📚 Quadratic Inequalities | 二次不等式
Quadratic inequalities are a core topic in IGCSE mathematics. They build on your knowledge of quadratic equations and graphs, but add an extra layer of reasoning about intervals and signs.
二次不等式是 IGCSE 数学的核心内容。它建立在二次方程和二次函数图像的基础上，同时增加了一层关于区间和符号的推理。
In this article, we will explore several methods for solving quadratic inequalities, including factorisation, the sign-pattern method, the graph method, completing the square, and the quadratic formula. We will also discuss common pitfalls and exam tips.
在本文中，我们将探讨求解二次不等式的多种方法，包括因式分解法、符号规律法、图像法、配方法和公式法。我们还会讨论常见错误与考试技巧。
1. Understanding Quadratic Inequalities | 理解二次不等式
Before solving, let us clarify what a quadratic inequality is. It is any statement of the form ax² + bx + c > 0, ax² + bx + c ≥ 0, ax² + bx + c < 0, or ax² + bx + c ≤ 0, where a ≠ 0.
在开始求解之前，我们先明确什么是二次不等式。它是指形如 ax² + bx + c > 0、ax² + bx + c ≥ 0、ax² + bx + c < 0 或 ax² + bx + c ≤ 0 的表达式，其中 a ≠ 0。
The solution set of a quadratic inequality is the range of x-values for which the inequality is true. Unlike linear inequalities, quadratic inequalities often have two separate intervals as their solution.
二次不等式的解集是使不等式成立的 x 的取值范围。与一次不等式不同，二次不等式的解通常包含两个不相交的区间。
2. Solving by Factorisation | 因式分解法
First, factor the quadratic expression as if you are solving a quadratic equation.
首先，像解一元二次方程那样对二次表达式进行因式分解。
Example: Solve x² − 5x + 6 > 0.
例：解不等式 x² − 5x + 6 > 0。
Factorising gives (x − 2)(x − 3) > 0.
因式分解得到 (x − 2)(x − 3) > 0。
The critical roots are x = 2 and x = 3. These divide the number line into three intervals:
临界根为 x = 2 和 x = 3。它们把数轴分成三个区间：
- x < 2
- 2 < x < 3
- x > 3
Test each interval by picking a representative number. For x < 2, choose x = 0: (0 − 2)(0 − 3) = 6 > 0, so this interval works.
选取一个代表数来检验每个区间。对于 x < 2，取 x = 0：(0 − 2)(0 − 3) = 6 > 0，所以该区间满足不等式。
For 2 < x < 3, choose x = 2.5: (2.5 − 2)(2.5 − 3) = −0.25 < 0, so this interval does not work.
对于 2 < x < 3，取 x = 2.5：(2.5 − 2)(2.5 − 3) = −0.25 < 0，所以该区间不满足不等式。
For x > 3, choose x = 4: (4 − 2)(4 − 3) = 2 > 0, so this interval works.
对于 x > 3，取 x = 4：(4 − 2)(4 − 3) = 2 > 0，所以该区间满足不等式。
Thus the solution is x < 2 or x > 3.
因此解为 x < 2 或 x > 3。
x < 2 or x > 3
3. The Sign-Pattern Method | 符号规律法
Instead of testing every interval, you can use the sign pattern of the factors. For a product of two linear factors, the sign changes at each root.
不需要逐个区间检验，你可以利用因子的符号规律。对于两个线性因子的乘积，符号在每个根处发生变化。
Consider (x − 2)(x − 3) > 0. Mark the roots 2 and 3 on a number line.
考虑 (x − 2)(x − 3) > 0。在数轴上标出根 2 和 3。
To the right of the largest root, both factors are positive, so the product is positive.
在最大根的右侧，两个因子均为正，因此乘积为正。
Between the roots, one factor is negative and the other positive, so the product is negative.
在两个根之间，一个因子为负，另一个为正，因此乘积为负。
To the left of the smallest root, both factors are negative, so the product is positive.
在最小根的左侧，两个因子均为负，因此乘积为正。
This confirms the solution: x < 2 or x > 3.
这证实了解为 x < 2 或 x > 3。
4. The Graph Method | 图像法
Consider the parabola y = x² − 5x + 6. Since a = 1 > 0, it opens upward.
考虑抛物线 y = x² − 5x + 6。由于 a = 1 > 0，抛物线开口向上。
The parabola crosses the x-axis at x = 2 and x = 3.
抛物线在 x = 2 和 x = 3 处与 x 轴相交。
For x < 2, the curve lies above the x-axis, so y > 0.
当 x < 2 时，曲线位于 x 轴上方，因此 y > 0。
For 2 < x < 3, the curve lies below the x-axis, so y < 0.
当 2 < x < 3 时，曲线位于 x 轴下方，因此 y < 0。
For x > 3, the curve is again above the x-axis, so y > 0.
当 x > 3 时，曲线再次位于 x 轴上方，因此 y > 0。
The graph method gives a clear visual confirmation of the algebraic result.
图像法为代数结果提供了清晰的视觉验证。
5. Solving with a Negative Leading Coefficient | 首项系数为负时的解法
When a < 0, the parabola opens downward, so the inequality sign pattern reverses.
当 a < 0 时，抛物线开口向下，因此不等式的符号规律发生反转。
Example: Solve −x² + 4x − 3 ≥ 0.
例：解不等式 −x² + 4x − 3 ≥ 0。
Multiply both sides by −1 and reverse the inequality sign:
两边同时乘以 −1 并反转不等号：
x² − 4x + 3 ≤ 0
Factorising gives (x − 1)(x − 3) ≤ 0.
因式分解得到 (x − 1)(x − 3) ≤ 0。
The roots are x = 1 and x = 3. For a ≤ sign, the product must be negative or zero.
根为 x = 1 和 x = 3。对于 ≤ 号，乘积必须为负或零。
Between the roots, (x − 1) is positive and (x − 3) is negative, so the product is negative.
在两个根之间，(x − 1) 为正而 (x − 3) 为负，因此乘积为负。
Thus the solution is 1 ≤ x ≤ 3.
因此解为 1 ≤ x ≤ 3。
1 ≤ x ≤ 3
Alternatively, you could solve the original inequality directly by observing that a downward-opening parabola is positive between its roots.
另一种方法是直接观察开口向下的抛物线在其两个根之间取正值，从而求解原不等式。
6. Inequalities with Two Distinct Roots | 具有两个不同实根的不等式
When a quadratic has two distinct real roots r₁ and r₂, the solution set depends on the direction of the inequality and the sign of a.
当二次方程具有两个不同的实根 r₁ 和 r₂ 时，其解集取决于不等号的方向以及 a 的符号。
For a > 0:
对于 a > 0：
- ax² + bx + c > 0 → x < r₁ or x > r₂
- ax² + bx + c ≥ 0 → x ≤ r₁ or x ≥ r₂
- ax² + bx + c < 0 → r₁ < x < r₂
- ax² + bx + c ≤ 0 → r₁ ≤ x ≤ r₂
For a < 0, the inequalities reverse:
对于 a < 0，不等式方向反转：
- ax² + bx + c > 0 → r₁ < x < r₂
- ax² + bx + c ≥ 0 → r₁ ≤ x ≤ r₂
- ax² + bx + c < 0 → x < r₁ or x > r₂
- ax² + bx + c ≤ 0 → x ≤ r₁ or x ≥ r₂
Knowing these patterns can save time in exams.
在考试中记住这些规律可以节省时间。
7. Inequalities with a Repeated Root | 具有重根的不等式
If the quadratic has a repeated root, the graph touches the x-axis at one point and does not cross it.
如果二次方程具有重根，则图像与 x 轴相切于一点而不穿过它。
Example: Solve x² − 4x + 4 > 0.
例：解不等式 x² − 4x + 4 > 0。
Factorising gives (x − 2)² > 0.
因式分解得到 (x − 2)² > 0。
The square of a real number is always non-negative. It equals zero only at x = 2.
一个实数的平方永远是非负的。它仅在 x = 2 时等于零。
Therefore the solution is all real numbers except x = 2:
因此解为除 x = 2 以外的所有实数：
x ∈ ℝ, x ≠ 2
If the inequality were x² − 4x + 4 ≥ 0, the solution would be all real numbers.
如果不等式为 x² − 4x + 4 ≥ 0，那么解集为所有实数。
If it were x² − 4x + 4 < 0, there would be no solution.
如果为 x² − 4x + 4 < 0，则无解。
If x² − 4x + 4 ≤ 0, the solution is exactly x = 2.
如果为 x² − 4x + 4 ≤ 0，则解恰为 x = 2。
8. Inequalities with No Real Roots | 无实根的不等式
When the discriminant b² − 4ac < 0, the quadratic does not touch or cross the x-axis.
当判别式 b² − 4ac < 0 时，二次函数图像与 x 轴不相交也不相切。
Example: Solve x² + 2x + 5 > 0.
例：解不等式 x² + 2x + 5 > 0。
Complete the square: (x + 1)² + 4 > 0.
配方：(x + 1)² + 4 > 0。
Since (x + 1)² is always non-negative, adding 4 makes the expression always positive.
由于 (x + 1)² 永远非负，加 4 后表达式恒为正。
Thus the solution is all real numbers.
因此解为所有实数。
x ∈ ℝ
If the inequality were x² + 2x + 5 < 0, there would be no solution, because the expression is always positive.
如果不等式为 x² + 2x + 5 < 0，则无解，因为该表达式恒为正。
9. Solving by Completing the Square | 配方法
Completing the square is useful when factorisation is not straightforward.
当因式分解不太直接时，配方法非常有用。
Example: Solve 2x² − 4x − 6 < 0.
例：解不等式 2x² − 4x − 6 < 0。
Divide by 2: x² − 2x − 3 < 0.
两边除以 2：x² − 2x − 3 < 0。
Complete the square:
配方：
(x − 1)² − 1 − 3 < 0
(x − 1)² < 4
Take the square root of both sides:
两边开平方：
−2 < x − 1 < 2
Add 1 throughout:
整体加 1：
−1 < x < 3
Therefore the solution is −1 < x < 3, which matches the factorisation result (x − 3)(x + 1) < 0.
因此解为 −1 < x < 3，与因式分解结果 (x − 3)(x + 1) < 0 一致。
10. The Quadratic Formula Method | 公式法
If the quadratic does not factorise easily, use the quadratic formula to find the roots.
如果二次式不易因式分解，可以使用二次公式求根。
x = (−b ± √(b² − 4ac)) / (2a)
Example: Solve x² + 2x − 4 > 0.
例：解不等式 x² + 2x − 4 > 0。
Here a = 1, b = 2, c = −4.
这里 a = 1，b = 2，c = −4。
Discriminant: b² − 4ac = 4 + 16 = 20.
判别式：b² − 4ac = 4 + 16 = 20。
Roots:
根为：
x = (−2 ± √20) / 2 = −1 ± √5
Thus the roots are x = −1 − √5 and x = −1 + √5.
因此根为 x = −1 − √5 和 x = −1 + √5。
Since a > 0 and the inequality is > 0, the solution is outside the interval between the roots:
由于 a > 0 且不等式为 > 0，解位于两根之外：
x < −1 − √5 or x >
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