📚 IGCSE Maths: Quadratic Equations, Inequalities & Algebraic Fractions | IGCSE 数学：二次方程、不等式与代数分式
Quadratic equations and algebraic fractions are central to the IGCSE Mathematics syllabus. They appear in pure algebra questions, in geometry problems, and in everyday applications such as area or speed calculations. This article covers factorising, completing the square, the quadratic formula, inequalities, and algebraic fraction simplification.
二次方程与代数分式是 IGCSE 数学课程中的核心内容。它们不仅出现在纯代数题中，还会出现在几何问题和面积、速度等实际应用中。本文涵盖因式分解、配方法、求根公式、二次不等式以及代数分式化简。
1. Recognising Quadratic Expressions | 识别二次表达式
A quadratic expression in one variable has the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. Common examples include x² + 5x + 6, 2x² - 3x + 1 and -x² + 4.
一元二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。常见例子包括 x² + 5x + 6、2x² - 3x + 1 和 -x² + 4。
Recognising the standard form is the first step because all solution methods rely on identifying a, b and c correctly. If an equation is not in standard form, collect all terms on one side first.
识别标准形式是第一步，因为所有解法都依赖于正确判断 a、b 和 c。如果方程不是标准形式，应先将所有项移到一边。
2. Factorising When a = 1 | 首项系数为 1 时的因式分解
c and whose sum is b. If these two numbers are p and q, then x² + bx + c = (x + p)(x + q).
对于形如 x² + bx + c 的二次式，我们需要找到两个数，它们的乘积为 c，和为 b。如果这两个数为 p 和 q，则 x² + bx + c = (x + p)(x + q)。
ers are 2 and 5 because 2 × 5 = 10 and 2 + 5 = 7. Therefore x² + 7x + 10 = (x + 2)(x + 5).
例如：分解 x² + 7x + 10。两个数为 2 和 5，因为 2 × 5 = 10 且 2 + 5 = 7。因此 x² + 7x + 10 = (x + 2)(x + 5)。
- Always check the signs: if c is positive, the two numbers have the same sign; if c is negative, they have opposite signs.
- 务必检查符号：若 c 为正，两数同号；若 c 为负，两数异号。
3. Factorising When a ≠ 1 | 首项系数不为 1 时的因式分解
When the coefficient of x² is not 1, we can use the ac method. Multiply a by c, then find two numbers whose product is ac and whose sum is b. Split the middle term and factor by grouping.
当 x² 的系数不为 1 时，可以使用 ac 法。先将 a 与 c 相乘，再找到两个数，使其乘积为 ac、和为 b。然后拆分中间项并分组分解。
ac = 6. The two numbers are 1 and 6 because 1 × 6 = 6 and 1 + 6 = 7. Split 7x into x + 6x:
例如：分解 2x² + 7x + 3。这里 a = 2，b = 7，c = 3，因此 ac = 6。两个数为 1 和 6，因为 1 × 6 = 6 且 1 + 6 = 7。将 7x 拆分为 x + 6x：
2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)
Factorising by grouping works for any quadratic with rational roots, but you must be careful when negative signs are involved.
分组分解适用于任何具有有理根的二次式，但涉及负号时必须特别小心。
4. Solving Quadratic Equations by Factorising | 通过因式分解解二次方程
zero product property: if AB = 0, then A = 0 or B = 0. This gives the solutions directly.
当二次方程写成乘积等于零的形式后，我们使用零积性质：若 AB = 0，则 A = 0 或 B = 0。这样可以直接得到解。
Example: Solve x² - 5x + 6 = 0. Factorise to get (x - 2)(x - 3) = 0, so x = 2 or x = 3.
例如：解方程 x² - 5x + 6 = 0。因式分解得 (x - 2)(x - 3) = 0，所以 x = 2 或 x = 3。
e of squares is a special case: x² - 9 = 0 becomes (x - 3)(x + 3) = 0, so x = 3 or x = -3.
平方差是一种特殊情况：x² - 9 = 0 可化为 (x - 3)(x + 3) = 0，因此 x = 3 或 x = -3。
5. Completing the Square | 配方法
in the form (x + p)² + q. For x² + bx, we use the identity x² + bx = (x + b/2)² - (b/2)².
配方法将二次式改写为 (x + p)² + q 的形式。对于 x² + bx，我们使用恒等式 x² + bx = (x + b/2)² - (b/2)²。
² + 6x as (x + 3)² - 9, so the equation becomes (x + 3)² - 9 + 5 = 0, giving (x + 3)² = 4.
例如：解方程 x² + 6x + 5 = 0。将 x² + 6x 写成 (x + 3)² - 9，方程变为 (x + 3)² - 9 + 5 = 0，即 (x + 3)² = 4。
x + 3 = ±2, so x = -1 or x = -5
This method is especially useful when the quadratic does not factorise neatly and when deriving the quadratic formula.
当二次式不能直接因式分解时，以及推导求根公式时，配方法尤其有用。
6. The Quadratic Formula | 求根公式
ovides the solutions of ax² + bx + c = 0 for any real coefficients a, b and c, with a ≠ 0:
求根公式给出了 ax² + bx + c = 0 在任意实系数 a、b、c 且 a ≠ 0 时的解：
x = (-b ± √b² - 4ac) / 2a
xample: Solve 3x² + 5x - 2 = 0. Here a = 3, b = 5 and c = -2. Substitute into the formula:
例如：解方程 3x² + 5x - 2 = 0。这里 a = 3，b = 5，c = -2。代入公式：
x = (-5 ± √5² - 4 × 3 × -2) / (2 × 3) = (-5 ± √49) / 6
Therefore x = (-5 + 7)/6 = 1/3 or x = (-5 - 7)/6 = -2.
因此 x = (-5 + 7)/6 = 1/3 或 x = (-5 - 7)/6 = -2。
In the IGCSE examination, the quadratic formula is given in the formula sheet, but you must know how to substitute values accurately.
在 IGCSE 考试中，求根公式通常出现在公式表中，但你必须能够准确代入数值。
7. The Discriminant | 判别式
written as D = b² - 4ac. It tells us the nature of the roots without solving the equation.
平方根内的表达式 b² - 4ac 称为判别式，通常写作 D = b² - 4ac。它可以让我们在不求解方程的情况下判断根的性质。
| Discriminant value | Nature of roots |
| D > 0 | Two distinct real roots |
| D = 0 | One repeated real root |
| D < 0 | No real roots |
x² - 4x + 5 = 0, D = (-4)² - 4 × 1 × 5 = 16 - 20 = -4, so the equation has no real roots.
例如：对于 x² - 4x + 5 = 0，D = (-4)² - 4 × 1 × 5 = 16 - 20 = -4，因此该方程没有实数根。
8. Quadratic Inequalities | 二次不等式
nding equation x² - 3x - 10 = 0. Factorising gives (x - 5)(x + 2) = 0, so x = 5 or x = -2.
要解二次不等式，例如 x² - 3x - 10 > 0，首先通过解对应方程 x² - 3x - 10 = 0 找到临界值。因式分解得 (x - 5)(x + 2) = 0，因此 x = 5 或 x = -2。
These critical values divide the number line into three intervals: x < -2, -2 < x < 5, and x > 5. Testing a value from each interval shows that the product is positive for x < -2 or x > 5.
这些临界值将数轴分为三个区间：x < -2、-2 < x < 5 和 x > 5。在每个区间中取一个检验值可知，当 x < -2 或 x > 5 时乘积为正。
Solution: x < -2 or x > 5
the inequality had been ≤ 0 instead, the solution would be the closed interval -2 ≤ x ≤ 5.
如果不等式是 ≤ 0，那么解就是闭区间 -2 ≤ x ≤ 5。
9. Algebraic Fractions with Quadratics | 含二次式的代数分式
Algebraic fractions often require factorising quadratic numerators or denominators before simplifying. For example, simplify (x² - 1) / (x² + 2x + 1).
代数分式通常需要先对分子或分母中的二次式进行因式分解，然后才能化简。例如，化简 (x² - 1) / (x² + 2x + 1)。
(x² - 1) / (x² + 2x + 1) = (x - 1)(x + 1) / (x + 1)² = (x - 1) / (x + 1)
This simplification is valid only when x ≠ -1, because the original denominator would be zero at that value.
这个化简仅在 x ≠ -1 时成立，因为当 x = -1 时原分式的分母为零。
When solving equations involving algebraic fractions, multiply through by the common denominator to obtain a quadratic equation, then solve using any suitable method.
在解含有代数分式的方程时，先乘以公分母得到二次方程，再使用合适的方法求解。
10. Exam-Style Problem Solving | 考试型问题解决
Quadratic equations often appear in word problems. Example: The product of two consecutive positive integers is 72. Find the integers.
二次方程经常出现在文字题中。例如：两个连续正整数的乘积为 72。求这两个整数。
0, so x = -9 or x = 8. Since the integer is positive, x = 8 and the integers are 8 and 9.
设较小的整数为 x。则 x(x + 1) = 72，即 x² + x - 72 = 0。因式分解得 (x + 9)(x - 8) = 0，所以 x = -9 或 x = 8。由于整数为正，因此 x = 8，两个整数为 8 和 9。
In geometry problems, the area of a rectangle may lead to a quadratic equation. Always define the variable clearly, form the equation, solve it, and interpret the answer in context.
在几何题中，矩形的面积也可能引出二次方程。务必清晰地设出变量、建立方程、求解，并结合题意解释答案。
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