📚 Mastering Quadratic Equations for IGCSE Mathematics | 掌握 IGCSE 数学二次方程
In this article you will learn how to recognise, solve, sketch and check quadratic equations using three main methods: factorising, completing the square and the quadratic formula. Each method has its own strengths, and a well-prepared IGCSE student knows which one to reach for in the exam.
本文将教你如何识别、求解、画出并检验二次方程，涵盖三种主要方法：因式分解法、配方法和求根公式法。每种方法各有优势，准备充分的 IGCSE 考生必须知道在考场上该用哪一种。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2. That means the highest power of the variable, usually x, is 2. In IGCSE Mathematics the standard form is written as ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. If a were equal to 0, the x² term would disappear and the equation would become linear.
二次方程是次数为 2 的多项式方程，也就是说变量（通常是 x）的最高次幂是 2。在 IGCSE 数学中，标准形式写作 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。如果 a = 0，x² 项就消失了，方程会退化成一次（线性）方程。
uch as 3x + 1 = 0 is not quadratic, because the highest power of x is 1. The word quadratic comes from the Latin quadratus, meaning square, which is exactly why the key feature is a squared term.
常见例子包括 x² - 9 = 0、x² + 3x - 4 = 0 以及 2x² - 5x + 3 = 0，注意 x² 的系数不一定等于 1。相反，像 3x + 1 = 0 这样的线性方程不是二次方程，因为 x 的最高次幂是 1。quadratic 一词源自拉丁语 quadratus，意思是"正方形"，这正是二次方程的核心特征在于平方项的原因。
Recognising a quadratic before you start solving is essential, because the method you choose, the number of solutions and the shape of the graph all depend on the values of a, b and c.
在动手求解之前先识别出二次方程非常关键，因为你选择的方法、解的个数以及图像的形状，都取决于 a、b、c 的取值。
2. Standard Form and Coefficients | 标准形式与系数
t be rewritten as 3x² - 5x + 2 = 0. Only then can you read off the coefficients correctly.
使用任何方法之前，都必须先把方程整理成一边恰好为 0 的形式。例如 3x² = 5x - 2 必须先改写成 3x² - 5x + 2 = 0，只有这样才能正确读出各个系数。
e have a = 1, b = 0 and c = -9. Remember that the sign belongs to the coefficient: in x² - 5x + 6 = 0, b = -5, not 5.
在 ax² + bx + c = 0 中：a 是 x² 的系数，b 是 x 的系数，c 是常数项。如果某一项缺失，它的系数就是 0，所以在 x² - 9 = 0 中 a = 1、b = 0、c = -9。要记住符号属于系数本身：在 x² - 5x + 6 = 0 中 b = -5，而不是 5。
| Equation 方程 | a | b | c |
|---|---|---|---|
| x² + 3x - 4 = 0 | 1 | 3 | -4 |
| 2x² - 5x + 3 = 0 | 2 | -5 | 3 |
| x² - 9 = 0 | 1 | 0 | -9 |
| 5x² = 20 | 5 | 0 | -20 |
3. Three Core Methods at a Glance | 三种核心方法概览
IGCSE questions on quadratics almost always expect one of three approaches. Factorising is the fastest when the roots are integers or simple fractions. Completing the square is needed when the question asks for the turning point or an exact form. The quadratic formula always works, even when factorising fails.
IGCSE 的二次方程题目几乎都要求使用以下三种方法之一。当根是整数或简单分数时，因式分解法最快；当题目要求求顶点或精确形式时，需要配方；而求根公式在任何情况下都能用，即使无法因式分解。
| Method 方法 | Best used when 最佳使用时机 | Extra benefit 额外用途 |
|---|---|---|
| Factorising 因式分解 | Roots are integers or simple fractions 根为整数或简单分数 | Fastest, good for graphs 最快，便于画图 |
| Completing the square 配方 | Question asks for vertex or exact values 题目要求顶点或精确值 | Gives minimum/maximum point 给出最值点 |
| Quadratic formula 求根公式 | Factorising is difficult or impossible 难以或无法分解 | Always works for real roots 有实根时总能适用 |
4. Method 1: Factorising | 方法一：因式分解法
least one of them must be 0. So if (x + 4)(x - 1) = 0, then either x + 4 = 0 or x - 1 = 0.
因式分解法依赖"零积性质"：如果两个表达式的乘积为 0，那么其中至少有一个为 0。所以若 (x + 4)(x - 1) = 0，则 x + 4 = 0 或 x - 1 = 0。
× (-1) = -4 and 4 + (-1) = 3. Hence x² + 3x - 4 = (x + 4)(x - 1), giving x = -4 or x = 1.
当 a = 1 时，找两个数，使它们的乘积等于 c、和等于 b。在 x² + 3x - 4 = 0 中，这两个数是 +4 和 -1，因为 4 × (-1) = -4，且 4 + (-1) = 3。因此 x² + 3x - 4 = (x + 4)(x - 1)，得到 x = -4 或 x = 1。
x² + bx + c = (x + p)(x + q), where p + q = b and p × q = c
Split the middle term: 2x² - 2x - 3x + 3 = 2x(x - 1) - 3(x - 1) = (2x - 3)(x - 1). So x = 3 ÷ 2 = 1.5 or x = 1.
当 a ≠ 1 时，使用"拆中项"法。对于 2x² - 5x + 3 = 0，先算 a × c = 6，再找两个数使乘积为 6、和为 -5，即 -2 和 -3。把中项拆开：2x² - 2x - 3x + 3 = 2x(x - 1) - 3(x - 1) = (2x - 3)(x - 1)，所以 x = 3 ÷ 2 = 1.5 或 x = 1。
ommon factor: 2x² - 8x = 0 becomes 2x(x - 4) = 0, so x = 0 or x = 4. Never divide both sides by x, because that destroys the root x = 0.
有两个特殊模式能节省大量时间：平方差 x² - 9 = (x + 3)(x - 3)，以及完全平方 x² + 6x + 9 = (x + 3)²。同时要注意提取公因式：2x² - 8x = 0 化为 2x(x - 4) = 0，所以 x = 0 或 x = 4。千万不要两边同时除以 x，那会丢掉 x = 0 这个根。
5. Method 2: Completing the Square | 方法二：配方法
Completing the square rewrites a quadratic as a perfect square plus a constant. For x² + bx + c, the rule is to take half of b, square it, and adjust the constant term accordingly. This form is ideal for finding exact roots and turning points.
配方法把二次式改写成"完全平方加常数"的形式。对于 x² + bx + c，法则是取 b 的一半再平方，并相应调整常数项。这种形式非常适合求精确根和顶点。
x² + bx + c = (x + b ÷ 2)² - (b ÷ 2)² + c
(x + 3)² - 9 + 1 = (x + 3)² - 8. Setting this equal to 0 gives (x + 3)² = 8, so x + 3 = ±√8 = ±2√2, and therefore x = -3 + 2√2 ≈ -0.17 or x = -3 - 2√2 ≈ -5.83.
例：解 x² + 6x + 1 = 0。6 的一半是 3，3² = 9，所以 x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8。令其等于 0，得 (x + 3)² = 8，故 x + 3 = ±√8 = ±2√2，因此 x = -3 + 2√2 ≈ -0.17 或 x = -3 - 2√2 ≈ -5.83。
3 = 2(x + 2)² - 8 + 3 = 2(x + 2)² - 5. Setting this to 0 gives (x + 2)² = 2.5, so x = -2 ± √2.5 ≈ -0.42 or x ≈ -3.58.
当 a ≠ 1 时，先把 a 只从 x² 项和 x 项中提出来。对于 2x² + 8x + 3 = 0：2(x² + 4x) + 3 = 2[(x + 2)² - 4] + 3 = 2(x + 2)² - 8 + 3 = 2(x + 2)² - 5。令其为 0，得 (x + 2)² = 2.5，所以 x = -2 ± √2.5 ≈ -0.42 或 x ≈ -3.58。
completed square immediately gives the vertex (h, k) and the line of symmetry x = h. If a
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