📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in algebra, graph sketching, problem solving, and many real-life applications. This article explains how to recognise quadratic equations, solve them by factorising, completing the square and using the quadratic formula, and how to interpret their roots, graphs and the discriminant.
二次方程是 IGCSE 数学中最重要的主题之一。它出现在代数、图像绘制、问题解决以及许多实际应用中。本文将解释如何识别二次方程，如何用因式分解法、配方法和求根公式求解，以及如何理解方程的根、图像与判别式。
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 written in the form ax² + bx + c = 0, where x is the unknown and a, b and c are constants with a ≠ 0.
二次方程是未知数的最高次数为 2 的方程。它通常可以写成 ax² + bx + c = 0 的形式，其中 x 是未知数，a、b、c 是常数，且 a ≠ 0。
ressions such as x³ + 2x - 1 or 5x - 3 are not quadratic, because in the first the highest power is 3, and in the second the equation is linear.
例如，x² - 5x + 6 = 0 和 2x² + 3x - 1 = 0 都是二次方程。像 x³ + 2x - 1 或 5x - 3 这样的式子不是二次方程，因为前者的最高次数不是 2，后者是一次方程。
The word "quadratic" comes from the Latin word for square, which reminds us that the squared term is the key feature. If a were 0, the x² term would disappear and the equation would become linear, so the condition a ≠ 0 is essential.
"二次"一词来自拉丁语中"平方"的意思，提醒我们平方项才是关键特征。如果 a 等于 0，x² 项就消失了，方程会变成一次方程，因此 a ≠ 0 这个条件非常必要。
2. Standard Form and Coefficients | 标准形式与系数
number a is the coefficient of x², b is the coefficient of x, and c is the constant term.
二次方程的标准形式是 ax² + bx + c = 0。其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。
belongs to the coefficient, so if the equation is x² + 4x - 7 = 0, then b = 4 and c = -7.
对于 3x² - 5x + 2 = 0，我们可以确定 a = 3，b = -5，c = 2。注意符号属于系数，因此如果方程是 x² + 4x - 7 = 0，那么 b = 4，c = -7。
ifying a, b and c. If the equation is given as 2x² = 3x + 5, first write 2x² - 3x - 5 = 0.
在识别 a、b、c 之前，一定要把方程先整理成标准形式。如果方程给出的是 2x² = 3x + 5，那么先写成 2x² - 3x - 5 = 0。
| Equation | 方程 | a | b | c |
|---|---|---|---|
| 3x² - 5x + 2 = 0 | 3 | -5 | 2 |
| x² + 4x - 7 = 0 | 1 | 4 | -7 |
| 2x² - 3x - 5 = 0 | 2 | -3 | -5 |
3. Solving by Factorising | 因式分解法
Factorising is often the fastest method when the quadratic can be written as the product of two linear brackets. The key is to find two numbers that multiply to give ac and add to give b, or directly to fit the factor form (px + q)(rx + s).
当二次式可以写成两个一次因式的乘积时，因式分解通常是最快的方法。关键是找到两个数，使它们的乘积等于 ac，和等于 b；或者直接写成 (px + q)(rx + s) 的因式形式。
ct is 12 and whose sum is -7. The numbers are -3 and -4, so x² - 7x + 12 = (x - 3)(x - 4).
对于 x² - 7x + 12 = 0，我们需要两个数，乘积为 12，和为 -7。这两个数是 -3 和 -4，所以 x² - 7x + 12 = (x - 3)(x - 4)。
bracket equal to zero gives x - 3 = 0 or x - 4 = 0, so the solutions are x = 3 and x = 4.
令每个括号等于零，得到 x - 3 = 0 或 x - 4 = 0，因此解为 x = 3 和 x = 4。
This works because of the zero product rule: if the product of two numbers is zero, at least one of them must be zero. Always expand your brackets back mentally to check the middle term and the sign of the constant term.
这样做依据的是零积法则：如果两个数的乘积为零，那么其中至少有一个数为零。做题后一定要在脑中把括号展开验算一次，检查一次项和常数项的符号。
ightly longer. For example 2x² - 3x - 5 = (2x - 5)(x + 1), which gives x = 2.5 and x = -1.
当 a 不等于 1 时，尝试的过程会更长一些。例如 2x² - 3x - 5 = (2x - 5)(x + 1)，由此得到 x = 2.5 和 x = -1。
4. Special Factorising Patterns | 特殊因式分解模式
Three patterns appear again and again in exams and are worth memorising, because they save a great deal of time.
有三种模式在考试中反复出现，非常值得记住，因为它们能节省大量时间。
Difference of two squares: x² - 9 = (x - 3)(x + 3), and more generally a²x² - b² = (ax - b)(ax + b).
平方差：x² - 9 = (x - 3)(x + 3)，更一般地 a²x² - b² = (ax - b)(ax + b)。
Perfect square: = (x + 3)², and x² - 10x + 25 = (x - 5)². The constant term is always half of b, squared.
完全平方：x² + 6x + 9 = (x + 3)²，x² - 10x + 25 = (x - 5)²。常数项总是 b 的一半的平方。
Common factor: 2x(x - 4) = 0, so x = 0 or x = 4. Never divide both sides by x, or the root x = 0 is lost.
公因式：2x² - 8x = 0 可写成 2x(x - 4) = 0，因此 x = 0 或 x = 4。千万不要两边同时除以 x，否则会丢失 x = 0 这个根。
0) or a missing linear term (b = 0) is a strong hint that one of these shortcuts applies.
缺少常数项（c = 0）或缺少一次项（b = 0）是使用上述快捷方法的重要提示。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form (x + p)² + q. This form shows the vertex of the parabola immediately and allows exact solutions to be found even when the expression does not factorise.
配方法把二次式改写为 (x + p)² + q 的形式。这种形式能直接看出抛物线的顶点，也能在表达式无法因式分解时求出精确解。
To complete the square for x² + bx + c, take half of b, square it, and add and subtract it:
对 x² + bx + c 配方时，取 b 的一半，平方后加上再减去：
x² + bx + c = (x + ½b)² + c - (½b)²
1 = (x + 3)² - 9 + 1 = (x + 3)² - 8, so the minimum point of y = x² + 6x + 1 is (-3, -8).
例如，x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8，因此 y = x² + 6x + 1 的最低点是 (-3, -8)。
To solve x² + 6x - 2 = 0 by completing the square, write (x + 3)² = 11, so x = -3 ± √11. These are exact surd answers; the decimal values are 0.317 and -6.317 to 3 s.f.
用配方法解 x² + 6x - 2 = 0，先写成 (x + 3)² = 11，所以 x = -3 ± √11。这是精确的根式答案；保留三位有效数字则为 0.317 和 -6.317。
ut as a factor first: 2x² + 8x + 3 = 2(x² + 4x) + 3 = 2[(x + 2)² - 4] + 3 = 2(x + 2)² - 5.
如果 a ≠ 1，先把 a 提取出来：2x² + 8x + 3 = 2(x² + 4x) + 3 = 2[(x + 2)² - 4] + 3 = 2(x + 2)² - 5。
6. Deriving the Quadratic Formula | 由配方法推导求根公式
: it is simply completing the square carried out on the general equation ax² + bx + c = 0.
求根公式并不神秘：它只是对一般式 ax² + bx + c = 0 做一次配方法而已。
. Completing the square on the first two terms gives (x + b ÷ 2a)² - b² ÷ 4a² + c ÷ a = 0.
假设 a ≠ 0，把每一项都除以 a：x² + (b ÷ a)x + (c ÷ a) = 0。对前两项配方得到 (x + b ÷ 2a)² - b² ÷ 4a² + c ÷ a = 0。
) ÷ 4a². Taking the square root of both sides and subtracting b ÷ 2a produces the formula.
移项并把右边通分，得到 (x + b ÷ 2a)² = (b² - 4ac) ÷ 4a²。两边开平方再减去 b ÷ 2a，就得到求根公式。
x = (-b ± √b² - 4ac) ÷ 2a
number has two square roots, one positive and one negative. That single symbol is what allows one formula to produce both roots of a quadratic equation.
出现 ± 号是因为一个正数有两个平方根，一正一负。正是这个符号让一个公式能够给出二次方程的两个根。
7. Using the Quadratic Formula | 使用求根公式
The formula works for every quadratic equation, even when the expression does not factorise with integers, which makes it the most reliable method in an exam.
求根公式对任何二次方程都适用，即使式子无法用整数因式分解，因此在考试中它是最可靠的方法。
- 5 = 0. Here a = 2, b = -3 and c = -5, so b² - 4ac = (-3)² - 4 × 2 × (-5) = 9 + 40 = 49.
解 2x² - 3x - 5 = 0。这里 a = 2，b = -3，c = -5，所以 b² - 4ac = (-3)² - 4 × 2 × (-5) = 9 + 40 = 49。
x = (3 ± √49) ÷ 4 = (3 ± 7) ÷ 4
ubstituting both values back into the original equation confirms that each one gives zero.
所以 x = 10 ÷ 4 = 2.5 或 x = -4 ÷ 4 = -1。把两个值代回原方程，可以验证它们都使方程等于零。
= -9. This single slip is one of the most common sources of lost marks in the whole topic.
平方前要把负数放进括号：(-3)² = 9，而 -3² = -9。这一处小失误是本主题中最常见的失分原因之一。
If the answer is a surd,
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