Solving Quadratic Inequalities | 解二次不等式

📚 Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities are a fundamental topic in A-Level Mathematics. They require you to find the set of values of \(x\) for which a quadratic expression is greater than, less than, or equal to zero. In this article, we will work through a detailed example — Example 1.8.1 — and explore the essential methods you need to master.

二次不等式是A-Level数学中的一个基础课题。你需要找出使二次表达式大于、小于或等于零的 \(x\) 的取值范围。在本文中，我们将详细讲解一个具体例题 —— 例题1.8.1 —— 并探讨你必须掌握的核心方法。

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

A quadratic inequality is an inequality that involves a quadratic expression. The general forms are:

二次不等式是涉及二次表达式的不等式。其一般形式为：

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

where \(a\), \(b\), and \(c\) are real numbers and \(a \neq 0\). Solving such an inequality means finding all real values of \(x\) that make the statement true.

其中 \(a\)、\(b\)、\(c\) 是实数，且 \(a \neq 0\)。解这样的不等式意味着找出所有使该不等式成立的实数 \(x\)。

2. The Sign Diagram Method | 符号图解法

One of the most reliable methods for solving quadratic inequalities is the sign diagram. First, factor the quadratic, then identify its roots. These roots split the real number line into intervals. In each interval, the quadratic has a constant sign — either positive or negative.

解二次不等式最可靠的方法之一是符号图解法。首先对二次式进行因式分解，然后找出它的根。这些根将实数轴分成若干个区间。在每个区间内，二次式的符号保持不变 —— 要么为正，要么为负。

For example, if a quadratic factors as \((x – r_1)(x – r_2)\), then the sign of the product depends on the signs of each factor. Sketching a sign diagram helps you visualise where the product is positive or negative.

例如，如果二次式可以分解为 \((x – r_1)(x – r_2)\)，那么积的符号取决于每个因式的符号。画符号图可以帮助你直观地看出积在哪些区间为正、哪些区间为负。

3. Example 1.8.1: Setting Up | 例题1.8.1：题目设定

Example 1.8.1. Solve the inequality \(x^2 – 5x + 6 > 0\).

例题1.8.1. 解不等式 \(x^2 – 5x + 6 > 0\)。

This is a classic quadratic inequality. Let us solve it step by step using the sign diagram method.

这是一个经典的二次不等式。下面我们使用符号图解法逐步求解。

4. Step 1: Factorise the Quadratic | 第一步：因式分解二次式

We look for two numbers that multiply to \(6\) and add to \(-5\). These numbers are \(-2\) and \(-3\). Therefore:

我们要找到两个数，它们相乘等于 \(6\)，相加等于 \(-5\)。这两个数是 \(-2\) 和 \(-3\)。因此：

x² − 5x + 6 = (x − 2)(x − 3)

So the inequality becomes \((x − 2)(x − 3) > 0\).

于是原不等式变为 \((x − 2)(x − 3) > 0\)。

5. Step 2: Find the Critical Values | 第二步：找到临界值

The critical values (also called roots or zeros) are found by setting the factorised expression equal to zero:

临界值（也称为根或零点）通过令因式分解后的表达式等于零来求得：

(x − 2)(x − 3) = 0

This gives \(x = 2\) or \(x = 3\). These two values divide the number line into three intervals:

由此得到 \(x = 2\) 或 \(x = 3\)。这两个值将数轴分成三个区间：

\(x < 2\)
\(2 < x < 3\)
\(x > 3\)

Note that the critical values themselves are not included because the inequality is strict (\(>\)).

注意，临界值本身不包括在内，因为这是严格不等式（\(>\)）。

6. Step 3: Draw a Sign Diagram | 第三步：绘制符号图

We now test the sign of each factor and the product in each interval. Choose a test value from each interval:

现在我们在每个区间内取一个测试值，检验每个因式以及乘积的符号：

Interval	Test value	(x − 2)	(x − 3)	Product
x < 2	x = 1	−	−	+
2 < x < 3	x = 2.5	+	−	−
x > 3	x = 4	+	+	+

The product is positive exactly when both factors have the same sign. From the table, this occurs when \(x < 2\) or \(x > 3\).

当两个因式同号时，乘积为正。从表中可以看出，这发生在 \(x < 2\) 或 \(x > 3\) 时。

7. Solution of Example 1.8.1 | 例题1.8.1的解

Because we need the product to be greater than zero, we select the intervals where the product is positive:

因为我们需要乘积大于零，所以我们选择乘积为正的区间：

x < 2 or x > 3

In set notation, the solution set is \(\{x : x < 2\} \cup \{x : x > 3\}\). This can also be written as \((-\infty, 2) \cup (3, \infty)\).

用集合符号表示，解集为 \(\{x : x < 2\} \cup \{x : x > 3\}\)。也可以写成 \((-\infty, 2) \cup (3, \infty)\)。

8. Alternative Method: Graphical Interpretation | 替代方法：图像解释

The graph of \(y = x^2 – 5x + 6\) is a parabola that opens upwards because the coefficient of \(x^2\) is positive. The parabola crosses the \(x\)-axis at \(x = 2\) and \(x = 3\).

函数 \(y = x^2 – 5x + 6\) 的图像是一条开口向上的抛物线，因为 \(x^2\) 的系数为正。抛物线与 \(x\) 轴交于 \(x = 2\) 和 \(x = 3\) 两点。

For values of \(x\) outside the roots, the parabola lies above the \(x\)-axis, so \(y > 0\). Between the roots, the parabola lies below the \(x\)-axis, so \(y < 0\). This confirms our algebraic solution.

当 \(x\) 取值在两根之外时，抛物线位于 \(x\) 轴上方，因此 \(y > 0\)。在两根之间时，抛物线位于 \(x\) 轴下方，因此 \(y < 0\)。这印证了我们的代数解。

Graphical rule: For an upward-opening parabola, the inequality \(ax² + bx + c > 0\) is satisfied outside the roots.

图像法则：对于开口向上的抛物线，不等式 \(ax² + bx + c > 0\) 在两根之外成立。

9. Working with Non-Strict Inequalities | 处理非严格不等式

If the inequality had been \(x^2 – 5x + 6 \geq 0\), the critical values \(x = 2\) and \(x = 3\) would also be included in the solution. The answer would then be \(x \leq 2\) or \(x \geq 3\).

如果原不等式是 \(x^2 – 5x + 6 \geq 0\)，那么临界值 \(x = 2\) 和 \(x = 3\) 也应包含在解集中。答案将是 \(x \leq 2\) 或 \(x \geq 3\)。

For a strict inequality, use open circles on the number line; for a non-strict inequality, use closed circles. This distinction is essential in examinations.

对于严格不等式，在数轴上使用空心圆圈；对于非严格不等式，使用实心圆圈。这种区分在考试中非常重要。

10. Special Cases: No Real Roots | 特殊情况：无实根

Some quadratic expressions do not factorise over the real numbers and have no real roots. In that case, the sign of the quadratic is the same for all \(x\).

有些二次式在实数范围内不能因式分解，没有实数根。在这种情况下，二次式的符号对所有 \(x\) 都相同。

For example, \(x^2 + 1 > 0\) is true for all real \(x\), because \(x^2 + 1\) is always positive. But \(x^2 + 1 < 0\) has no real solutions.

例如，\(x^2 + 1 > 0\) 对所有实数 \(x\) 都成立，因为 \(x^2 + 1\) 总是正的。而 \(x^2 + 1 < 0\) 没有实数解。

Always check the discriminant \(Δ = b² − 4ac\). If \(Δ < 0\), the parabola does not cross the \(x\)-axis, and its sign is determined by the coefficient of \(x^2\).

始终检查判别式 \(Δ = b² − 4ac\)。如果 \(Δ < 0\)，抛物线不与 \(x\) 轴相交，其符号由 \(x^2\) 的系数决定。

11. Common Mistakes | 常见错误

Multiplying or dividing both sides of an inequality by a negative number without reversing the inequality sign. This is a crucial error.
忽略乘以或除以负数时需要改变不等号方向。这是一个严重错误。

Forgetting to factorise or finding incorrect roots due to sign errors.
忘记因式分解或因符号错误而求错根。

Including the roots when the inequality is strict, or excluding them when it is non-strict.
当不等式是严格时误将根包含进去，或非严格时误将根排除。

Writing the answer incorrectly, such as \(2 < x < 3\) when the solution is actually two separate regions.
写出错误的答案，例如把解为两个分开区间的情况写成 \(2 < x < 3\)。

Remember that a quadratic inequality often has a solution set consisting of two disjoint intervals, not a single interval.

记住，二次不等式的解集通常由两个不相交的区间组成，而不是一个单一区间。

12. Practice and Further Questions | 练习与进阶问题

To master this topic, try solving the following inequalities:

为了掌握这一主题，请尝试求解以下不等式：

\(x^2 + 2x – 3 > 0\)
\(2x^2 – 4x ≤ 6\)
\(x^2 – 6x + 9 > 0\)
\(-x^2 + 4x – 3 < 0\)

For the last one, remember that if the coefficient of \(x^2\) is negative, the parabola opens downwards. You may multiply through by \(-1\) and reverse the inequality sign, or deal with the signs directly.

对于最后一个，注意如果 \(x^2\) 的系数为负，抛物线开口向下。你可以两边乘以 \(-1\) 并改变不等号方向，或者直接处理符号。

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Solving Quadratic Inequalities | 解二次不等式