📚 IGCSE Maths: Solving Quadratic Equations | IGCSE 数学：二次方程求解
Quadratic equations are central to IGCSE Mathematics. They appear in algebra, coordinate geometry and real-world problems such as area, projectile motion and number patterns. This guide covers the three main solving methods - factorisation, completing the square and the quadratic formula - and shows you how the discriminant decides how many solutions exist and how an equation links to its graph. Every method is illustrated with worked examples written the way examiners expect to see them.
二次方程是 IGCSE 数学的核心内容。它们出现在代数、坐标几何和现实问题中，例如面积、抛体运动和数字规律。本指南涵盖三种主要求解方法——因式分解法、配方法和求根公式法，并说明判别式如何决定解的个数，以及方程如何与其图像相互联系。每个方法都配有按考官期望格式书写的例题。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation in which the highest power of the unknown is 2. Every quadratic equation can be rearranged into the standard form a x² + b x + c = 0, where a, b and c are constants and a ≠ 0. The unknown is usually written as x, but letters such as t, n or y are used when they stand for time, an integer or a coordinate.
二次方程是未知数最高次数为 2 的方程。每个二次方程都可以整理成标准形式 a x² + b x + c = 0，其中 a、b、c 为常数，且 a ≠ 0。未知数通常写作 x，但当它表示时间、整数或坐标时，也会使用 t、n、y 等字母。
a x² + b x + c = 0, a ≠ 0
equation becomes linear. The condition a ≠ 0 is therefore essential, because it is exactly what makes the equation quadratic rather than a straight-line equation with one solution.
如果 a = 0，x² 项就会消失，方程变成一次方程。因此条件 a ≠ 0 至关重要，因为正是它使方程成为二次方程，而不是只有一个解的一次方程。
two answers: a positive root and a negative root. Writing only x = 3 loses half of the solution.
例如，x² - 9 = 0 是二次方程，因为 x 的最高次数为 2。它可以快速求解：整理为 x² = 9，得 x = ±3。注意这里有两个答案：一个正根和一个负根。只写 x = 3 会丢掉一半的解。
In IGCSE, a quadratic equation has either two distinct real solutions, one repeated real solution, or no real solutions at all. Later sections show how to tell which case applies before you do any heavy algebra.
在 IGCSE 中，二次方程要么有两个不同的实数解，要么有一个重复的实数解，要么完全没有实数解。后文会说明如何在繁重计算之前就判断出属于哪种情况。
2. Standard Form and Coefficients | 标准形式与系数
ackets, collecting like terms, and moving every term to the left-hand side so that the right-hand side is exactly zero.
在选择方法之前，一定要先把方程化为标准形式 a x² + b x + c = 0。也就是说，展开所有括号、合并同类项，并把所有项移到等号左边，使等号右边正好为 0。
Once the equation is in standard form, identify the coefficients carefully, including their signs. The coefficient a is the number in front of x², b is the number in front of x, and c is the constant term. If a term is missing, its coefficient is 0, not 1 and not "nothing".
方程化为标准形式后，要仔细写出各系数，包括它们的符号。系数 a 是 x² 前面的数，b 是 x 前面的数，c 是常数项。如果某一项缺失，它的系数是 0，而不是 1，更不是"没有"。
| Given equation 给出的方程 | Standard form 标准形式 | a | b | c |
| 3x² = 12 | 3x² - 12 = 0 | 3 | 0 | -12 |
| x(x - 4) = 5 | x² - 4x - 5 = 0 | 1 | -4 | -5 |
| (x + 1)² = 9 | x² + 2x - 8 = 0 | 1 | 2 | -8 |
| 2x² + 3x = x² - 7 | x² + 3x + 7 = 0 | 1 | 3 | 7 |
= 0. When b = 0 the equation is a x² + c = 0, which is solved by rearranging and taking square roots, as in x² - 9 = 0.
有两种特殊情形值得记牢。当 c = 0 时，方程为 a x² + b x = 0，它总能分解为 x(a x + b) = 0，因此必定有一个根 x = 0。当 b = 0 时，方程为 a x² + c = 0，可通过移项后开平方求解，如 x² - 9 = 0。
irst gives 2x² + x - 5 = 0 with a = 2, b = 1, c = -5, which is the form you actually need.
一个非常常见的考试陷阱是像 5 - 2x² = x 这样的方程。有些学生过早地因式分解，结果符号出错。先整理得 2x² + x - 5 = 0，此时 a = 2、b = 1、c = -5，这才是真正需要的形式。
3. Method 1: Factorisation | 方法一：因式分解法
Factorisation is usually the fastest method and is always worth trying first. It relies on the zero-product property: if the product of two factors is zero, then at least one of the factors must be zero. So after factorising, you simply set each bracket equal to zero and solve two short linear equations.
因式分解通常是最快的方法，值得优先尝试。它依据的是零积性质：若两个因式的乘积为 0，则至少有一个因式为 0。因此分解之后，只需令每个括号等于 0，解两个简单的一次方程即可。
For a monic quadratic x² + b x + c, look for two numbers that multiply to give c and add to give b. If those two numbers are p and q, the factorised form is (x + p)(x + q).
对于二次项系数为 1 的方程 x² + b x + c，要寻找两个数：它们的乘积等于 c，和等于 b。若这两个数为 p 和 q，则分解形式为 (x + p)(x + q)。
stituting either value back into the original equation gives zero, which confirms both answers.
例：解 x² - 5x + 6 = 0。需要两个数，乘积为 6、和为 -5，这两个数是 -2 和 -3，所以 x² - 5x + 6 = (x - 2)(x - 3) = 0，得 x = 2 或 x = 3。把任一根代回原方程都得到 0，从而验证了两个答案。
Split the middle term, then factorise in pairs: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3), so x = -1/2 or x = -3.
当 a 不为 1 时，寻找过程稍长。对于 2x² + 7x + 3 = 0，先算 a 与 c 的乘积得 6，再找乘积为 6、和为 7 的两个数：6 和 1。拆开中间项后分组分解：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)，故 x = -1/2 或 x = -3。
Do not assume every quadratic factorises neatly. If no integer pair works, the roots are irrational or complex-looking surds, and you should switch to completing the square or the quadratic formula instead of wasting time guessing.
不要以为每个二次式都能整齐地分解。如果找不到整数对，根就是无理数（根式），此时应改用配方法或求根公式，而不是浪费时间继续猜。
4. Method 2: Completing the Square | 方法二：配方法
Completing the square rewrites a quadratic expression as a perfect square plus a constant. It works for every quadratic, gives exact surd answers, and reveals the turning point of the graph, which is why examiners love it.
配方法把二次式改写为"完全平方 + 常数"的形式。它对所有二次方程都适用，能给出精确的根式答案，还能直接看出图像的顶点，因此深受考官青睐。
x² + b x + c = (x + b/2)² + c - b²/4
ing the expression equal to zero then gives (x + b/2)² = b²/4 - c, and you finish by taking square roots of both sides.
规则很简单：取 x 的系数的一半放进括号内，再减去它的平方，使整个式子数值不变。令该式等于 0，即得 (x + b/2)² = b²/4 - c，最后对两边开平方即可完成求解。
, so write (x + 3)² - 9 + 2 = 0, that is (x + 3)² = 7. Taking square roots gives x + 3 = ±√7, so x = -3 + √7 or x = -3 - √7. In decimal form these are approximately -0.35 and -5.65.
例：解 x² + 6x + 2 = 0。6 的一半是 3，于是写成 (x + 3)² - 9 + 2 = 0，即 (x + 3)² = 7。开平方得 x + 3 = ±√7，故 x = -3 + √7 或 x = -3 - √7，小数形式约为 -0.35 和 -5.65。
leted-square form tells you immediately that the curve has a minimum value of -5 at x = 2.
当 a 不为 1 时，先把 a 从前两项中提出。对于 2x² - 8x + 3，写成 2(x² - 4x) + 3 = 2[(x - 2)² - 4] + 3 = 2(x
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