📚 Quadratic Inequalities | 二次不等式
Quadratic inequalities are a key topic in IGCSE Mathematics, combining algebraic manipulation with graphical reasoning. In this article, we will break down the methods step by step, using clear examples and visual explanations to help you master this topic.
二次不等式是 IGCSE 数学中的一个重要考点，它将代数运算与图像推理结合在一起。本文将分步骤为你讲解解题方法，并通过清晰的例题和图像化解释，帮助你彻底掌握这一知识点。
1. What Is a Quadratic Inequality | 什么是二次不等式
A quadratic inequality is an inequality that involves a quadratic expression. The general forms are:
二次不等式是包含二次表达式的不等式，其一般形式为：
ax² + bx + c > 0 , ax² + bx + c ≥ 0 , ax² + bx + c < 0 , ax² + bx + c ≤ 0
where a, b and c are constants and a ≠ 0. Unlike a linear inequality, which gives a simple range of x values, a quadratic inequality often gives two separate intervals or a single continuous interval.
其中 a、b、c 为常数，且 a ≠ 0。与只给出单一 x 取值范围的一次不等式不同，二次不等式通常给出两个分离的区间或一个连续的区间。
2. Solving by Factorising | 通过因式分解求解
When the quadratic expression can be factorised, we first find the roots (also called critical values) by treating the inequality as an equation. Then we use a number line or a sign table to determine where the expression is positive or negative.
当二次表达式可以因式分解时，我们先将不等式看作方程，求出根（也称为临界值）。然后利用数轴或符号表来判断表达式在何处为正、何处为负。
Example 1: Solve x² − 5x + 6 > 0.
例 1：解不等式 x² − 5x + 6 > 0。
Step 1: Factorise the quadratic expression:
第一步：对二次表达式进行因式分解：
x² − 5x + 6 = (x − 2)(x − 3)
Step 2: Find the roots by setting (x − 2)(x − 3) = 0. The roots are x = 2 and x = 3.
第二步：令 (x − 2)(x − 3) = 0，求出根为 x = 2 和 x = 3。
Step 3: Draw a number line and test the sign of the product in each interval: x < 2, 2 < x < 3, and x > 3.
第三步：画数轴，并在每个区间 x < 2、2 < x < 3、x > 3 中检验乘积的符号。
For x = 0 (x < 2): (0 − 2)(0 − 3) = (−)(−) = + .
取 x = 0（属于 x < 2）：(0 − 2)(0 − 3) = (−)(−) = +。
For x = 2.5 (2 < x < 3): (2.5 − 2)(2.5 − 3) = (+)(−) = − .
取 x = 2.5（属于 2 < x < 3）：(2.5 − 2)(2.5 − 3) = (+)(−) = −。
For x = 4 (x > 3): (4 − 2)(4 − 3) = (+)(+) = + .
取 x = 4（属于 x > 3）：(4 − 2)(4 − 3) = (+)(+) = +。
Because we want the expression to be greater than zero, the solution is the union of the intervals where the sign is positive:
因为我们要求表达式大于零，所以解为正号所在的区间之并集：
x < 2 or x > 3
3. The Sign Table Method | 符号表法
A sign table is a compact way to organise the signs of each factor across the intervals. It is especially useful when the quadratic has more than two factors or when you want to avoid mistakes.
符号表是一种将各因子在不同区间内的符号进行整理的紧凑方法。当二次式含有两个以上因子，或你想避免出错时，符号表尤其有用。
For the example above, the sign table looks like this:
对于上面的例子，符号表如下：
| Interval | x < 2 | 2 < x < 3 | x > 3 |
| (x − 2) | − | + | + |
| (x − 3) | − | − | + |
| Product (x − 2)(x − 3) | + | − | + |
Since we need the product > 0, we select the intervals with the + sign.
由于我们需要乘积 > 0，因此选择带 + 号的区间。
4. Solving by Completing the Square | 通过配方法求解
If the quadratic cannot be factorised easily, we can complete the square to express the inequality in the form (x + p)² > q or (x + p)² < q. This method also works when the roots are irrational.
如果二次式不容易因式分解，我们可以通过配方法将不等式化为 (x + p)² > q 或 (x + p)² < q 的形式。当根为无理数时，这种方法同样适用。
Example 2: Solve x² + 6x + 4 > 0.
例 2：解不等式 x² + 6x + 4 > 0。
Complete the square:
配方：
x² + 6x + 4 = (x + 3)² − 5
So the inequality becomes:
因此不等式变为：
(x + 3)² − 5 > 0 ⇒ (x + 3)² > 5
Taking the square root of both sides gives:
两边开平方得：
x + 3 > √5 or x + 3 < −√5
Therefore:
因此：
x > −3 + √5 or x < −3 − √5
5. Using the Quadratic Formula | 使用二次公式
When factorisation is not possible and completing the square seems too lengthy, the quadratic formula can be used to find the roots exactly:
当无法因式分解且配方法显得冗长时，可以使用二次公式精确求出根：
x = (−b ± √(b² − 4ac)) / (2a)
Once the roots are known, the sign of the quadratic expression depends on the coefficient of x². If a > 0, the parabola opens upward; if a < 0, it opens downward. This is essential for finding the correct intervals.
一旦求出根，二次表达式的符号便取决于 x² 的系数 a。若 a > 0，抛物线开口向上；若 a < 0，抛物线开口向下。这一点对于确定正确区间至关重要。
Example 3: Solve 2x² − 3x − 2 < 0.
例 3：解不等式 2x² − 3x − 2 < 0。
Using the quadratic formula with a = 2, b = −3, c = −2:
使用二次公式，a = 2，b = −3，c = −2：
x = (3 ± √(9 + 16)) / 4 = (3 ± 5) / 4
So x = 2 or x = −0.5. Since a = 2 > 0, the parabola opens upward. The expression is negative between the two roots.
因此 x = 2 或 x = −0.5。由于 a = 2 > 0，抛物线开口向上，表达式在两个根之间为负。
The solution is:
解为：
−0.5 < x < 2
6. Graphical Interpretation | 图像解释
Graphing the quadratic function y = ax² + bx + c helps visualise the solution set. The roots correspond to the x-intercepts of the parabola. The inequality asks us to find the x-values where the parabola lies above or below the x-axis.
绘制二次函数 y = ax² + bx + c 的图像有助于直观理解解集。根对应于抛物线与 x 轴的交点。不等式要求我们找出抛物线位于 x 轴上方或下方的 x 值。
If a > 0 and the discriminant is positive, the curve is a “U” shape that crosses the x-axis at two points. The region above the x-axis is outside the roots, and the region below the x-axis is between the roots.
若 a > 0 且判别式为正，曲线呈“U”形并与 x 轴交于两点。x 轴上方的区域位于两根之外，而 x 轴下方的区域位于两根之间。
If a < 0, the curve is an upside-down "U". The region above the x-axis is between the roots, and the region below the x-axis is outside the roots.
若 a < 0，曲线呈倒“U”形。x 轴上方的区域位于两根之间，而 x 轴下方的区域位于两根之外。
This graphical reasoning is a powerful tool for solving inequalities quickly and checking your algebraic answers.
这种图像推理是快速求解不等式以及检验代数答案的有力工具。
7. Special Cases: No Real Roots | 特殊情况：无实根
When the discriminant b² − 4ac is negative, the quadratic equation has no real roots. The parabola never crosses the x-axis. Therefore, the sign of the expression is always the same as the sign of a.
当判别式 b² − 4ac 为负时，二次方程没有实根，抛物线永远不会与 x 轴相交。因此，表达式的符号始终与 a 的符号相同。
Example 4: Solve x² + 2x + 5 > 0.
例 4：解不等式 x² + 2x + 5 > 0。
Complete the square:
配方：
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