📚 Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
stants and a ≠ 0. Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebraic manipulation to graphs, geometry and real-world modelling. Mastering three core methods - factorisation, completing the square and the quadratic formula - gives you the flexibility to solve any quadratic confidently in examinations.
二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。二次方程贯穿 IGCSE 数学大纲，从代数变形到图像、几何和实际建模都会出现。掌握因式分解、配方法和求根公式这三种核心方法，你就能在考试中灵活、自信地求解任何二次方程。
1. Standard Form and Key Terms | 标准形式与关键术语
condition a ≠ 0 is essential; if a = 0, the equation becomes linear rather than quadratic.
在 IGCSE 数学中，二次方程通常需要先整理成标准形式 ax² + bx + c = 0。ax² 是二次项，bx 是一次项，c 是常数项。a ≠ 0 这一条件非常重要；如果 a = 0，方程就变成一次方程而不是二次方程。
wever, 2x² = 5x - 3 must be rearranged to 2x² - 5x + 3 = 0 before you attempt to solve it.
例如，3x² - 7x + 2 = 0 已经是标准形式，其中 a = 3，b = -7，c = 2。但 2x² = 5x - 3 需要先整理为 2x² - 5x + 3 = 0，然后才能求解。
When rearranging, always move all terms to one side so that the other side equals zero. Keep the sign of each term correct, especially when moving terms that are subtracted.
整理方程时，务必把所有项移到一边，使另一边等于零。尤其要注意移项时各项的符号是否正确，尤其是被减去的项。
2. The Zero Product Property | 零乘积性质
q = 0. This simple rule is the logical foundation of solving quadratics by factorisation.
零乘积性质指出：如果两个表达式的乘积为零，那么其中至少一个表达式必须为零。用符号表示：若 pq = 0，则 p = 0 或 q = 0。这条简单规则是因式分解法解二次方程的逻辑基础。
et equal to zero separately, giving x - 2 = 0 or x + 3 = 0, and therefore x = 2 or x = -3.
当二次方程写成 (x - 2)(x + 3) = 0 时，根据零乘积性质，我们可以令每个括号分别等于零，得到 x - 2 = 0 或 x + 3 = 0，因此 x = 2 或 x = -3。
It is important to remember that this property only works when the product equals zero. If the product equals a non-zero number, you cannot set each factor equal to that number; you must first bring the equation to standard form with zero on one side.
重要的是要记住，这个性质只在乘积等于零时成立。如果乘积等于一个非零数，你不能让每个因式都等于这个数；必须先把方程整理成一边为零的标准形式。
3. Solving by Factorisation | 因式分解法
two numbers that multiply to give ac and add to give b. Use these numbers to split the middle term, factorise by grouping, and finally apply the zero product property.
用因式分解法解二次方程时，首先将方程整理成标准形式 ax² + bx + c = 0。然后找到两个数，使它们相乘等于 ac，相加等于 b。用这两个数拆分一次项，再分组分解，最后应用零乘积性质。
and add to -5: -2 and -3. Therefore, x² - 5x + 6 = (x - 2)(x - 3) = 0, so x = 2 or x = 3.
例如，解 x² - 5x + 6 = 0。我们需要两个数，相乘等于 6，相加等于 -5：-2 和 -3。因此，x² - 5x + 6 = (x - 2)(x - 3) = 0，所以 x = 2 或 x = 3。
hen factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0, so x = -1/2 or x = -3.
当 a ≠ 1 时，先找公因式或使用 ac 法。解 2x² + 7x + 3 = 0：ac = 6，b = 7，所以两个数是 6 和 1。拆分一次项：2x² + 6x + x + 3 = 0，然后分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0，所以 x = -1/2 或 x = -3。
4. Completing the Square | 配方法
Completing the square transforms a quadratic expression into the form a(x + p)² + q. This is particularly useful when a quadratic does not factorise easily, and it also reveals the vertex of the corresponding parabola.
配方法将二次表达式转化为 a(x + p)² + q 的形式。当二次方程不容易因式分解时，这个方法尤其有用，而且它还能揭示对应抛物线的顶点。
tract 9: x² + 6x + 9 - 9 + 1 = 0, so (x + 3)² - 8 = 0. Then (x + 3)² = 8, giving x + 3 = ±√8, so x = -3 ± 2√2.
对于 x² + 6x + 1 = 0，取 x 系数的一半 3，平方得 9。加上并减去 9：x² + 6x + 9 - 9 + 1 = 0，所以 (x + 3)² - 8 = 0。然后 (x + 3)² = 8，得到 x + 3 = ±√8，所以 x = -3 ± 2√2。
x² + bx = (x + b/2)² - (b/2)²
对于一般形式，x² + bx = (x + b/2)² - (b/2)²。这是配方法的核心恒等式。
de the bracket: 2[(x + 2)² - 4] + 5 = 0, then 2(x + 2)² - 8 + 5 = 0, giving 2(x + 2)² = 3.
当 x² 的系数不是 1 时，先把它从前两项中提出来。对于 2x² + 8x + 5 = 0，写成 2(x² + 4x) + 5 = 0。在括号内完成配方：2[(x + 2)² - 4] + 5 = 0，然后 2(x + 2)² - 8 + 5 = 0，得到 2(x + 2)² = 3。
5. The Quadratic Formula | 求根公式
hen factorisation is difficult or impossible. It is derived from completing the square and should be memorised accurately.
求根公式可以直接给出 ax² + bx + c = 0 的解，即使因式分解很难或不可能。它由配方法推导而来，需要准确记忆。
x = (-b ± √b² - 4ac) / 2a
求根公式为：x = (-b ± √b² - 4ac) / 2a。
3x - 5 = 0, we have a = 2, b = -3 and c = -5. Substituting into the formula gives x = (3 ± √9 + 40) / 4 = (3 ± √49) / 4, so x = (3 + 7)/4 = 2.5 or x = (3 - 7)/4 = -1.
对于方程 2x² - 3x - 5 = 0，a = 2，b = -3，c = -5。代入公式得 x = (3 ± √9 + 40) / 4 = (3 ± √49) / 4，所以 x = (3 + 7)/4 = 2.5 或 x = (3 - 7)/4 = -1。
When the roots are not whole numbers, use a calculator to evaluate the square root and round to the required number of decimal places or significant figures. Always write down the substitution step to earn method marks in the exam.
当根不是整数时，用计算器计算平方根，并按题目要求四舍五入到指定的小数位数或有效数字。考试中务必写出代入步骤，以获得方法分。
6. The Discriminant | 判别式
tells you how many real solutions a quadratic equation has without solving it completely.
判别式是求根公式中平方根号下的部分：D = b² - 4ac。它可以在不解出全部解的情况下，告诉你二次方程有多少个实数解。
| Discriminant | 判别式 | Number of real roots | 实数根个数 | Graph behaviour | 图像特征 |
|---|---|---|
| D > 0 | Two distinct real roots | Cuts x-axis twice |
