📚 Example 1.6.2: Solving Quadratic Inequalities by Sketching | 例 1.6.2：通过草图法求解一元二次不等式
In AQA A-Level Mathematics, solving quadratic inequalities is a core skill that combines factorisation, curve sketching and interval notation. Example 1.6.2 focuses on using a rough sketch of the parabola to determine where a quadratic expression is positive or negative. This approach avoids sign errors and helps you express solutions clearly in set notation or interval form.
在 AQA A-Level 数学中，解一元二次不等式是一项核心技能，它综合了因式分解、函数图像草图和区间表示法。例 1.6.2 的重点是利用抛物线的草图来判断二次函数在哪些区间取正值或负值。这种方法可以避免符号错误，并帮助你用集合符号或区间形式清晰地表示解集。
The full method has four fixed stages: rearrange into standard form, find the roots, sketch the parabola, then read the answer off the sketch. Once you trust the sketch, you never have to test random values again, and the final answer almost writes itself. This article works through every stage slowly, then extends the same idea to harder cases such as negative leading coefficients, repeated roots and quadratics that will not factorise.
完整的方法有四个固定步骤：整理成标准形式、求出根、画出抛物线草图、再从草图上读出答案。一旦你信任草图，就不必再反复代入试值，最终答案几乎可以自动写出。本文将逐步讲解每个环节，并把同一思路推广到更复杂的情形，例如首项系数为负、出现重根，以及无法因式分解的二次式。
1. Recognising a Quadratic Inequality | 识别一元二次不等式
c > 0, ax² + bx + c ≥ 0, ax² + bx + c < 0 or ax² + bx + c ≤ 0, where a ≠ 0. The word 'inequality' e or union of ranges of x. In other words, the answer is a set of numbers, not one number.
一元二次不等式是指形如 ax² + bx + c > 0、ax² + bx + c ≥ 0、ax² + bx + c < 0 或 ax² + bx + c ≤ 0 的不等式，其中 a ≠ 0。这里的不等式意味着我们不是求 x 的单个值，而是求 x 的一个区间或多个区间的并集。换句话说，答案是数的集合，而不是单独一个数。
Before doing anything else, rearrange so that the quadratic is on one side and zero is on the other. For instance, 3x + 2 > x² should be rewritten as x² − 3x − 2 < 0. Only after this rearrangement can you correctly read off a, b and c, because a sketch of the wrong expression gives the wrong answer.
在做任何运算之前，先把二次式移到一边、零留在另一边。例如，3x + 2 > x² 应改写为 x² − 3x − 2 < 0。只有整理之后，你才能正确读出 a、b、c 的值，因为画出错误表达式的草图只会得到错误答案。
For example, the statement x² − 5x + 6 > 0 asks: for which real values of x is the quadratic expression strictly greater than zero? Such questions appear frequently in AQA Pure Mathematics, especially in the first module.
例如，表达式 x² − 5x + 6 > 0 的问题是：当 x 取哪些实数时，这个二次表达式的值严格大于零？这类问题在 AQA Pure Mathematics 中经常出现，尤其是在第一模块中。
2. The Role of the Roots | 根的关键作用
e the expression equal zero. These roots act as boundary points that split the real number line into intervals. Within each interval, the sign of the quadratic expression stays constant: it cannot jump from positive to negative without passing through zero.
二次方程 ax² + bx + c = 0 的根是使表达式等于零的 x 值。这些根作为边界点，将实数轴分割成若干区间。在每一个区间内，二次表达式的符号保持不变：它不可能在不经过零的情况下从正跳到负。
d positive everywhere outside them. If a < 0 the parabola opens downwards, and the pattern reverses. The roots themselves always give equality, so they are included in the solution only when the inequality is ≥ or ≤.
这个事实正是草图法成立的唯一理由。若 a > 0，抛物线开口向上，因此它只在两根之间为负，在两根之外处处为正。若 a < 0，抛物线开口向下，上述规律正好相反。根本身对应等号成立的情形，因此只有当不等式为 ≥ 或 ≤ 时，根才被包含在解集中。
How many roots there are is controlled by the discriminant:
根的个数由判别式决定：
Δ = b² − 4ac
and the curve only touches the axis; if Δ < 0 there are no real roots and the curve never meets the axis at all.
若 Δ > 0，有两个不同的根，草图与 x 轴相交两次；若 Δ = 0，有一个重根，曲线只与 x 轴相切；若 Δ < 0，没有实根，曲线根本不与 x 轴相交。
3. Factorising the Quadratic: Finding the Critical Values | 因式分解：求临界值
are −2 and −3, so we write x² − 5x + 6 = (x − 2)(x − 3). Setting each bracket to zero gives the roots x = 2 and x = 3.
要求根，首先尝试对二次式因式分解。对于 x² − 5x + 6，我们要找两个数，它们的乘积为 6、和为 −5。这两个数是 −2 和 −3，于是 x² − 5x + 6 = (x − 2)(x − 3)。令每个括号等于零，得到根 x = 2 与 x = 3。
ly and gives roots x = ½ and x = −3. A useful check is to multiply the brackets back out and confirm you recover the original expression term by term.
当 x² 的系数不是 1 时，同样的逻辑仍然适用，只是要更细心。对于 2x² + 5x − 3，我们得到 (2x − 1)(x + 3)，展开后完全正确，根为 x = ½ 与 x = −3。一个实用的检查是把括号重新展开，逐项确认能还原出原来的表达式。
rmula instead. The roots do not have to be neat integers, and surd answers such as x = 1 ± √5 are perfectly acceptable in AQA examinations.
如果尝试合理时间后仍看不出因式分解，不要浪费时间：直接使用求根公式。根不必是漂亮的整数，像 x = 1 ± √5 这样的根式答案在 AQA 考试中完全可接受。
4. Sketching the Parabola | 画出抛物线草图
² is positive, the parabola is a U-shape opening upwards. Mark the x-axis intercepts at x = 2 and x = 3, and optionally the y-intercept at (0, 6), obtained by substituting x = 0.
现在画出 y = x² − 5x + 6 的草图。它不需要按比例绘制，只需形状正确、截距正确。由于 x² 的系数为正，抛物线是开口向上的 U 形。在 x 轴上标出交点 x = 2 与 x = 3，也可以标出 y 轴截距 (0, 6)，它由代入 x = 0 得到。
d to the right of x = 3 the curve is above the axis, so there the expression is positive. Labelling the sketch with + and − signs takes ten seconds and removes almost all possibility of a sign error later.
接着把两个区域标注出来。在 x = 2 与 x = 3 之间，曲线落到 x 轴下方，所以该区间内表达式为负。在 x = 2 左侧和 x = 3 右侧，曲线位于 x 轴上方，所以表达式为正。在草图上标出 + 与 − 只需十秒钟，却能几乎消除之后出现符号错误的可能。
A common error is to sketch the curve as a straight line or to draw it opening the wrong way. Always check the sign of a first: a > 0 opens upwards, a < 0 opens downwards. Everything else about the sketch is negotiable; the direction of opening is not.
一个常见错误是把曲线画成直线，或把开口方向画反。永远先检查 a 的符号：a > 0 开口向上，a < 0 开口向下。草图的其它细节都可以将就，唯独开口方向不能出错。
5. Reading the Solution from the Sketch | 从草图读出解集
The inequality you are solving tells you which region to read. For x² − 5x + 6 > 0 you want the parts above the axis, which are the two outer branches, so the solution is x < 2 or x > 3. For x² − 5x + 6 < 0 you want the part below the axis, which is the single middle interval 2 < x < 3.
你要求解的不等式决定该读哪个区域。对于 x² − 5x + 6 > 0，需要的是轴上方部分，也就是外侧的两支，因此解为 x < 2 或 x > 3。对于 x² − 5x + 6 < 0，需要的是轴下方部分，也就是中间那一段，因此解为 2 < x < 3。
Notice how the answer changes shape. When the parabola opens upwards, a '>' inequality gives two separate tails joined by 'or', while a '<' inequality gives one single interval joined by 'and' or written as a double inequality. The word you use matters: 'or' means union and 'and' means intersection.
注意答案的形状如何变化。当抛物线开口向上时，'>' 型不等式的解是两段分离的尾部，用「或」连接；而 '<' 型不等式的解是一个完整区间，用「且」连接，或写成双边不等式。用词很重要：「或」表示并集，「且」表示交集。
Always verify the shape of your answer before writing it down. A solution such as 2 > x > 3 is impossible, because no number is simultaneously greater than 3 and less than 2. If your working produces that, you have almost certainly read the sketch the wrong way round.
写下答案之前，务必先检查答案的形状。像 2 > x > 3 这样的解是不可能存在的，因为没有数能同时大于 3 又小于 2。如果你的推导得出这种结果，几乎可以肯定你把草图读反了。
6. Worked Example: x² − 5x + 6 > 0 | 完整例题：x² − 5x + 6 > 0
Let us now write the full solution exactly as it should appear in an examination. Each step earns method marks, so show all of them, even the ones that feel obvious.
下面按照考试中应有的样子写出完整解答。每一步都有过程分，所以都要写出来，即使看起来显而易见。
- Step 1 — Standard form. 6 is already in standard form with a = 1, b = −5, c = 6, and the right-hand side is zero. 第一步——标准形式：x² − 5x + 6 已经是标准形式，其中 a = 1、b = −5、c = 6，右边为零。
- Step 2 — Solve the related equation. x² − 5x + 6 = 0 factorises as (x − 2)(x − 3) = 0, so x = 2 or x = 3. 第二步——解对应方程：x² − 5x + 6 = 0 分解为 (x − 2)(x − 3) = 0，因此 x = 2 或 x = 3。
- Step 3 — Sketch. Draw an upward parabola crossing the x-axis at 2 and 3, and mark that the curve is above the axis outside these values and below it between them. 第三步——画草图：画一条开口向上的抛物线，与 x 轴交于 2 和 3，并标出在这两点之外曲线位于轴上方、在两点之间位于轴下方。
- Step 4 — Read the answer. The inequality requires the expression to be positive, so take the outer branches. 第四步——读出答案：不等式要求表达式为正，因此取外侧两支。
The solution can be written in several equivalent ways:
解可以写成几种等价的形式：
x < 2 or x > 3
{x ∈ ℝ : x < 2} ∪ {x ∈ ℝ : x > 3}
(−∞, 2) ∪ (3, ∞)
is negative; at x = 4 it equals 2, which is positive. This agrees exactly with the sketch.
为了确认答案，从每个区域各取一个值检验。当 x = 0 时表达式的值为 6，为正；当 x = 2.5 时值为 −0.25，为负；当 x = 4 时值为 2，为正。这与草图完全一致。
| Test value x | x² − 5x + 6 | Sign | Region |
| 0 | 6 | positive | x < 2 |
| 2.5 | −0.25 | negative | 2 < x < 3 |
| 4 | 2 | positive | x > 3 |
7. Strict and Non-Strict Inequalities | 严格不等式与非严格不等式
s open brackets or strict symbols. The symbols ≥ and ≤ are non-strict: the roots do satisfy them, so the endpoints are included.
符号 > 与 < 称为严格不等式：根本身不满足不等号，因此端点被排除，答案使用开括号或严格符号。符号 ≥ 与 ≤ 是非严格的：根本身满足不等号，因此端点被包含在解集中。
For the same quadratic x² − 5x + 6:
对同一个二次式 x² − 5x + 6：
x² − 5x + 6 ≥ 0 gives x ≤ 2 or x ≥ 3, that is (−∞, 2] ∪ [3, ∞)
x² − 5x + 6 ≤ 0 gives 2 ≤ x ≤ 3, that is [2, 3]
The difference between a round bracket and a square bracket is often worth a mark, so be deliberate about it. Sketch the parabola, decide whether the curve at the roots should be counted, and only then write the brackets.
圆括号与方括号的差别常常值一分，所以要刻意留意。先画抛物线，判断根处的点是否应计入，然后再写括号。
mber you can reach, so it always takes a round bracket: (−∞, 2) and (3, ∞), never [−∞, 2].
另外请记住，∞ 永远不是一个能取到的数，因此总是用圆括号：(−∞, 2) 与 (3, ∞)，绝不写 [−∞, 2]。
8. Negative Leading Coefficients | 首项系数为负
Suppose the inequality is −x² + 5x − 6 > 0. There are two safe routes. The first is to multiply through by −1, remembering that multiplying an inequality by a negative number reverses the inequality sign: this gives x² − 5x + 6 < 0, which factorises to (x − 2)(x − 3) < 0 and therefore has solution 2 < x < 3.
假设不等式为 −x² + 5x − 6 > 0。有两条安全的路径。第一条是两边同乘 −1，同时记住乘以负数会改变不等号方向：得到 x² − 5x + 6 < 0，分解为 (x − 2)(x − 3) < 0，因此解为 2 < x < 3。
wards, crossing the axis at 2 and 3. A downward parabola is above the axis between its roots and below the axis outside them, which gives the same answer, 2 < x < 3. Both methods are valid; the sketch alone is safer if you are tired or rushed.
第二条路径是照常画草图。此时 a = −1，抛物线开口向下，与 x 轴交于 2 和 3。开口向下的抛物线在两根之间位于 x 轴上方、在两根之外位于轴下方，同样得到 2 < x < 3。两种方法都正确；但在疲劳或时间紧张时，单靠草图更稳妥。
The table below summarises the two shapes and is worth memorising:
下表总结了两种情形，值得记住：
| Shape | Expression > 0 | style="border:1px solid #
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