📚 Solving Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程求解：因式分解、配方法与求根公式
Quadratic equations run through the whole of the IGCSE and A Level mathematics syllabus, from algebraic manipulation to coordinate geometry and real-world modelling. This revision guide covers the three main algebraic methods for solving quadratics, the role of the discriminant, and how the solutions connect to the graph of a parabola.
二次方程贯穿 IGCSE 与 A Level 数学大纲，从代数运算到坐标几何、再到实际建模都会出现。本复习指南涵盖求解二次方程的三种主要代数方法、判别式的作用，以及方程的解与抛物线图像之间的联系。
1. Recognising a Quadratic Equation | 识别二次方程
A quadratic equation is any equation that can be rearranged into the form ax² + bx + c = 0, where x is the variable and a, b, c are constants with a ≠ 0. The defining feature is the x² term, which makes the relationship non-linear.
二次方程是任何可以整理为 ax² + bx + c = 0 形式的方程，其中 x 是变量，a、b、c 是常数且 a ≠ 0。它的决定性特征是含有 x² 项，这使关系成为非线性。
This means that equations such as 2x² − 7 = 0, x² + 4x = 1 and (x − 3)(x + 2) = 0 are all quadratics, while 5x + 3 = 0 is linear because the highest power of x is 1.
这意味着 2x² − 7 = 0、x² + 4x = 1 以及 (x − 3)(x + 2) = 0 都是二次方程，而 5x + 3 = 0 是一次方程，因为 x 的最高次数是 1。
In exam questions, quadratics often appear only after you have expanded brackets, cleared fractions or moved every term to one side. Identifying the equation type early helps you choose the most efficient solving strategy instead of guessing.
在考试题目中，二次方程常常在你展开括号、去分母或把各项移到一边之后才显现出来。尽早识别方程类型有助于你选择最高效的求解策略，而不是凭感觉硬算。
Three algebraic methods dominate the syllabus: factorising, completing the square, and the quadratic formula. All three give exactly the same roots - they simply suit different situations.
大纲中最主要的三种代数方法是：因式分解、配方法与求根公式。三种方法得到的根完全相同，只是适用于不同情形。
2. The Standard Form ax² + bx + c = 0 | 标准形式 ax² + bx + c = 0
Before you factorise or substitute into a formula, the equation must be in standard form with zero on the right-hand side. Solving from a rearranged or partially expanded form is the single most common source of lost marks.
在因式分解或代入公式之前，方程必须化为右侧为 0 的标准形式。在未整理或未完全展开的形式下求解，是最常见的失分原因。
ax² + bx + c = 0, a ≠ 0
3x² = 5x − 2 becomes 3x² − 5x + 2 = 0, so a = 3, b = −5, c = 2.
3x² = 5x − 2 化为 3x² − 5x + 2 = 0，因此 a = 3，b = −5，c = 2。
x(x + 4) = 5 becomes x² + 4x − 5 = 0, so a = 1, b = 4, c = −5.
x(x + 4) = 5 化为 x² + 4x − 5 = 0，因此 a = 1，b = 4，c = −5。
2x² + 8 = 10x becomes 2x² − 10x + 8 = 0, which can be divided by 2 to give x² − 5x + 4 = 0.
2x² + 8 = 10x 化为 2x² − 10x + 8 = 0，两边除以 2 可得 x² − 5x + 4 = 0。
Dividing through by a common factor never changes the roots, and it makes factorising much easier. However, never divide by x itself, because x = 0 may be a valid solution.
约去公因数不会改变方程的根，却能让因式分解容易得多。但绝不要除以 x，因为 x = 0 可能本身就是有效解。
3. Method 1: Factorising | 方法一：因式分解
Factorising reverses the process of expanding brackets. When a = 1, you look for two numbers that multiply to give c and add to give b, then write the quadratic as (x + p)(x + q).
因式分解是展开括号的逆过程。当 a = 1 时，你需要找两个数，使它们的乘积等于 c、和等于 b，然后把二次式写成 (x + p)(x + q)。
The method works because of the null factor law: if the product of two expressions is zero, then at least one of them must be zero. Setting each bracket equal to zero gives the two roots separately.
这个方法成立的原因是零积定律：若两个表达式的乘积为零，则其中至少有一个为零。令每个括号分别等于零，就能独立求出两个根。
If (x + p)(x + q) = 0 then x = −p or x = −q
Common factor: x² − 9x = 0 gives x(x − 9) = 0, so x = 0 or x = 9.
提公因式：x² − 9x = 0 化为 x(x − 9) = 0，所以 x = 0 或 x = 9。
Difference of two squares: x² − 16 = 0 gives (x − 4)(x + 4) = 0, so x = ±4.
平方差：x² − 16 = 0 化为 (x − 4)(x + 4) = 0，所以 x = ±4。
When a ≠ 1: split the middle term or use trial and error, for example 2x² + 7x + 3 = (2x + 1)(x + 3).
当 a ≠ 1 时：可拆分中间项或逐步试算，例如 2x² + 7x + 3 = (2x + 1)(x + 3)。
Factorising is the fastest method, but it only works neatly when the roots are rational. If the numbers refuse to cooperate, switch to the formula rather than spending minutes hunting for a non-existent pair.
因式分解是最快的方法，但只在根为有理数时才干净利落。如果怎么都凑不出来，应果断改用求根公式，而不是花几分钟去找根本不存在的数对。
4. Factorising in Practice: Worked Examples | 因式分解实例：典型例题
Example 1. Solve x² − 5x + 6 = 0. The numbers −2 and −3 multiply to 6 and add to −5, so (x − 2)(x − 3) = 0 and x = 2 or x = 3.
例 1. 解 x² − 5x + 6 = 0。数 −2 与 −3 的乘积为 6、和为 −5，所以 (x − 2)(x − 3) = 0，得 x = 2 或 x = 3。
Example 2. Solve x² + 2x − 15 = 0. Here +5 and −3 work, giving (x + 5)(x − 3) = 0, so x = −5 or x = 3.
例 2. 解 x² + 2x − 15 = 0。这里 +5 与 −3 满足条件，得到 (x + 5)(x − 3) = 0，所以 x = −5 或 x = 3。
Example 3. Solve 2x² + 7x . Since 2 × 3 = 6 and we need two numbers multiplying to 6 and adding to 7, use 6 and 1: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Hence x = −½ or x = −3.
例 3. 解 2x² + 7x + 3 = 0。因为 2 × 3 = 6，需要两个数乘积为 6、和为 7，取 6 与 1：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。因此 x = −½ 或 x = −3。
Example 4. Solve x² = 9x. Rearranging gives x² − 9x = 0, then x(x − 9) = 0, so x = 0 or x = 9. Dividing both sides by x would lose the root x = 0 and cost you marks.
例 4. 解 x² = 9x。移项得 x² − 9x = 0，再因式分解为 x(x − 9) = 0，所以 x = 0 或 x = 9。若两边同时除以 x，就会丢掉根 x = 0，直接丢分。
Always check your factorisation by expanding it back out mentally. A quick expansion of two brackets takes five seconds and catches sign errors immediately.
务必在脑中把因式分解结果重新展开检验一遍。展开两个括号只需五秒，却能立刻发现符号错误。
5. Method 2: Completing the Square | 方法二：配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. It always works, even when the roots are irrational, and it also reveals the turning point of the parabola.
配方法把二次式改写为 (x + p)² + q 的形式。它总是可行，即使根是无理数也不例外，而且还能直接揭示抛物线的顶点。
The key identity is that x² + bx is almost (x + b/2)², but the squaring introduces an extra term (b/2)², which must be subtracted to keep the expression unchanged.
关键恒等式是：x² + bx 几乎就是 (x + b/2)²，但平方展开后会多出一项 (b/2)²，必须把它减掉才能保持原式不变。
x² + bx = (x + b/2)² − (b/2)²
x² + 10x = (x + 5)² − 25
x² + 10x = (x + 5)² − 25
x² − 6x + 2 = (x − 3)² − 9 + 2 = (x − 3)² − 7
x² − 6x + 2 = (x − 3)² − 9 + 2 = (x − 3)² − 7
When a ≠ 1, first take a out as a factor of the first two terms: ax² + bx + c = a(x² + (b/a)x) + c.
当 a ≠ 1 时，先把前两项中的 a 提出来：ax² + bx + c = a(x² + (b/a)x) + c。
s, and remember the ± sign. Forgetting the negative square root is the classic error here.
求解时，先把常数移到等号另一边，再对两边开平方，并记得加上 ± 号。忘记负的平方根是这里最典型的错误。
6. Completing the Square: Worked Examples | 配方法：典型例题
Example 1. Solve x² + 6x + 4 = 0. Complete the square: (x + 3)² − 9 + 4 = 0, so (x + 3)² = 5 and x = −3 ± √5, giving x ≈ 0.764 or x ≈ −6.764.
例 1. 解 x² + 6x + 4 = 0。配方得 (x + 3)² − 9 + 4 = 0，即 (x + 3)² = 5，故 x = −3 ± √5，得到 x ≈ 0.764 或 x ≈ −6.764。
Example 2. Solve 2x² − 8x + 3 = 0. Factor out 2: 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5. Setting this to zero gives (x − 2)² = 5/2, so x = 2 ± √10/2.
例 2. 解 2x² − 8x + 3 = 0。提出 2：2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5。令其为零得 (x − 2)² = 5/2，所以 x = 2 ± √10/2。
Example 3. Solve x² − 5x + 1 = 0. Here (x − 5/2)² − 25/4 + 1 = 0, so (x − 5/2)² = 21/4 and x = (5 ± √21)/2, approximately 4.791 or 0.209.
例 3. 解 x² − 5x + 1 = 0。此时 (x − 5/2)² − 25/4 + 1 = 0，即 (x − 5/2)² = 21/4，故 x = (5 ± √21)/2，约为 4.791 或 0.209。
The completed-square form also gives the vertex directly. For y = 2(x − 2)² − 5 the minimum point is (2, −5), which examiners frequently ask for in the same question as the roots.
配方的结果还能直接给出顶点。对 y = 2(x − 2)² − 5，其最低点为 (2, −5)，考官常在同一道题中既问根又问顶点。
7. Method 3: The Quadratic Formula | 方法三：求根公式
The quadratic formula solves any quadratic equation, whatever the coefficients, as long as the equation is first written in standard form. It should be memorised, since it is not always given in the formula booklet.
只要先把方程写成标准形式，求根公式可以解任何二次方程，无论系数如何。由于公式手册中不一定提供，务必牢记。
x = (−b ± √b² − 4ac) / (2a)
Notice that the whole numerator is divided by 2a, and the square root covers the entire expression b² − 4ac. When b is negative, −b becomes positive, so write out the substitution line before simplifying.
注意整个分子都要除以 2a，而根号覆盖的是 b² − 4ac 整体。当 b 为负时，−
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