📚 Example 3.2.1: Solving Quadratic Inequalities by Sketching Graphs | 例 3.2.1：通过画图解二次不等式
In AQA A-Level Mathematics, quadratic inequalities often appear in pure topics such as equations, inequalities, and curve sketching. This worked example shows how to solve a quadratic inequality without guessing, by using factorisation, critical values, and a clear sketch of the parabola.
在 AQA A-Level 数学中，二次不等式常出现在纯数学的方程、不等式和曲线作图等主题中。本例展示如何利用因式分解、临界值和清晰的抛物线图像，而不是靠猜测来解二次不等式。
1. Understanding the Problem | 理解题意
The inequality chosen for Example 3.2.1 is x² − 4x + 3 > 0. Our task is to find all real values of x for which the quadratic expression is strictly positive, meaning the graph lies above the x-axis.
例 3.2.1 选用的不等式是 x² − 4x + 3 > 0。我们的任务是找出所有使该二次式严格大于零的实数 x，也就是图像位于 x 轴上方时对应的 x 值。
Since this is a strict inequality, the boundary values where the expression equals zero will not be included in the final solution.
由于这是严格不等式，使表达式等于零的边界值不会包含在最终解集中。
2. Factorising the Quadratic Expression | 将二次式因式分解
The first step is to factorise the quadratic x² − 4x + 3. We look for two numbers that multiply to give the constant term 3 and add to give the coefficient of x, which is −4.
第一步是将二次式 x² − 4x + 3 进行因式分解。我们需要找到两个数，它们相乘得到常数项 3，相加得到 x 的系数 −4。
x² − 4x + 3 = (x − 1)(x − 3)
The numbers −1 and −3 multiply to give 3 and add to give −4, so the factorised form is (x − 1)(x − 3). Always check your factorisation by expanding the brackets.
数字 −1 和 −3 相乘得 3，相加得 −4，因此因式分解后的形式为 (x − 1)(x − 3)。请始终通过展开括号来检查你的因式分解是否正确。
3. Finding the Critical Values | 求临界值
To find the critical values, set each factor equal to zero. This gives the points where the quadratic expression changes sign.
为了求临界值，令每个因式等于零。这样就得到二次式可能变号的点。
(x − 1)(x − 3) = 0
x − 1 = 0 → x = 1
x − 3 = 0 → x = 3
oots of the equation x² − 4x + 3 = 0 and they divide the number line into three intervals.
临界值为 x = 1 和 x = 3。它们是方程 x² − 4x + 3 = 0 的根，并把数轴分成三个区间。
4. Sketching the Graph | 画出图像
nd the corresponding y-value is −1, but we do not need the vertex to solve the inequality.
x² 的系数为正，所以抛物线开口向上。它的两个 x 轴截距位于 x = 1 和 x = 3。顶点位于两根中点 x = 2 处，对应的 y 值为 −1，但解这个不等式并不需要顶点。
| Feature | Value |
|---|---|
| Leading coefficient | 1 (positive) |
| Direction | Opens upward |
| Roots / x-intercepts | x = 1, x = 3 |
| Vertex | (2, −1) |
A rough sketch is enough in an exam: mark the roots on the x-axis, draw a U-shaped curve crossing the axis at those roots, and label the intervals.
考试中画出一个简图即可：在 x 轴上标出两根，画一条 U 形曲线穿过这些根，并标出区间。
5. Determining the Sign in Each Interval | 判断各区间的符号
The graph shows that the quadratic expression is positive outside the interval between the roots and negative inside it. We can confirm this by testing one value in each interval.
图像显示，二次式在两根之间的区间内为负，在两根之外的区间为正。我们可以通过在每个区间测试一个值来验证。
< 3, and x = 4 for x > 3. Substitute these into the factorised form (x − 1)(x − 3).
在区间 x < 1 中取 x = 0，在 1 < x < 3 中取 x = 2，在 x > 3 中取 x = 4。将它们代入因式分解式 (x − 1)(x − 3)。
| Interval | Test value | (x−1)(x−3) | Sign |
|---|---|---|---|
| x < 1 | x = 0 | (−1)(−3) = 3 | Positive |
| 1 < x < 3 | x = 2 | (1)(−1) = −1 | Negative |
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