📚 Solving Quadratic Equations: Factorising, Completing the Square and the Formula | IGCSE数学：解二次方程——因式分解、配方法与公式法
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from pure algebra to geometry, measurement and word problems. This article brings together the three main algebraic methods - factorising, completing the square and the quadratic formula - and shows how each method connects to the graph of a parabola, so that you can choose the fastest and safest route in the exam.
二次方程贯穿 IGCSE 数学大纲，从纯代数到几何、测量和应用题都会用到。本文汇总三种主要代数方法——因式分解、配方法和二次公式法，并说明每种方法与抛物线图像之间的联系，帮助你在考场上选择最快、最稳妥的解法。
1. Recognising Quadratic Equations | 识别二次方程
re constants and a ≠ 0. The highest power of the unknown x is 2, which is why it is called 'quadratic' or second-degree.
二次方程是任何可以整理成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。未知数 x 的最高次数为 2，因此称为二次方程或二阶方程。
x - 4 = 0 are quadratic equations, but x³ + x² = 0 is not, because its highest power is 3.
例如 x² - 5x + 6 = 0 和 2x² + 3x - 4 = 0 是二次方程，而 x³ + x² = 0 不是，因为它的最高次数是 3。
Many equations are quadratic in disguise: the equation only becomes clear after expanding brackets, multiplying by x, or moving every term to one side.
很多方程是"伪装"的二次方程：展开括号、两边乘以 x，或把所有项移到一边之后，二次结构才会显现。
- x(x + 4) = 12 expands to x² + 4x - 12 = 0, a quadratic equation. | x(x + 4) = 12 展开后为 x² + 4x - 12 = 0，是二次方程。
- x + 3/x = 5 becomes x² - 5x + 3 = 0 after multiplying by x (x ≠ 0). | x + 3/x = 5 两边乘以 x 后变为 x² - 5x + 3 = 0（x ≠ 0）。
- (x - 2)² = 7 expands to x² - 4x - 3 = 0. | (x - 2)² = 7 展开后为 x² - 4x - 3 = 0。
- x³ + x² = 0 is cubic, not quadratic. | x³ + x² = 0 是三次方程，不是二次方程。
2. Standard Form and Core Terms | 标准形式与核心术语
are real numbers, and a cannot be zero because otherwise the equation would become linear.
二次方程的标准形式为 ax² + bx + c = 0。其中 a、b、c 为实数，且 a 不能为零，否则方程就退化为一次方程。
The term ax² is called the quadratic term, bx is the linear term, and c is the constant term. Naming them correctly helps you avoid sign errors when you substitute into the formula.
ax² 称为二次项，bx 称为一次项，c 称为常数项。准确辨认这三项，可以在代入公式时避免符号错误。
| Equation | a | b | c |
|---|---|---|---|
| x² - 5x + 6 = 0 | 1 | -5 | 6 |
| 2x² + 3x - 4 = 0 | 2 | 3 | -4 |
| x² = 9, rewritten as x² - 9 = 0 | 1 | 0 | -9 |
| 4 - 3x² = x, rewritten as 3x² + x - 4 = 0 | 3 | 1 | -4 |
Before solving, always rearrange the equation into standard form and collect like terms. Solving a quadratic that is not equal to zero means you cannot use the zero-product property, and substituting the wrong c into the formula is one of the most common exam errors.
解题前务必先把方程整理成标准形式并合并同类项。若方程右边不是零，就不能使用零积性质；而把符号错的 c 代入公式，是考试中最常见的失分点之一。
3. Solving by Factorising | 用因式分解法求解
Factorising reverses the process of expanding brackets. The method relies on the zero-product property: if two factors multiply to give zero, then at least one of them must be zero.
因式分解是展开括号的逆过程。该方法依赖零积性质：若两个因式的乘积为零，则其中至少有一个因式为零。
If AB = 0, then A = 0 or B = 0
p and q such that p × q = c and p + q = b. The factorised form is then (x + p)(x + q) = 0.
对于二次项系数为 1 的方程 x² + bx + c，要找两个数 p 和 q，满足 p × q = c 且 p + q = b，因式分解形式即为 (x + p)(x + q) = 0。
to 6 and add to -5, namely -2 and -3, so (x - 2)(x - 3) = 0 and therefore x = 2 or x = 3.
例：解 x² - 5x + 6 = 0。需要两数相乘得 6、相加得 -5，即 -2 与 -3，所以 (x - 2)(x - 3) = 0，从而 x = 2 或 x = 3。
r 3 and 7 multiplies to 21 and adds to 10, giving (x + 3)(x + 7) = 0, so x = -3 or x = -7.
例：解 x² + 10x + 21 = 0。3 与 7 相乘得 21、相加得 10，得到 (x + 3)(x + 7) = 0，所以 x = -3 或 x = -7。
Sign patterns are worth memorising. If c is positive and b is positive, both numbers are positive; if c is positive and b is negative, both numbers are negative; if c is negative, the two numbers have opposite signs.
符号规律值得记牢：c 为正、b 为正，两数同为正；c 为正、b 为负，两数同为负；c 为负，则两数异号。
4. Factorising When a ≠ 1 | 二次项系数不为 1 时的因式分解
, where m × n = a and p × q = c. A reliable technique is the split-the-middle-term method.
当 x² 的系数不为 1 时，分解形式为 (mx + p)(nx + q)，其中 m × n = a，p × q = c。一个稳妥的技巧是"拆中间项"法。
d add to b. Step 3: split bx into those two terms. Step 4: factorise by grouping in pairs.
第一步：计算 a × c；第二步：找两个数，相乘得 a × c、相加得 b；第三步：把 bx 拆成这两项；第四步：分组提取公因式。
groups as 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0. Hence x = -1/2 or x = -3.
例：解 2x² + 7x + 3 = 0。此处 a × c = 6，6 与 1 相乘得 6、相加得 7。于是 2x² + 6x + x + 3 = 0，分组为 2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0，所以 x = -1/2 或 x = -3。
Always check your answer by expanding the brackets mentally or on paper. If the middle term or the constant term does not match the original equation, the factorisation is wrong.
务必通过展开括号来检验答案。若一次项或常数项与原来的方程不符，说明分解有误。
- If a × c has many factor pairs, test them systematically rather than guessing. | 若 a × c 的因数对很多，应系统尝试而不是乱猜。
- out a common numerical factor first, e.g. 2x² + 4x - 6 = 2(x² + 2x - 3) = 2(x + 3)(x - 1). | 先提取公因数，例如 2x² + 4x - 6 = 2(x² + 2x - 3) = 2(x + 3)(x - 1)。
- If no integer pair works, the equation may need the formula instead. | 若找不到整数因数对，就该改用公式法。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression as a perfect square plus a constant. It is the key technique for finding the vertex of a parabola and for solving equations that do not factorise.
配方法把二次式改写为"完全平方加常数"的形式。它是求抛物线顶点、以及解无法因式分解的方程的关键技巧。
x² + bx = (x + b/2)² - (b/2)²
)² = x² + 8x + 16. Subtracting the extra 16 and adding the original 3 gives (x + 4)² - 13.
例：把 x² + 8x + 3 写成配方形式。8 的一半是 4，所以完全平方为 (x + 4)² = x² + 8x + 16。减去多出的 16 再加上原来的 3，得到 (x + 4)² - 13。
When the coefficient of x² is a, first factor a out of the first two terms, then complete the square inside the bracket.
当 x² 的系数为 a 时，先从二次项和一次项中提取 a，再在括号内配方。
ax² + bx + c = a(x + b/2a)² + c - b²/4a
Example: 3x² - 12x + 7 = 3(x² - 4x) + 7 = 3[(x - 2)² - 4] + 7 = 3(x - 2)² - 5.
例：3x² - 12x + 7 = 3(x² - 4x) + 7 = 3[(x - 2)² - 4] + 7 = 3(x - 2)² - 5。
ex (h, k) and the line of symmetry x = h, which is why it is so useful in graph questions.
配方形式 a(x - h)² + k 能直接读出顶点 (h, k) 和对称轴 x = h，这正是它在图像题中特别有用的原因。
6. Solving Equations by Completing the Square | 用配方法解方程
To solve a quadratic equation by completing the square, isolate the squared bracket and then take square roots of both sides. Remember that taking a square root produces both a positive and a negative value.
用配方法解二次方程时，先把完全平方项单独留在一边，然后两边开平方。注意开平方会产生正、负两个值。
8x + 3 = 0. Completing the square gives (x + 4)² - 13 = 0, so (x + 4)² = 13 and x + 4 = ±√13. Therefore x = -4 + √13 or x = -4 - √13, that is x ≈ -0.39 or x ≈ -7.61.
例：解 x² + 8x + 3 = 0。配方得 (x + 4)² - 13 = 0，所以 (x + 4)² = 13，x + 4 = ±√13。因此 x = -4 + √13 或 x = -4 - √13，即 x ≈ -0.39 或 x ≈ -7.61。
Example: solve 3x² - 12x + 7 = 0. From 3(x - 2)² - 5 = 0 we get (x - 2)² = 5/3, so x = 2 ± √5/3, which simplifies to x = 2 ± √15/3, approximately 3.29 or 0.71.
例：解 3x² - 12x + 7 = 0。由 3(x - 2
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