A-Level Maths: The Complete Guide to Solving Quadratic Inequalities | A-Level 数学：二次不等式解法全攻略

📚 A-Level Maths: The Complete Guide to Solving Quadratic Inequalities | A-Level 数学：二次不等式解法全攻略

Quadratic inequalities appear regularly in A-Level Mathematics, especially in pure maths and even in some applied contexts. Mastering them is not only about algebraic manipulation; it also requires a clear visual understanding of parabolas and sign changes.

二次不等式是 A-Level 数学中的常考内容，无论是在纯数学还是部分应用问题中都会出现。掌握它不仅仅需要代数变形能力，更需要对抛物线和符号变化有清晰的图像理解。

1. Understanding Quadratic Inequalities | 理解二次不等式

A quadratic inequality is any inequality that can be written in one of the four standard forms: ax² + bx + c > 0, ax² + bx + c < 0, ax² + bx + c ≥ 0, or ax² + bx + c ≤ 0, where a ≠ 0.

二次不等式是指可以写成以下四种标准形式之一的不等式：ax² + bx + c > 0，ax² + bx + c < 0，ax² + bx + c ≥ 0，或 ax² + bx + c ≤ 0，其中 a ≠ 0。

Unlike a linear inequality, the solution of a quadratic inequality is usually an interval, the union of two intervals, or sometimes the empty set or all real numbers.

与一次不等式不同，二次不等式的解通常是一个区间、两个区间的并集，有时也可能是空集或全体实数。

2. Standard Form and Basic Definitions | 标准形式与基本定义

Before solving, always make sure the quadratic expression is arranged in descending powers of x: ax² + bx + c. The coefficient a is especially important because its sign determines whether the parabola opens upward or downward.

在求解之前，请确保二次表达式按 x 的降幂排列：ax² + bx + c。系数 a 特别重要，因为它的正负决定了抛物线开口向上还是向下。

If a > 0, the parabola opens upward, like a U shape.

如果 a > 0，抛物线开口向上，形状类似字母 U。
If a < 0, the parabola opens downward, like an inverted U shape.

如果 a < 0，抛物线开口向下，形状类似倒过来的 U。
The roots of ax² + bx + c = 0 are the x-intercepts of the corresponding parabola.

方程 ax² + bx + c = 0 的根就是对应抛物线与 x 轴的交点。

The discriminant, Δ = b² − 4ac, tells us how many real roots exist:

判别式 Δ = b² − 4ac 告诉我们存在多少个实根：

Δ > 0: two distinct real roots.

Δ > 0：两个不同的实根。
Δ = 0: one repeated real root.

Δ = 0：一个重根。
Δ < 0: no real roots.

Δ < 0：没有实根。

3. Step 1: Rewrite in Standard Form | 第一步：化为标准形式

If the inequality is not yet in standard form, move all terms to one side of the inequality sign. For example, x² < 4 − 2x should first be rewritten as x² + 2x − 4 < 0.

如果不等式还不是标准形式，先把所有项移到不等号的一侧。例如，x² < 4 − 2x 应首先改写为 x² + 2x − 4 < 0。

Be careful when multiplying or dividing by a negative number: you must reverse the inequality sign. However, for quadratics, it is usually cleaner to keep the coefficient of x² positive.

注意：当乘以或除以一个负数时，必须改变不等号方向。不过对于二次不等式，通常保持 x² 的系数为正会更加简洁。

Example: −x² + 4x − 3 > 0 ⇒ x² − 4x + 3 < 0

In the example above, both sides were multiplied by −1, so the direction of the inequality was reversed.

在上面的例子中，两边同时乘以 −1，所以不等号方向发生了反转。

4. Step 2: Solve the Corresponding Quadratic Equation | 第二步：解对应二次方程

To find the critical values, solve the equation ax² + bx + c = 0. This can be done by factorising, completing the square, or using the quadratic formula.

要求出临界值，需要先解方程 ax² + bx + c = 0。可以使用因式分解、配方法或二次公式来求解。

x = (−b ± √(b² − 4ac)) / (2a)

If the quadratic factorises easily, factorising is usually the fastest method. For example, x² − 5x + 6 = 0 can be written as (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

如果二次式容易因式分解，因式分解通常是最快的方法。例如，x² − 5x + 6 = 0 可以写成 (x − 2)(x − 3) = 0，因此得到 x = 2 或 x = 3。

These roots divide the real number line into distinct intervals. Inside each interval, the sign of the quadratic expression does not change.

这些根将实数轴分成若干个区间。在每个区间内，二次表达式的正负号不会改变。

5. Step 3: Sketch the Graph or Use a Sign Table | 第三步：画图或使用符号表

There are two reliable methods for determining the sign of the quadratic on each interval: drawing a quick sketch of the parabola, or constructing a sign table.

有两种可靠的方法可以判断二次式在每个区间内的符号：快速画出抛物线草图，或者构造符号表。

When sketching, pay attention to the coefficient of x². If a > 0, the graph is U-shaped, so it is positive outside the roots and negative between the roots. If a < 0, the graph is ∩-shaped, so the signs are reversed.

画图时，注意 x² 的系数。如果 a > 0，图像是 U 形，因此在两根之外为正，在两根之间为负。如果 a < 0，图像是 ∩ 形，符号正好相反。

For example, solve x² − 5x + 6 > 0:

例如，解 x² − 5x + 6 > 0：

Roots are x = 2 and x = 3.

根是 x = 2 和 x = 3。
The graph of y = x² − 5x + 6 opens upward.

函数 y = x² − 5x + 6 的图像开口向上。
It is above the x-axis when x < 2 or x > 3.

当 x < 2 或 x > 3 时，图像在 x 轴上方。

Solution: x < 2 or x > 3

6. The Sign Table Method | 符号表法详解

A sign table is a systematic alternative to sketching. It shows the sign of each linear factor on every interval.

符号表是画图之外的一种系统化方法。它展示每个线性因子在各个区间内的符号。

Let us use the same example: solve x² − 5x + 6 > 0.

我们使用同样的例子：解 x² − 5x + 6 > 0。

x² − 5x + 6 = (x − 2)(x − 3)

Interval	x < 2	2 < x < 3	x > 3
x − 2	−	+	+
x − 3	−	−	+
(x − 2)(x − 3)	+	−	+

The product is positive for x < 2 and for x > 3, so the solution is x < 2 or x > 3.

乘积在 x < 2 和 x > 3 时为正，因此解为 x < 2 或 x > 3。

For a strict inequality such as > or <, the roots themselves are not included. For ≥ or ≤, the roots are included and should be written as closed intervals.

对于严格不等式如 > 或 <，根本身不包含在解集内。对于 ≥ 或 ≤，根包含在解集内，应写成闭区间。

7. Dealing with “No Real Roots” Cases | 处理无实数根的情形

When Δ < 0, the quadratic equation ax² + bx + c = 0 has no real roots. In this case the parabola does not cross or touch the x-axis at all.

当 Δ < 0 时，方程 ax² + bx + c = 0 没有实根。此时抛物线完全不与 x 轴相交或相切。

If a > 0 and Δ < 0, the entire parabola lies above the x-axis, so the quadratic expression is positive for every real x.

如果 a > 0 且 Δ < 0，整条抛物线都在 x 轴上方，因此二次表达式对任意实数 x 都为正。

x² + 2x + 3 > 0 is true for all real x.

x² + 2x + 3 > 0 对所有实数 x 都成立。
x² + 2x + 3 < 0 has no real solutions.

x² + 2x + 3 < 0 没有实数解。
x² + 2x + 3 ≤ 0 also has no real solutions.

x² + 2x + 3 ≤ 0 同样没有实数解。
x² + 2x + 3 ≥ 0 is true for all real x.

x² + 2x + 3 ≥ 0 对所有实数 x 都成立。

If a < 0 and Δ < 0, the entire parabola lies below the x-axis, so the quadratic expression is negative for every real x.

如果 a < 0 且 Δ < 0，整条抛物线都在 x 轴下方，因此二次表达式对任意实数 x 都为负。

8. Special Cases: Perfect Squares and Linear Factors | 特殊情况：完全平方与线性因子

When the discriminant equals zero, the quadratic has a repeated root. The expression is either always non-negative or always non-positive, depending on the sign of a.

当判别式等于零时，二次方程有一个重根。此时表达式要么恒为非负，要么恒为非正，具体取决于 a 的符号。

Consider x² − 4x + 4 ≥ 0. Notice that x² − 4x + 4 = (x − 2)².

考虑 x² − 4x + 4 ≥ 0。注意 x² − 4x + 4 = (x − 2)²。

Since a square is never negative, (x − 2)² ≥ 0 is true for all real x. Therefore the solution is all real numbers.

由于平方永远不会为负，(x − 2)² ≥ 0 对一切实数 x 都成立。因此解集为全体实数。

However, for the strict inequality (x − 2)² > 0, the solution is x ≠ 2, because zero is not greater than zero.

然而，对于严格不等式 (x − 2)² > 0，解集是 x ≠ 2，因为零不大于零。

Another special case occurs when the quadratic can be written as a product involving a repeated factor. The sign table will still work, but zero appears at the repeated root only once. The sign does not change as x passes through a repeated root.

另一种特殊情况是二次式可以写成含有重因子的乘积形式。符号表仍然适用，但零只在重根处出现一次。当 x 经过重根时，符号不会发生变化。

9. Common Mistakes to Avoid | 常见错误避免

Many students lose marks on quadratic inequalities because of small but avoidable mistakes. Here are the most common ones.

许多学生在二次不等式上失分，是因为犯了一些细小但可以避免的错误。以下是最常见的几个。

Forgetting to reverse the inequality sign when multiplying or dividing by a negative number.

在乘以或除以负数时忘记反转不等号方向。
Jumping from (x − 2)(x − 3) > 0 directly to a one-sided inequality such as x > 3. You must consider both intervals.

从 (x − 2)(x − 3) > 0 直接跳到 x > 3 这样的单边不等式。你必须同时考虑两个区间。
Including or excluding the roots incorrectly. Strict inequalities exclude roots; non-strict inequalities include them.

错误地包含或排除根。严格不等式排除根；非严格不等式包含根。
Using “and” instead of “or” between two solution intervals. In cases like x < 2 or x > 3, these two sets must be written with “or”.

在两个解区间之间错误地使用“且”而不是“或”。在类似 x < 2 或 x > 3 的情况下，两个集合必须用“或”连接。
Forgetting to check the sign of a when graphing the parabola.

画抛物线时忘记检查 a 的符号。

Always test one value from each interval if you are unsure. This is a quick and reliable way to check your final answer.

如果不确定，可以在每个区间内选择一个数进行检验。这是检查最终答案的快速而可靠的方法。

10. Worked Examples | 典型例题精讲

Let us work through two further examples in detail.

下面我们详细完成两个进一步的例子。

Example 1: Solve 2x² − 5x − 3 ≤ 0.

例 1：解 2x² − 5x − 3 ≤ 0。

2x² − 5x − 3 = (2x + 1)(x − 3)

Setting each factor to zero gives x = −1/2 and x = 3. The coefficient 2 is positive, so the parabola opens upward.

令每个因式为零，得到 x = −1/2 和 x = 3。系数 2 为正，所以抛物线开口向上。

For an upward-opening parabola, the expression is negative or zero between the roots. Therefore the solution is:

对于开口向上的抛物线，表达式在两根之间小于或等于零。因此解为：

−1/2 ≤ x ≤ 3

Example 2: Solve −x² + 4x − 4 < 0.

例 2：解 −x² + 4x − 4 < 0。

First multiply by −1 to make the leading coefficient positive. This reverses the inequality:

首先乘以 −1 使首项系数为正。这会反转不等号方向：

x² − 4x + 4 > 0

Notice that x² − 4x + 4 = (x − 2)², so the expression is zero at x = 2 and positive everywhere else. Hence the solution is x ≠ 2.

注意 x² − 4x + 4 = (x − 2)²，因此表达式在 x = 2 处为零，在其他任何位置都为正。因此解为 x ≠ 2。

11. Practice Questions | 练习题

Try these questions on your own before checking the answers.

请先独立完成以下题目，再核对答案。

1. Solve x² − 4x − 12 > 0.

1. 解 x² − 4x − 12 > 0。
2. Solve x² − 6x + 9 ≤ 0.

2. 解 x² − 6x + 9 ≤ 0。
3. Solve 3x² + x + 2 > 0.

3. 解 3x² + x + 2 > 0。
4. Solve −2x² + 3x + 2 ≥ 0.

4. 解 −2x² + 3x + 2 ≥ 0。

Answers:

答案：

x < −2 or x > 6.

x < −2 或 x > 6。
x = 3 is the only solution.

x = 3 是唯一解。
All real x, because Δ < 0 and the coefficient of x² is positive.

全体实数 x，因为 Δ < 0 且 x² 的系数为正。
−1/2 ≤ x ≤ 2.

−1/2 ≤ x ≤ 2。

12. Summary | 总结

To solve any quadratic inequality, follow the same reliable procedure:

要解任何二次不等式，都可以遵循以下可靠的步骤：

Rewrite the inequality in standard form with all terms on one side.

将不等式化为标准形式，把所有项放到同一侧。
Solve the corresponding quadratic equation to find the critical roots.

解对应的二次方程，求出临界根。
Sketch the parabola or construct a sign table to determine the sign in each interval.

画出抛物线草图或构造符号表，判断每个区间内的符号。
Write the final solution using correct inequality notation or interval notation.

用正确的不等式记号或区间记号写出最终解集。

With practice, quadratic inequalities become a reliable source of marks. Always check your answer by substituting a test point from each interval.

只要多加练习，二次不等式会成为稳定的得分点。永远记得在每个区间代入一个测试点来检验答案。

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A-Level Maths: The Complete Guide to Solving Quadratic Inequalities | A-Level 数学：二次不等式解法全攻略