📚 Quadratic Equations: From Factorising to the Quadratic Formula | 二次方程：从因式分解到求根公式
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from simple factorising questions to problems involving the discriminant and real-life modelling. This revision guide walks through the standard form, three main solution methods, the discriminant, graphical links, common mistakes, and exam-style applications.
二次方程贯穿 IGCSE 数学课程，从简单的因式分解题到涉及判别式和实际建模的问题。本复习指南将带你梳理标准形式、三种主要解法、判别式、图像联系、常见错误以及考试风格的应用题。
1. What is a Quadratic Equation? | 什么是二次方程？
est power of the unknown x is 2, which is why the graph of y = ax² + bx + c is a parabola.
二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。未知数 x 的最高次数为 2，因此 y = ax² + bx + c 的图像是一条抛物线。
so the condition a ≠ 0 is essential. Quadratic equations can have two real solutions, one repeated real solution, or no real solutions depending on the value of the discriminant.
若 a = 0，方程就变成一次方程，因此 a ≠ 0 的条件至关重要。二次方程可能有两个实数解、一个重复实数解或没有实数解，具体取决于判别式的值。
2. Standard Form and Coefficients | 标准形式与系数
refully, including their signs. For example, in 3x² - 5x + 2 = 0, a = 3, b = -5 and c = 2.
在应用任何方法之前，先把方程整理为标准形式 ax² + bx + c = 0。仔细识别系数 a、b、c 及其符号。例如，在 3x² - 5x + 2 = 0 中，a = 3, b = -5, c = 2。
after multiplying by -1. Changing all signs is optional but keeps a positive if preferred.
如果方程给出的是 2x - x² = 4，把所有项移到一边得到 -x² + 2x - 4 = 0，或乘以 -1 得到 x² - 2x + 4 = 0。改变所有符号是可选的，但若希望 a 为正可以这样做。
ax² + bx + c = 0, a ≠ 0
3. Solving by Factorisation | 因式分解法
Factorisation is usually the fastest method when the quadratic can be expressed as a product of two linear factors. We look for two numbers that multiply to give ac and add to give b, then split the middle term or use direct factorising.
当二次方程可以表示为两个一次因式的乘积时，因式分解通常是最快的方法。我们要找两个数，它们相乘等于 ac、相加等于 b，然后拆分中间项或直接因式分解。
5) = 0, giving x = -2 or x = -5. Always set each factor equal to zero to obtain the roots.
对于 x² + 7x + 10 = 0，两个数是 2 和 5，因为 2 + 5 = 7 且 2 × 5 = 10。因此 (x + 2)(x + 5) = 0，得到 x = -2 或 x = -5。始终令每个因式等于零来求根。
the middle term is split as 6x - x, then factor by grouping to obtain (2x - 1)(x + 3) = 0.
当 a ≠ 1 时，例如 2x² + 5x - 3 = 0，需要两个数乘积为 2 × (-3) = -6、和为 5。这两个数是 6 和 -1，因此中间项拆为 6x - x，然后分组分解得到 (2x - 1)(x + 3) = 0。
If (px + q)(rx + s) = 0, then x = -q/p or x = -s/r.
4. Solving by Completing the Square | 配方法
ch can then be solved by taking square roots. This method is especially useful for deriving the quadratic formula and for finding the vertex of a parabola.
配方法将二次方程转化为 (x + p)² = q 的形式，然后通过开平方求解。这种方法在推导求根公式以及求抛物线顶点时特别有用。
d add to both sides: x² + 6x + 9 = 7, so (x + 3)² = 7. Taking square roots gives x + 3 = ±√7, hence x = -3 ± √7.
对于 x² + 6x + 2 = 0，先改写为 x² + 6x = -2。取 6 的一半，平方得 9，两边同时加上：x² + 6x + 9 = 7，所以 (x + 3)² = 7。开平方得 x + 3 = ±√7，因此 x = -3 ± √7。
8x + 3 = 0, divide by 2 to get x² + 4x + 3/2 = 0, then proceed with completing the square.
如果 a ≠ 1，先将整个方程除以 a，使 x² 的系数为 1。对于 2x² + 8x + 3 = 0，除以 2 得到 x² + 4x + 3/2 = 0，然后继续配方。
x² + bx = (x + b/2)² - (b/2)²
5. The Quadratic Formula | 求根公式
standard form. It is derived by completing the square on ax² + bx + c = 0 and is given by:
求根公式为任何标准形式的二次方程提供了直接解法。它由对 ax² + bx + c = 0 配方推导而来，公式如下：
x = (-b ± √b² - 4ac) / 2a
stitute carefully. For 3x² - 4x - 2 = 0, a = 3, b = -4, c = -2. The formula gives x = (4 ± √16 + 24) / 6 = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3.
使用公式时，先识别 a、b、c，然后仔细代入。对于 3x² - 4x - 2 = 0，a = 3, b = -4, c = -2。公式给出 x = (4 ± √16 + 24) / 6 = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3。
The quadratic formula is particularly helpful when factorisation is difficult or when coefficients are non-integer. Make sure to simplify surds fully unless the question asks for decimal answers.
当因式分解
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