📚 IGCSE Maths: Solving Quadratic Equations | IGCSE 数学：二次方程求解
A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in algebra, graphs, problem-solving and even in real-world modelling. This revision guide explains how to solve quadratic equations using factorising, completing the square and the quadratic formula, with clear examples and exam tips.
二次方程是 IGCSE 数学中最重要的主题之一。它出现在代数、函数图像、应用题甚至现实建模中。本复习指南将介绍如何用因式分解法、配方法和求根公式解二次方程，并配有清晰的例题和考试技巧。
1. What is a Quadratic Equation? | 什么是二次方程？
re a, b and c are constants and a ≠ 0. The highest power of the unknown x is 2, which is why it is called 'quadratic'.
二次方程是一个次数为 2 的多项式方程。它可以写成一般形式 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。未知数 x 的最高次数是 2，因此称为 'quadratic'。
+ c = 0. Therefore, the condition a ≠ 0 is essential for the equation to remain quadratic.
如果 a = 0，方程就变成一次方程 bx + c = 0。因此，a ≠ 0 是保证方程仍为二次方程的关键条件。
2. The General Form ax² + bx + c = 0 | 一般形式 ax² + bx + c = 0
c is the constant term. For example, in 3x² - 5x + 2 = 0, we have a = 3, b = -5 and c = 2.
二次方程的标准形式为 ax² + bx + c = 0。其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。例如在 3x² - 5x + 2 = 0 中，a = 3，b = -5，c = 2。
Sometimes an equation is not given in this order. Before solving, you should rearrange the terms so that one side of the equation equals zero.
有时方程并不按这个顺序给出。求解前应先把各项移到一边，使另一边等于零。
3. Solving by Factorising | 因式分解法
If a quadratic expression can be written as the product of two linear factors, we can solve the equation by setting each factor equal to zero. This method is fast when the factors are simple integers.
如果一个二次式可以写成两个一次因式的乘积，我们就可以通过令每个因式等于零来解方程。当因式为简单整数时，这种方法非常快捷。
x² + 5x + 6 = (x + 2)(x + 3), so the equation x² + 5x + 6 = 0 becomes (x + 2)(x + 3) = 0.
例如，x² + 5x + 6 = (x + 2)(x + 3)，因此方程 x² + 5x + 6 = 0 变为 (x + 2)(x + 3) = 0。
4. The Zero Product Property | 零乘积性质
o, then at least one of the numbers must be zero. That is, if AB = 0, then A = 0 or B = 0.
零乘积性质指出，如果两个数的乘积为零，那么至少有一个数为零。也就是说，如果 AB = 0，则 A = 0 或 B = 0。
Applying this to (x + 2)(x + 3) = 0 gives x + 2 = 0 or x + 3 = 0, so x = -2 or x = -3.
将此性质应用于 (x + 2)(x + 3) = 0，得到 x + 2 = 0 或 x + 3 = 0，所以 x = -2 或 x = -3。
(x + 2)(x + 3) = 0 ⇒ x = -2 or x = -3
5. Factorising When a = 1 | 首项系数为 1 的因式分解
When the coefficient of x² is 1, we look for two numbers that multiply to give c and add to give b. These two numbers become the constants in the factors.
当 x² 的系数为 1 时，我们需要寻找两个数，它们的乘积等于 c，和等于 b。这两个数就成为因式中的常数。
e numbers 2 and 5 work because 2 × 5 = 10 and 2 + 5 = 7. So x² + 7x + 10 = (x + 2)(x + 5).
对于 x² + 7x + 10 = 0，我们需要两个数，乘积为 10，和为 7。2 和 5 符合条件，因为 2 × 5 = 10 且 2 + 5 = 7。所以 x² + 7x + 10 = (x + 2)(x + 5)。
If c is negative, the two numbers have opposite signs. If b is positive, the number with the larger absolute value is positive; if b is negative, the number with the larger absolute value is negative.
如果 c 为负，两个数异号。如果 b 为正，绝对值较大的那个数为正；如果 b 为负，绝对值较大的那个数为负。
6. Factorising When a ≠ 1 | 首项系数不为 1 的因式分解
When a ≠ 1, factorising is slightly harder. You can use the 'ac method': multiply a and c, find two numbers with that product and sum b, then split the middle term and factor by grouping.
当 a ≠ 1 时，因式分解稍难一些。可以使用 'ac 法'：先求 a 与 c 的乘积，找出两个数使其乘积等于 ac、和等于 b，然后拆中项并分组分解。
ith product 6 and sum 7: 6 and 1. Split 7x as 6x + x: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x +
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