Example 4.2.1: Solving a Quadratic Inequality | 例题4.2.1：解二次不等式

📚 Example 4.2.1: Solving a Quadratic Inequality | 例题4.2.1：解二次不等式

In this example, we will solve the quadratic inequality x² − 5x + 4 ≤ 0 step by step, following the standard AQA A-level Mathematics approach. This type of question frequently appears in the pure mathematics paper and requires a clear understanding of factorisation, critical points, and sign analysis.

在本例题中，我们将一步一步地求解二次不等式 x² − 5x + 4 ≤ 0，遵循 AQA A-Level 数学的标准方法。这类问题在纯数学试卷中经常出现，要求我们清晰地理解因式分解、临界点和符号分析。

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1. Problem Statement | 问题陈述

We are given the inequality x² − 5x + 4 ≤ 0. We must find the set of real numbers x that satisfy this condition.

我们被给定不等式 x² − 5x + 4 ≤ 0。我们必须找出满足此条件的实数 x 的集合。

This is a quadratic inequality because the highest power of x is 2. The symbol ≤ means the expression on the left may equal zero as well as be less than zero.

这是一个二次不等式，因为 x 的最高次数是 2。符号 ≤ 表示左边的表达式可以等于零，也可以小于零。

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2. Factorising the Quadratic | 因式分解二次式

To solve a quadratic inequality, we first try to factorise the quadratic expression. We look for two numbers that multiply to 4 and add to −5.

为了解二次不等式，我们首先尝试对二次表达式进行因式分解。我们寻找两个数，它们相乘得 4，相加得 −5。

These numbers are −1 and −4, so we can write:

这两个数是 −1 和 −4，因此我们可以写出：

x² − 5x + 4 = (x − 1)(x − 4)

Therefore the inequality becomes (x − 1)(x − 4) ≤ 0.

因此不等式变为 (x − 1)(x − 4) ≤ 0。

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3. Finding Critical Points | 求临界点

Critical points are the values of x for which the expression equals zero. We set each factor equal to zero.

临界点是使表达式等于零的 x 的值。我们令每个因式等于零。

From x − 1 = 0, we get x = 1. From x − 4 = 0, we get x = 4.

由 x − 1 = 0，得到 x = 1。由 x − 4 = 0，得到 x = 4。

These points split the real number line into three intervals: x < 1, 1 < x < 4, and x > 4.

这些点将实数轴分为三个区间：x < 1，1 < x < 4 和 x > 4。

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4. Sign Analysis | 符号分析

We choose a test point in each interval and evaluate the sign of (x − 1)(x − 4).

我们在每个区间内选择一个测试点，并判断 (x − 1)(x − 4) 的符号。

For x < 1, take x = 0. Then (0 − 1)(0 − 4) = (−1)(−4) = 4 > 0, positive.

对于 x < 1，取 x = 0。则 (0 − 1)(0 − 4) = (−1)(−4) = 4 > 0，为正。

For 1 < x < 4, take x = 2. Then (2 − 1)(2 − 4) = (1)(−2) = −2 ≤ 0, negative.

对于 1 < x < 4，取 x = 2。则 (2 − 1)(2 − 4) = (1)(−2) = −2 ≤ 0，为负。

For x > 4, take x = 5. Then (5 − 1)(5 − 4) = (4)(1) = 4 > 0, positive.

对于 x > 4，取 x = 5。则 (5 − 1)(5 − 4) = (4)(1) = 4 > 0，为正。

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5. Graphical Interpretation | 图形解释

The quadratic y = x² − 5x + 4 is a parabola opening upwards because the coefficient of x² is positive.

二次函数 y = x² − 5x + 4 是一条开口向上的抛物线，因为 x² 的系数为正。

It crosses the x-axis at x = 1 and x = 4. The part of the curve below or on the x-axis corresponds to y ≤ 0.

它与 x 轴在 x = 1 和 x = 4 处相交。曲线位于 x 轴下方或与轴相交的部分对应于 y ≤ 0。

This graphical view confirms that the solution is the interval between the two roots, including the endpoints.

这种图形视角证实了解集为两根之间的区间，包括端点。

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6. Writing the Solution Set | 写出解集

Because the inequality is ≤, the endpoints are included in the solution. We use square brackets to denote a closed interval.

因为不等式是 ≤，端点包含在解集中。我们用方括号表示闭区间。

Thus the solution set is:

因此解集为：

1 ≤ x ≤ 4

In interval notation, we write x ∈ [1, 4].

用区间表示法，我们写作 x ∈ [1, 4]。

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7. Checking by Substitution | 代入验证

We can verify the answer by substituting a few values. For x = 1, the original expression equals 0, satisfying ≤.

我们可以通过代入一些数值来验证答案。对于 x = 1，原表达式等于 0，满足 ≤。

For x = 4, the expression also equals 0. For x = 2, the expression equals −2, which is less than 0.

对于 x = 4，表达式也等于 0。对于 x = 2，表达式等于 −2，小于 0。

For x = 0 or x = 5, the expression is positive, so these values do not belong to the solution set.

对于 x = 0 或 x = 5，表达式为正，因此这些值不属于解集。

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8. Common Mistakes | 常见错误

A common mistake is to write x < 1 and x < 4 as the solution. This is wrong because the product of two negative numbers is positive, and the product of two positive numbers is positive.

一个常见错误是将解写为 x < 1 且 x < 4。这是错误的，因为两个负数相乘为正，两个正数相乘也为正。

Another mistake is to forget that the ≤ symbol includes the boundary points. If the inequality were strict (<), the endpoints would be excluded.

另一个错误是忘记 ≤ 符号包含边界点。如果不等式是严格的 (<)，则端点将被排除。

Always test signs in each interval rather than relying on memory.

务必在每个区间测试符号，而不是凭记忆。

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9. Variations | 变式

If the inequality were x² − 5x + 4 > 0, the solution would be the union of the two outer intervals: x < 1 or x > 4.

如果不等式是 x² − 5x + 4 > 0，则解集为两个外侧区间的并集：x < 1 或 x > 4。

If the leading coefficient were negative, for example −x² + 5x − 4 ≥ 0, we could multiply both sides by −1 and reverse the inequality sign before solving.

如果首项系数为负，例如 −x² + 5x − 4 ≥ 0，我们可以先将两边乘以 −1 并反转不等号，然后再求解。

In some exam questions, the quadratic may not factorise with integers; then we use the quadratic formula to find critical points.

在某些考试题中，二次式可能无法用整数因式分解；这时我们用二次公式求临界点。

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10. Summary | 总结

The key steps to solve a quadratic inequality are: factorise, find the critical points, test intervals, and state the solution using the correct notation.

求解二次不等式的关键步骤如下：因式分解、求临界点、测试区间，并用正确的符号写出解集。

For x² − 5x + 4 ≤ 0, the solution is x ∈ [1, 4]. Understanding this example prepares you for more complex inequalities involving higher-order polynomials.

对于 x² − 5x + 4 ≤ 0，解集为 x ∈ [1, 4]。理解这个例题能帮助你为更复杂的涉及高阶多项式的不等式做好准备。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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