📚 Algebraic Fractions and Quadratic Equations | 代数分式与二次方程
Algebraic fractions and quadratic equations form the backbone of IGCSE Mathematics. They appear in almost every exam paper, from simplifying expressions to solving complex simultaneous equations. Mastery of these topics not only secures top grades in algebra but also builds essential skills for coordinate geometry, calculus preparation, and problem-solving.
代数分式与二次方程构成了 IGCSE 数学的基石。几乎每份试卷都会出现它们的身影，从简化表达式到求解复杂的联立方程。掌握这些主题不仅能在代数部分获得高分，还能为坐标几何、微积分预备和问题解决能力奠定坚实基础。
1. Expanding Brackets: The Foundation | 展开括号：基础技能
Before tackling fractions and quadratics, you must be completely fluent in expanding brackets. The most common method is FOIL: First, Outer, Inner, Last. For example, (x + 5)(x + 2) expands to x² + 2x + 5x + 10, which simplifies to x² + 7x + 10. Speed and accuracy in expanding directly determine your ability to factorise later.
在处理分式和二次方程之前，你必须完全熟练地掌握去括号。最常用的方法是 FOIL 法则：首项、外项、内项、末项。例如，(x + 5)(x + 2) 展开为 x² + 2x + 5x + 10，化简后得到 x² + 7x + 10。展开的速度和准确性直接决定了你后续因式分解的能力。
Expanding perfect squares is a common exam favourite. You must remember that (a + b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b². A frequent mistake is writing (a + b)² = a² + b², which is completely incorrect. The middle term, 2ab, must never be forgotten.
完全平方展开是考试中的常客。你必须牢记 (a + b)² = a² + 2ab + b² 以及 (a – b)² = a² – 2ab + b²。一个常见错误是写成 (a + b)² = a² + b²，这是完全错误的。中间项 2ab 绝不能忘记。
The difference of two squares formula, (a – b)(a + b) = a² – b², is another powerful tool. It is particularly useful for quickly simplifying certain algebraic fractions, as we will see in Section 4.
平方差公式 (a – b)(a + b) = a² – b² 是另一个强大的工具。它在快速简化某些代数分式时特别有用，我们将在第 4 节中看到。
2. Factorising Quadratics (a = 1) | 因式分解二次式（首项系数为 1）
Factorising is the reverse of expanding. To factorise x² + bx + c, you must find two numbers that multiply to give c and add to give b. For example, factorise x² – 5x + 6. The numbers -2 and -3 multiply to give 6 and add to give -5. Therefore, the factorised form is (x – 2)(x – 3).
因式分解是展开的逆过程。要分解 x² + bx + c，需要找到两个数，它们相乘等于 c，相加等于 b。例如，分解 x² – 5x + 6。数字 -2 和 -3 相乘等于 6，相加等于 -5。因此，分解后的形式是 (x – 2)(x – 3)。
Always check if you can factorise out a common number first. Take 2x² + 6x + 4. Although the coefficient of x² is not 1, all terms share a factor of 2. This becomes 2(x² + 3x + 2), which simplifies to 2(x + 1)(x + 2). Ignoring common factors is a leading cause of lost marks.
始终检查是否能先行提取公因数。以 2x² + 6x + 4 为例。虽然 x² 的系数不是 1，但所有项都有公因数 2。这变为 2(x² + 3x + 2)，进一步简化为 2(x + 1)(x + 2)。忽略公因数是导致失分的主要原因之一。
| Type | Form | Example |
| General quadratic | x² + bx + c | x² + 7x + 12 = (x + 3)(x + 4) |
| Perfect square | (a ± b)² | x² – 6x + 9 = (x – 3)² |
| Difference of squares | a² – b² | x² – 25 = (x – 5)(x + 5) |
3. Factorising Quadratics (a ≠ 1) | 因式分解二次式（首项系数不为 1）
When the coefficient of x² is not 1, the process requires more care. For the general form ax² + bx + c, you must find two numbers that multiply to give a × c and add to give b. Consider 2x² + 7x + 3. Here, a × c = 6 and b = 7. The numbers 1 and 6 satisfy this. Rewrite the middle term: 2x² + x + 6x + 3. Group and factorise: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
当 x² 的系数不为 1 时，因式分解的过程需要更加小心。对于一般形式 ax² + bx + c，你需要找到两个数，它们相乘等于 a × c，相加等于 b。考虑 2x² + 7x + 3。这里，a × c = 6，b = 7。数字 1 和 6 满足条件。重写中间项：2x² + x + 6x + 3。分组并因式分解：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。
The “cross method” is a popular alternative. List the factors of a and c, then test combinations until the outer and inner products sum to b. This method requires practice but is faster once mastered. The “grouping” method shown above is more systematic and less prone to guess-and-check errors.
“十字相乘法”是另一种常用方法。列出 a 和 c 的因数，然后测试组合，直到外项积与内项积之和等于 b。这个方法需要练习，但一旦掌握就会更快。上面展示的“分组分解法”更系统化，且不易出现尝试性错误。
4. Simplifying Algebraic Fractions | 简化代数分式
Algebraic fractions follow the exact same rules as numerical fractions. The golden rule for simplification is to factorise the numerator and denominator completely, then cancel common factors. For example, simplify (x² – 9) / (x² – 3x). The numerator factorises to (x – 3)(x + 3). The denominator factorises to x(x – 3). Both the numerator and denominator share the factor (x – 3). Cancelling gives (x + 3) / x.
代数分式遵循与数值分数完全相同的法则。简化的黄金法则是将分子和分母完全因式分解，然后约去公因式。例如，化简 (x² – 9) / (x² – 3x)。分子分解为 (x – 3)(x + 3)。分母分解为 x(x – 3)。分子和分母都含有公因式 (x – 3)。约去后得到 (x + 3) / x。
It is crucial to remember that you can only cancel factors, never terms. For instance, in the fraction (x + 5) / (x + 2), you cannot cancel the x terms. The x is a term, not a factor, because the numerator is a sum, not a product. Cancelling terms is one of the most heavily penalised errors in IGCSE algebra.
必须记住，你只能约去因式，绝不能约去项。例如，在分式 (x + 5) / (x + 2) 中，你不能约去 x。这里的 x 是项而非因式，因为分子是和而非乘积。约项是 IGCSE 代数中最严重的失分错误之一。
Also, be mindful of the domain. When you cancel (x – 3), you are implicitly stating that x ≠ 3. In the original fraction, x cannot be 0 or 3 because these values would make the denominator zero. Including this restriction is often required in higher-level questions.
同时，要注意定义域。当你约去 (x – 3) 时，实际上隐含了 x ≠ 3 的条件。在原分式中，x 不能是 0 或 3，因为这些值会使分母为零。在更高水平的问题中，通常需要注明这一限制条件。
5. Multiplying and Dividing Algebraic Fractions | 代数分式的乘除
Multiplication of algebraic fractions is straightforward: multiply the numerators together and multiply the denominators together. It is often easier to factorise and cancel before multiplying to avoid large numbers. For example, (x² – 1) / (x + 2) × (x + 2) / (x – 1). First, factorise (x² – 1) into (x – 1)(x + 1). The (x + 2) terms cancel, and the (x – 1) terms cancel. The final answer is simply (x + 1).
代数分式的乘法很直接：分子相乘作为结果的分子，分母相乘作为结果的分母。通常为避免大数字计算，可以先因式分解并约分再相乘。例如，(x² – 1) / (x + 2) × (x + 2) / (x – 1)。首先，将 (x² – 1) 分解为 (x – 1)(x + 1)。(x + 2) 项相互约去，(x – 1) 项也相互约去。最终答案就是 (x + 1)。
Division follows the rule of “keep, change, flip”: keep the first fraction, change the division sign to multiplication, and flip the second fraction upside down. For instance, (x² + 5x + 6) / (x + 1) ÷ (x + 3) / (x² – 1). Flip the second fraction and multiply. Factorise everything: (x + 2)(x + 3) / (x + 1) × (x – 1)(x + 1) / (x + 3). Cancel the common factors (x + 3) and (x + 1). The simplified result is (x + 2)(x – 1).
除法遵循“保持、变号、翻转”法则：保持第一个分式不变，将除号变为乘号，并将第二个分式分子分母颠倒。例如，(x² + 5x + 6) / (x + 1) ÷ (x + 3) / (x² – 1)。翻转第二个分式并相乘。将一切因式分解：(x + 2)(x + 3) / (x + 1) × (x – 1)(x + 1) / (x + 3)。约去公因式 (x + 3) 和 (x + 1)。简化结果为 (x + 2)(x – 1)。
6. Adding and Subtracting Algebraic Fractions | 代数分式的加减
Adding and subtracting algebraic fractions requires a common denominator. The most efficient denominator is the Lowest Common Multiple (LCM) of the denominators. Take the expression 1 / x + 1 / (x + 2). The LCM of x and (x + 2) is x(x + 2). Rewrite each fraction with this denominator: (x + 2) / [x(x + 2)] + x / [x(x + 2)]. Add the numerators: (2x + 2) / [x(x + 2)].
代数分式的加减需要公分母。最有效的分母是各分母的最小公倍数（LCM）。以表达式 1 / x + 1 / (x + 2) 为例。x 与 (x + 2) 的 LCM 是 x(x + 2)。将每个分式重写为该分母：(x + 2) / [x(x + 2)] + x / [x(x + 2)]。将分子相加：(2x + 2) / [x(x + 2)]。
When subtracting, use brackets around the second numerator to avoid sign errors. For example, subtract 2 / (x – 1) from 3 / (x + 1). The LCM is (x + 1)(x – 1). The calculation is 3(x – 1) / [(x + 1)(x – 1)] – 2(x + 1) / [(x + 1)(x – 1)]. This gives [3(x – 1) – 2(x + 1)] / [(x + 1)(x – 1)]. Expanding and simplifying the numerator yields (3x – 3 – 2x – 2) = (x – 5). The final answer is (x – 5) / [(x + 1)(x – 1)].
当进行减法时，请在第二个分子周围使用括号，以避免符号错误。例如，从 3 / (x + 1) 中减去 2 / (x – 1)。LCM 为 (x + 1)(x – 1)。计算过程为 3(x – 1) / [(x + 1)(x – 1)] – 2(x + 1) / [(x + 1)(x – 1)]。结果为 [3(x – 1) – 2(x + 1)] / [(x + 1)(x – 1)]。展开并简化分子得到 (3x – 3 – 2x – 2) = (x – 5)。最终答案是 (x – 5) / [(x + 1)(x – 1)]。
7. Solving Quadratic Equations by Factorising | 因式分解法解二次方程
To solve a quadratic equation by factorising, the first step is always to rearrange the equation so that one side equals zero. Once you have the form ax² + bx + c = 0, factorise the quadratic expression. Then, use the Zero Product Property: if the product of two factors is zero, then one or both of the factors must be zero.
用因式分解法解二次方程，第一步总是将方程重新排列，使一边等于零。当得到 ax² + bx + c = 0 的形式后，对二次表达式进行因式分解。然后，运用零乘积性质：如果两个因式的乘积为零，则其中一个或两个因式必须为零。
Solve the equation 2x² – 5x – 3 = 0. Factorising (2x + 1)(x – 3) = 0. Therefore, either 2x + 1 = 0 or x – 3 = 0. Solving these linear equations gives x = -½ or x = 3. Always check your solutions by substituting them back into the original equation.
解方程 2x² – 5x – 3 = 0。因式分解得到 (2x + 1)(x – 3) = 0。因此，要么 2x + 1 = 0，要么 x – 3 = 0。解这些线性方程得到 x = -½ 或 x = 3。务必通过代回原方程来检验你的解。
Be aware of the common “divide by x” trap. If the equation is x² = 4x, students often divide both sides by x to get x = 4,
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