📚 Mastering Quadratic Equations: Completing the Square & Quadratic Formula | 掌握二次方程：配方法与求根公式
Quadratic equations form the backbone of IGCSE algebra and appear in everything from projectile motion to profit optimisation. In this article we explore three core solution methods – factorisation, completing the square and the quadratic formula – and show how they connect. You will learn step‑by‑step techniques, see worked examples and discover how to choose the most efficient method for any problem.
二次方程是 IGCSE 代数的核心内容，从抛体运动到利润优化问题中处处可见。本文将深入探讨三种核心解法——因式分解法、配方法和求根公式——并揭示它们之间的内在联系。你将逐步学习相关技巧，通过例题加深理解，并学会如何在任何题目中选出最高效的方法。
1. What Is a Quadratic Equation? | 什么是二次方程？
stants and a ≠ 0. The highest power of the variable x is 2, which is why the graph of a quadratic function is a parabola. Every quadratic equation has two solutions (roots), although they may be real and distinct, real and equal, or complex.
二次方程是能够写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。变量 x 的最高次幂是 2，这也是二次函数图像呈抛物线的原因。每个二次方程都有两个解（根），这些根可能为两相异实数、两相等实数，或复数。
2. Standard Form and Coefficients | 标准形式与系数
coefficient a determines the parabola’s width and direction of opening (if a > 0 it opens upwards; if a < 0 it opens downwards). b influences the position of the axis of symmetry, and c is the y‑intercept. Before solving, always rearrange the equation so that one side equals zero.
标准形式为 ax² + bx + c = 0。系数 a 决定抛物线的宽度和开口方向（a > 0 时向上开口，a < 0 时向下开口）。b 影响对称轴的位置，c 则是 y 轴截距。在求解前，务必先将方程整理成一边等于零的形式。
3. Solving by Factorisation | 因式分解法
2)(x + 3) = 0, giving x = −2 or x = −3. This method is quick when the roots are integers.
当二次方程能表示为两个一次因式的乘积时，可使用因式分解法：(px + q)(rx + s) = 0。根据零积性质，要么 px + q = 0，要么 rx + s = 0。例如 x² + 5x + 6 = 0 可化为 (x + 2)(x + 3) = 0，解得 x = −2 或 x = −3。当根为整数时，此法最快。
4. When Factorisation Is Not Straightforward | 因式分解不直接的情况
atic has a coefficient a ≠ 1 or does not factorise neatly into integer pairs, simple trial‑and‑ res practice. For messy coefficients or irrational roots, we need more systematic methods.
当二次项系数 a ≠ 1 或无法用整数对轻松分解时，简单的试错因式分解会变得繁琐。例如 2x² + 3x − 2 = 0 可分解为 (2x − 1)(x + 2) = 0，但找到正确的组合需要一定训练。对于系数复杂或根为无理数的情况，则需要更系统的方法。
5. Completing the Square – The Concept | 配方法的概念
Completing the square rewrites ax² + bx + c into the form a(x + p)² + q, where p and q are constants. This form reveals the vertex of the parabola and provides a direct way to solve any quadratic equation. The key idea is to create a perfect square trinomial from the x² and x terms, then adjust the constant term accordingly.
配方法将 ax² + bx + c 转化为 a(x + p)² + q 的形式，其中 p 和 q 为常数。这一形式不仅揭示了抛物线的顶点，还能直接求解任何二次方程。其核心思路是利用 x² 和 x 项构造一个完全平方三项式，再对常数项进行相应调整。
6. Step‑by‑Step Completing the Square | 配方法步骤
For x² + bx + c = 0 (where a = 1):
1. Move the constant term: x² + bx = −c.
2. Take half of b, square it: (b/2)².
3. Add this to both sides: x² + bx + (b/2)² = −c + (b/2)².
4. Factor the left side as (x + b/2)².
5. Take the square root of both sides and solve for x.
For a ≠ 1, first divide every term by a.
对于 x² + bx + c = 0（a = 1 时）：
1. 将常数项移到右侧：x² + bx = −c。
2. 取 b 的一半，再平方：(b/2)²。
3. 将该平方加到等式两边：x² + bx + (b/2)² = −c + (b/2)²。
4. 左边因式分解为 (x + b/2)²。
5. 两边开平方，然后解出 x。
若 a ≠ 1，则需先用 a 除以每一项。
7. Deriving the Quadratic Formula | 推导求根公式
are: (x + b/(2a))² = (b/(2a))² − c/a = (b² − 4ac)/(4a²). Take square roots: x + b/(2a) = ±√b² − 4ac/(2a). Finally, isolate x to obtain the quadratic formula: x = (−b ± √b² − 4ac) / (2a). This single formula solves every quadratic equation.
从标准形式 ax² + bx + c = 0 出发，除以 a 得 x² + (b/a)x + c/a = 0。配方得 (x + b/(2a))² = (b/(2a))² − c/a = (b² − 4ac)/(4a²)。两边开平方：x + b/(2a) = ±√b² − 4ac/(2a)。最后移项即得求根公式：x = (−b ± √b² − 4ac) / (2a)。这一公式可求解所有二次方程。
8. The Quadratic Formula in Practice | 求根公式的应用
Identify a, b and c from the standard form and substitute into x = (−b ± √b² − 4ac . For example, 2x² − 4x − 6 = 0 gives a = 2, b = −4, c = −6. Then b² − 4ac = 16 + 48 = 64, √64 = 8, so x = (4 ± 8) / 4, yielding x = 3 or x = −1.
从标准形式中确定 a、b 和 c，代入 x = (−b ± √b² − 4ac) / (2a) 中。先化简根号内的部分，再计算出两个可能的根。例如 2x² − 4x − 6 = 0，其中 a = 2，b = −4，c = −6。计算 b² − 4ac = 16 + 48 = 64，√64 = 8，于是 x = (4 ± 8) / 4，得 x = 3 或 x = −1。
9. Discriminant and Nature of Roots | 判别式与根的性质
lled the discriminant. It determines the nature of the roots without solving the equation:
| Discriminant | Nature of roots |
| Δ > 0 | Two distinct real roots |
| Δ = 0 | One repeated real root (or two equal roots) |
| Δ < 0 | No real roots (two complex roots) |
This is a quick check that often saves time in IGCSE problems.
根号内的表达式 Δ = b² − 4ac 称为判别式，它能在不解方程的情况下判断根的性质：
| 判别式 | 根的性质 |
| Δ > 0 | 两个相异实根 |
| Δ = 0 | 一个重实根（两个相等实根） |
| Δ < 0 | 无实根（两个复根） |
在 IGCSE 题目中，这一快速判断常能节省时间。
10. Choosing the Right Method | 选择合适的方法
For simple quadratics with integer factor pairs, factorisation is fastest. If factorisation is not obvious, completing the square is elegant and directly gives the vertex form. The quadratic formula is the most general – it works for every case, but involves more arithmetic. In exams, check the discriminant first: if Δ is a perfect square, factorisation is likely possible; if not, use the formula or complete the square.
对于具有整数因式对的简单二次方程，因式分解法最快。若因式分解不易看出，配方法更为优雅，还能直接给出顶点式。求根公式最为通用——适用于所有情况，但计算量稍大。考试时，先判断判别式：若 Δ 为完全平方数，很可能可用因式分解；否则采用求根公式或配方法。
11. Worked Examples | 例题解析
Example 1 (Factorisation): hat multiply to 10 and add to −7: −5 and −2. So (x − 5)(x − 2) = 0, giving x = 5 or x = 2.
例1（因式分解法）：解 x² − 7x + 10 = 0。找出积为 10，和为 −7 的两个数：−5 和 −2。于是 (x − 5)(x − 2) = 0，解得 x = 5 或 x = 2。
Example 2 (Completing the square): is 3, squared is 9. Add 9: x² + 6x + 9 = 5, so (x + 3)² = 5. Take square roots: x + 3 = ±√5, so x = −3 ± √5.
例2（配方法）：解 x² + 6x + 4 = 0。移常数项：x² + 6x = −4。6 的一半是 3，平方得 9。两边加 9：x² + 6x + 9 = 5，即 (x + 3)² = 5。开平方：x + 3 = ±√5，故 x = −3 ± √5。
Example 3 (Quadratic formula): − 4(3)(−8) = 4 + 96 = 100. √100 = 10. x = (2 ± 10)/(6) → x = 12/6 = 2 or x = −8/6 = −4/3.
例3（求根公式）：解 3x² − 2x − 8 = 0。a = 3, b = −2, c = −8。Δ = (−2)² − 4(3)(−8) = 4 + 96 = 100。√100 = 10。x = (2 ± 10)/6 → x = 12/6 = 2 或 x = −8/6 = −4/3。
12. Common Mistakes and How to Avoid Them | 常见错误与规避方法
Forgetting to set the equation equal to zero before factorising is a frequent slip. Always rearrange first. With the quadratic formula, misidentifying the sign of b or c leads to wrong answers – write down a, b, c explicitly. When completing the square, many forget to add the same value to both sides or mishandle the coefficient a. Finally, check your solutions by substituting back into the original equation.
忘记先使方程等于零就进行因式分解是常见失误，务必先移项。使用求根公式时，误判 b 或 c 的符号会导致错误——请明确写出 a、b、c。配方法中，很多人忘记在两边同时加上同一个值，或未正确处理系数 a。最后，将所得解代入原方程验算即可避免大部分失分。
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