Solving Quadratic Inequalities | 解二次不等式

📚 Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities are a key topic in the Edexcel IGCSE Mathematics syllabus. They test not only your algebraic manipulation but also your understanding of the graphical behaviour of quadratic functions. In this article, we will explore a systematic method for solving quadratic inequalities of the form ax² + bx + c > 0, < 0, ≥ 0, or ≤ 0.

二次不等式是 Edexcel IGCSE 数学大纲中的一个核心主题。它不仅考查代数运算能力，还考查对二次函数图像性质的理解。在本文中，我们将系统学习如何求解形如 ax² + bx + c > 0、< 0、≥ 0 或 ≤ 0 的二次不等式。

1. What is a Quadratic Inequality? | 什么是二次不等式？

A quadratic inequality is an inequality that contains a quadratic expression. The general form is ax² + bx + c compared to zero, where a ≠ 0. The comparison can be greater than (>), less than (<), greater than or equal to (≥), or less than or equal to (≤).

二次不等式是含有二次表达式的不等式。其一般形式是将 ax² + bx + c 与零进行比较，其中 a ≠ 0。比较关系可以是大于（>）、小于（<）、大于或等于（≥）、小于或等于（≤）。

For example, x² − 4x + 3 > 0 and 2x² + 5x − 3 ≤ 0 are both quadratic inequalities.

例如，x² − 4x + 3 > 0 和 2x² + 5x − 3 ≤ 0 都是二次不等式。

The key difference between a quadratic equation and a quadratic inequality is that an equation gives specific solutions (roots), while an inequality gives a range of values, often expressed as intervals of x.

二次方程与二次不等式之间的关键区别在于：方程给出确定的解（根），而不等式则给出一个取值范围，通常以 x 的区间形式表达。

2. The Quadratic Graph Approach | 二次函数图像法

To understand quadratic inequalities deeply, we must first recall the shape of a quadratic graph. When a > 0, the parabola opens upwards (U-shaped); when a < 0, it opens downwards (∩-shaped).

要深入理解二次不等式，我们必须首先回顾二次函数图像的形状。当 a > 0 时，抛物线开口朝上（U 形）；当 a < 0 时，开口朝下（∩ 形）。

Consider the function f(x) = x² − 3x − 10. Factorising gives f(x) = (x − 5)(x + 2). The roots are x = 5 and x = −2. These are the points where the curve crosses the x-axis.

考虑函数 f(x) = x² − 3x − 10。因式分解得 f(x) = (x − 5)(x + 2)。根为 x = 5 和 x = −2。这些是曲线与 x 轴相交的点。

Between the roots, the curve lies below the x-axis (f(x) < 0). Outside the roots, the curve lies above the x-axis (f(x) > 0). This simple observation is the foundation of our solution strategy.

在两根之间，曲线位于 x 轴下方（f(x) < 0）。在两根之外，曲线位于 x 轴上方（f(x) > 0）。这一简单观察就是我们解题策略的基础。

3. Finding the Critical Values | 求关键值（临界值）

The critical values are the roots of the corresponding quadratic equation. To find them, we set the quadratic expression equal to zero and solve:

关键值就是对应二次方程的根。要求它们，我们将二次表达式设为零并求解：

ax² + bx + c = 0

This is achieved by factorising, completing the square, or using the quadratic formula:

这可以通过因式分解、配方法或求根公式来实现：

x = (−b ± √(b² − 4ac)) ⁄ (2a)

For the Edexcel IGCSE exam, factorising is the most common and preferred method, as most exam questions are designed with integer roots.

对于 Edexcel IGCSE 考试而言，因式分解是最常用且首选的方法，因为大多数考题都设计为整数根。

Worked Example: Solve x² − 3x − 10 = 0 to find critical values.

示例：解方程 x² − 3x − 10 = 0 以求出关键值。

Factorise: (x − 5)(x + 2) = 0. Therefore x = 5 or x = −2.

因式分解：(x − 5)(x + 2) = 0。因此 x = 5 或 x = −2。

4. Solving x² − 3x − 10 > 0 | 求解 x² − 3x − 10 > 0

Now let us solve the inequality x² − 3x − 10 > 0 step by step.

现在我们逐步求解不等式 x² − 3x − 10 > 0。

Step 1: Factorise the quadratic expression.

步骤 1：对二次表达式进行因式分解。

x² − 3x − 10 = (x − 5)(x + 2)

Step 2: Find the critical values by setting each factor to zero: x = 5 and x = −2.

步骤 2：令每个因式为零，求关键值：x = 5 和 x = −2。

Step 3: Mark these two values on a number line. They divide the line into three regions:

步骤 3：在数轴上标出这两个值。它们将数轴分为三个区域：

Region A: x < −2
Region B: −2 < x < 5
Region C: x > 5

Step 4: Test a value from each region in the original inequality.

步骤 4：从每个区域取一个值代入原不等式进行检验。

Region A: try x = −3. Then (−3)² − 3(−3) − 10 = 9 + 9 − 10 = 8 > 0 ✓
Region B: try x = 0. Then 0² − 0 − 10 = −10 < 0 ✗
Region C: try x = 6. Then 6² − 18 − 10 = 36 − 18 − 10 = 8 > 0 ✓

Step 5: The solution is the union of the regions where the inequality is satisfied:

步骤 5：解为满足不等式的区域的并集：

x < −2 or x > 5

In interval notation: x ∈ (−∞, −2) ∪ (5, ∞).

用区间表示：x ∈ (−∞, −2) ∪ (5, ∞)。

5. Solving x² − 3x − 10 < 0 | 求解 x² − 3x − 10 < 0

Using the same critical values, we now solve the opposite inequality x² − 3x − 10 < 0.

使用相同的关键值，我们现在求解相反的不等式 x² − 3x − 10 < 0。

From our region testing above, we saw that the expression is negative only in Region B (−2 < x < 5).

从上述区域检验中，我们看到表达式仅在区域 B（−2 < x < 5）中为负。

Therefore the solution is:

因此解为：

−2 < x < 5

This corresponds to the interval x ∈ (−2, 5).

这对应于区间 x ∈ (−2, 5)。

Key Insight: For a positive quadratic (a > 0):
– The inequality > 0 (or ≥ 0) gives solutions “outside” the roots.
– The inequality < 0 (or ≤ 0) gives solutions "between" the roots.

关键要点：对于正系数二次函数（a > 0）：
– 不等式 > 0（或 ≥ 0）的解在两根之外。
– 不等式 < 0（或 ≤ 0）的解在两根之间。

Remember the mnemonic: “First outside, then between” — or think of the U-shaped smile of the parabola: the smile is above the x-axis at the edges.

记住口诀：「外大于、内小于」——或者想象抛物线 U 形的微笑：微笑的两端在 x 轴上方。

6. The Sign Diagram Method | 符号图表法

The sign diagram is a more formal and efficient alternative to region testing. It is particularly useful in the exam because it is quick and less prone to arithmetic errors.

符号图表是区域检验法的一种更正式且更高效的方法。它在考试中特别有用，因为速度快且不容易出现计算错误。

Step 1: Factorise and identify critical values.

步骤 1：因式分解并找出关键值。

Step 2: List the critical values in increasing order on a number line.

步骤 2：在数轴上按递增顺序列出关键值。

Step 3: For each factor, determine its sign in each region. A linear factor (x − k) is negative when x < k, zero when x = k, and positive when x > k.

步骤 3：对每个因式，判断其在每个区域的符号。线性因式 (x − k) 在 x < k 时为负，在 x = k 时为零，在 x > k 时为正。

Step 4: Multiply the signs together to determine the sign of the whole expression in each region.

步骤 4：将符号相乘，确定整个表达式在每个区域的符号。

Let us construct a sign diagram for (x − 5)(x + 2):

让我们为 (x − 5)(x + 2) 构建符号图表：

Region	x < −2	−2 < x < 5	x > 5
(x + 2)	Negative (−)	Positive (+)	Positive (+)
(x − 5)	Negative (−)	Negative (−)	Positive (+)
Product	(+) — above axis	(−) — below axis	(+) — above axis

This table immediately reveals that the product is positive for x < −2 or x > 5, and negative for −2 < x < 5.

该表立即揭示出：乘积在 x < −2 或 x > 5 时为正，在 −2 < x < 5 时为负。

With practice, you can draw this sign diagram in under a minute, making it ideal for the exam environment.

通过练习，你可以在一分钟内画出这个符号图表，非常适合考试环境。

7. When the Coefficient a < 0 | 当系数 a < 0 时

What happens when the quadratic has a negative leading coefficient, such as −x² + 4x + 5 ≥ 0? We have two options.

当二次函数首项系数为负时，例如 −x² + 4x + 5 ≥ 0，该怎么办？我们有两种选择。

Option 1 — Multiply by −1: Multiply both sides by −1 and reverse the inequality sign. This is usually the cleanest approach.

方法 1 — 乘以 −1：将不等式两边同时乘以 −1，并反转不等号。这通常是最简洁的方法。

−x² + 4x + 5 ≥ 0 ⟹ x² − 4x − 5 ≤ 0

Option 2 — Work with the negative quadratic: The parabola opens downwards, so the graph is above the x-axis between the roots and below the x-axis outside the roots. This reverses the relationship from the a > 0 case.

方法 2 — 直接处理负二次函数：抛物线开口朝下，因此图像在两根之间位于 x 轴上方，在两根之外位于 x 轴下方。这与 a > 0 的情况正好相反。

Let us complete Option 1. Factorise x² − 4x − 5 = (x − 5)(x + 1) ≤ 0. The critical values are −1 and 5. Since the inequality is ≤ 0, we take the region between the roots:

我们完成方法 1。因式分解 x² − 4x − 5 = (x − 5)(x + 1) ≤ 0。关键值为 −1 和 5。由于不等式为 ≤ 0，我们取两根之间的区域：

−1 ≤ x ≤ 5

Important: Never forget to reverse the inequality sign when multiplying or dividing by a negative number. This is one of the most common careless errors in IGCSE examinations.

重要：在乘以或除以负数时，切勿忘记反转不等号。这是 IGCSE 考试中最常见的粗心错误之一。

8. Inequalities with ≥ and ≤ | 含 ≥ 和 ≤ 的不等式

The difference between strict inequalities (>, <) and non-strict inequalities (≥, ≤) lies in whether the critical values themselves are included in the solution.

严格不等式（>、<）与非严格不等式（≥、≤）之间的区别在于关键值本身是否包含在解中。

For ≥ 0 or ≤ 0, the roots are included because the expression equals zero at those points, and zero satisfies both ≥ 0 and ≤ 0.

对于 ≥ 0 或 ≤ 0，根被包含在解中，因为在那些点上表达式等于零，而零同时满足 ≥ 0 和 ≤ 0。

For example, the solution of x² − 3x − 10 ≥ 0 is x ≤ −2 or x ≥ 5, using square brackets in interval notation:

例如，x² − 3x − 10 ≥ 0 的解为 x ≤ −2 或 x ≥ 5，在区间表示中使用方括号：

x ∈ (−∞, −2] ∪ [5, ∞)

And the solution of x² − 3x − 10 ≤ 0 is −2 ≤ x ≤ 5, or x ∈ [−2, 5].

而 x² − 3x − 10 ≤ 0 的解为 −2 ≤ x ≤ 5，即 x ∈ [−2, 5]。

The graph makes this clear: at the roots, the parabola touches the x-axis, so those points satisfy the equality condition. Use a closed dot ● on the number line for included values, and an open dot ○ for excluded values.

图像清楚地说明了这一点：在根处，抛物线接触 x 轴，因此这些点满足等号条件。数轴上用实心点 ● 表示包含该值，用空心点 ○ 表示不包含该值。

9. Perfect Squares and No Real Roots | 完全平方与无实数根

Not all quadratic inequalities factorise into two distinct linear factors. Two special cases require extra attention.

并非所有二次不等式都能分解为两个不同的线性因式。两种特殊情况需要特别注意。

Case 1 — Perfect square: Suppose x² − 6x + 9 > 0. This factorises to (x − 3)² > 0.

情形 1 — 完全平方：假设 x² − 6x + 9 > 0。它分解为 (x − 3)² > 0。

A perfect square is always greater than or equal to zero. It equals zero only when x = 3. Therefore the solution to (x − 3)² > 0 is all real numbers except x = 3:

完全平方总是大于或等于零。它仅在 x = 3 时等于零。因此 (x − 3)² > 0 的解是所有实数除了 x = 3：

x ∈ ℝ, x ≠ 3

If the inequality were (x − 3)² ≥ 0, the solution would simply be all real numbers, since the square is never negative.

如果不等式为 (x − 3)² ≥ 0，解就是所有实数，因为平方永远不会为负。

Case 2 — No real roots: Consider x² + 2x + 5 > 0. The discriminant is b² − 4ac = 4 − 20 = −16 < 0, so the quadratic has no real roots. The parabola never touches the x-axis. Since a = 1 > 0, the parabola opens upwards and lies entirely above the x-axis. Therefore the inequality is true for all real x.

情形 2 — 无实数根：考虑 x² + 2x + 5 > 0。判别式为 b² − 4ac = 4 − 20 = −16 < 0，因此该二次函数没有实数根。抛物线不与 x 轴相交。由于 a = 1 > 0，抛物线开口朝上且完全位于 x 轴上方。因此该不等式对所有实数 x 均成立。

In contrast, x² + 2x + 5 < 0 would have no solution at all, because a positive quadratic can never be negative if it never crosses the x-axis.

相反，x² + 2x + 5 < 0 将完全没有解，因为一个不穿越 x 轴的正系数二次函数永远不可能为负。

10. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Here are the most frequent errors students make in the Edexcel IGCSE exam when dealing with quadratic inequalities, along with strategies to avoid them.

以下是学生在 Edexcel IGCSE 考试中处理二次不等式时最常见的错误，以及相应的避免策略。

Forgetting the direction of the parabola. Always check the sign of a. If a < 0, the 'outside-between‘ rule reverses. Solution: multiply by −1 first.
错误一：忘记抛物线的开口方向。始终检查 a 的符号。如果 a < 0，则「外大于、内小于」的规律会反转。解决办法：先将不等式两边乘以 −1。
Mixing up the inequality sign after multiplying by a negative. Example: −x² + 4 > 0 must become x² − 4 < 0, not still > 0.
错误二：乘以负数后没有反转不等号。例如：−x² + 4 > 0 必须变为 x² − 4 < 0，而不是保持不变。
Writing the answer as two separate intervals without the word ‘or’. For > 0 with two roots, write x < a or x > b — never write x < a and x > b as a single statement.
错误三：将两个分离区间未用「或」连接。对于 > 0 且有两个根的情况，应写为 x < a 或 x > b——绝不能将其作为单一语句写成 x < a 且 x > b。
Forgetting to include roots when the sign is ≥ or ≤. Check whether the critical values should use solid dots (included) or hollow dots (excluded).
错误四：当符号为 ≥ 或 ≤ 时忘记包含根。检查关键值应使用实心点（包含）还是空心点（不包含）。
Not testing the regions. Test one value from each region to confirm your algebraic conclusion matches the actual sign pattern.
错误五：没有检验各区域。从每个区域取一个值代入检验，以确认代数结论与实际符号模式一致。

11. Worked Exam-Style Examples | 考试风格例题

Let us work through three full examples in the format you will encounter in the Edexcel IGCSE examination.

让我们以 Edexcel IGCSE 考试中会遇到的形式，完成三个完整例题。

Example A: Solve 2x² − 5x − 3 < 0.

例题 A：求解 2x² − 5x − 3 < 0。

Factorise: 2x² − 5x − 3 = (2x + 1)(x − 3). Critical values satisfy 2x + 1 = 0 ⟹ x = −½, and x − 3 = 0 ⟹ x = 3.

因式分解：2x² − 5x − 3 = (2x + 1)(x − 3)。关键值满足 2x + 1 = 0 ⟹ x = −½，以及 x − 3 = 0 ⟹ x = 3。

Since a = 2 > 0, the parabola opens upwards. The inequality is ‘less than zero‘, so we take the region between the roots:

由于 a = 2 > 0，抛物线开口朝上。不等式为「小于零」，因此我们取两根之间的区域：

−½ < x < 3

Example B: Solve x² + 6x + 9 ≤ 0.

例题 B：求解 x² + 6x + 9 ≤ 0。

Factorise: x² + 6x + 9 = (x + 3)² ≤ 0. A perfect square is always non-negative. The only value that makes it equal to zero is x = −3. Since it can never be negative, the solution is simply:

因式分解：x² + 6x + 9 = (x + 3)² ≤ 0。完全平方始终非负。唯一使其等于零的值是 x = −3。由于它永远不能为负，解法就是：

x = −3

Example C: Solve −x² + 3x + 4 > 0.

例题 C：求解 −x² + 3x + 4 > 0。

Multiply by −1 and reverse the sign: x² − 3x − 4 < 0. Factorise: (x − 4)(x + 1) < 0. Roots are 4 and −1. The solution lies between the roots:

乘以 −1 并反转符号：x² − 3x − 4 < 0。因式分解：(x − 4)(x + 1) < 0。根为 4 和 −1。解在两根之间：

−1 < x < 4

12. Summary of Key Points | 要点总结

Let us consolidate everything we have learned into a concise revision checklist.

让我们将所学内容浓缩为一份简洁的复习清单。

Always rearrange the inequality so that one side is zero.
总是将不等式整理为一边为零的形式。
Factorise the quadratic and find the critical values (roots).
对二次式进行因式分解并求出关键值（根）。
If necessary, multiply by −1 to make a > 0, remembering to reverse the inequality sign.
如有必要，乘以 −1 使 a > 0，并记得反转不等号。
Sketch the parabola or construct a sign diagram.
画出抛物线草图或构建符号图表。
For a > 0: > 0 gives x < root₁ or x > root₂; < 0 gives root₁ < x < root₂.
对于 a > 0：> 0 给出 x < 根₁ 或 x > 根₂；< 0 给出根₁ < x < 根₂。
Use ≥/≤ to include the roots; use >/< to exclude them.
使用 ≥/≤ 时包含根；使用 >/< 时排除根。
Handle special cases: perfect squares touch the axis at one point; positive quadratics with a negative discriminant are always positive.
处理特殊情况：完全平方在一点处接触 x 轴；判别式为负的正系数二次函数恒为正。

Mastery of quadratic inequalities comes from practice. Work through every question in your Edexcel IGCSE past papers, and draw a sketch for every problem until the graphical intuition becomes second nature. This skill will serve you well both in Paper 1 and Paper 2.

掌握二次不等式需要大量练习。仔细完成 Edexcel IGCSE 历年真题中的每一道题目，并为每道题画草图，直到图形直觉成为你的第二天性。这一技能在 Paper 1 和 Paper 2 中都将使你受益良多。

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Solving Quadratic Inequalities | 解二次不等式