📚 Worked Example 1.6.1: Solving a Quadratic Inequality by Sketching | 例题 1.6.1：借助图像求解二次不等式
This worked example is taken from the AQA A-Level Pure Mathematics algebra strand. It demonstrates a reliable method for solving a quadratic inequality by factorising the quadratic, identifying critical values, and using a sketch of the curve to interpret where the function is negative, zero, or positive. Mastering this approach is essential because quadratic inequalities appear regularly in both pure and applied contexts at A-Level.
本题选自 AQA A-Level 纯数学代数部分。它演示了一种可靠的方法来求解二次不等式：先对二次式进行因式分解，确定临界值，再利用曲线图像判断函数在何处为负、为零或为正。掌握这种方法至关重要，因为二次不等式在 A-Level 的纯数学和应用数学中经常出现。
1. The Problem | 题目
s is a quadratic inequality because the highest power of x is 2 and the expression is compared to zero using an inequality sign. Unlike a linear inequality, the solution is not simply a single interval obtained by isolating x, because the graph of a quadratic has a turning point and can cross the x-axis twice.
求解不等式 x² − 6x + 5 ≤ 0。这是一个二次不等式，因为 x 的最高次数为 2，并且该表达式通过不等号与零进行比较。与线性不等式不同，它的解并不是通过简单地移项得到的一个区间，因为二次函数的图像具有一个转折点，并且可能与 x 轴相交两次。
We need a method that considers the shape of the parabola and the sign of the quadratic expression over each portion of the number line defined by the roots.
我们需要一种方法，结合抛物线的形状以及由根所划分的实数轴各区间上二次式的符号来求解。
2. Factorising the Quadratic | 因式分解二次式
We first factorise the quadratic expression x² − 6x + 5. To do this, we need two numbers whose product is +5 and whose sum is −6. The numbers −1 and −5 satisfy both conditions, so we can write:
我们首先对二次式 x² − 6x + 5 进行因式分解。为此，我们需要找到两个数，使它们的乘积为 +5，和为 −6。数 −1 和 −5 满足这两个条件，因此可以写成：
x² − 6x + 5 = (x − 1)(x − 5)
ly visible. It is often quicker to factorise than to use the quadratic formula, but the quadratic formula is an acceptable alternative if you do not immediately see the factors.
这种因式分解形式让方程 x² − 6x + 5 = 0 的根一目了然。通常情况下，因式分解比使用求根公式更快，但如果你不能立即看出因式，使用求根公式也是可以接受的一种替代方法。
ly: (x − 1)(x − 5) = x² − 5x − x + 5 = x² − 6x + 5, which matches the original expression.
在因式分解后，一定要通过展开括号来检查：(x − 1)(x − 5) = x² − 5x − x + 5 = x² − 6x + 5，这与原式一致。
3. Locating Critical Values | 确定临界值
tical values because the sign of the quadratic expression can only change at these points.
临界值就是使二次式等于零的 x 值。令每个因式等于零，得到 x − 1 = 0，即 x = 1；以及 x − 5 = 0，即 x = 5。我们称它们为临界值，因为二次式的符号只能在这些点处发生改变。
These two values divide the real number line into three intervals: x < 1, 1 < x < 5, and x > 5. On each interval, the sign of (x − 1)(x − 5) remains constant, so we can test one value from each interval to determine that sign.
这两个值将实数轴分成三个区间：x < 1、1 < x < 5 和 x > 5。在每个区间上，(x − 1)(x − 5) 的符号保持不变，因此我们可以从每个区间中选取一个值进行
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