Solving Quadratic Inequalities Using Graphs | 二次不等式解法：结合图像判断范围

📚 Solving Quadratic Inequalities Using Graphs | 二次不等式解法：结合图像判断范围

Quadratic inequalities often appear in IGCSE Edexcel Mathematics papers. The most reliable method is to sketch the graph of the quadratic function and then read the required range of x-values directly from the sketch. This article explains that approach step by step.

二次不等式是 IGCSE Edexcel 数学考试中的常见题型。最可靠的方法是先画出二次函数的图像，然后从图像中直接读出 x 的取值范围。本文将逐步解释这一方法。

1. The General Form of a Quadratic Inequality | 二次不等式的一般形式

A quadratic inequality can be written in one of these forms:

二次不等式通常可以写成以下形式之一：

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

Here a, b and c are real numbers, and a ≠ 0. The sign of a determines the shape of the graph: if a > 0 the parabola opens upwards (a “smile”); if a < 0 it opens downwards (a "frown").

其中 a、b、c 为实数，且 a ≠ 0。a 的正负决定抛物线开口方向：a > 0 时开口向上（”笑脸”形）；a < 0 时开口向下（"倒笑脸"形）。

2. Sketching the Quadratic Graph | 画出二次函数图像

To solve a quadratic inequality, first solve the corresponding quadratic equation ax² + bx + c = 0. The solutions are the x-coordinates where the graph crosses the x-axis. These points are called the roots or x-intercepts.

解二次不等式时，首先解对应的二次方程 ax² + bx + c = 0。方程的解是图像与 x 轴交点的 x 坐标，这些点称为方程的根或 x 轴的截距。

For example, to solve x² – 4x + 3 > 0, first solve x² – 4x + 3 = 0. This factorises as (x – 1)(x – 3) = 0, so x = 1 or x = 3. The graph crosses the x-axis at x = 1 and x = 3.

例如，解 x² – 4x + 3 > 0，先解 x² – 4x + 3 = 0。因式分解得 (x – 1)(x – 3) = 0，所以 x = 1 或 x = 3。图像与 x 轴交于 x = 1 和 x = 3。

x² – 4x + 3 = (x – 1)(x – 3)

The leading coefficient is positive, so the parabola opens upwards. Its y-intercept is at (0, 3) if needed, but the main features for the inequality are the roots and the direction of the parabola.

由于二次项系数为正，抛物线开口向上。如果需要，y 截距为 (0, 3)，但解不等式时主要关注的是根和开口方向。

3. The “Above or Below the Axis” Rule | “在轴上方或下方”法则

Once the sketch is drawn, look at the sign of the quadratic expression. For any x-value, the y-coordinate of the graph equals ax² + bx + c. Therefore:

画出草图后，观察二次表达式的符号。对于任意 x 值，图像的 y 坐标等于 ax² + bx + c。因此：

ax² + bx + c > 0 means the graph is above the x-axis.
ax² + bx + c > 0 表示图像在 x 轴上方。
ax² + bx + c < 0 means the graph is below the x-axis.
ax² + bx + c < 0 表示图像在 x 轴下方。

This rule converts an algebra problem into a visual one. You never need to test random points if your sketch is correct.

这个法则把代数问题转化为视觉问题。只要草图正确，就不需要测试随机点。

4. Solving (x – a)(x – b) > 0 | 解 (x – a)(x – b) > 0

Consider a quadratic with two distinct roots a and b, where a < b. If the coefficient of x² is positive, the graph is a "smile" that crosses the x-axis at a and b. It is above the x-axis on the left of a, below between a and b, and above again on the right of b.

考虑具有两个不同根 a 和 b 的二次函数，其中 a < b。若 x² 的系数为正，图像形如"笑脸"，在 a 和 b 处穿过 x 轴。它在 a 的左侧位于 x 轴上方，在 a 与 b 之间位于下方，在 b 的右侧又位于上方。

x < a or x > b

Therefore the solution to (x – a)(x – b) > 0 is the union of two separate intervals: x < a or x > b. In set notation this is {x : x < a} ∪ {x : x > b}.

因此 (x – a)(x – b) > 0 的解是两个独立区间的并集：x < a 或 x > b。用集合符号表示为 {x : x < a} ∪ {x : x > b}。

5. Solving (x – a)(x – b) < 0 | 解 (x - a)(x - b) < 0

Using the same “smile” graph, the portion below the x-axis is exactly the interval between the two roots. Hence:

使用同样的”笑脸”图像，x 轴下方的部分正好是两根之间的区间。因此：

a < x < b

This single interval is the solution to (x – a)(x – b) < 0. For example, x² - 4x + 3 < 0 has roots 1 and 3, so its solution is 1 < x < 3.

这个单一区间就是 (x – a)(x – b) < 0 的解。例如 x² - 4x + 3 < 0 的根为 1 和 3，所以解为 1 < x < 3。

Remember that for a quadratic opening upwards, the “outside” regions are positive and the “inside” region is negative. This memory aid works for any two real roots.

记住：对开口向上的二次函数，”外侧”区域为正，”内侧”区域为负。这个记忆方法适用于任意两个实数根。

6. Strict vs Non-Strict Inequalities | 严格不等式与非严格不等式

If the inequality is strict (> or <), the roots themselves are not included, so we use open circles on a number line and write < or >. If the inequality is non-strict (≥ or ≤), the roots make the expression equal to zero, so they are included.

如果不等式是严格的（> 或 <），根本身不包含在解集中，数轴上用空心圆圈，书写时用 < 或 >。如果不等式是非严格的（≥ 或 ≤），在根处表达式等于 0，因此根包含在解集中。

For example, x² – 4x + 3 ≥ 0 has solution x ≤ 1 or x ≥ 3. And x² – 4x + 3 ≤ 0 has solution 1 ≤ x ≤ 3.

例如 x² – 4x + 3 ≥ 0 的解为 x ≤ 1 或 x ≥ 3。x² – 4x + 3 ≤ 0 的解为 1 ≤ x ≤ 3。

Inequality	Solution for a “smile” graph	Number line notation
(x – a)(x – b) > 0	x < a or x > b	Open circles at a and b
(x – a)(x – b) < 0	a < x < b	Open circles at a and b
(x – a)(x – b) ≥ 0	x ≤ a or x ≥ b	Closed circles at a and b
(x – a)(x – b) ≤ 0	a ≤ x ≤ b	Closed circles at a and b

7. When a Is Negative | 当 a 为负数时

If the coefficient of x² is negative, the parabola opens downwards. For example, solve -x² + 5x – 6 > 0. Multiply the whole inequality by -1. This reverses the inequality sign: x² – 5x + 6 < 0. Then factorise: (x - 2)(x - 3) < 0, so the solution is 2 < x < 3.

如果 x² 的系数为负，抛物线开口向下。例如解 -x² + 5x – 6 > 0。将整个不等式乘以 -1，不等号方向反转：x² – 5x + 6 < 0。因式分解得 (x - 2)(x - 3) < 0，所以解为 2 < x < 3。

-x² + 5x – 6 > 0 ⇔ x² – 5x + 6 < 0

Alternatively, you can sketch the downward-opening parabola directly. It is above the x-axis only between its two roots at 2 and 3. Both approaches give the same answer.

另一种方法是直接画出开口向下的抛物线。它仅在两根 2 和 3 之间位于 x 轴上方。两种方法得到相同结果。

8. Quadratic with No Real Roots | 没有实数根的二次函数

When the discriminant Δ = b² – 4ac is negative, the quadratic has no real roots. The graph never crosses the x-axis.

当判别式 Δ = b² – 4ac 为负时，二次方程没有实数根，图像与 x 轴无交点。

If a > 0, the graph is always above the x-axis. Thus ax² + bx + c > 0 is true for all real x, and ax² + bx + c < 0 has no solution.
如果 a > 0，图像始终在 x 轴上方。因此 ax² + bx + c > 0 对所有实数 x 都成立，而 ax² + bx + c < 0 无解。
If a < 0, the graph is always below the x-axis. Thus ax² + bx + c < 0 is true for all real x, and ax² + bx + c > 0 has no solution.
如果 a < 0，图像始终在 x 轴下方。因此 ax² + bx + c < 0 对所有实数 x 都成立，而 ax² + bx + c > 0 无解。

For example, x² + 2x + 5 > 0. The discriminant is 4 – 20 = -16 < 0, and a = 1 > 0. So the solution is all real numbers: x ∈ ℝ.

例如 x² + 2x + 5 > 0。判别式为 4 – 20 = -16 < 0，且 a = 1 > 0。所以解为全体实数：x ∈ ℝ。

9. Quadratic with a Repeated Root | 有重根的二次函数

If Δ = 0, the quadratic has one repeated root. The graph touches the x-axis at that point but does not cross it. The y-coordinate is zero at the root and positive everywhere else if a > 0.

若 Δ = 0，二次方程有一个重根。图像在该点与 x 轴相切但不穿过。在根处 y 坐标为 0；若 a > 0，其他地方的 y 坐标均为正。

Example: solve (x – 2)² > 0. The only root is x = 2. The graph touches the axis at x = 2 and is above elsewhere. Therefore the solution is all real numbers except x = 2.

例如：解 (x – 2)² > 0。唯一根为 x = 2。图像在 x = 2 处与轴相切，其余位置在轴上方。因此解为除 x = 2 以外的所有实数。

x ∈ ℝ, x ≠ 2

For (x – 2)² < 0 there is no solution, because a square is never negative. For (x - 2)² ≥ 0 the solution is all real numbers, and for (x - 2)² ≤ 0 the solution is only x = 2.

对于 (x – 2)² < 0，无解，因为平方永不为负。对于 (x - 2)² ≥ 0，解为全体实数；对于 (x - 2)² ≤ 0，解仅为 x = 2。

10. Common Mistakes and Exam Tips | 常见错误与应试建议

Many students lose marks on quadratic inequalities because of careless errors. Avoid these common pitfalls:

许多学生在二次不等式上失分是因为粗心错误。避免以下常见陷阱：

Forgetting to reverse the inequality sign when multiplying or dividing by a negative number.
忘记在乘以或除以负数时反转不等号方向。
Sketching the graph without checking whether a is positive or negative.
画草图时未检查 a 的正负。
Writing “1 < x > 3″ instead of two separate inequalities. The correct form for > is x < 1 or x > 3.
把解写成 “1 < x > 3″ 而非两个独立不等式。对于 > 正确写法是 x < 1 或 x > 3。
Using the roots from the equation but failing to test which side is positive.
解出方程的根但没有判断哪一侧为正。

Always sketch a simple graph in the working space. Even a rough sketch shows the correct regions and prevents sign errors. In exam questions, marks are often awarded for the correct set of x-values, so present your final answer clearly as an inequality or in set notation.

务必在答题区域画一个简单草图。即使粗略的草图也能显示正确的区域，避免符号错误。考试中，分数通常取决于最终的 x 取值范围，因此请用不等式或集合符号清晰地写出答案。

Finally, check your answer by substituting a test x-value from each region back into the original inequality. This is a fast and reliable verification.

最后，从每个区域选取一个测试 x 值代回原不等式进行验证。这是一种快速而可靠的检查方法。

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Quadratic Inequalities Using Graphs | 二次不等式解法：结合图像判断范围