📚 Example 4.7.2: Solving Quadratic Inequalities | 例题 4.7.2：解二次不等式
In AQA A-Level Mathematics, Example 4.7.2 is a standard question on quadratic inequalities. It requires you to find the interval of x values that makes a quadratic expression less than or equal to zero, and then to write that interval using correct mathematical notation. Although the algebra is short, the question tests several skills at once: factorising a quadratic, interpreting the sign of a product, reading a parabola, and communicating a solution set precisely.
在 AQA A-Level 数学中，例题 4.7.2 是二次不等式的标准题型。它要求你找出使二次函数表达式小于或等于零的 x 值区间，并用正确的数学符号写出该区间。虽然代数运算并不长，但这道题同时考查多项能力：因式分解二次式、判断乘积的符号、读懂抛物线图像，以及准确表达解集。
This article expands the original example into a complete revision lesson. We will state the problem, explain the graphical meaning of the inequality, work through the factorisation and sign analysis step by step, compare the sketch method with the algebra method, list the mistakes that cost marks in exams, and finish with a set of exam-style variants so that you can apply the same reasoning to any quadratic inequality.
本文将原例题扩展为一节完整的复习课。我们会给出题目、解释不等式在图像上的含义、逐步完成因式分解与符号分析、比较草图法与代数法、列出考试中容易失分的错误，最后给出一组考题风格的变式，帮助你把同样的思路应用到任意二次不等式上。
1. The Problem Statement | 题目陈述
trict about one thing only: the symbol is ≤, not <, so the boundary values themselves are included in the answer.
本例题要求学生解一个二次不等式，并用区间符号或集合符号表示解集。这个不等式有一个关键细节：符号是 ≤ 而不是 <，因此边界值本身也包含在答案中。
Problem: f x for which 2x² − 5x − 3 ≤ 0. Give your answer in set notation and in interval notation.
题目：求满足 2x² − 5x − 3 ≤ 0 的所有实数 x，并用集合符号和区间符号表示答案。
2x² − 5x − 3 ≤ 0
The expression on the left is a quadratic with leading coefficient 2, linear coefficient −5 and constant term −3. Because the leading coefficient is positive, its graph opens upwards, and the expression is negative only between its two roots. That single observation already tells us the shape of the answer before any algebra is done.
左侧表达式是一个二次式，二次项系数为 2，一次项系数为 −5，常数项为 −3。由于二次项系数为正，其图像开口向上，表达式只在两个根之间取负值。仅凭这一点，在做任何代数运算之前，我们就已经知道答案的大致形状了。
2. Why Quadratic Inequalities Matter | 为什么二次不等式如此重要
Quadratic inequalities appear throughout the A-Level specification. They are needed when finding the domain of a function, when determining where a curve lies above or below a line, when solving inequalities arising from the discriminant, and when working with kinematics problems in mechanics where a particle is above a certain height. In pure mathematics, the same technique reappears in the study of polynomial inequalities, rational inequalities and the location of stationary points.
二次不等式贯穿整个 A-Level 考纲。求函数定义域时需要它，判断曲线位于某条直线上方还是下方时需要它，处理由判别式引出的不等式时需要它，在力学中研究质点高于某一高度的时间区间时也需要它。在纯数学部分，同样的技巧还会出现在多项式不等式、分式不等式以及驻点位置的讨论中。
Examiners also value this topic because it links algebra to graphs. A student who can only factorise may find the roots but choose the wrong region; a student who understands the parabola will rarely make that error. The mark scheme usually rewards both the correct critical values and the correct final interval, so method and interpretation both matter.
考官重视这一题型，还因为它把代数与图像联系起来。只会因式分解的学生可能求出根，却选错区间；而理解抛物线的学生极少犯这种错误。评分标准通常既给临界值的分，也给最终区间的分，因此方法与理解同样重要。
3. Understanding Quadratic Inequalities Graphically | 从图像理解二次不等式
ent, the graph is U-shaped. The inequality 2x² − 5x − 3 ≤ 0 asks for the parts of the parabola on or below the x-axis, that is, where the y-value is zero or negative.
二次不等式的解法需要将代数表达式与抛物线图像联系起来。对于二次项系数为正的二次函数，图像呈 U 形。不等式 2x² − 5x − 3 ≤ 0 要找的是抛物线位于 x 轴及其下方的部分，也就是 y 值为零或为负的部分。
If the graph is below the x-axis, the quadratic expression is negative; if the graph is above the x-axis, the expression is positive. Since the parabola opens upwards, it can only be below the axis between its two roots. Outside that interval, the arms of the parabola rise above the axis and the expression becomes positive again.
如果图像在 x 轴下方，二次表达式为负；如果在 x 轴上方，则表达式为正。由于抛物线开口向上，它只能在两个根之间位于 x 轴下方。在该区间之外，抛物线的两支向上扬起，表达式重新变为正值。
Negative region: between the roots | 负值区间：两根之间
crosses the x-axis, and positive on the two slopes climbing away from the valley. For the inequality ≤ 0, we want the valley floor plus its two edges.
一个有用的心理图像是山谷。二次表达式在山谷底部取负值，在山谷穿过 x 轴的两个点处恰好为零，而在向两侧升起的坡面上取正值。对于 ≤ 0 的不等式，我们要的是谷底以及两侧的边缘点。
4. Step 1: Finding the Critical Values by Factorising | 步骤 1：因式分解求临界值
The first step is to find the critical values where the quadratic expression equals zero. These are the boundary points that separate positive and negative regions, and they will later become the endpoints of the solution interval.
第一步是求临界值，即二次表达式等于零的点。这些是区分正负区间的边界点，随后会成为解区间的端点。
2x² − 5x − 3 = 0
= −6 and whose sum is −5. Those numbers are −6 and +1. Split the middle term accordingly.
对二次式进行因式分解。我们需要两个数，它们的乘积为 2 × (−3) = −6，和为 −5。这两个数是 −6 与 +1。据此拆分中间项。
2x² − 6x + x − 3 = 0
Now factorise in pairs. From the first two terms take out 2x, and from the last two terms take out 1.
现在分组分解。前两项提出 2x，后两项提出 1。
2x(x − 3) + 1(x − 3) = 0
(2x + 1)(x − 3) = 0
ese two critical values divide the number line into three regions, and the solution to the inequality must lie in exactly one of them (or possibly two, depending on the inequality symbol).
令每个因式等于零：2x + 1 = 0 得 x = −1/2；x − 3 = 0 得 x = 3。这两个临界值把数轴分成三个区域，不等式的解必定落在其中之一（也可能落在其中两个，取决于不等号方向）。
ginal expression. Always expand mentally to confirm the signs are right, because a single sign slip here changes the entire answer.
快速检验因式分解：展开 (2x + 1)(x − 3) 得 2x² − 6x + x − 3 = 2x² − 5x − 3，与原式一致。务必在脑中展开确认符号正确，因为这里一个符号错误就会改变整个答案。
5. Step 2: Sign Analysis of the Intervals | 步骤 2：区间符号分析
The critical values −1/2 and 3 divide the real number line into three intervals: x < −1/2, −1/2 < x < 3, and x > 3. Choose a test point from each interval and substitute it into the factorised form to determine the sign of the expression.
临界值 −1/2 和 3 将实数轴分成三个区间：x < −1/2、−1/2 < x < 3、x > 3。从每个区间选一个测试点，代入因式分解后的形式，判断表达式的符号。
| Interval | Test point | Value of 2x² − 5x − 3 | Sign | Satisfies ≤ 0? |
| x < −1/2 | x = −1 | 4 | Positive | No |
| −1/2 < x < 3 | x = 0 | −3 | Negative | Yes |
| x > 3 | x = 4 | 9 | Positive | No |
− 3 = 32 − 20 − 3 = 9, which is positive. Only the middle interval produces a negative value, so only the middle interval satisfies the inequality.
在 x = −1 处代入得 2(1) − 5(−1) − 3 = 2 + 5 − 3 = 4，为正；在 x = 0 处代入得 0 − 0 − 3 = −3，为负；在 x = 4 处代入得 2(16) − 5(4) − 3 = 32 − 20 − 3 = 9，为正。只有中间区间给出负值，因此只有中间区间满足该不等式。
You can also read the signs directly from the factorised form (2x + 1)(x − 3). For x < −1/2, both factors are negative, so their product is positive. For −1/2 < x < 3, the first factor is positive while the second is negative, so the product is negative. For x > 3, both factors are positive, so the product is positive. This factor-sign reasoning is faster and less error-prone than substituting numbers.
也可以直接从因式分解形式 (2x + 1)(x − 3) 读出符号。当 x < −1/2 时，两个因式都为负，乘积为正；当 −1/2 < x < 3 时，第一个因式为正、第二个为负，乘积为负；当 x > 3 时，两个因式都为正，乘积为正。这种"因式符号"推理比逐一代入数字更快，也更不易出错。
6. Step 3: Writing the Solution Set | 步骤 3：写出解集
<). The endpoints x = −1/2 and x = 3 make the expression equal to zero, and zero is allowed by ≤ 0.
解集就是表达式取负值的区间，再加上两个端点，因为不等式是非严格的（≤ 而不是 <）。端点 x = −1/2 与 x = 3 使表达式等于零，而 ≤ 0 允许取零。
−1/2 ≤ x ≤ 3
s are included. In set notation it is written as {x : −1/2 ≤ x ≤ 3}, where the colon means "such that" ation, but the question may specify which one it wants, so read the instruction carefully.
用区间符号表示为 [−1/2, 3]，方括号表示端点包含在内。用集合符号表示为 {x : −1/2 ≤ x ≤ 3}，其中冒号表示"使得"。在 AQA 考试中两种写法都接受，但题目可能指定其中一种，因此要仔细阅读要求。
If the inequality had been strict, 2x² − 5x − 3 < 0, the solution would be −1/2 < x < 3 and the interval notation would use round brackets, (−1/2, 3). The difference between including and excluding endpoints is a very common source of lost marks, so decide deliberately which brackets to use.
如果不等式是严格的，即 2x² − 5x − 3 < 0，那么解为 −1/2 < x < 3，区间符号应使用圆括号 (−1/2, 3)。端点包含与否的区别是极常见的失分点，因此要有意识地决定使用哪种括号。
7. The Sketch Method: Faster and Safer | 草图法：更快也更稳妥
In an examination, the quickest reliable method is to sketch the parabola. Mark the two roots on the x-axis, note that the parabola opens upwards because the coefficient of x² is positive, and shade the region below the axis.
在考试中，最快且可靠的方法是画出抛物线草图。在 x 轴上标出两个根，注意由于 x² 的系数为正，抛物线开口向上，然后把位于 x 轴下方的部分涂上阴影。
er to ≤ 0 is the shaded central region, and the answer to ≥ 0 would be the two outer arms.
草图立即显示：曲线在 −1/2 与 3 之间下探到 x 轴之下，在该区间之外则升到 x 轴之上。因此 ≤ 0 的答案是中间涂阴影的区域，而 ≥ 0 的答案则是两侧向外伸展的两支。
A good sketch takes about fifteen seconds and protects you from the classic error of choosing the wrong region. Even when the question says "solve algebraically", a small sketch in the margin is accepted as working and can earn method marks if the final answer is wrong.
一幅好的草图大约只需十五秒，却能避免"选错区间"这一经典错误。即使题目要求"用代数方法求解"，在页边画一幅小草图也会被视为有效过程，万一最终答案出错，仍可能拿到方法分。
8. Common Mistakes and Exam Warnings | 常见错误与考试提醒
ts are only the boundaries; the answer must be a region, not two isolated numbers. Writing "x = −1/2 or x = 3" answers the wrong question entirely.
最常见的错误是把不等式当作方程，求出根后就停下。根只是边界，答案必须是一个区间，而不是两个孤立的数。写"x = −1/2 或 x = 3"完全答非所问。
x ≥ 3 for the ≤ 0 inequality. A sketch prevents this. Remember: for a positive quadratic, "less than zero" means between the roots, and "greater than zero" means outside the roots.
第二个常见错误是选错区域。忘记抛物线形状的学生有时会对 ≤ 0 的不等式写出 x ≤ −1/2 或 x ≥ 3。画草图可以避免这一点。记住：对于开口向上的二次函数，"小于零"位于两根之间，"大于零"位于两根之外。
rks. The fourth mistake is failing to check the factorisation by expanding, which lets a sign error pass unnoticed. The fifth mistake is forgetting to change the direction of the inequality when multiplying or dividing by a negative number, which matters in variants such as −x² + 4x − 3 ≥ 0.
第三个错误是端点处理不当：不等式为 ≤ 时用了圆括号，或为 < 时用了方括号，都会丢分。第四个错误是不展开检验因式分解，让符号错误蒙混过关。第五个错误是在乘以或除以负数时忘记改变不等号方向，这在 −x² + 4x − 3 ≥ 0 这类变式中至关重要。
9. Alternative Methods: Formula and Completing the Square | 其他方法：求根公式与配方法
not rational. For 2x² − 5x − 3 = 0, substitute a = 2, b = −5 and c = −3 into the formula.
因式分解并非唯一途径。求根公式永远适用，即使因式不明显或不是有理数。对于 2x² − 5x − 3 = 0，把 a = 2、b = −5、c = −3 代入公式。
x = (5 ± √25 + 24) ÷ 4 = (5 ± 7) ÷ 4
is positive and a perfect square, so the roots are rational and the factorisation exists.
由此得 x = 12 ÷ 4 = 3 或 x = −2 ÷ 4 = −1/2，与因式分解结果一致。此处的判别式 Δ = b² − 4ac = 25 + 24 = 49，为正且是完全平方数，因此根是有理数，因式分解存在。
= 2(x − 5/4)² − 49/8, the inequality 2(x − 5/4)² − 49/8 ≤ 0 becomes (x − 5/4)² ≤ 49/16, so |x − 5/4| e interval. This method is especially useful when the quadratic does not factorise nicely.
配方法提供了另一种视角。把 2x² − 5x − 3 写成 2(x − 5/4)² − 49/8，不等式 2(x − 5/4)² − 49/8 ≤ 0 化为 (x − 5/4)² ≤ 49/16，即 |x − 5/4| ≤ 7/4。各项加 5/4 得 −1/2 ≤ x ≤ 3，与之前结果相同。当二次式不易分解时，这种方法尤其有用。
10. Inequalities with ≥, > and < Signs | 含 ≥、> 与 < 号的不等式
Once the roots are known, the sign of the inequality determines which region to choose. For a quadratic with a positive leading coefficient and two distinct
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图