📚 Mastering Quadratic Equations for IGCSE Mathematics | 掌握 IGCSE 数学二次方程
Quadratic equations are a central topic in IGCSE Mathematics, appearing in algebra, graphs, problem-solving and even in applied questions. This revision guide builds the essential skills from factorising to the quadratic formula, with clear bilingual explanations and exam-style tips.
二次方程是 IGCSE 数学的核心主题，出现在代数、图像、应用题甚至实际问题中。本复习指南从因式分解到求根公式，提供清晰的双语讲解和应试技巧。
1. Quadratic Expressions and Equations: Definitions | 二次表达式与方程：定义
A quadratic expression has the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. If the expression is set equal to zero, ax² + bx + c = 0, it becomes a quadratic equation.
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。如果将该表达式设为零，即 ax² + bx + c = 0，它就成为二次方程。
The highest power of the variable x is 2, which is why the equation is called ‘quadratic’. Examples include x² – 5x + 6 = 0 and 2x² + 3x – 4 = 0.
变量 x 的最高次数为 2，因此该方程称为“二次”。例子包括 x² – 5x + 6 = 0 和 2x² + 3x – 4 = 0。
In IGCSE examinations, you are usually expected to give solutions in exact form, such as fractions or surds, unless the question asks for decimal answers. A quadratic equation can have two real solutions, one repeated solution, or no real solutions depending on the discriminant, which is covered later in this guide.
在 IGCSE 考试中，通常要求给出精确形式的解，如分数或根式，除非题目要求写出小数答案。根据判别式的不同，二次方程可以有两个实数解、一个重根或没有实数解，本指南稍后会讨论判别式。
2. Expanding and Factorising Quadratic Expressions | 展开与因式分解二次表达式
Expanding two binomials, for example (x + p)(x + q), produces a quadratic expression: x² + (p + q)x + pq. This pattern is the key to factorising many quadratics.
展开两个二项式，例如 (x + p)(x + q)，会得到二次表达式：x² + (p + q)x + pq。这一规律是许多二次式因式分解的关键。
To factorise x² + bx + c, find two numbers that multiply to c and add to b. For x² + 5x + 6, the numbers 2 and 3 give factors (x + 2)(x + 3).
要因式分解 x² + bx + c，需要找到两个数，它们的积为 c、和为 b。对于 x² + 5x + 6，数 2 和 3 给出因式 (x + 2)(x + 3)。
When a ≠ 1, such as in 2x² + 7x + 3, use inspection or the ‘ac method’: multiply a and c to get 6, find 6 and 1 whose sum is 7, then split the middle term and factorise by grouping.
当 a ≠ 1 时，例如 2x² + 7x + 3，可以使用观察法或 ‘ac 法’：先将 a 与 c 相乘得 6，找到 6 和 1 其和为 7，然后拆中项并按组分解。
- Example: 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
- 示例：2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
You should also recognise special forms: a difference of squares a² – b² = (a – b)(a + b), and perfect square trinomials such as x² + 10x + 25 = (x + 5)². These appear frequently in IGCSE papers.
你还应该识别特殊形式：平方差 a² – b² = (a – b)(a + b)，以及完全平方三项式，如 x² + 10x + 25 = (x + 5)²。这些在 IGCSE 试卷中经常出现。
3. Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程
If a product equals zero, then at least one factor must equal zero. Therefore, after factorising ax² + bx + c = 0 into (px + q)(rx + s) = 0, set each bracket equal to zero and solve.
如果乘积等于零，那么至少有一个因式为零。因此，将 ax² + bx + c = 0 因式分解为 (px + q)(rx + s) = 0 后，令每个括号等于零并求解。
For x² – 7x + 12 = 0, factorise to (x – 3)(x – 4) = 0. The solutions are x = 3 and x = 4.
对于 x² – 7x + 12 = 0，因式分解为 (x – 3)(x – 4) = 0。解为 x = 3 和 x = 4。
Always rearrange the equation to have zero on one side before factorising. For example, x² – 3x = 10 becomes x² – 3x – 10 = 0, then (x – 5)(x + 2) = 0, giving x = 5 or x = -2.
在因式分解之前，一定要先把方程整理为一边为零。例如，x² – 3x = 10 变为 x² – 3x – 10 = 0，然后 (x – 5)(x + 2) = 0，得到 x = 5 或 x = -2。
Factorisation is usually the fastest method when the quadratic has integer roots. If you cannot find integer factors quickly, move to completing the square or the quadratic formula.
当二次方程有整数根时，因式分解通常是最快的方法。如果无法快速找到整数因式，则改用配方法或求根公式。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q, which makes the turning point of the graph visible and allows solving when factorising is difficult.
配方法将二次式改写为 a(x + p)² + q 的形式，使图像的顶点可见，并且在难以因式分解时可用于求解。
For x² + 6x + 2, take half of 6 to get 3, square it to get 9, and write: x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7.
对于 x² + 6x + 2，取 6 的一半得 3，平方得 9，写成：x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7。
To solve x² + 6x + 2 = 0, use (x + 3)² – 7 = 0, so (x + 3)² = 7, giving x + 3 = ±√7, hence x = -3 ± √7.
要解 x² + 6x + 2 = 0，用 (x + 3)² – 7 = 0，所以 (x + 3)² = 7，得到 x + 3 = ±√7，因此 x = -3 ± √7。
x² + bx = (x + b/2)² – (b/2)²
If the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 5, write 2[x² + 4x] + 5 = 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3.
如果 x² 的系数不是 1，先将其提出。对于 2x² + 8x + 5，写成 2[x² + 4x] + 5 = 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3。
5. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation ax² + bx + c = 0:
求根公式可解任意二次方程 ax² + bx + c = 0：
x = (-b ± √(b² – 4ac)) ÷ (2a)
For 2x² – 4x – 3 = 0, substitute a = 2, b = -4, c = -3 into the formula: x = (4 ± √(16 + 24)) ÷ 4 = (4 ± √40) ÷ 4 = (2 ± √10) ÷ 2.
对于 2x² – 4x – 3 = 0，将 a = 2、b = -4、c = -3 代入公式：x = (4 ± √(16 + 24)) ÷ 4 = (4 ± √40) ÷ 4 = (2 ± √10) ÷ 2。
Use the formula when the quadratic does not factorise easily, or when the coefficients are not integers. Quote the formula accurately and show substitution steps to earn method marks.
当二次式不易因式分解或系数不是整数时，使用求根公式。准确写出公式并展示代入步骤，以获得方法分。
Be careful with signs: if b is negative, -b becomes positive; if c is negative, -4ac becomes positive. Many marks are lost through sign errors in the square root part.
注意符号：如果 b 为负，则 -b 为正；如果 c 为负，则 -4ac 变为正。许多分数因平方根部分的符号错误而丢失。
6. Discriminant and Nature of Roots | 判别式与根的性质
The discriminant is the expression inside the square root of the quadratic formula: Δ = b² – 4ac. It tells you how many real solutions the quadratic equation has.
判别式是求根公式平方根内的表达式：Δ = b² – 4ac。它可以告诉你二次方程有多少个实数解。
- If b² – 4ac > 0, there are two distinct real roots.
- 如果 b² – 4ac > 0，则有两个不同的实数根。
- If b² – 4ac = 0, there is one repeated real root.
- 如果 b² – 4ac = 0，则有一个重根。
- If b² – 4ac < 0, there are no real roots.
- 如果 b² – 4ac < 0，则没有
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