📚 Mastering Quadratic Inequalities | 掌握二次不等式
Quadratic inequalities are a fundamental topic in AQA A-Level Mathematics, often appearing in both pure mathematics and applied contexts. They test your ability to manipulate expressions, interpret graphs, and solve problems systematically.
二次不等式是 AQA A-Level 数学的基础内容，常出现在纯数学和应用题中。它考察你化简表达式、解读图像以及系统化解题的能力。
1. What is a Quadratic Inequality? | 什么是二次不等式？
e a ≠ 0. Unlike a quadratic equation, the solution is usually an interval or a union of intervals, not a single value.
二次不等式是形如 ax² + bx + c < 0、ax² + bx + c > 0、ax² + bx + c ≤ 0 或 ax² + bx + c ≥ 0 的表达式，其中 a ≠ 0。与二次方程不同，它的解通常是一个区间或几个区间的并集，而不是单个值。
For example, the inequality x² − 4 < 0 has the solution −2 < x < 2, meaning all values of x between −2 and 2.
例如，不等式 x² − 4 < 0 的解为 −2 < x < 2，即 −2 和 2 之间的所有 x 值。
2. Solving by Factorisation | 因式分解法
When the quadratic expression factorises neatly, we can solve the inequality by finding the roots and then testing intervals. The standard approach is to rearrange the inequality to have zero on one side, factorise, and then use a sign diagram or test points.
当二次表达式可以轻松因式分解时，我们可以先求出根，再通过区间测试来解不等式。标准做法是将不等式整理为一边为零，分解因式，然后使用符号图或测试点。
x² − x − 6 > 0 → (x − 3)(x + 2) > 0
tic opens upwards, so the expression is positive outside the interval between the roots. Thus the solution is x < −2 or x > 3.
根为 x = −2 和 x = 3。该二次函数开口向上，因此在两根之外的区间上表达式为正。所以解为 x < −2 或 x > 3。
3. Solving by Completing the Square | 配方法
If the quadratic does not factorise easily, completing the square is a reliable algebraic method. It also gives a direct idea of the range of the quadratic function.
如果二次式不易因式分解，配方法是一种可靠的代数方法。它还能直接给出二次函数的取值范围。
Take x² + 4x + 1 ≥ 0. Complete the square:
以 x² + 4x + 1 ≥ 0 为例。配方：
(x + 2)² − 3 ≥ 0 → (x + 2)² ≥ 3
Taking square roots gives |x + 2| ≥ √3, so x ≤ −2 − √3 or x ≥ −2 + √3. This method is especially useful when roots are surds.
两边开平方得 |x + 2| ≥ √3，因此 x ≤ −2 − √3 或 x ≥ −2 + √3。这种方法在根为无理数时尤其有用。
4. Graphical Interpretation | 图像解释
. Solving an inequality like ax² + bx + c > 0 means finding the x-values for which the curve lies above the x-axis. This visual interpretation helps avoid sign errors.
y = ax² + bx + c 的图像是抛物线。解形如 ax² + bx + c > 0 的不等式，就是求曲线位于 x 轴上方的 x 取值范围。这种直观理解有助于避免符号错误。
- If a > 0, the parabola is U-shaped, and the region above the x-axis is outside the roots.
- 如果 a > 0，抛物线为 U 形，位于 x 轴上方的区域在两根之外。
- If a < 0, the parabola is n-shaped, and the region above the x-axis is between the roots.
- 如果 a < 0，抛物线为倒 U 形，位于 x 轴上方的区域在两根之间。
Always sketch a quick graph or write down the sign of the quadratic in each interval.
务必快速画出草图，或写出二次式在每个区间的符号。
5. The Role of the Discriminant | 判别式的作用
determines whether the quadratic crosses the x-axis. This affects the inequality solution:
判别式 Δ = b² − 4ac 决定二次函数是否与 x 轴相交。这会影响不等式的解：
| Δ | Roots | Example: ax² + bx + c > 0 (a > 0) |
| Δ > 0 | Two distinct real roots | x < α or x > β |
| Δ = 0 | One repeated root | All x except the root |
| Δ < 0 | No real roots | All real x (if a > 0) |
For Δ < 0 and a > 0, the quadratic is always positive, so the inequality holds for all real x. For a < 0, it is always negative, so the inequality never holds.
当 Δ < 0 且 a > 0 时，二次式恒为正，因此不等式对所有实数 x 成立。若 a < 0，则恒为负，不等式无解。
6. Rearranging to Standard Form | 整理为标准形式
Many inequalities do not initially appear as quadratics. They may involve fractions, exponentials, or trigonometric functions that reduce to quadratics after substitution or manipulation.
许多不等式最初并非二次形式。它们可能包含分式、指数函数或三角函数，经过代换或变形后可化为二次式。
For example, solve 2/x < x + 1 for x ≠ 0. Multiply by x² (positive) to avoid sign reversal:
例如，解 2/x < x + 1（x ≠ 0）。两边乘以 x²（正数）以避免变号：
2x < x³ + x² → x³ + x² − 2x > 0 → x(x + 2)(x − 1) > 0
The critical points are −2, 0, and 1. Using a sign chart, the solution is −2 < x < 0 or x > 1. Remember to exclude any values that make the original denominator zero.
临界点为 −2、0 和 1。利用符号表，解为 −2 < x < 0 或 x > 1。注意排除使原分母为零的值。
7. Inequalities in the Form of a Quotient | 商形式的不等式
When an inequality involves a rational function, it is often necessary to find the critical values where the numerator or denominator is zero. Then test sign changes across those intervals.
当不等式涉及有理函数时，通常需要找到分子或分母为零的临界值。然后测试这些区间上的符号变化。
intervals gives −3 < x ≤ 2. The value x = −3 is excluded because the denominator is zero.
考虑 (x − 2)/(x + 3) ≤ 0。临界值为 x = 2 和 x = −3。测试区间得 −3 < x ≤ 2。由于分母为零，x = −3 被排除。
It is useful to write the quotient as a single algebraic fraction and keep the denominator positive by multiplying by the square of the denominator.
一个技巧是先将分式化为单个代数分式，并通过乘以分母的平方来保证分母为正。
8. Solving (a): A Classic Example | 第 (a) 部分：经典例题
n asks you to solve a quadratic inequality directly. For instance, solve 2x² − 7x + 3 ≤ 0.
在考试题中，第 (a) 部分常要求直接解一个二次不等式。例如，解 2x² − 7x + 3 ≤ 0。
wards, so the expression is negative between the roots. Hence the solution is 1/2 ≤ x ≤ 3.
首先分解因式：(2x − 1)(x − 3) ≤ 0。根为 x = 1/2 和 x = 3。因为 a = 2 > 0，抛物线开口向上，所以表达式在两根之间为负。因此解为 1/2 ≤ x ≤ 3。
Include the endpoints when the inequality symbol is ≤ or ≥.
当不等号包含等号时，应包含端点值。
9. Solving (b): Application of the Solution | 第 (b) 部分：解的应用
function such as f(x) = 2x² − 7x + 3 for a given domain, or to solve a related inequality.
第 (b) 部分通常扩展前面的结果。它可能要求你求函数 f(x) = 2x² − 7x + 3 在给定定义域上的值域，或解一个相关的不等式。
f x for which the function is below or equal to the x-axis. You might also need to find where f(x) crosses the x-axis, which gives the roots.
例如，在解出 2x² − 7x + 3 ≤ 0 后，可能要求你求出函数位于 x 轴下方或与 x 轴相交的 x 值。你也可能需要求 f(x) 与 x 轴的交点，即根。
= k does not intersect the curve y = 2x² − 7x + 3. This becomes a discriminant inequality.
另一种常见扩展是解带参数的不等式，例如求 k 的取值范围，使得直线 y = k 与曲线 y = 2x² − 7x + 3 不相交。这转化为判别式不等式。
10. Common Mistakes and How to Avoid Them | 常见错误与避免方法
- Multiplying both sides by a variable expression without considering its sign. Always multiply by a positive quantity, such as the square.
- 两边乘以含变量的表达式而不考虑其符号。务必乘以正数，例如平方项。
- Forgetting to flip the inequality sign when multiplying or dividing by a negative number.
- 乘以或除以负数时忘记改变不等号方向。
- Writing the solution as a single compound inequality when the intervals are disjoint, e.g., x < 1 or x > 3 is not 1 > x > 3.
- 当区间不相连时，写成单一复合不等式，例如 x < 1 或 x > 3 不能写成 1 > x > 3。
- gnoring the roots when the inequality is strict (< or >) versus non-strict (≤ or ≥).
- 忽视严格不等式（< 或 >）与非严格不等式（≤ 或 ≥）在根值处的区别。
Always check your interval by substituting a test value into the original inequality.
务必通过代入测试值到原不等式来检验你的区间。
11. Practice Questions | 练习题
1. Solve x² − 5x − 14 > 0.
1. 解 x² − 5x − 14 > 0。
2. Solve −x² + 6x − 8 ≥ 0.
2. 解 −x² + 6x − 8 ≥ 0。
3. Find the range of values of m for which the quadratic mx² + 3x + m > 0 for all real x.
3. 求 m 的取值范围，使得二次式 mx² + 3x + m > 0 对所有实数 x 恒成立。
or x > 7. 2. 2 ≤ x ≤ 4. 3. For m > 0 and discriminant < 0, giving 9 − 4m² < 0, so m > 3/2.
答案：1. x < −2 或 x > 7。2. 2 ≤ x ≤ 4。3. 要求 m > 0 且判别式 < 0，即 9 − 4m² < 0，所以 m > 3/2。
12. Summary | 总结
Quadratic inequalities are solved by factorising, completing the square, or using the quadratic formula, then interpreting the graph's position relative to the x-axis. Always consider the sign of a and the discriminant to determine the structure of the solution.
解二次不等式可以通过因式分解、配方或求根公式，然后结合图像与 x 轴的相对位置来解释。务必考虑 a 的符号和判别式，以确定解的结构。
In AQA A-Level exams, part (a) typically asks for a straightforward inequality solution, while part (b) applies it to a curve, a function range, or a parameter problem. Practice is essential to gain speed and accuracy.
在 AQA A-Level 考试中，第 (a) 部分通常直接要求解不等式，第 (b) 部分则将其应用于曲线、函数值域或参数问题。多做练习是提高速度和准确性的关键。
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