📚 Mastering Quadratics and Inequalities for Cambridge IGCSE Mathematics | 剑桥 IGCSE 数学：掌握二次函数与不等式
Quadratic expressions and inequalities appear in almost every Cambridge IGCSE Mathematics paper, especially in the Extended syllabus. Being fluent in factorising, completing the square, using the quadratic formula, and reading quadratic graphs is essential for top marks.
二次表达式与二次不等式几乎出现在每一份剑桥 IGCSE 数学试卷中，尤其是 Extended 课程。熟练因式分解、配方法、使用二次公式以及阅读二次函数图像，是取得高分的必备能力。
1. Understanding Quadratic Expressions | 认识二次表达式
A quadratic expression has the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. The highest power of x is 2, which is why it is called ‘quadratic’. Common examples include x² + 5x + 6, 2x² − 7x + 3 and −x² + 4x + 1.
二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 是常数且 a ≠ 0。x 的最高次数为 2，因此称为 ‘二次’。常见例子包括 x² + 5x + 6、2x² − 7x + 3 和 −x² + 4x + 1。
In the Cambridge specification, you must be able to recognise the coefficients a, b and c. These values are used directly in the quadratic formula and the discriminant, so misreading them is a frequent source of error.
在剑桥考试大纲中，你必须能够识别系数 a、b 和 c。这些值会直接代入二次公式和判别式，因此读错它们是常见的错误来源。
For example, in 3x² − 2x + 5, the coefficient a = 3, b = −2 and c = 5. Always include the sign before each term when identifying b and c.
例如，在 3x² − 2x + 5 中，系数 a = 3、b = −2、c = 5。识别 b 和 c 时一定要包含每项前面的符号。
2. Factorising Quadratics | 因式分解二次式
Factorising is often the fastest way to solve a quadratic equation when the roots are simple. For a monic quadratic such as x² + 7x + 10, look for two numbers that add to 7 and multiply to 10. These are 2 and 5, so x² + 7x + 10 = (x + 2)(x + 5).
当根较为简单时，因式分解通常是解二次方程最快的方法。对于首一二次式如 x² + 7x + 10，找出两个数，使它们相加得 7、相乘得 10。这两个数是 2 和 5，因此 x² + 7x + 10 = (x + 2)(x + 5)。
When a ≠ 1, use a trial method or split the middle term. For 2x² − 5x − 3, multiply a and c: 2 × (−3) = −6. Find two numbers that add to −5 and multiply to −6: −6 and 1. Rewrite: 2x² − 6x + x − 3, then group: 2x(x − 3) + 1(x − 3) = (2x + 1)(x − 3).
当 a ≠ 1 时，可使用尝试法或裂项法。对于 2x² − 5x − 3，先计算 a × c：2 × (−3) = −6。找出两个数，使它们相加得 −5、相乘得 −6：−6 和 1。重写为 2x² − 6x + x − 3，然后分组：2x(x − 3) + 1(x − 3) = (2x + 1)(x − 3)。
Factorising also helps with quadratic inequalities. The expression x² − 4x + 3 becomes (x − 1)(x − 3), which tells you the critical values are x = 1 and x = 3.
因式分解也有助于解二次不等式。表达式 x² − 4x + 3 可化为 (x − 1)(x − 3)，由此可知临界值为 x = 1 和 x = 3。
3. Completing the Square | 配方法
Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This is useful for finding the vertex of a quadratic graph and for solving equations when factorising is difficult.
配方法可将 ax² + bx + c 写成 a(x + p)² + q 的形式。这在求二次函数图像的顶点以及因式分解困难时解方程非常有用。
For x² + 6x + 5, take half of the coefficient of x, which is 6 ÷ 2 = 3, and write (x + 3)². Expand (x + 3)² = x² + 6x + 9, so subtract 9 and add 5: x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4.
对于 x² + 6x + 5，取 x 系数的一半，即 6 ÷ 2 = 3，写成 (x + 3)²。展开 (x + 3)² = x² + 6x + 9，因此减去 9 再加上 5：x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4。
If the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 1, write 2(x² + 4x) + 1. Complete the square inside: x² + 4x = (x + 2)² − 4, so 2[(x + 2)² − 4] + 1 = 2(x + 2)² − 8 + 1 = 2(x + 2)² − 7.
如果 x² 的系数不是 1，先将其提出来。对于 2x² + 8x + 1，写成 2(x² + 4x) + 1。在括号内配方：x² + 4x = (x + 2)² − 4，因此 2[(x + 2)² − 4] + 1 = 2(x + 2)² − 8 + 1 = 2(x + 2)² − 7。
This form immediately shows the vertex. For y = (x + 3)² − 4, the vertex is (−3, −4). Remember the sign changes for the x-coordinate.
这个形式可以直接读出顶点。对于 y = (x + 3)² − 4，顶点为 (−3, −4)。注意 x 坐标的符号要变号。
4. The Quadratic Formula | 二次公式
The quadratic formula solves any equation ax² + bx + c = 0. It is especially useful when the quadratic does not factorise neatly. The formula is:
二次公式可以解任何形如 ax² + bx + c = 0 的方程。当二次式不能直接因式分解时，这个公式尤其有用。公式为：
x = (−b ± √(b² − 4ac)) / (2a)
For x² + 4x − 5 = 0, identify a = 1, b = 4, c = −5. Substitute into the formula: x = (−4 ± √(4² − 4 × 1 × (−5))) / (2 × 1) = (−4 ± √(16 + 20)) / 2 = (−4 ± √36) / 2 = (−4 ± 6) / 2. This gives x = (−4 + 6) / 2 = 1 or x = (−4 − 6) / 2 = −5.
对于 x² + 4x − 5 = 0，确定 a = 1、b = 4、c = −5。代入公式：x = (−4 ± √(4² − 4 × 1 × (−5))) / (2 × 1) = (−4 ± √(16 + 20)) / 2 = (−4 ± √36) / 2 = (−4 ± 6) / 2。由此得 x = (−4 + 6) / 2 = 1 或 x = (−4 − 6) / 2 = −5。
When the discriminant b² − 4ac is negative, the equation has no real roots. For x² + 2x + 5 = 0, b² − 4ac = 4 − 20 = −16, so there are no real solutions; the graph does not cross the x-axis.
当判别式 b² − 4ac 为负时，方程无实数根。例如 x² + 2x + 5 = 0，b² − 4ac = 4 − 20 = −16，所以无实数解；图像不与 x 轴相交。
5. Discriminant and Nature of Roots | 判别式与根的性质
The discriminant Δ = b² − 4ac tells you how many real roots a quadratic equation has. This is a very common exam topic and can be tested without asking you to actually solve the equation.
判别式 Δ = b² − 4ac 可以告诉你二次方程有多少个实数根。这是非常常见的考点，有时不需要真正解方程也会考查。
If Δ > 0, there are two distinct real roots. If Δ = 0, there is one repeated real root, meaning the graph touches the x-axis at one point. If Δ < 0, there are no real roots, so the graph never crosses the x-axis.
如果 Δ > 0，有两个不相等的实数根。如果 Δ = 0，有一个重复实根，即图像与 x 轴只有一个交点。如果 Δ < 0，没有实数根，图像不与 x 轴相交。
For example, for x² − 4x + 4 = 0, Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0, so there is one repeated root. Factorising confirms this: (x − 2)² = 0, giving x = 2 only.
例如，对于 x² − 4x + 4 = 0，Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0，所以有一个重复实根。因式分解也可以验证：(x − 2)² = 0，仅得 x = 2。
Exam questions often ask: ‘Find the value of k for which the equation has equal roots.’ This means you must set Δ = 0 and solve for k.
考试题经常问：’求 k 的值，使方程有两个相等的实根。’ 这意味着你需要令 Δ = 0，然后解出 k。
6. Graphs of Quadratic Functions | 二次函数的图像
The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有一个最小值点。如果 a < 0，开口向下，有一个最大值点。
The vertex can be found by completing the square or by using the symmetry line x = −b / (2a). Once you know the x-coordinate of the vertex, substitute it back into the equation to find the y-coordinate.
顶点可以通过配方法或使用对称轴 x = −b / (2a) 来求。知道顶点的 x 坐标后，将其代回方程即可求出 y 坐标。
For y = 2x² − 8x + 3, a = 2 and b = −8, so the vertex x-coordinate is −(−8) / (2 × 2) = 8 / 4 = 2. Substitute x = 2: y = 2(2)² − 8(2) + 3 = 8 − 16 + 3 = −5. The vertex is (2, −5).
对于 y = 2x² − 8x + 3，a = 2、b = −8，所以顶点的 x 坐标为 −(−8) / (2 × 2) = 8 / 4 = 2。代入 x = 2：y = 2(2)² − 8(2) + 3 = 8 − 16 + 3 = −5。顶点为 (2, −5)。
The roots are where y = 0. The graph crosses the x-axis at the solutions of ax² + bx + c = 0. The y-intercept is always the constant c, because when x = 0, y = c.
根是 y = 0 时的位置。图像在 ax² + bx + c = 0 的解处与 x 轴相交。y 轴截距始终是常数 c，因为当 x = 0 时，y = c。
7. Solving Quadratic Inequalities | 解二次不等式
To solve a quadratic inequality such as x² − 4x + 3 < 0, first factorise: (x − 1)(x − 3) < 0. The critical values are x = 1 and x = 3, where the expression equals zero.
要解二次不等式如 x² − 4x + 3 < 0，首先因式分解：(x − 1)(x − 3) < 0。临界值为 x = 1 和 x = 3，此时表达式等于零。
Draw a quick sketch or use a sign table. For a parabola opening upwards, the expression is below zero between the roots. Therefore the solution is 1 < x < 3.
画一个简图或使用符号表。对于开口向上的抛物线，表达式在两根之间小于零。因此解为 1 < x < 3。
If the inequality is x² − 4x + 3 ≥ 0, the solution is the opposite region: x ≤ 1 or x ≥ 3. Always check whether the inequality includes equality, because ≤ and ≥ require closed endpoints.
如果不等式为 x² − 4x + 3 ≥ 0，解则为相反的区域：x ≤ 1 或 x ≥ 3。一定要检查不等式是否包含等号，因为 ≤ 和 ≥ 需要使用闭区间端点。
When the coefficient of x² is negative, the parabola opens downwards, so you must reverse the interval logic. For −x² + 2x + 3 > 0, multiply through by −1 to get x² − 2x − 3 < 0, remembering to flip the inequality sign.
当 x² 的系数为负时，抛物线开口向下，因此区间逻辑要反过来。对于 −x² + 2x + 3 > 0，两边乘以 −1 得到 x² − 2x − 3 < 0，注意要翻转不等号方向。
8. Applications to Word Problems | 应用题
Quadratic word problems often involve area, projectile motion, or profit maximisation. The key steps are: define the variable, write an equation, solve it, and reject meaningless solutions such as negative lengths.
二次应用题通常涉及面积、抛体运动或利润最大化。关键步骤是：设出变量、列出方程、解方程，并舍去无意义的解，如负长度。
Example: A rectangle has length 3 cm longer than its width. Its area is 40 cm². Let width be x, so length is x + 3. The area equation is x(x + 3) = 40, which gives x² + 3x − 40 = 0. Factorise: (x + 8)(x − 5) = 0. The solutions are x = −8 or x = 5. Since width cannot be negative, width = 5 cm and length = 8 cm.
例题：一个矩形的长比宽多 3 cm，面积为 40 cm²。设宽为 x，则长为 x + 3。面积方程为 x(x + 3) = 40，即 x² + 3x − 40 = 0。因式分解：(x + 8)(x − 5) = 0。解为 x = −8 或 x = 5。由于宽度不能为负，宽 = 5 cm，长 = 8 cm。
Projectile problems often give height h(t) = −5t² + 20t + 1. To find when the object hits the ground, set h(t) = 0 and solve. To find maximum height, complete the square or use t = −b / (2a) and substitute back.
抛体问题常给出高度 h(t) = −5t² + 20t + 1。要求物体何时落地，令 h(t) = 0 并解方程。要求最大高度，可配方或使用 t = −b / (2a) 代回求值。
9. Common Mistakes and Exam Tips | 常见错误与应试技巧
One common mistake is forgetting to divide by 2a correctly in the quadratic formula. Another is ignoring the sign of b, especially when b is negative. Always write the formula first and substitute carefully with brackets around negative numbers.
一个常见错误是在二次公式中没有正确除以 2a。另一个常见错误是忽略 b 的符号，尤其是当 b 为负时。一定要先写出公式，并在代入负数时使用括号。
When solving inequalities, students often write 1 < x > 3, which is incorrect. The correct notation separates the two regions or uses a single interval: x < 1 or x > 3.
解不等式时，学生常写成 1 < x > 3，这是不正确的。正确的表示方法要将两个区域分开，或使用单个区间：x < 1 或 x > 3。
In completing the square, remember to adjust the constant term after squaring. For x² − 8x + 10, half of −8 is −4, and (−4)² = 16, so write (x − 4)² − 16 + 10 = (x − 4)² − 6, not (x − 4)² + 10.
在配方时，记得平方后要调整常数项。对于 x² − 8x + 10，−8 的一半是 −4，(−4)² = 16，因此写成 (x − 4)² − 16 + 10 = (x − 4)² − 6，而不是 (x − 4)² + 10。
10. Practice Questions and Summary | 练习题与总结
The best way to master quadratics is to practise a variety of styles. Try these quick questions: (1) Solve x² − 7x + 12 = 0. (2) Solve 3x² − 5x − 2 = 0. (3) Find the vertex of y = −x² + 4x − 3. (4) Solve x² − 2x − 3 > 0.
掌握二次函数的最佳方法是练习多种题型。试试这些快速练习：(1) 解 x² − 7x + 12 = 0。(2) 解 3x² − 5x − 2 = 0。(3) 求 y = −x² + 4x − 3 的顶点。(4) 解 x² − 2x − 3 > 0。
- x² − 7x + 12 = (x − 3)(x − 4), so x = 3 or x = 4.
- 3x² − 5x − 2 = (3x + 1)(x − 2), so x = −1/3 or x = 2.
- y = −(x − 2)² + 1, so the vertex is (2, 1).
- x² − 2x − 3 = (x − 3)(x + 1) > 0, so x < −1 or x > 3.
In each topic, focus on linking the algebraic method to the graph. This will help you check your answers and avoid sign errors in the exam.
在每个主题中，都要把代数方法与图像联系起来。这将帮助你检查答案，避免考试中的符号错误。
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