📚 Mastering Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 掌握二次方程：因式分解、配方法与求根公式
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebraic manipulation and graph sketching to real-life modelling. This revision guide covers the standard form, three main solution methods, the discriminant, graph interpretation and common exam techniques you need to secure high marks.
二次方程贯穿 IGCSE 数学大纲，从代数运算、函数图像到实际建模都有涉及。本复习指南涵盖标准形式、三种主要求解方法、判别式、图像解读以及考试中冲击高分所需的常见技巧。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation in one variable is any equation that can be written as ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The condition a ≠ 0 is essential, because if a = 0 the equation becomes linear rather than quadratic.
一元二次方程是任何可以写成 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。a ≠ 0 的条件至关重要，因为如果 a = 0，方程就变成了一次方程而不是二次方程。
Before solving, you should always rearrange the equation so that all terms are on one side and zero is on the other. For example, 3x² = 5x – 2 becomes 3x² – 5x + 2 = 0, and 2x – x² = 1 becomes -x² + 2x – 1 = 0.
在求解前，你应当先将所有项移到一边，使另一边为零。例如，3x² = 5x – 2 可化为 3x² – 5x + 2 = 0；2x – x² = 1 可化为 -x² + 2x – 1 = 0。
2. Solving by Factorising | 因式分解法
Factorising is often the fastest method when the quadratic has simple integer roots. Write the quadratic as a product of two linear brackets, such as (px + q)(rx + s) = 0, then solve each bracket equal to zero separately.
当二次方程具有简单的整数根时，因式分解通常是最快的方法。将二次式写成两个一次括号的乘积，如 (px + q)(rx + s) = 0，然后分别令每个括号等于零求解。
Example: x² – 7x + 12 = 0 factorises to (x – 3)(x – 4) = 0, giving x = 3 or x = 4. Always expand your factors back to check that they match the original quadratic.
示例：x² – 7x + 12 = 0 因式分解为 (x – 3)(x – 4) = 0，得到 x = 3 或 x = 4。务必通过展开来检查因式是否与原二次式一致。
- When a = 1, find two numbers that multiply to c and add to b.
- 当 a = 1 时，找出两个数，使其乘积为 c，和为 b。
- When a ≠ 1, use the ac method: find two numbers that multiply to ac and add to b, then split the middle term and factor by grouping.
- 当 a ≠ 1 时，使用 ac 方法：找出乘积为 ac 且和为 b 的两个数，然后拆分中间项并进行分组分解。
3. Zero Product Property | 零乘积性质
The zero product property states that if A × B = 0, then A = 0 or B = 0, or both. This is the key reason factorising works: once a product of brackets equals zero, at least one bracket must be zero.
零乘积性质指出，如果 A × B = 0，则 A = 0 或 B = 0，或两者都为零。这是因式分解法有效的关键原因：一旦括号的乘积等于零，至少有一个括号必须为零。
Do not apply this property when the product equals a non-zero number. For (x – 1)(x + 2) = 3, you cannot simply set each bracket equal to 3; you must first expand or rearrange so the right-hand side is zero.
当乘积等于非零数字时，不能使用这一性质。对于 (x – 1)(x + 2) = 3，你不能简单令每个括号等于 3；必须先展开或整理，使右边为零。
Correct approach: expand to x² + x – 2 = 3, then subtract 3 to get x² + x – 5 = 0, which can be solved using the quadratic formula.
正确做法：展开得到 x² + x – 2 = 3，然后减去 3 得到 x² + x – 5 = 0，再使用求根公式求解。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form is valuable for finding the vertex, solving equations that do not factorise neatly, and proving the quadratic formula.
配方法将二次式改写为 a(x + p)² + q 的形式。这种形式对于求顶点、求解无法整齐因式分解的方程以及证明求根公式都非常有价值。
For x² + 6x + 5, take half of the coefficient of x: 6/2 = 3, square it to get 9, then write x² + 6x + 5 = (x + 3)² – 9 + 5 = (x + 3)² – 4.
对于 x² + 6x + 5，取 x 项系数的一半：6/2 = 3，平方得到 9，然后写成 x² + 6x + 5 = (x + 3)² – 9 + 5 = (x + 3)² – 4。
x² + bx = (x + b/2)² – (b/2)²
If the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 3, write 2(x² + 4x) + 3 = 2[(x + 2)² – 4] + 3 = 2(x + 2)² – 5.
如果 x² 的系数不是 1，先将其提出。对于 2x² + 8x + 3，写成 2(x² + 4x) + 3 = 2[(x + 2)² – 4] + 3 = 2(x + 2)² – 5。
5. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation ax² + bx + c = 0, including cases where factorising is difficult or impossible over integers. It is derived by completing the square on the general quadratic.
求根公式可以求解任何二次方程 ax² + bx + c = 0，包括难以或无法在整数范围内因式分解的情况。它是由对一般二次式进行配方推导出来的。
x = [-b ± √(b² – 4ac)] / 2a
To use the formula, identify a, b and c carefully, including negative signs. Substitute them into the formula and simplify step by step, paying attention to the order of operations.
使用公式时，要仔细确定 a、b 和 c，包括负号。将其代入公式并逐步化简，注意运算顺序。
Example: 2x² – 3x – 5 = 0 has a = 2, b = -3, c = -5. Then x = [3 ± √(9 + 40)] / 4 = [3 ± 7] / 4, giving x = 10/4 = 2.5 or x = -4/4 = -1.
示例：2x² – 3x – 5 = 0 中 a = 2, b = -3, c = -5。因此 x = [3 ± √(9 + 40)] / 4 = [3 ± 7] / 4，得到 x = 10/4 = 2.5 或 x = -4/4 = -1。
6. The Discriminant and Nature of Roots | 判别式与根的性质
The discriminant D = b² – 4ac tells you how many real solutions a quadratic equation has without solving it fully. It is the expression under the square root in the quadratic formula.
判别式 D = b² – 4ac 可以在不完全求解的情况下告诉你二次方程有多少个实数解。它就是求根公式中平方根号下的表达式。
- If D > 0: two distinct real roots. / 如果 D > 0：两个不相等的实根。
- If D = 0: one repeated real root. / 如果 D = 0：一个重根（两个相等的实根）。
- If D < 0: no real roots. / 如果 D < 0：没有实数根。
This concept is commonly tested through questions that ask you to find the range of k for which a quadratic has equal roots, two roots or no roots. For example, x² + kx + 9 = 0 has equal roots when D = k² – 36 = 0, so k = ±6.
这一概念常见于考题，要求你求出使二次方程具有重根、两个根或无实根的 k 的取值范围。例如，x² + kx + 9 = 0 当 D = k² – 36 = 0 时有重根，因此 k = ±6。
7. Graphs of Quadratic Functions | 二次函数图像
The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point; if a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上且有最低点；如果 a < 0，抛物线开口向下且有最高点。
The y-intercept is at (0, c). The x-intercepts, if they exist, are the real roots of the equation ax² + bx + c = 0. The axis of symmetry passes through the turning point.
y 轴截距位于 (0, c)。x 轴截距（如果存在）就是方程 ax² + bx + c = 0 的实根。对称轴穿过顶点。
| Feature / 特征 | Formula / 公式 |
|---|---|
| Axis of symmetry / 对称轴 | x = -b / 2a |
| Turning point / 顶点 | (-b / 2a, f(-b / 2a)) |
| y-intercept / y 轴截距 | (0, c) |
8. Turning Point and Vertex Form | 顶点与顶点式
Writing a quadratic in completed-square form y = a(x + p)² + q reveals the turning point directly: (-p, q). This is extremely useful for sketching graphs and for optimisation problems.
将二次函数写成配方式 y = a(x + p)² + q 可直接展示顶点：(-p, q)。这对画图和最优化问题非常有用。
For y = (x + 3)² – 4, the turning point is (-3, -4). Since a = 1 > 0, this is a minimum point. The graph crosses the y-axis at y = 5 and the x-axis at the roots x = -1 and x = -5.
对于 y = (x + 3)² – 4，顶点为 (-3, -4)。由于 a = 1 > 0，这是一个最低点。图像与 y 轴交于 y = 5，与 x 轴交于根 x = -1 和 x = -5。
You can also find the turning point from standard form using x = -b / 2a, then substituting this x-value back into the equation to find the y-coordinate.
你也可以从标准形式出发，使用 x = -b / 2a 求顶点的 x 坐标，然后将该 x 值代回方程求出 y 坐标。
9. Applications and Word Problems | 应用题与文字题
IGCSE questions often model area, projectile motion, profit or product relationships with quadratic equations. Translate the words into an equation, solve it, and check the context for valid answers.
IGCSE 题目常用二次方程来模拟面积、抛体运动、利润或乘积关系。将文字转化为方程，求解并结合实际
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