📚 Solving Quadratic Inequalities | 解一元二次不等式
Quadratic inequalities are a key topic in IGCSE Mathematics. They combine the familiar techniques of solving quadratic equations with the extra step of determining which ranges of x satisfy the inequality. This guide will walk you through the essential methods, common pitfalls, and exam-style examples.
一元二次不等式是 IGCSE 数学中的一个核心考点。它将我们熟悉的解二次方程的方法，与判断哪些 x 的取值范围满足不等式的额外步骤结合起来。本指南将带你系统地掌握关键方法、常见错误以及考试风格的例题。
1. What Is a Quadratic Inequality? | 什么是一元二次不等式
A quadratic inequality is an expression involving a quadratic polynomial and one of the inequality symbols: <, >, ≤, or ≥. In standard form, it can be written as ax² + bx + c < 0, ax² + bx + c > 0, ax² + bx + c ≤ 0, or ax² + bx + c ≥ 0, where a ≠ 0.
一元二次不等式是包含二次多项式以及不等号（<, >, ≤ 或 ≥）的表达式。其标准形式可写成 ax² + bx + c < 0、ax² + bx + c > 0、ax² + bx + c ≤ 0 或 ax² + bx + c ≥ 0，其中 a ≠ 0。
For example, x² − 5x + 6 > 0 is a quadratic inequality. The goal is to find all real values of x that make the inequality true.
例如，x² − 5x + 6 > 0 就是一个一元二次不等式。我们的目标是找出所有能使不等式成立的实数 x。
Unlike a quadratic equation, which usually has two specific solutions (or one repeated, or none), a quadratic inequality typically has a range of solutions, often expressed as intervals on the number line.
与通常有两个具体解（或一个重根、或无解）的二次方程不同，一元二次不等式通常有一个解集范围，通常用数轴上的区间来表示。
2. Step-by-Step Method | 分步解法
There are two common approaches to solving quadratic inequalities: the number-line method (also called the sign diagram method) and the graphical method. Both rely on the same first two steps.
解一元二次不等式有两种常用方法：数轴法（也称符号图法）和图解法。这两种方法的前两步是相同的。
Step 1: Rearrange the inequality so that one side is zero. For example, x² < 4 becomes x² − 4 < 0.
第一步：将不等式整理为一边为零的形式。例如，x² < 4 可化为 x² − 4 < 0。
Step 2: Solve the corresponding quadratic equation, i.e., set the quadratic expression equal to zero and find the roots. These roots are called critical values or boundary points.
第二步：解对应的二次方程，即令二次表达式等于零，求出根。这些根称为临界值或边界点。
Step 3: Use a sign diagram or a sketch of the parabola to decide which intervals between the roots satisfy the inequality.
第三步：利用符号图或抛物线草图来判断根之间的哪些区间满足不等式。
Let us illustrate this method with a worked example.
下面用一个完整的例子来说明这个方法。
3. Worked Example 1: x² − 5x + 6 > 0 | 例题 1：x² − 5x + 6 > 0
Solve the inequality x² − 5x + 6 > 0.
解不等式 x² − 5x + 6 > 0。
Step 1: The expression is already on the left with zero on the right.
第一步：不等式左边已经是二次表达式，右边为零。
Step 2: Solve x² − 5x + 6 = 0. Factorising gives (x − 2)(x − 3) = 0, so x = 2 or x = 3.
第二步：解 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。
Step 3: Draw a number line with the critical values 2 and 3. These two points split the line into three intervals: x < 2, 2 < x < 3, and x > 3. Test a value from each interval in the original inequality.
第三步：在数轴上标出临界值 2 和 3。这两个点把数轴分成三个区间：x < 2、2 < x < 3 和 x > 3。在每个区间内取一个测试值代入原不等式。
| Interval | Test value | (x−2)(x−3) | Sign |
| x < 2 | x = 0 | (−2)(−3) = 6 | Positive |
| 2 < x < 3 | x = 2.5 | (0.5)(−0.5) = −0.25 | Negative |
| x > 3 | x = 4 | (2)(1) = 2 | Positive |
We want x² − 5x + 6 > 0, so we choose the intervals where the product is positive: x < 2 or x > 3.
因为我们需要 x² − 5x + 6 > 0，所以选择乘积为正的区间：x < 2 或 x > 3。
x < 2 or x > 3
In interval notation, the solution is (−∞, 2) ∪ (3, ∞). Note that the endpoints are not included because the inequality is strictly greater than zero.
用区间表示，解集为 (−∞, 2) ∪ (3, ∞)。注意端点不包含在内，因为不等式是严格大于零。
4. Worked Example 2: x² − 5x + 6 ≤ 0 | 例题 2：x² − 5x + 6 ≤ 0
Now solve the same quadratic with ≤ 0 instead of > 0.
现在解同一个二次表达式，但不等号改为 ≤ 0。
The critical values are still x = 2 and x = 3. The sign of the expression is positive for x < 2, negative for 2 < x < 3, and positive for x > 3.
临界值仍然是 x = 2 和 x = 3。表达式的符号在 x < 2 时为正，2 < x < 3 时为负，x > 3 时为正。
Since we want the expression to be less than or equal to zero, we take the interval where it is negative, and we include the endpoints because the inequality allows equality.
因为我们要求表达式小于或等于零，所以取符号为负的区间，并且因为不等号允许相等，所以端点要包含在内。
2 ≤ x ≤ 3
In interval notation: [2, 3].
用区间表示：[2, 3]。
5. The Graphical Approach | 图形方法
Another way to visualise quadratic inequalities is by sketching the graph of y = ax² + bx + c. The parabola opens upward if a > 0 and downward if a < 0.
另一种理解一元二次不等式的方法是通过绘制 y = ax² + bx + c 的图像。当 a > 0 时抛物线开口向上，当 a < 0 时开口向下。
For x² − 5x + 6, since a = 1 > 0, the parabola opens upward and crosses the x-axis at x = 2 and x = 3. The graph lies above the x-axis when x < 2 or x > 3, and below the x-axis when 2 < x < 3.
对于 x² − 5x + 6，因为 a = 1 > 0，抛物线开口向上，并在 x = 2 和 x = 3 处与 x 轴相交。当 x < 2 或 x > 3 时，图像在 x 轴上方；当 2 < x < 3 时，图像在 x 轴下方。
Thus, the inequality ax² + bx + c > 0 corresponds to the portions of the graph above the x-axis, while ax² + bx + c < 0 corresponds to the portions below the x-axis.
因此，ax² + bx + c > 0 对应图像位于 x 轴上方的部分，而 ax² + bx + c < 0 对应图像位于 x 轴下方的部分。
This graphical intuition is very useful, especially when the quadratic does not factor easily. However, for IGCSE, factorisation is the most common method.
这种图形直觉非常有用，特别是当二次式不容易因式分解时。不过在 IGCSE 中，因式分解是最常用的方法。
6. Dealing with a Negative Leading Coefficient | 处理首项系数为负数的情况
If the coefficient of x² is negative, it is often helpful to multiply the entire inequality by −1. Remember that when you multiply or divide an inequality by a negative number, you must reverse the inequality sign.
如果 x² 的系数为负，通常可以先在整个不等式两边乘以 −1。记住：当不等式两边乘以或除以一个负数时，不等号方向必须反转。
For example, to solve −x² + 4x − 3 > 0, multiply by −1 to get x² − 4x + 3 < 0. The sign has reversed.
例如，解 −x² + 4x − 3 > 0，两边乘以 −1 得到 x² − 4x + 3 < 0。不等号方向发生了反转。
Now solve x² − 4x + 3 < 0. The equation x² − 4x + 3 = 0 factors as (x − 1)(x − 3) = 0, so the critical values are x = 1 and x = 3.
现在解 x² − 4x + 3 < 0。方程 x² − 4x + 3 = 0 因式分解为 (x − 1)(x − 3) = 0，所以临界值为 x = 1 和 x = 3。
The expression (x − 1)(x − 3) is negative for 1 < x < 3. Therefore the solution is 1 < x < 3.
表达式 (x − 1)(x − 3) 在 1 < x < 3 时为负。因此解集为 1 < x < 3。
1 < x < 3
Always double-check which way the inequality sign points after multiplying by a negative number.
乘以负数后，一定要重新检查不等号的方向是否正确。
7. Inequalities with a Perfect Square or No Real Roots | 完全平方式或无实数根的不等式
Sometimes the quadratic equation has a repeated root, meaning the parabola touches the x-axis at exactly one point. For example, x² − 6x + 9 = (x − 3)² has a double root at x = 3.
有时二次方程有重根，即抛物线与 x 轴恰好相切于一点。例如，x² − 6x + 9 = (x − 3)² 在 x = 3 处有重根。
Consider x² − 6x + 9 > 0. The expression is zero at x = 3 and positive everywhere else (since (x − 3)² ≥ 0 for all real x). Therefore the solution is all real numbers except x = 3, written as x < 3 or x > 3.
考虑 x² − 6x + 9 > 0。表达式在 x = 3 时等于零，在其他所有地方都为正（因为 (x − 3)² ≥ 0 对所有实数 x 成立）。因此解集为除了 x = 3 之外的所有实数，写作 x < 3 或 x > 3。
What if the quadratic has no real roots, for example x² + 1 > 0? Since x² + 1 ≥ 1 for all x, the inequality is true for all real values of x.
如果二次方程没有实数根呢？例如 x² + 1 > 0。因为 x² + 1 ≥ 1 对所有 x 都成立，所以这个不等式对所有实数 x 都成立。
However, x² + 1 < 0 has no real solution because the expression is always positive. Be careful with these special cases in the exam.
然而，x² + 1 < 0 没有实数解，因为表达式总是正的。考试中要特别注意这类特殊情况。
8. Using the Discriminant | 使用判别式
The discriminant, Δ = b² − 4ac, tells us whether the quadratic equation ax² + bx + c = 0 has two, one, or no real roots. This is useful when solving quadratic inequalities as well.
判别式 Δ = b² − 4ac 告诉我们二次方程 ax² + bx + c = 0 有两个、一个还是没有实数根。这在解一元二次不等式时同样有用。
If Δ > 0, there are two distinct critical values. If Δ = 0, there is one repeated critical value. If Δ < 0, there are no real critical values, and the sign of the quadratic is the same everywhere (the sign of a).
如果 Δ > 0，有两个不同的临界值。如果 Δ = 0，有一个重根临界值。如果 Δ < 0，没有实数临界值，二次表达式的符号处处相同（与 a 的符号相同）。
For example, consider x² + x + 1 > 0. Here a = 1, b = 1, c = 1, so Δ = 1 − 4 = −3 < 0. Since a > 0, the expression is always positive, so the inequality holds for all real x.
例如，考虑 x² + x + 1 > 0。这里 a = 1，b = 1，c = 1，所以 Δ = 1 − 4 = −3 < 0。由于 a > 0，表达式总是正的，因此该不等式对所有实数 x 都成立。
This method is particularly helpful when the quadratic cannot be factorised using simple integers.
当二次式无法用简单整数因式分解时，这种方法尤其有用。
9. Common Mistakes to Avoid | 避免常见错误
Mistake 1: Reversing the inequality sign incorrectly. This often happens when multiplying or dividing by a negative number. Always reverse the sign when doing so.
错误一：错误地反转不等号。 这通常发生在乘以或除以负数时。此时必须反转不等号。
Mistake 2: Including endpoints in a strict inequality. For < and >, the critical values themselves are not part of the solution. For ≤ and ≥, they are included.
错误二：在严格不等式中包含端点。 对于 < 和 >，临界值本身不属于解集。对于 ≤ 和 ≥，端点应包含在内。
Mistake 3: Writing the solution as a single interval when it should be two intervals. For example, x² − 5x + 6 > 0 has solution x < 2 or x > 3, not 2 < x > 3.
错误三：把本应是两个区间的解写成一个区间。 例如，x² − 5x + 6 > 0 的解为 x < 2 或 x > 3，而不是 2 < x > 3。
Mistake 4: Forgetting to gather all terms on one side first. For instance, the inequality x² < 4 must be rewritten as x² − 4 < 0 before solving the corresponding equation.
错误四：忘记先把所有项移到一边。 例如，不等式 x² < 4 必须先改写为 x² − 4 < 0，再解对应的方程。
Mistake 5: Confusing the sign of the product with the sign of the sum. In a sign diagram, always test actual values from each interval rather than guessing.
错误五：混淆乘积的符号与和的符号。 在符号图中，务必在每一个区间取实际值进行测试，而不是凭感觉猜测。
10. Exam-Style Practice Questions | 考试风格练习题
Try these questions to test your understanding. Write down your solutions before checking the answers below.
尝试做下面这些题来检验你的理解。先写下你的解答，再核对下面的答案。
Question 1: Solve x² − 7x + 10 > 0.
练习 1：解 x² − 7x + 10 > 0。
Question 2: Solve 2x² − 3x − 2 ≤ 0.
练习 2：解 2x² − 3x − 2 ≤ 0。
Question 3: Solve x² < 9.
练习 3：解 x² < 9。
Question 4: Solve −x² + 6x − 8 ≥ 0.
练习 4：解 −x² + 6x − 8 ≥ 0。
Question 5: Find the set of values of k for which x² + kx + 4 > 0 for all real x.
练习 5：求使 x² + kx + 4 > 0 对所有实数 x 都成立的 k 的取值范围。
11. Solutions to Practice Questions | 练习题答案
Solution 1: x² − 7x + 10 = (x − 2)(x − 5) = 0 gives x = 2 or x = 5. The product is positive for x < 2 or x > 5.
答案 1：x² − 7x + 10 = (x − 2)(x − 5) = 0 得到 x = 2 或 x = 5。乘积在 x < 2 或 x > 5 时为正。
x < 2 or x > 5
Solution 2: Solve 2x² − 3x − 2 = 0. Factoring gives (2x + 1)(x − 2) = 0, so x = −½ or x = 2. The expression is negative between these values.
答案 2：解 2x² − 3x − 2 = 0。因式分解得 (2x + 1)(x − 2) = 0，所以 x = −½ 或 x = 2。表达式在这两个值之间为负。
−½ ≤ x ≤ 2
Solution 3: x² < 9 is equivalent to x² − 9 < 0, i.e., (x − 3)(x + 3) < 0. The solution is −3 < x < 3.
答案 3：x² < 9 等价于 x² − 9 < 0，即 (x − 3)(x + 3) < 0。解为 −3 < x < 3。
−3 < x < 3
Solution 4: Multiply by −1 and reverse the sign: x² − 6x + 8 ≤ 0. Factorising gives (x − 2)(x − 4) ≤ 0. The solution is 2 ≤ x ≤ 4.
答案 4：两边乘以 −1 并反转不等号：x² − 6x + 8 ≤ 0。因式分解得 (x − 2)(x − 4) ≤ 0。解为 2 ≤ x ≤ 4。
2 ≤ x ≤ 4
Solution 5: For x² + kx + 4 > 0 to hold for all real x, the discriminant must be negative. So k² − 16 < 0, which means k² < 16, giving −4 < k < 4.
答案 5：要使 x² + kx + 4 > 0 对所有实数 x 都成立，判别式必须小于零。因此 k² − 16 < 0，即 k² < 16，得到 −4 < k < 4。
−4 < k < 4
12. Summary | 总结
To solve a quadratic inequality, first rearrange it so that one side is zero, then solve the corresponding quadratic equation to find the critical values. Use a sign diagram or a parabola sketch to determine which intervals satisfy the inequality. Remember to reverse the inequality sign when multiplying or dividing by a negative number, and be careful about whether endpoints are included.
解一元二次不等式时，首先将不等式整理为一边为零的形式，然后解对应的二次方程求出临界值。接着利用符号图或抛物线草图判断哪些区间满足不等式。记住：当乘以或除以负数时，不等号要反转；同时注意端点是否包含在解集中。
With regular practice, quadratic inequalities become straightforward. Always draw a quick sign diagram, even if you think you can see the answer immediately. This reduces the chance of careless errors.
通过规律练习，一元二次不等式会变得简单易懂。即使你觉得自己能立刻看出答案，也建议快速画一个符号图。这样可以减少粗心错误的发生。
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