📚 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 the IGCSE Maths syllabus, from algebra and graphs to problem-solving in areas such as area and projectile motion. In this revision guide, you will learn four reliable methods: factorising, completing the square, the quadratic formula, and using graphs. Each section pairs a key concept with a worked example so you can build exam-ready accuracy.
二次方程贯穿 IGCSE 数学大纲，从代数、图像到面积和抛体运动等应用题中都会出现。在本复习指南中，你将学习四种可靠的方法：因式分解、配方法、二次公式和图像法。每节都把一个关键概念与实例配对，帮助你达到考试要求的准确性。
1. What is a Quadratic Equation? | 什么是二次方程？
ion becomes linear. The graph of a quadratic function is a parabola, and solving the equation means finding the x-values where this parabola crosses the x-axis.
二次方程是一个二次多项式方程。它的一般形式为 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。如果 a = 0，方程就变成一次方程。二次函数的图像是一条抛物线，解方程意味着找到这条抛物线与 x 轴交点的 x 值。
2. Standard Form and Key Terms | 标准形式与关键术语
nt a is the leading coefficient. If a = 1, the quadratic is monic and easier to factorise.
解题前总是先把二次方程整理成标准形式 ax² + bx + c = 0。ax² 是二次项，bx 是一次项，c 是常数项。系数 a 是首项系数。如果 a = 1，这个二次式是首一多项式，更容易因式分解。
In standard form, the terms are usually written in descending powers of x. The quadratic term ax² controls the shape of the parabola, the linear term bx controls its tilt, and the constant c gives the y-intercept. Always check that the equation is set equal to 0 before choosing a method.
在标准形式中，各项通常按 x 的降幂排列。二次项 ax² 控制抛物线的形状，一次项 bx 控制倾斜，常数项 c 给出 y 轴截距。选择方法前，一定要检查方程是否已设为零。
3. Solving by Factorising: The Zero Product Property | 因式分解法：零乘积性质
oth). After writing the quadratic as a product of two linear brackets, set each bracket equal to zero and solve separately. This method is fastest when the quadratic has rational roots.
因式分解法使用零乘积性质：如果 p × q = 0，那么 p = 0 或 q = 0（或两者都为零）。把二次式写成两个一次因式的乘积后，令每个括号分别为零并分别求解。当二次方程有有理根时，这种方法最快。
If (x - m)(x - n) = 0, then x = m or x = n
如果 (x - m)(x - n) = 0，那么 x = m 或 x = n
4. Worked Example: Factorising with a = 1 | 实例：首项系数为1的因式分解
(x - 3) = 0. Setting each bracket to zero gives x - 2 = 0 or x - 3 = 0, so x = 2 or x = 3.
解方程 x² - 5x + 6 = 0。我们需要两个数，它们的乘积为 +6，和为 -5。这两个数是 -2 和 -3。因此 x² - 5x + 6 = (x - 2)(x - 3) = 0。令每个括号为零，得到 x - 2 = 0 或 x - 3 = 0，所以 x = 2 或 x = 3。
x² - 5x + 6 = 0 → (x - 2)(x - 3) = 0 → x = 2 or x = 3
5. Factorising when a ≠ 1 | 首项系数不为1的因式分解
Factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Then set each factor to zero.
对于像 2x² + 7x + 3 这样的二次式，找到两个数，它们的乘积为 a × c = 2 × 3 = 6，和为 b = 7。这两个数是 6 和 1。拆分中间项：2x² + 6x + x + 3。分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。然后令每个因式为零。
x = -3. Always substitute back into the original equation to confirm both roots are valid.
解 2x + 1 = 0 得到 x = -1/2，解 x + 3 = 0 得到 x = -3。总是代入原方程以确认两个根都有效。
6. Completing the Square | 配方法
ct (b/2)²: x² + bx + (b/2)² - (b/2)² + c = 0, which becomes (x + b/2)² + (c - (b/2)²) = 0.
配方法把二次式写成 a(x + p)² + q 的形式。即使二次式不能整齐地因式分解，这种方法仍然有效。从 x² + bx + c = 0 开始。关键步骤是加上并减去 (b/2)²：x² + bx + (b/2)² - (b/2)² + c = 0，整理为 (x + b/2)² + (c - (b/2)²) = 0。
x² + bx + c = (x + b/2)² + (c - b²/4)
配方法的目的是把 x 项集中在一个完全平方内，然后通过开平方求解。如果 a ≠ 1，先提取 a。
7. Worked Example: Completing the Square | 实例：配方法
- 9 + 5 = 0, hence (x + 3)² - 4 = 0. So (x + 3)² = 4. Taking square roots gives x + 3 = ±√4 = ±2. Therefore x = -3 ± 2, so x = -1 or x = -5.
用配方法解 x² + 6x + 5 = 0。x 的系数是 6，所以 (6/2)² = 9。重写：x² + 6x + 9 - 9 + 5 = 0，因此 (x + 3)² - 4 = 0。所以 (x + 3)² = 4。开方得到 x + 3 = ±√4 = ±2。因此 x = -3 ± 2，所以 x = -1 或 x = -5。
x² + 6x + 5 = 0 → (x + 3)² - 4 = 0 → x = -1 or x = -5
8. The Quadratic Formula | 二次公式
The quadratic formula solves any quadratic equation ax² + bx + c = 0: x = (-b ± √b² - 4ac) / 2a. Substitute a, b and c carefully, and simplify the square root if possible. This formula is derived by completing the square and is given in IGCSE exams.
二次公式可解任何二次方程 ax² + bx + c = 0：x = (-b ± √b² - 4ac) / 2a。仔细代入 a、b、c，如果可能就化简根号。这个公式由配方法推导而来，IGCSE 考试中会提供。
x = (-b ± √b² - 4ac) / 2a
For example, solve 2x² + 3x - 2 = 0. Here a = 2, b = 3, c = -2. Then x = (-3 ± √3² - 4 × 2 × -2) / (2 × 2) = (-3 ± √9 + 16) / 4 = (-3 ± √25) / 4 = (-3 ± 5) / 4. This gives x = 1/2 or x = -2.
例如，解 2x² + 3x - 2 = 0。这里 a = 2，b = 3，c = -2。那么 x = (-3 ± √3² - 4 × 2 × -2) / (2 × 2) = (-3 ± √9 + 16) / 4 = (-3 ± √25) / 4 = (-3 ± 5) / 4。得到 x = 1/2 或 x = -2。
9. Using the Discriminant | 判别式的使用
s one repeated real root. If Δ < 0, there are no real roots, only complex roots. In IGCSE, 'no real roots' is the expected answer.
判别式 Δ = b² - 4ac 让你不必解方程就能判断根的性质。如果 Δ > 0，有两个不同的实数根。如果 Δ = 0，有一个重根。如果 Δ < 0，没有实数根，只有复数根。在 IGCSE 中，预期答案是 '没有实数根'。
| Discriminant | 判别式 | Nature of roots | 根的性质 |
| Δ > 0 | Two distinct real roots | 两个不同的实数根 |
| Δ = 0 | One repeated real root | 一个重根 |
| Δ < 0 | No real roots | 没有实数根 |
10. Quadratic Graphs and Solutions | 二次函数图像与解
x-intercepts of the graph. If the parabola touches the x-axis at one point, the quadratic has one repeated root. If it never touches the x-axis, there are
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