📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解法、配方法与求根公式
Quadratic equations are one of the most important algebra topics in IGCSE Mathematics. They appear in pure algebra, graph sketching, geometry and real-life modelling. This guide explains how to solve quadratic equations using factorising, completing the square and the quadratic formula, and how to use the discriminant to understand the nature of the roots.
二次方程是 IGCSE 数学中最重要的代数主题之一，广泛出现在纯代数、函数图像、几何和实际建模中。本指南将讲解如何用因式分解法、配方法和求根公式解二次方程，以及如何利用判别式判断根的性质。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation of degree two, which means the highest power of the unknown is 2. Its standard form is shown below, where a, b and c are constants and a ≠ 0. If a were zero, the equation would become linear, not quadratic.
一元二次方程是次数为 2 的方程，也就是说未知数的最高次数是 2。它的标准形式如下，其中 a、b、c 为常数，且 a ≠ 0。如果 a 等于零，方程就变成了一次方程，而不是二次方程。
ax² + bx + c = 0, a ≠ 0
values of x that make the equation true are called the roots or the solutions, and when the equation is drawn as a graph they are the x-intercepts.
例如，x² - 5x + 6 = 0 和 2x² + 3x - 2 = 0 都是二次方程。使方程成立的 x 值称为根或解，从图像角度看也就是 x 轴截距。
thods in this guide. Quadratic means the highest power of x is 2, so an equation that looks complicated at first may still turn out to be quadratic after rearranging and simplifying.
任何能化简为 ax² + bx + c = 0 且 a ≠ 0 的方程都可以用本指南中的方法求解。二次意味着 x 的最高次数为 2，因此有些方程看起来复杂，但整理化简之后仍然是二次方程。
2. Standard Form and Key Terms | 标准形式与关键术语
all the terms to one side, simplify and arrange them in descending powers of x. This makes the coefficients a, b and c clear and helps you avoid sign errors.
求解前，先把方程整理为 ax² + bx + c = 0 的形式。将所有项移到一边，化简并按 x 的降幂排列。这样可以清楚地看出系数 a、b、c，避免符号错误。
c = 2. Notice that the sign belongs to the coefficient: the term is -7x, so b = -7, not 7.
系数 a 称为首项系数，b 是 x 的系数，c 是常数项。对于 3x² - 7x + 2 = 0，有 a = 3、b = -7、c = 2。注意符号属于系数本身：这一项是 -7x，所以 b = -7，而不是 7。
| Term | 项 | Name | 名称 | Example in 3x² - 7x + 2 = 0 |
| 3x² | quadratic term 二次项 | a = 3 |
| -7x | linear term 一次项 | b = -7 |
| 2 | constant term 常数项 | c = 2 |
2x(x - 3) = 8 becomes 2x² - 6x - 8 = 0, which can be divided by 2 to give x² - 3x - 4 = 0.
有些方程必须先整理。例如 x² = 5x - 6 化为 x² - 5x + 6 = 0；2x(x - 3) = 8 化为 2x² - 6x - 8 = 0，两边同除以 2 得 x² - 3x - 4 = 0。
3. Solving by Factorising | 因式分解法求解
Factorising is the fastest method when the roots are integers or simple fractions. It relies on the zero product rule: if two numbers multiply to give zero, then at least one of them must be zero.
当根是整数或简单分数时，因式分解是最快的方法。它的依据是零积法则：如果两个数的乘积为零，那么其中至少有一个数为零。
If pq = 0, then p = 0 or q = 0
numbers -2 and -3 work, so the equation becomes (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
用因式分解解 x² - 5x + 6 = 0 时，要找两个数，使它们的乘积等于 c = 6，和等于 b = -5。数 -2 与 -3 符合条件，于是方程变为 (x - 2)(x - 3) = 0，得到 x = 2 或 x = 3。
The general working steps are listed below. Always check your factorisation by expanding the brackets back, because a single sign mistake changes the answer completely.
通用的解题步骤如下。因式分解后一定要展开回去检查，因为一个符号错误就会让答案完全不同。
- Step 1 - Rearrange: move every term to one side so the equation equals zero.
- Step 2 - Factorise: take out common factors first, then split into two brackets.
- Step 3 - Apply the zero product rule: set each bracket equal to zero.
- Step 4 - Solve and check: solve the two small linear equations and substitute back.
- 第一步 - 整理：把所有项移到一边，使方程等于零。
- 第二步 - 分解因式：先提取公因式，再拆成两个括号。
- 第三步 - 用零积法则：令每个括号等于零。
- 第四步 - 求解并检验：解出两个一次方程并代回原式检验。
4. Factorising Harder Cases and Special Patterns | 更难的因式分解与特殊结构
re 4 and -1. Split the middle term, then group: 2x² + 4x - x - 2 = 2x(x + 2) - 1(x + 2) = (2x - 1)(x + 2). So x = ½ or x = -2.
当 a ≠ 1 时，因式分解需要更细心。对于 2x² + 3x - 2 = 0，乘积 ac = 2 × (-2) = -4，需要找两个数乘积为 -4、和为 3，即 4 与 -1。把中间项拆开再分组：2x² + 4x - x - 2 = 2x(x + 2) - 1(x + 2) = (2x - 1)(x + 2)，所以 x = ½ 或 x = -2。
ect square gives a² ± 2ab + b² = (a ± b)², so x² - 6x + 9 = 0 becomes (x - 3)² = 0, giving a repeated root x = 3.
有两种特殊结构能节省大量时间。平方差公式 a² - b² = (a + b)(a - b)，所以 x² - 9 = 0 化为 (x - 3)(x + 3) = 0，得到 x = ±3。完全平方公式 a² ± 2ab + b² = (a ± b)²，所以 x² - 6x + 9 = 0 化为 (x - 3)² = 0，得到重根 x = 3。
2) = 0, so x = ±2. Never divide both sides by x, because that would delete the root x = 0.
在进一步分解前，一定要先提取公因式。例如 2x² - 8 = 0 化为 2(x² - 4) = 0，再化为 2(x - 2)(x + 2) = 0，所以 x = ±2。千万不要两边同除以 x，那样会丢掉 x = 0 这个根。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form (x + p)² + q, which shows the turning point of the graph immediately. The key identity for a monic quadratic is given below.
配方法把二次式改写成 (x + p)² + q 的形式，可以直接看出图像的顶点。首项系数为 1 时的关键恒等式如下。
x² + bx + c = (x + b/2)² - (b/2)² + c
ssion is -8, and it occurs when x = -3. This is exactly the minimum point of the parabola.
例如 x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8。由于平方不可能为负，该式的最小值就是 -8，在 x = -3 时取得。这正是抛物线的最低点。
[(x + 2)² - 4] + 3 = 2(x + 2)² - 8 + 3 = 2(x + 2)² - 5. The minimum value is -5 at x = -2.
当 a ≠ 1 时，先把前两项的系数 a 提出来。对于 2x² + 8x + 3，写成 2(x² + 4x) + 3 = 2[(x + 2)² - 4] + 3 = 2(x + 2)² - 8 + 3 = 2(x + 2)² - 5。最小值是 -5，在 x = -2 时取得。
6. Solving Equations by Completing the Square | 用配方法解方程
red bracket, and finally take the square root of both sides. Remember the ± sign when you take a square root.
用配方法解方程时，先把常数项移到右边，再对左边配方，把平方括号单独留下，最后两边开平方。开平方时不要忘记 ± 号。
he square gives (x - 2)² - 4 - 1 = 0, so (x - 2)² = 5. Taking square roots gives x - 2 = ±√5, so x = 2 ± √5, which is approximately 4.236 or -0.236.
以 x² - 4x - 1 = 0 为例。配方得 (x - 2)² - 4 - 1 = 0，即 (x - 2)² = 5。两边开平方得 x - 2 = ±√5，所以 x = 2 ± √5，约等于 4.236 或 -0.236。
Unless the question asks for decimal answers to a given accuracy, leave your answer in surd form. Completing the square is also the method used to derive the quadratic formula, and it is the cleanest way to find the exact maximum or minimum value of a quadratic function.
除非题目要求给出指定精度的近似值，否则答案应保留根号形式。配方法也是推导求根公式所用的方法，并且是求二次函数精确最大值或最小值最简洁的途径。
7. The Quadratic Formula | 求根公式
is written below, and it works for every equation in the form ax² + bx + c = 0 with a ≠ 0.
求根公式可以解任何二次方程，即使该式无法因式分解。公式如下，对任何形如 ax² + bx + c = 0（a ≠ 0）的方程都适用。
x = (-b ± √b² - 4ac) ÷ (2a)
To use it: write the equation in standard form, identify a, b and c with their signs, compute b² - 4ac first, then substitute everything carefully. Keep the whole expression under the square root, and remember that 2a divides the entire numerator.
使用时：先把方程化为标准形式，带符号写出 a、b、c，先算 b² - 4ac，然后仔细代入。注意根号要覆盖整个式子，而且分母 2a 除以的是整个分子。
= 0 has a = 2, b = 3, c = -2, so b² - 4ac = 9 - 4 × 2 × (-2) = 9 + 16 = 25. Then x = (-3 ± √25) ÷ 4 = (-3 ± 5) ÷ 4, giving x = ½ or x = -2, which matches the factorising result.
检验前面的例子：2x² + 3x - 2 = 0 中 a = 2、b = 3、c = -2，所以 b² - 4ac = 9 - 4 × 2 × (-2) = 9 + 16 = 25。于是 x = (-3 ± √25) ÷ 4 = (-3 ± 5) ÷ 4，得到 x = ½ 或 x = -2，与因式分解的结果一致。
If the number under the square root is not a perfect square, simplify the surd. For instance, √20 = √4 × 5 = 2√5, which keeps the answer exact before you round it to 2 or 3 significant figures.
如果根号内的数不是完全平方数，就化简根式。例如 √20 = √4 × 5 = 2√5，这样在四舍五入到 2 位或 3 位有效数字前，答案仍是精确形式。
8. The Discriminant and the Nature of the Roots | 判别式与根的性质
The part of the formula under the square root is called the discriminant, written Δ. It tells you how many real roots an equation has without solving it fully, which is a very common exam question.
公式中根号内的部分称为判别式，记作 Δ。它可以在不完整求解的情况下告诉你方程有多少个实根，这是考试中非常常见的考点。
Δ = b² - 4ac
| Value of Δ | Nature of roots 根的性质 | Graph 图像 |
| Δ > 0 | two distinct real roots 两个不同的实根 | crosses the x-axis twice 与 x 轴相交两次 |
| Δ = 0 | one repeated real root 一个重实根 | touches the x-axis 与 x 轴相切 |
| Δ < 0 | no real roots 没有实根 | never meets the x-axis 与 x 轴不相交 |
-4, which is negative, so the equation has no real roots. If Δ is a perfect square, the roots
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