Quadratic Inequalities: Solving, Sketching and Notation | 二次不等式：解法、图像与记号

📚 Quadratic Inequalities: Solving, Sketching and Notation | 二次不等式：解法、图像与记号

Quadratic inequalities are a core topic in the Edexcel A-Level Pure Mathematics specification. A question of this type asks you to find the complete set of x-values for which a quadratic expression is positive, negative, zero, non-negative or non-positive. Success depends on three habits: factorising accurately, sketching a clear parabola or sign diagram, and writing the final answer in exactly the notation the question demands.

二次不等式是 Edexcel A-Level 纯数学考试大纲的核心考点。这类题目要求你找出使某个二次式的值为正、为负、为零、非负或非正的全部 x 值。解题成功取决于三个习惯：准确因式分解、画出清晰的抛物线草图或符号图，以及严格按题目要求的形式书写最终答案。

---

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

A quadratic inequality is any inequality that can be rearranged into one of the four standard forms below, where a, b and c are real constants and a ≠ 0:

二次不等式是指可以整理成下列四种标准形式之一的不等式，其中 a、b、c 为实常数，且 a ≠ 0：

• ax² + bx + c > 0
• ax² + bx + c < 0
• ax² + bx + c ≥ 0
• ax² + bx + c ≤ 0

• ax² + bx + c > 0（大于零）
• ax² + bx + c < 0（小于零）
• ax² + bx + c ≥ 0（大于或等于零）
• ax² + bx + c ≤ 0（小于或等于零）

Unlike a quadratic equation, which has at most two real roots, a quadratic inequality normally has infinitely many solutions. The solution set is one interval, two separate intervals, the whole real line, or the empty set.

二次方程最多只有两个实根，而二次不等式通常有无穷多个解。其解集可能是一个区间、两个分离的区间、整个实数集，或者空集。

abola lies above the x-axis (for > 0 and ≥ 0) or below the x-axis (for < 0 and ≤ 0).

函数 y = ax² + bx + c 的图像是一条抛物线。解不等式就是在寻找使这条抛物线位于 x 轴上方（对应 > 0 和 ≥ 0）或位于 x 轴下方（对应 < 0 和 ≤ 0）的所有 x 值。

---

2. Critical Values from Factorisation | 因式分解求临界值

e roots you obtain are called critical values, because they are the only points at which the quadratic expression can change sign.

解二次不等式的第一步，永远是解对应的二次方程 ax² + bx + c = 0。求得的根称为临界值，因为它们是二次式唯一可能改变符号的位置。

o values split the number line into three regions: x < 2, 2 < x < 3 and x > 3.

例如，x² − 5x + 6 可以因式分解为 (x − 2)(x − 3)，因此临界值为 x = 2 和 x = 3。这两个值把数轴分成三个区域：x < 2、2 < x < 3 和 x > 3。

x² − 5x + 6 = (x − 2)(x − 3)

Always take out any common factor first, and check whether the quadratic is a difference of two squares or a perfect square. If the expression does not factorise neatly with integers, use the quadratic formula or complete the square instead.

务必先提取公因式，并检查该二次式是否为平方差或完全平方。如果无法用整数干净地因式分解，就改用求根公式或配方法。

Two special situations are worth remembering. If the quadratic is a perfect square such as (x − 2)², there is only one critical value, and the expression never changes sign. If the discriminant b² − 4ac is negative, there are no critical values at all.

有两种特殊情形值得牢记。如果二次式是完全平方，例如 (x − 2)²，则只有一个临界值，且该表达式永远不变号。如果判别式 b² − 4ac 为负，则根本不存在临界值。

---

3. Sketches and Sign Diagrams | 抛物线草图与符号图

he roots on the x-axis and pay attention to the sign of a, because this decides whether the parabola opens upwards or downwards.

求出临界值后，画出抛物线 y = ax² + bx + c 的草图。把根标在 x 轴上，并特别留意 a 的符号，因为它决定了抛物线开口向上还是向下。

• If a > 0, the parabola opens upwards (a U-shape). It lies above the x-axis outside the roots and below the x-axis between the roots.
• If a < 0, the parabola opens downwards (an inverted U). It lies below the x-axis outside the roots and above the x-axis between the roots.

• 若 a > 0，抛物线开口向上（U 形）。它在两根之外位于 x 轴上方，在两根之间位于 x 轴下方。
• 若 a < 0，抛物线开口向下（倒 U 形）。它在两根之外位于 x 轴下方，在两根之间位于 x 轴上方。

A sign diagram is a faster alternative. Write the critical values in increasing order along a number line, then mark the sign of the expression in each region. For (x − 2)(x − 3), the signs read + , − , + from left to right, which immediately gives the solution of x² − 5x + 6 > 0 as x < 2 or x > 3.

符号图是更快捷的替代方法。把临界值按从小到大的顺序标在数轴上，再标出每个区域内表达式的符号。对于 (x − 2)(x − 3)，从左到右的符号依次为 +、−、+，由此可直接得出 x² − 5x + 6 > 0 的解为 x < 2 或 x > 3。

---

4. Solving the Two Basic Cases | 两种基本情形的求解

Consider the inequality x² − 5x + 6 > 0. Factorising gives (x − 2)(x − 3) > 0. The product of two factors is positive when both factors are positive, or when both are negative.

先看不等式 x² − 5x + 6 > 0。因式分解得 (x − 2)(x − 3) > 0。两个因式的乘积为正，当且仅当两个因式同为正，或同为负。

• If x < 2 then both (x − 2) and (x − 3) are negative, so the product is positive.
• If 2 < x < 3 then (x − 2) is positive and (x − 3) is negative, so the product is negative.
• If x > 3 then both factors are positive, so the product is positive.

• 若 x < 2，则 (x − 2) 与 (x − 3) 均为负，乘积为正。
• 若 2 < x < 3，则 (x − 2) 为正而 (x − 3) 为负，乘积为负。
• 若 x > 3，则两个因式均为正，乘积为正。

Therefore x² − 5x + 6 > 0 has the solution x < 2 or x > 3. Notice that the answer is two separate regions joined by the word "or", which corresponds to the union symbol ∪ in set notation.

因此 x² − 5x + 6 > 0 的解为 x < 2 或 x > 3。注意答案是两个分离的区域，用"或"字连接，在集合记号中对应并集符号 ∪。

Now consider x² − 5x + 6 < 0. The product is negative only when the two factors have opposite signs, which happens between the roots. The solution is therefore 2 < x < 3, a single interval.

再来看 x² − 5x + 6 < 0。乘积为负，仅当两个因式异号，而这只会发生在两根之间。因此解为 2 < x < 3，是一个单独的区间。

x² − 5x + 6 > 0 ⟺ x < 2 or x > 3

x² − 5x + 6 < 0 ⟺ 2 < x < 3

广告 / Sponsored

---

5. Handling the Symbols ≥ and ≤ | 处理 ≥ 与 ≤

itical values themselves belong to the solution set, because at those points the expression equals zero. Everything else is identical to the strict case.

当不等号是 ≥ 或 ≤ 时，临界值本身属于解集，因为在这些点上表达式等于零。除此之外，解法与严格不等式完全一致。

• x² − 5x + 6 ≥ 0 gives x ≤ 2 or x ≥ 3.
• x² − 5x + 6 ≤ 0 gives 2 ≤ x ≤ 3.

• x² − 5x + 6 ≥ 0 的解为 x ≤ 2 或 x ≥ 3。
• x² − 5x + 6 ≤ 0 的解为 2 ≤ x ≤ 3。

rval uses round brackets: (2, 3). A half-open case such as 2 < x ≤ 3 is written (2, 3].

用区间记号表示时，闭区间的端点用方括号书写：[2, 3]。开区间则用圆括号：(2, 3)。像 2 < x ≤ 3 这样的半开情形写作 (2, 3]。

estion actually says ≥ or ≤. Read the symbol carefully before you write your final answer.

本考点最常见的错误，就是题目本来说的是 ≥ 或 ≤，答题时却写成了严格不等号。写最终答案之前，务必重新确认题目所用的符号。

---

6. Negative Leading Coefficients | 首项系数为负的情形

If the coefficient of x² is negative, you have two safe strategies. Either sketch the inverted parabola, or multiply the whole inequality by −1 and reverse the inequality sign.

如果 x² 的系数为负，你有两种稳妥策略：画出开口向下的抛物线草图，或者把整个不等式乘以 −1 并反转不等号方向。

Take −x² + 5x − 6 > 0. Multiplying both sides by −1 reverses the sign and gives x² − 5x + 6 < 0.

以 −x² + 5x − 6 > 0 为例。两边同乘 −1 并反转不等号，得到 x² − 5x + 6 < 0。

−x² + 5x − 6 > 0 ⟺ x² − 5x + 6 < 0 ⟺ 2 < x < 3

s and crosses the x-axis at 2 and 3, so it is above the x-axis between the roots. This confirms 2 < x < 3.

用草图验证一下。抛物线 y = −x² + 5x − 6 开口向下，在 2 和 3 处与 x 轴相交，因此在两根之间位于 x 轴上方。这与 2 < x < 3 一致。

Never divide an inequality by an expression whose sign is unknown, and never multiply by a negative number without reversing the direction of the inequality sign.

切勿除以符号未知的表达式，也切勿在乘以负数时忘记反转不等号方向。

---

7. Quadratics That Do Not Factorise | 无法因式分解的二次式

When a quadratic does not factorise with integers, find the critical values with the quadratic formula. The logic that follows is unchanged.

当二次式无法用整数因式分解时，就用求根公式求出临界值。后续的逻辑完全不变。

x = (−b ± √b² − 4ac) ÷ (2a)

mple, take x² − 4x + 1 < 0. Here a = 1, b = −4 and c = 1, so b² − 4ac = 16 − 4 = 12 and √12 = 2√3.

例如，解 x² − 4x + 1 < 0。此处 a = 1、b = −4、c = 1，故 b² − 4ac = 16 − 4 = 12，√12 = 2√3。

x = (4 ± 2√3) ÷ 2 = 2 ± √3

Since a > 0, the expression is negative between the roots, so the solution is 2 − √3 < x < 2 + √3. Numerically this is approximately 0.268 < x < 3.732, which is a useful sanity check.

由于 a > 0，该表达式在两根之间为负，所以解为 2 − √3 < x < 2 + √3。数值上约为 0.268 < x < 3.732，可作为快速检验。

Completing the square gives the same answer, and it is often the quicker route when the numbers are simple:

配方也能得到同样的答案，而且当数字比较简单时，这往往是更快的路径：

x² − 4x + 1 = (x − 2)² − 3

Setting (x − 2)² − 3 < 0 gives (x − 2)² < 3, so |x − 2| < √3, which rearranges to 2 − √3 < x < 2 + √3. Surd answers should normally be left in exact form unless the question asks for decimals.

由 (x − 2)² − 3 < 0 得 (x − 2)² < 3，即 |x − 2| < √3，整理得 2 − √3 < x < 2 + √3。除非题目要求取小数，否则根式答案通常应保留精确形式。

<

更多咨询请联系16621398022（同微信）