📚 Quadratic Equations: Solving by Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式解法
Quadratic equations are central to IGCSE and A-Level Mathematics. This article explains how to solve them by factorising, completing the square, and using the quadratic formula. You will also learn how the discriminant reveals the nature of the roots, how the completed-square form gives the turning point of a parabola, and how to choose the best method for each exam question.
二次方程是 IGCSE 与 A-Level 数学的核心内容。本文将讲解如何通过因式分解、配方法和求根公式求解二次方程。你还会学习判别式如何揭示根的性质、配方形式如何给出抛物线的顶点，以及如何为每道考题选择最佳方法。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation of degree 2, meaning the highest power of the unknown variable x is 2. Quadratic equations appear in problems involving area, projectile motion, profit maximisation, and many other optimisation tasks.
二次方程是次数为 2 的方程，即未知数 x 的最高次数为 2。二次方程出现在涉及面积、抛体运动、利润最大化以及许多其他优化任务的问题中。
General form: ax² + bx + c = 0, a ≠ 0
not quadratic. For example, x² − 5x + 6 = 0 and 2x² + 3x − 2 = 0 are quadratic equations.
常数 a、b、c 称为系数。条件 a ≠ 0 非常重要，因为如果 a = 0，方程就变成一次方程，而不是二次方程。例如 x² − 5x + 6 = 0 和 2x² + 3x − 2 = 0 都是二次方程。
- x² − 5x + 6 = 0 — x squared minus 5x plus 6 equals 0 / x² 减 5x 加 6 等于 0
- 2x² + 3x − 2 = 0 — 2x squared plus 3x minus 2 equals 0 / 2x² 加 3x 减 2 等于 0
- x² − 9 = 0 — a quadratic with no linear term / 没有一次项的二次方程
e equation are the x-values where the graph crosses the x-axis. These values are also called roots, zeros, or x-intercepts.
当二次式绘制为 y = ax² + bx + c 时，方程的解就是图像与 x 轴相交处的 x 值。这些值也称为根、零点或 x 轴截距。
2. Standard Form and Key Terms | 标准形式与关键术语
Before solving any quadratic equation, rearrange all terms onto one side so that the other side is exactly zero. The standard form makes it easy to identify the coefficients a, b and c correctly before you substitute them anywhere.
在求解任何二次方程之前，先把所有项移到一边，使另一边恰好为零。标准形式有助于在代入任何公式之前正确识别系数 a、b、c。
ax² + bx + c = 0
- Example: 3x² = 5x + 2 becomes 3x² − 5x − 2 = 0, so a = 3, b = −5, c = −2. / 示例：3x² = 5x + 2 移项后成为 3x² − 5x − 2 = 0，因此 a = 3，b = −5，c = −2。
- Example: 4x² + 7 = 12x becomes 4x² − 12x + 7 = 0. / 示例：4x² + 7 = 12x 移项后成为 4x² − 12x + 7 = 0。
| Term | Meaning | 中文 |
| a | leading coefficient (of x²) | 二次项系数 |
| b | linear coefficient (of x) | 一次项系数 |
| c | constant term | 常数项 |
| roots | solutions of the equation | 方程的根 |
| x-intercepts | where the graph crosses the x-axis | 图像与 x 轴交点 |
Understanding this link between algebra and the graph is essential for IGCSE paper questions, because the same quadratic can be asked about as an equation to solve and as a curve to sketch.
理解代数与图像之间的这种联系，对 IGCSE 试卷题目至关重要，因为同一个二次式既可能作为方程要求求解，也可能作为曲线要求作图。
3. Solving by Factorising | 因式分解法求解
Factorising is the fastest method when the quadratic factorises neatly with integer coefficients. It relies on the zero product property: if the product of two factors is zero, then at least one factor must be zero.
当二次式能用整数系数干净地分解时，因式分解是最快的方法。它依赖零积性质：如果两个因子的乘积为零，那么至少有一个因子必须为零。
If (x − p)(x − q) = 0, then x = p or x = q
For a monic quadratic x² + bx + c, you need two numbers that multiply to give c and add to give b. Write them inside the brackets and solve each bracket separately.
对于首项系数为 1 的二次式 x² + bx + c，你需要找到两个数，它们相乘等于 c、相加等于 b。把它们写进括号中，再分别令每个括号为零求解。
- Step 1: Rearrange into standard form ax² + bx + c = 0. / 步骤 1：整理成标准形式 ax² + bx + c = 0。
- Step 2: Factorise the expression completely. / 步骤 2：将二次式完全因式分解。
- Step 3: Set each factor equal to zero. / 步骤 3：令每个因子等于零。
- Step 4: Solve each linear equation and state both roots. / 步骤 4：解每个一次方程并写出两个根。
Worked example: mbers −2 and −3 multiply to 6 and add to −5, so (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
例题：解 x² − 5x + 6 = 0。−2 与 −3 相乘得 6、相加得 −5，所以 (x − 2)(x − 3) = 0，得 x = 2 或 x = 3。
Worked example: ² + 7x + 12 = 0. The numbers 3 and 4 work, so (x + 3)(x + 4) = 0, giving x = −3 or x = −4.
例题：解 x² + 7x + 12 = 0。3 与 4 符合条件，所以 (x + 3)(x + 4) = 0，得 x = −3 或 x = −4。
4. Factorising When a ≠ 1 | 当 a ≠ 1 时的因式分解
When the leading coefficient is not 1, the factorising is still possible but needs a systematic method. The most reliable technique is to split the middle term using the product ac and the sum b.
当二次项系数不为 1 时，仍然可以因式分解，但需要系统的方法。最可靠的做法是利用乘积 ac 与和 b 来拆分中间项。
- Step 1: Find two numbers whose product is ac and whose sum is b. / 步骤 1：找到两个数，乘积为 ac，和为 b。
- Step 2: Rewrite bx as the sum of those two terms. / 步骤 2：把 bx 改写为这两项之和。
- Step 3: Factorise in pairs (grouping). / 步骤 3：成对提取公因式（分组分解）。
- Step 4: Take out the common bracket and solve. / 步骤 4：提出公因式括号并求解。
Worked example: So 2x² + 4x − x − 2 = 2x(x + 2) − 1(x + 2) = (x + 2)(2x − 1) = 0, giving x = −2 or x = ½.
例题：解 2x² + 3x − 2 = 0。此处 ac = −4，而 4 与 −1 的乘积为 −4、和为 3。故 2x² + 4x − x − 2 = 2x(x + 2) − 1(x + 2) = (x + 2)(2x − 1) = 0，得 x = −2 或 x = ½。
Worked example: 15x − 4x − 10 = 3x(2x + 5) − 2(2x + 5) = (2x + 5)(3x − 2) = 0, giving x = −5/2 or x = 2/3.
例题：解 6x² + 11x − 10 = 0。此处 ac = −60，15 与 −4 符合条件。故 6x² + 15x − 4x − 10 = 3x(2x + 5) − 2(2x + 5) = (2x + 5)(3x − 2) = 0，得 x = −5/2 或 x = 2/3。
5. Special Cases: Difference of Two Squares and Perfect Squares | 特殊情形：平方差与完全平方
Two special patterns appear constantly in exams and can be factorised instantly once you recognise them. Spotting them saves a great deal of time.
有两种特殊模式在考试中频繁出现，一旦识别出来就能立刻分解，可以节省大量时间。
a² − b² = (a − b)(a + b) a² ± 2ab + b² = (a ± b)²
- Difference of two squares: x² − 9 = (x − 3)(x + 3), so x = 3 or x = −3. / 平方差：x² − 9 = (x − 3)(x + 3)，所以 x = 3 或 x = −3。
- Difference of two squares: 4x² − 25 = (2x − 5)(2x + 5), so x = 5/2 or x = −5/2. / 平方差：4x² − 25 = (2x − 5)(2x + 5)，所以 x = 5/2 或 x = −5/2。
- Perfect square: x² − 6x + 9 = (x − 3)² = 0, so x = 3 (a repeated root). / 完全平方：x² − 6x + 9 = (x − 3)² = 0，所以 x = 3（重根）。
- Perfect square: 9x² + 24x + 16 = (3x + 4)² = 0, so x = −4/3. / 完全平方：9x² + 24x + 16 = (3x + 4)² = 0，所以 x = −4/3。
A repeated root means the curve touches the x-axis at exactly one point, which is a useful check when you sketch the graph. A common error is to write only one root when the equation has two, or to forget the negative root in a difference of two squares.
重根意味着曲线与 x 轴恰好只接触一点，这在作图时是很有用的检验。常见错误是在方程有两个根时只写一个，或在平方差中漏掉负根。
6. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form is not only useful for solving equations, but also immediately reveals the coordinates of the vertex of the parabola.
配方法把二次式改写成 a(x + p)² + q 的形式。这种形式不仅便于解方程，还能直接揭示抛物线顶点的坐标。
x² + bx + c = (x + b/2)² − (b/2)² + c
- Step 1: Take half of the coefficient of x. / 步骤 1：取 x 的系数的一半。
- Step 2: Square the bracket (x + b/2)². / 步骤 2：写出平方括号 (x + b/2)²。
- Step 3: Subtract the extra (b/2)² and add back the original constant. / 步骤 3：减去多出的 (b/2)²，再加上原来的常数项。
- Step 4: If a ≠ 1, first factor a out of the x² and x terms only. / 步骤 4：若 a ≠ 1，先从 x² 与 x 项中提出 a。
Worked example: − 9 + 1 = (x + 3)² − 8. The vertex is at (−3, −8) and the curve has a minimum value of −8.
例题：x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。顶点为 (−3, −8)，曲线的最小值为 −8。
Worked example: 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3. The vertex is at (2, −3) and the minimum value is −3.
例题：2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3。顶点为 (2, −3)，最小值为 −3。
7. Solving Equations by Completing the Square | 用配方法解方程
emember that every positive number has two square roots, so you must include the ± symbol.
用配方法解方程时，先把方程改写成 (x + p)² = k 的形式，然后对两边开平方。记住每个正数都有两个平方根，因此必须写上 ± 符号。
- Step 1: Divide by a if needed so the coefficient of x² is 1. / 步骤 1：如有需要先除以 a，使 x² 的系数为 1。
- Step 2: Move the constant to the right-hand side. / 步骤 2：把常数项移到等号右边。
- Step 3: Complete the square on the left, adding the same amount on the right. / 步骤 3：在左边配方，并在右边加上同样的量。
- Step 4: Take square roots with ± and simplify the surd. / 步骤 4：开平方并带 ±，再化简根式。
Worked example: Solve x² + 6x + 1 = 0. Then (x + 3)² = 8, so x + 3 = ±2√2 and x = −3 ± 2√2, i.e. x ≈ −0.172 or x ≈ −5.83.
例题：解 x² + 6x + 1 = 0。得 (x + 3)² = 8，所以 x + 3 = ±2√2，x = −3 ± 2√2，即 x ≈ −0.172 或 x ≈ −5.83。
Worked example: Solve x² − 4x − 1 = 0. Then (x − 2)² = 5, so x = 2 ± √5, i.e. x ≈ 4.24 or x ≈ −0.236 (3 significant figures).
例题：解 x² − 4x − 1 = 0。得 (x − 2)² = 5，所以 x = 2 ± √5，即 x ≈ 4.24 或 x ≈ −0.236（保留 3 位有效数字）。
This method is compulsory when a question says "solve by completing the square" or "give your answer in surd form". It is also the way the quadratic formula itself is derived.
当题目要求"用配方法求解"或"答案用根式表示"时，必须使用此方法。求根公式本身也正是通过配方推导出来的。
8. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation, whether or not it factorises. Substitute the values of a, b and c directly from standard form, keeping the signs exactly as they appear.
求根公式可解任何二次方程，无论它能否因式分解。直接把标准形式中的 a、b、c 代入，符号要完全按原样保留。
x = (−b ± √b² − 4ac) / (2a)
Worked example: b = 3, c = −2, so b² − 4ac = 9 + 16 = 25. Hence x = (−3 ± 5) / 4, giving x = ½ or x = −2.
例题：解 2x² + 3x − 2 = 0。此处 a = 2，b = 3，c = −2，所以 b² − 4ac = 9 + 16 = 25。因此 x = (−3 ± 5) / 4，得 x = ½ 或 x = −2。
Worked example: Solve x² − 3x − 1 = 0. Here b² − 4ac = 9 + 4 = 13, so x = (3 ± √13) / 2, i.e. x ≈ 3.30 or x ≈ −0.303 (3 s.f.).
例题：解 x² − 3x − 1 = 0。此处 b² − 4ac = 9 +
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