📚 Exercise 4D: Quadratic Inequalities | 二次不等式
s on factorisation and graph sketching to identify intervals on the x-axis where a quadratic expression is positive or negative.
在 AQA A Level 数学中，练习 4D 聚焦二次不等式。这些不等式形如 ax² + bx + c > 0、ax² + bx + c < 0、ax² + bx + c ≥ 0 或 ax² + bx + c ≤ 0。本练习建立在因式分解和函数草图的基础上，用于确定二次式在 x 轴上哪些区间为正或为负。
1. What Exercise 4D Covers | 练习 4D 涵盖内容
ritical values, then use the shape of the parabola to decide which intervals satisfy the inequality.
主要技能是将二次不等式转化为读图问题。先解对应的二次方程 ax² + bx + c = 0 求出临界值，再利用抛物线的形状判断哪些区间满足不等式。
AQA may ask for your answer in set notation, interval notation, or as an inequality on a number line. You must be comfortable with all three forms and be able to convert between them quickly.
AQA 可能要求用集合记号、区间记号或数轴上的不等式表示答案。你必须熟悉这三种形式，并能在它们之间快速转换。
2. The Core Rule: Sketch Before Solving | 核心原则：先画草图再解
Always begin by rearranging the inequality so that 0 is on one side. For a quadratic with a positive leading coefficient, the graph is a ∪-shaped parabola. If it has two distinct real roots, the graph is below the x-axis between the roots and above the x-axis outside the roots.
始终先将不等式整理成一边为 0 的形式。对于首项系数为正的二次函数，其图像为 ∪ 形抛物线。若有两个不同实根，图像在两根之间位于 x 轴下方，在两根之外位于 x 轴上方。
For ax² + bx + c > 0 with a > 0 and roots α < β, the solution is x < α or x > β. For ax² + bx + c < 0, the solution is α < x < β. This sign rule is the foundation of Exercise 4D.
对于 a > 0 且两根 α < β 的 ax² + bx + c > 0，解为 x < α 或 x > β；对于 ax² + bx + c < 0，解为 α < x < β。这条符号规则是练习 4D 的基础。
3. Standard Method: Factorisation | 标准方法：因式分解
Step 1: Rearrange to get 0 on one side of the inequality. Step 2: Factorise the quadratic expression fully. Step 3: Set each factor equal to zero to find the critical values. Step 4: Place these values on a number line and test a value from each interval, or use the sketch. Step 5: Write the final solution using correct notation.
步骤一：移项使不等式一边为 0。步骤二：将二次式完整因式分解。步骤三：令每个因式等于 0 求出临界值。步骤四：将这些值标在数轴上，并在每个区间取一个试值，或直接使用草图。步骤五：用正确记号写出最终解。
Solve x² − 5x + 6 < 0 → (x − 2)(x − 3) < 0 → critical values x = 2, x = 3
ph is below the x-axis between x = 2 and x = 3. Therefore the solution is 2 < x < 3.
画出 y = (x − 2)(x − 3) 的草图后，图像在 x = 2 与 x = 3 之间位于 x 轴下方，因此解为 2 < x < 3。
4. Worked Example: Positive Leading Coefficient | 例题：首项系数为正
ns upwards, the expression is positive outside the roots and zero at the roots themselves.
解不等式 x² − 5x + 6 ≥ 0。因式分解得 (x − 2)(x − 3) ≥ 0。临界值为 x = 2 和 x = 3。由于抛物线开口向上，函数在两根外为正，在两个根处等于零。
Therefore the solution is x ≤ 2 or x ≥ 3. In interval notation: (−∞, 2] ∪ tice the square brackets include the critical values because the inequality is non-strict.
因此解为 x ≤ 2 或 x ≥ 3。用区间记号表示为 (−∞, 2] ∪ [3, ∞)。注意使用方括号包含临界值，因为不等式带有等号。
If the question required x² − 5x + 6 > 0, the solution would be (−∞, 2) ∪ (3, ∞) with round brackets only.
如果题目要求 x² − 5x + 6 > 0，解则为 (−∞, 2) ∪ (3, ∞)，只使用圆括号。
5. Handling a Negative x² Coefficient | 处理负二次项系数
If the coefficient of x² is negative, do not simply flip the inequality sign without care. The safest method is to multiply both sides by −1 and reverse the inequality, or to factorise out −1 and rewrite the quadratic with a positive leading coefficient.
如果 x² 的系数为负，不要随意翻转不等号。最安全的方法是两边乘以 −1 并反转不等式方向，或者因式分解提出 −1，将其改写为首项系数为正的二次式。
For example, solve −x² + 4x − 3 > 0. Multiply by −1: x² − 4x + 3 < 0. Factorise: (x − 1)(x − 3) < 0, giving 1 < x < 3.
例如解 −x² + 4x − 3 > 0。两边乘以 −1 得 x² − 4x + 3 < 0。因式分解为 (x − 1)(x − 3) < 0，解得 1 < x < 3。
Always multiply by −1 before sketching. If you sketch the original negative parabola, the sign regions are reversed: it is above the x-axis between the roots and below outside them, which is a common source of error.
始终先乘以 −1 再画草图。如果你直接画原负抛物线的草图，其正负区域正好相反：在两根基之间位于 x 轴上方，在两根之外位于下方，这是常见错误来源。
6. Using the Discriminant | 使用判别式
how many real roots the quadratic has. If Δ < 0, the graph does not cross the x-axis, so the expression is either always positive when a > 0, or always negative when a < 0.
判别式 Δ = b² − 4ac 告诉你二次函数有多少个实根。若 Δ < 0，图像不与 x 轴相交，因此当 a > 0 时函数恒正，当 a < 0 时函数恒负。
nce a = 1 > 0, the quadratic is always positive, so the solution is all real numbers: x ∈ ℝ.
例如 x² + 2x + 5 > 0 的判别式为 Δ = 4 − 20 = −16 < 0。由于 a = 1 > 0，该二次式恒正，所以解为全体实数：x ∈ ℝ。
If the inequality were x² + 2x + 5 < 0, there would be no solution, written as ∅ or { }. This case appears in Exercise 4D to test whether you check the discriminant before attempting to factorise.
如果不等式是 x² + 2x + 5 < 0，则无解，记作 ∅ 或 { }。这种情形在练习 4D 中出现，是为了考查你是否在尝试因式分解前先检查判别式。
7. Double Roots and Strict Inequalities | 重根与严格不等式
d root. The parabola touches the x-axis at that point but does not cross it. For example, x² − 6x + 9 > 0 can be written as (x − 3)² > 0.
当 Δ = 0 时，二次函数有重根。抛物线在该点与 x 轴相切但不穿过 x 轴。例如 x² − 6x + 9 > 0 可写成 (x − 3)² > 0。
ative. The inequality (x − 3)² > 0 is true for all x except x = 3, so the solution is x ∈ ℝ, x ≠ 3. In interval notation: (−∞, 3) ∪ (3, ∞).
一个平方数总是非负的。不等式 (x − 3)
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