Quadratic Inequalities | 二次不等式

📚 Quadratic Inequalities | 二次不等式

Quadratic inequalities are expressions involving a quadratic polynomial set with an inequality sign, such as ax² + bx + c > 0, < 0, ≥ 0 or ≤ 0, where a ≠ 0. In A-Level Edexcel Mathematics, solving these inequalities requires a solid understanding of quadratic graphs, factorisation, completing the square, and the discriminant. The solution sets describe ranges of x-values that satisfy the given condition and are often expressed using interval or set notation.

二次不等式是指含有二次多项式的不等式，例如 ax² + bx + c > 0、< 0、≥ 0 或 ≤ 0，其中 a ≠ 0。在 A-Level Edexcel 数学中，求解这类不等式需要牢固掌握二次函数图像、因式分解、配方法以及判别式的知识。解集描述了满足条件的 x 的取值范围，通常用区间或集合符号表示。

This article systematically covers the methods needed to solve quadratic inequalities, from sketching parabolas and using critical values to handling parameters and avoiding common mistakes. Each section provides a clear step-by-step approach, followed by practical examples and examination insights.

本文系统梳理了求解二次不等式所需的各种方法，从绘制抛物线草图、利用临界值，到处理参数和避免常见错误。每个部分都提供了清晰的分步思路，并配有实例和考试要点分析。

1. Introduction to Quadratic Inequalities | 二次不等式简介

A quadratic inequality compares a quadratic expression to zero or another expression using one of the signs <, >, ≤ or ≥. The general form is ax² + bx + c > 0, where the quadratic can be factorised, completed to a square, or have no real roots. The solution is not a single number but an interval or a union of intervals on the real line.

二次不等式是用 <、>、≤ 或 ≥ 将二次式与零（或其他式子）进行比较的不等式。一般形式为 ax² + bx + c > 0，其中的二次式可以因式分解、配方，也可能无实根。解通常不是一个孤立的数，而是实数轴上一个区间或几个区间的并集。

To understand the solution, we must consider the shape of the parabola y = ax² + bx + c. When a > 0, the parabola opens upwards; when a < 0, it opens downwards. The points where the parabola crosses the x-axis are the roots of the equation ax² + bx + c = 0, and they serve as critical boundaries for the inequality.

要理解解集，我们需要考虑抛物线 y = ax² + bx + c 的形状。当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。抛物线与 x 轴的交点就是方程 ax² + bx + c = 0 的根，它们构成了不等式的临界边界。

If the inequality is strict ( > or < ), the boundary points are not included; if it is non‑strict ( ≥ or ≤ ), they are included. The solution can be visualised by looking at the parts of the parabola that lie above or below the x‑axis.

如果不等式是严格不等号（> 或 <），边界点不包含在解内；若为非严格不等号（≥ 或 ≤），则包含边界。通过观察抛物线在 x 轴上方或下方的部分，可以直观地获得解集。

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2. Solving Quadratic Inequalities Graphically | 用图像法解二次不等式

Graphical method is fundamental: sketch the graph of y = ax² + bx + c, identify the x-intercepts, and then read off the x-values for which y is positive or negative. For example, solve x² – x – 6 < 0. Factorising gives (x – 3)(x + 2) = 0, so roots are –2 and 3. The parabola y = x² – x – 6 opens upwards (a = 1 > 0).

图像法是基础方法：画出 y = ax² + bx + c 的草图，找出与 x 轴的交点，然后读出使 y 为正或负的 x 值。例如，求解 x² – x – 6 < 0。分解因式得 (x – 3)(x + 2) = 0，因此根为 –2 和 3。抛物线 y = x² – x – 6 开口向上（a = 1 > 0）。

Since the parabola opens upwards, y is negative between the roots because the curve dips below the x-axis. Hence the solution is –2 < x < 3. If the inequality had been x² – x – 6 > 0, the solution would be x < –2 or x > 3, corresponding to the two outer regions where y is positive.

由于抛物线开口向上，两根之间的部分曲线低于 x 轴，因此 y 为负。解集为 –2 < x < 3。如果不等式是 x² – x – 6 > 0，则解为 x < –2 或 x > 3，对应曲线在 x 轴上方的两个外侧区域。

When the quadratic has a repeated root (discriminant = 0), the parabola touches the x-axis. For (x – 2)² > 0, the expression is positive everywhere except at x = 2, where it equals zero. Thus the solution is all real x except x = 2, which can be written as x ∈ ℝ, x ≠ 2.

当二次式有重根（判别式 = 0）时，抛物线与 x 轴相切。对于 (x – 2)² > 0，除 x = 2 处为零外，该式处处为正。因此解集为所有实数 x，但 x ≠ 2，可写成 x ∈ ℝ, x ≠ 2。

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3. The Critical Values Method | 临界值法

The critical values method provides a systematic algebraic technique. Begin by rearranging the inequality so that one side is zero. Then solve the corresponding quadratic equation to find the critical values. These values divide the number line into intervals. Test a point from each interval in the original inequality to determine which intervals satisfy it.

临界值法是一种系统的代数技巧。首先将不等式整理为一边为零的形式，然后求解相应的二次方程得到临界值。这些临界值将数轴分成若干个区间。在每个区间内选取一个检验点，代入原不等式，判断该区间是否满足不等式。

Example: solve 2x² + 3x – 5 ≥ 0. Equation 2x² + 3x – 5 = 0 factorises to (2x + 5)(x – 1) = 0, giving critical values x = –5/2 and x = 1. The intervals are x < –5/2, –5/2 < x < 1, and x > 1. Testing x = –3 gives positive, x = 0 gives negative, x = 2 gives positive. Since the inequality is ≥ 0, we include the critical values. Solution: x ≤ –5/2 or x ≥ 1.

例题：求解 2x² + 3x – 5 ≥ 0。方程 2x² + 3x – 5 = 0 分解为 (2x + 5)(x – 1) = 0，得临界值 x = –5/2 和 x = 1。划分区间：x < –5/2、–5/2 < x < 1 和 x > 1。检验 x = –3 得正，x = 0 得负，x = 2 得正。因为不等式是 ≥ 0，故包含临界值。解集：x ≤ –5/2 或 x ≥ 1。

This method works even when factorisation is not obvious, as the critical values can be found using the quadratic formula or completing the square. It also helps avoid sign mistakes when multiplying or dividing inequalities by negative numbers.

即使因式分解不明显，此方法也行之有效，因为临界值可用求根公式或配方法求出。它还有助于避免在不等式两边乘除负数时出现符号错误。

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4. Working with Factorised Quadratics | 使用因式分解的二次式

When a quadratic is already given in factorised form, or can be easily factorised, solving the inequality becomes very straightforward. For (x + 3)(2x – 1) ≤ 0, the critical values are –3 and ½. Sketch a sign table or a quick parabola: since the coefficient of x² is positive (2), the graph opens upwards, so the negative region lies between the roots.

当二次式已给出因式分解形式或容易分解时，求解不等式就变得非常直接。对于 (x + 3)(2x – 1) ≤ 0，临界值为 –3 和 ½。列出符号表或快速画出抛物线：因为 x² 的系数为正（2），图像开口向上，所以两根之间为负值区域。

Thus the solution is –3 ≤ x ≤ ½. Note the use of ≤ signs because the inequality includes equality. If the inequality were strictly less than zero, we would write –3 < x < ½.

因此解集为 –3 ≤ x ≤ ½。注意因为包含等号，所以使用了 ≤。若不等式是严格小于零，则应写作 –3 < x < ½。

For an expression like (4 – x)(x + 1) > 0, it is safer to rewrite it with a positive leading coefficient first. Multiply by –1, giving (x – 4)(x + 1) < 0 (the inequality sign reverses). Now the critical values are –1 and 4, and with a positive quadratic coefficient the interior gives the negative region. Hence solution: –1 < x < 4.

对于 (4 – x)(x + 1) > 0 这样的式子，最好先将首项系数化为正。乘以 –1 得 (x – 4)(x + 1) < 0（不等号方向改变）。此时临界值为 –1 和 4，且二次系数为正，两根之间为负值区域。故解为 –1 < x < 4。

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5. Completing the Square Approach | 配方法

Sometimes a quadratic cannot be factorised easily, or the question explicitly requires completing the square. Rewrite ax² + bx + c in the form a(x + p)² + q. Then solve the inequality by isolating the squared term and applying square root properties. This method also reveals the vertex and the range of the quadratic function.

有时二次式不容易分解，或题目明确要求使用配方法。将 ax² + bx + c 改写为 a(x + p)² + q 的形式，然后通过分离平方项并运用平方根的性质求解不等式。配方法还能揭示抛物线的顶点和二次函数的值域。

Example: solve x² + 4x + 1 < 0 by completing the square. Write (x + 2)² – 3 < 0, so (x + 2)² < 3. Taking the square root (and remembering |x+2| < √3) gives –√3 < x + 2 < √3. Subtracting 2 yields –2 – √3 < x < –2 + √3. This is the exact solution, which can be approximated if requested.

例题：用配方法求解 x² + 4x + 1 < 0。配方得 (x + 2)² – 3 < 0，因此 (x + 2)² < 3。开平方（注意 |x+2| < √3）得 –√3 < x + 2 < √3。减去 2 得到 –2 – √3 < x < –2 + √3。这是精确解，若题目要求可给出近似值。

Another common form is inequalities like 2x² – 8x + 5 ≥ 0. Factor out the 2: 2(x² – 4x) + 5 ≥ 0. Complete the square inside: 2[(x – 2)² – 4] + 5 ≥ 0 → 2(x – 2)² – 8 + 5 ≥ 0 → 2(x – 2)² ≥ 3 → (x – 2)² ≥ 1.5. Then x – 2 ≤ –√1.5 or x – 2 ≥ √1.5, leading to two separate intervals.

另一种常见形式为 2x² – 8x + 5 ≥ 0。将 2 提出：2(x² – 4x) + 5 ≥ 0。括号内配方：2[(x – 2)² – 4] + 5 ≥ 0 → 2(x – 2)² – 8 + 5 ≥ 0 → 2(x – 2)² ≥ 3 → (x – 2)² ≥ 1.5。于是 x – 2 ≤ –√1.5 或 x – 2 ≥ √1.5，得到两个独立的区间。

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6. Discriminant and No Real Roots | 判别式与无实根

The discriminant Δ = b² – 4ac determines the nature of the roots. If Δ < 0, the quadratic has no real roots and the parabola does not cross the x‑axis. In this case, the quadratic expression is either always positive or always negative, depending on the sign of a. This simplifies the inequality greatly.

判别式 Δ = b² – 4ac 决定了根的性质。若 Δ < 0，二次式无实根，抛物线不与 x 轴相交。此时二次式的值要么恒正，要么恒负，仅取决于 a 的符号。这使得不等式的求解大大简化。

Consider 2x² + x + 3 > 0. Here a = 2 > 0, and Δ = 1² – 4(2)(3) = –23 < 0. Since the parabola opens upwards and never touches the x‑axis, the expression is positive for all real x. Thus the solution is x ∈ ℝ. Conversely, if we had –x² + 2x – 5 < 0 with Δ < 0 and a < 0, the expression is always negative, so the inequality holds for all real x.

考虑 2x² + x + 3 > 0。此时 a = 2 > 0，Δ = 1² – 4(2)(3) = –23 < 0。由于抛物线开口向上且永不相交于 x 轴，该式对所有实数 x 恒为正。故解为 x ∈ ℝ。反之，对于 –x² + 2x – 5 < 0，Δ < 0 且 a < 0，表达式恒负，因此不等式对所有实数成立。

If the inequality were 2x² + x + 3 < 0 with Δ < 0 and a > 0, there would be no solution because the quadratic is never negative. Thus understanding the discriminant prevents incorrectly attempting to find non‑existent critical values.

若不等式为 2x² + x + 3 < 0，同样 Δ < 0 且 a > 0，则无解，因为二次式永不为负。因此，理解判别式可以避免错误地试图找出根本不存在的临界值。

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7. Inequalities with a Negative Leading Coefficient | 首项系数为负的不等式

When the coefficient a is negative, the parabola opens downwards. Students often mishandle this by sketching the graph incorrectly or by forgetting to reverse the inequality when multiplying by –1. A safe approach is to multiply the entire inequality by –1, flipping the sign, so that the quadratic term has a positive coefficient.

当二次项系数 a 为负时，抛物线开口向下。学生常因画错草图或在乘以 –1 时忘记改变不等号方向而出错。稳妥的方法是将整个不等式乘以 –1，并反转不等号，使二次项系数变为正。

Solve –x² + 4x – 3 > 0. Multiply by –1 to obtain x² – 4x + 3 < 0. Factorising gives (x – 1)(x – 3) < 0. The roots are 1 and 3, and with a positive a, the solution is 1 < x < 3. If we had kept the negative a and sketched the downward parabola, we would also see that y > 0 between the roots, but multiplying by –1 is a foolproof algebraic method.

求解 –x² + 4x – 3 > 0。乘以 –1 得 x² – 4x + 3 < 0。分解因式得 (x – 1)(x – 3) < 0。根为 1 和 3，由于 a 为正，解集为 1 < x < 3。如果保留负系数并画出开口向下的抛物线，同样会看到两根之间 y > 0，但乘以 –1 的代数方法万无一失。

Be careful with non‑strict inequalities: –x² – 2x ≥ 0 becomes x² + 2x ≤ 0 after multiplying by –1, so x(x + 2) ≤ 0, giving –2 ≤ x ≤ 0. Always remember the sign change applies to the entire inequality, including any non‑zero constant on the opposite side.

处理非严格不等式时要小心：–x² – 2x ≥ 0 乘以 –1 后变为 x² + 2x ≤ 0，即 x(x + 2) ≤ 0，解为 –2 ≤ x ≤ 0。始终要记住变号适用于整个不等式，包括另一边非零的常数。

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8. Quadratic Inequalities Involving Parameters | 含参数的二次不等式

Edexcel exam questions sometimes include an unknown constant k in the quadratic, asking for the range of k such that the inequality holds for all real x, or such that the solution set satisfies certain conditions. These problems combine quadratic inequalities with the discriminant and properties of parabolas.

Edexcel 考题有时会在二次式中包含未知常数 k，要求找出使不等式对所有实数 x 成立、或使解集满足特定条件的 k 的取值范围。这类题目将二次不等式与判别式及抛物线性质结合在一起。

For example, find k such that x² + kx + 9 > 0 for all real x. The quadratic has a = 1 > 0, so it opens upwards. To be always positive, it must have no real roots, meaning discriminant Δ < 0. Δ = k² – 4(1)(9) = k² – 36 < 0, so k² < 36, thus –6 < k < 6.

例如，求 k 使得 x² + kx + 9 > 0 对所有实数 x 成立。这里 a = 1 > 0，抛物线开口向上。要使它恒正，必须无实根，即判别式 Δ < 0。Δ = k² – 4(1)(9) = k² – 36 < 0，因此 k² < 36，即 –6 < k < 6。

Another typical problem: determine the values of k for which the inequality (k – 1)x² + 4x + (k + 2) has a solution set that is all real numbers except perhaps a single point. This requires a > 0 and Δ ≤ 0, but also ensures that the quadratic is not degenerate (a ≠ 0). Such analysis merges inequality theory with algebraic conditions.

另一类典型问题是：求 k 使得不等式 (k – 1)x² + 4x + (k + 2) 的解集为全体实数（至多扣除一个点）。这要求 a > 0 且 Δ ≤ 0，同时要确保二次式不退化（a ≠ 0）。这种分析将不等式理论与代数条件融合在一起。

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9. Set Notation and Interval Notation | 集合符号与区间符号

In Edexcel mark schemes, correct notation is essential. Solutions are usually expressed using set notation with inequalities, such as {x: x < –2} ∪ {x: x > 3}, or interval notation using parentheses for open ends and brackets for closed ends: (–∞, –2) ∪ (3, ∞). For inclusive boundaries, use square brackets: [–3, ½].

在 Edexcel 评分标准中，正确的符号至关重要。解通常用带有不等号的集合符号表示，如 {x: x < –2} ∪ {x: x > 3}；或用区间符号表示，开区间用圆括号，闭区间用方括号： (–∞, –2) ∪ (3, ∞)。包含边界时使用方括号：[–3, ½]。

When two intervals are disjoint, the union symbol ∪ must be used. Never write an answer like 2 < x > 5 — that notation is illogical. Instead, write x < 2 or x > 5, or in set notation {x: x < 2} ∪ {x: x > 5}. For infinite intervals, always use parentheses next to ∞ or –∞, because infinity is not a real number and cannot be included.

当两个区间不相连时，必须使用并集符号 ∪。绝对不要写成 2 < x > 5 之类的错误表示。应写为 x < 2 或 x > 5，或用集合符号 {x: x < 2} ∪ {x: x > 5}。对于无穷区间，在 ∞ 或 –∞ 旁始终使用圆括号，因为无穷并不是实数，不可包含。

A small summary table for interval forms:

Inequality	Interval Notation	Set Notation
a < x < b	(a, b)	{x: a < x < b}
a ≤ x ≤ b	[a, b]	{x: a ≤ x ≤ b}
x < a or x > b	(–∞, a) ∪ (b, ∞)	{x: x < a} ∪ {x: x > b}
x ≥ a	[a, ∞)	{x: x ≥ a}

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10. Applications and Word Problems | 应用与文字题

Quadratic inequalities appear frequently in applied contexts. A typical problem gives a model for revenue, cost, or height as a quadratic function of a variable, and asks for the range of input for which the output is above or below a certain threshold. The steps are: set up the inequality, bring all terms to one side, and solve using standard methods.

二次不等式经常出现在应用情境中。典型问题会给出收入、成本或高度关于某个变量的二次函数模型，要求找出使输出高于或低于某一阈值的自变量范围。解题步骤为：建立不等式，将所有项移至一边，然后使用标准方法求解。

Example: The height of a firework is modelled by h(t) = –5t² + 40t + 2, where t ≥ 0 is time in seconds. Find the time interval for which the firework is above 50 metres. Inequality: –5t² + 40t + 2 > 50 → –5t² + 40t – 48 > 0. Multiply by –1: 5t² – 40t + 48 < 0. Solve 5t² – 40t + 48 = 0 → t² – 8t + 9.6 = 0 → t = [8 ± √(64 – 38.4)]/2 = [8 ± √25.6]/2 ≈ (8 ± 5.06)/2, giving t ≈ 1.47 and t ≈ 6.53. With positive a after sign change, the quadratic is negative between the roots. Thus the firework is above 50 m for 1.47 < t < 6.53 seconds.

例题：烟花高度模型为 h(t) = –5t² + 40t + 2，其中 t ≥ 0 为时间（秒）。求烟花高于 50 米的时间区间。不等式：–5t² + 40t + 2 > 50 → –5t² + 40t – 48 > 0。乘以 –1：5t² – 40t + 48 < 0。解 5t² – 40t + 48 = 0 → t² – 8t + 9.6 = 0 → t = [8 ± √(64 – 38.4)]/2 = [8 ± √25.6]/2 ≈ (8 ± 5.06)/2，得 t ≈ 1.47 和 t ≈ 6.53。变号后首项系数正，二次式在两根之间为负。因此烟花在 1.47 < t < 6.53 秒期间高于 50 米。

Always interpret the answer in the context of the problem, checking that the domain restrictions (e.g. t ≥ 0) do not cut off part of the solution. In this example both roots are positive, so the whole interval applies.

务必结合实际问题解释答案，并检查定义域限制（如 t ≥ 0）是否会截去部分解。此例中两根均为正，故整个区间有效。

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11. Common Errors to Avoid | 常见错误与避免方法

Many students lose marks because of sign errors when multiplying or dividing by a negative number. When you change –x² + 3x + 10 ≥ 0 to x² – 3x – 10 ≤ 0, the inequality sign must flip. Another frequent mistake is misinterpreting the inequality when a < 0 without re‑scaling. Always make the coefficient of x² positive before applying graphical or critical‑value methods, or sketch very carefully.

很多学生因乘除负数时符号错误而失分。将 –x² + 3x + 10 ≥ 0 变为 x² – 3x – 10 ≤ 0 时，不等号必须翻转。另一个常见错误是未调整 a < 0 的情况就作图或直接应用临界值法。务必先将 x² 系数化为正，或者非常仔细地画出草图。

A third error is miswriting the final solution. For strict inequalities like > 0, do not include the critical values; for ≥ 0, do include them. Also, avoid

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