 轴之上的部分，而 < 0 对应位于 x 轴之下的部分。
The table below summarises every case when the roots r₁ and r₂ are written with r₁ < r₂.
下表在 r₁ < r₂ 的前提下，总结了全部情形。
| Inequality | a > 0 (∪ shape) | a < 0 (∩ shape) |
| ax² + bx + c > 0 | x < r₁ or x > r₂ | r₁ < x < r₂ |
| ax² + bx + c < 0 | r₁ < x < r₂ | x < r₁ or x > r₂ |
e the two roots: x ≤ −4 or x ≥ 2. Change the sign to ≤ and the answer flips to −4 ≤ x ≤ 2.
例如 x² + 2x − 8 ≥ 0 的根是 −4 与 2，抛物线开口向上，因此曲线在两根之外位于 x 轴上方：x ≤ −4 或 x ≥ 2。若把符号改成 ≤，答案就翻转为 −4 ≤ x ≤ 2。
8. Step 5: Present the Answer on a Number Line | 第五步：用数轴规范呈现答案
dot for ≥ or ≤ and a hollow dot for > or <. Draw arrows or thicken the relevant sections to show exactly which parts of the line are included.
数轴图不是装饰，它往往正是最后一个方法分的所在。按从小到大标出两个临界值，≥ 或 ≤ 用实心圆点，> 或 < 用空心圆点，再用箭头或加粗线段标出数轴上被包含的部分。
Then translate the diagram into formal notation. Interval notation and set notation say exactly the same thing, but examiners sometimes specify one of them, so be ready to write either.
然后把图转写成规范记号。区间记法与集合记法表达的是同一件事，但考官有时会指定其中一种，所以两种都要会写。
x ≤ 4 ↔ [2, 4] x < −3 or x > ½ ↔ (−∞, −3) ∪ (½, ∞)
included, while round brackets and strict inequality signs mean it is excluded. Mixing the two up is one of the most common ways to lose a mark on an otherwise perfect solution.
注意：方括号与 ≥、≤ 一样，表示端点被包含；圆括号与严格不等号一样，表示端点被排除。把这两者混淆，是在一份本可完美的解答中丢分最常见的方式之一。
9. Special Cases: No Real Roots, Double Roots and Negative a | 特殊情形：无实根、重根与负系数
If the discriminant is negative, the parabola never touches the x-axis, so the quadratic expression keeps the same sign for every value of x. When a > 0 it is always positive, which means ax² + bx + c > 0 is true for all real x, while ax² + bx + c < 0 has no solutions at all.
若判别式为负，抛物线与 x 轴永不相交，二次表达式的符号对任意 x 都保持不变。当 a > 0 时它恒为正，也就是说 ax² + bx + c > 0 对所有实数 x 成立，而 ax² + bx + c < 0 则完全无解。
(x − 3)² > 0 excludes only the single value x = 3, and (x − 1)² ≤ 0 is satisfied only at x = 1.
重根会产生完全平方式。因为 (x + 2)² ≥ 0 对一切 x 成立，所以 (x + 2)² ≥ 0 的解集是全体实数；而 (x − 3)² > 0 只排除 x = 3 这一个值，(x − 1)² ≤ 0 则仅在 x = 1 时成立。
rough by −1 and reverse the inequality sign to work with an upward parabola, or sketch the ∩ nt is 4 − 20 = −16, so it is always positive and the original inequality has no solutions.
当 a 为负时，有两条稳妥路线：要么两边同乘 −1 并把不等号反向，改用开口向上的抛物线处理；要么直接画出 ∩ 形草图，对 > 0 的情形去读 x 轴下方的区域。例如 −x² + 2x − 5 > 0 可改写为 x² − 2x + 5 < 0，其判别式为 4 − 20 = −16，说明该式恒为正，故原不等式无解。
10. Common Exam Mistakes to Avoid | 常见失分点
s the final answer. A quadratic inequality expects a range or a pair of ranges, and a bare "x = −4, x = 2" scores nothing.
代价最高的错误是把不等式当成方程，最后只写两个临界值了事。二次不等式要的是一个范围或两个范围的并集，孤零零地写 "x = −4, x = 2" 是拿不到分的。
sing the sign, forgetting to reverse the inequality after multiplying by −1, and confusing "and" with "or". Between the two roots you need "and", written as a single interval such as −4 ≤ x ≤ 2; outside the roots you need "or", written as two separate inequalities.
其他高频失误还包括：两边同时除以 x 或除以负数却没有反向不等号；同乘 −1 后忘记反向；以及把 "且" 与 "或" 用错。两根之间用 "且"，写成一个区间如 −4 ≤ x ≤ 2；两根之外用 "或"，写成两条独立的不等式。
substitute it into the original expression: the sign you get tells you whether that region belongs in the answer.
最后一道保险是"取点检验"。在每一段区间里各取一个值代回原式，例如取 x = 0 或 x = 1000，得到的正负号就告诉你这一段是否属于答案。
11. Worked Exam-Style Examples | 真题式例题精讲
means below the axis, which happens between the roots. The answer is 2 ≤ x ≤ 4, or [2, 4].
例 1：解 x² − 6x + 8 ≤ 0。因式分解得 (x − 2)(x − 4) = 0，临界值为 2 与 4。抛物线开口向上，≤ 0 表示位于 x 轴下方，即两根之间。答案：2 ≤ x ≤ 4，或写成 [2, 4]。
e critical values are −3 and ½. Since a > 0 and the condition is > 0, take the regions outside the roots. The answer is x < −3 or x > ½.
例 2：解 2x² + 5x − 3 > 0。因式分解得 (2x − 1)(x + 3) = 0，临界值为 −3 与 ½。因 a > 0 且条件为 > 0，取两根之外的区域。答案：x < −3 或 x > ½。
e 1 and 3, the parabola opens upward and the condition is ≤ 0, so the answer is 1 ≤ x ≤ 3.
例 3：解 −x² + 4x − 3 ≥ 0。两边同乘 −1 并反向，得 x² − 4x + 3 ≤ 0，可分解为 (x − 1)(x − 3) = 0。临界值为 1 与 3，开口向上且条件为 ≤ 0，故答案为 1 ≤ x ≤ 3。
Example 4 (hidden quadratic): solve x⁴
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