📚 Solving Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 解二次方程：因式分解、配方法与求根公式
Quadratic equations appear throughout IGCSE Mathematics, from algebra problem solving to geometry and real-life modelling. This article explains the three main algebraic methods for solving quadratics, how the discriminant tells you about the roots, and how to avoid common errors in exams.
二次方程贯穿 IGCSE 数学，从代数问题求解到几何和实际生活建模中都会出现。本文将讲解求解二次方程的三种主要代数方法、判别式如何说明根的性质，以及如何避免考试中的常见错误。
1. What is a Quadratic Equation? | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2. Its general form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. If a were zero, the equation would become linear. The highest power of the variable x is 2.
二次方程是一个次数为 2 的多项式方程。它的一般形式是 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。如果 a 为零，方程就变成一次方程。变量 x 的最高次数是 2。
dratic equations. Some equations may need rearranging before they show this standard form.
例如，x² - 5x + 6 = 0 和 2x² + 3x - 2 = 0 都是二次方程。有些方程需要先整理才能呈现标准形式。
2. Standard Form and Key Terms | 标准形式和关键术语
m ax² is the quadratic term, bx is the linear term, and c is the constant term. The coefficient a must not be zero.
求解前，一定要先把方程写成 ax² + bx + c = 0 的形式。ax² 是二次项，bx 是一次项，c 是常数项。系数 a 不能为零。
The word 'root' means a solution of the equation. A quadratic equation can have two real roots, one repeated real root, or no real roots depending on the discriminant b² - 4ac.
"根"是指方程的解。根据判别式 b² - 4ac，二次方程可以有两个实根、一个重复实根或没有实根。
3. Solving by Factorising | 因式分解法求解
Factorising is often the fastest method when the quadratic expression factorises into two brackets. You look for two numbers that multiply to give ac and add to give b, then split the middle term, or you use direct inspection.
当二次式可以分解成两个括号时，因式分解通常是最快的方法。你要找到两个数，使它们的乘积等于 ac，和等于 b，然后拆分中项，或直接观察分解。
e product is +6 and sum is -5. The numbers are -2 and -3, so x² - 5x + 6 = (x - 2)(x - 3).
示例：解 x² - 5x + 6 = 0。我们需要两个数，乘积为 +6，和为 -5。这两个数是 -2 和 -3，因此 x² - 5x + 6 = (x - 2)(x - 3)。
Then set each bracket equal to zero: x - 2 = 0 gives x = 2, and x - 3 = 0 gives x = 3.
然后令每个括号等于零：x - 2 = 0 得 x = 2，x - 3 = 0 得 x = 3。
4. The Zero Product Property | 零乘积性质
e zero. This is why factorising works: if (x + p)(x + q) = 0, then x + p = 0 or x + q = 0.
零乘积性质指出，如果两个式子的乘积为零，那么至少有一个式子必须为零。这就是因式分解有效的原因：如果 (x + p)(x + q) = 0，那么 x + p = 0 或 x + q = 0。
pply this property. For example, (x - 2)(x - 3) = 5 does not imply x - 2 = 5 or x - 3 = 5.
重要提示：不能把每个因式等于一个非零数。必须先把方程化为等于零，再应用这一性质。例如，(x - 2)(x - 3) = 5 并不意味着 x - 2 = 5 或 x - 3 = 5。
5. Solving by Completing the Square | 配方法求解
This method is useful when the quadratic does not factorise easily, and it is essential for deriving the quadratic formula.
配方法把二次式转化为 a(x + p)² + q = 0 的形式。当二次式不容易因式分解时，这个方法很有用，而且它是推导求根公式的基础。
and subtracting (b/2)². That gives x² + bx = (x + b/2)² - (b/2)². Then add the constant c.
对于 x² + bx + c，你可以通过加上并减去 (b/2)² 来改写 x² + bx。这就得到 x² + bx = (x + b/2)² - (b/2)²。然后再加上常数 c。
+ 2 = 0. Half of 6 is 3, and 3² = 9. So x² + 6x + 2 = (x + 3)² - 9 + 2 = (x + 3)² - 7 = 0.
示例：解 x² + 6x + 2 = 0。6 的一半是 3，3² = 9。因此 x² + 6x + 2 = (x + 3)² - 9 + 2 = (x + 3)² - 7 = 0。
Then (x + 3)² = 7, so x + 3 = ±√7, giving x = -3 ± √7.
然后 (x + 3)² = 7，所以 x + 3 = ±√7，得到 x = -3 ± √7。
6. Deriving the Quadratic Formula | 推导求根公式
+ bx + c = 0, divide by a, then complete the square. The result is the quadratic formula:
从 ax² + bx + c = 0 开始，两边除以 a，然后配方。结果就是求根公式：
x = (-b ± √b² - 4ac) ÷ (2a)
This formula gives the solutions of any quadratic equation, provided you first identify a, b and c correctly from the standard form.
只要你先从标准形式中正确识别出 a、b、c，这个公式就能给出任何二次方程的解。
ans there are usually two solutions: one using the plus sign and one using the minus sign.
符号 ± 表示通常有两个解：一个用加号，一个用减号。
7. Using the Quadratic Formula | 使用求根公式
ample: Solve 2x² - 3x - 2 = 0. Here a = 2, b = -3 and c = -2. Substitute into the formula:
例：解 2x² - 3x - 2 = 0。这里 a = 2，b = -3，c = -2。代入公式：
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