📚 Solving Quadratic Equations | 解一元二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics, appearing in both Core and Extended papers. They show up as direct algebra questions and are also woven into geometry, trigonometry and problem-solving contexts. Mastering the main strategies – factorisation, the quadratic formula, completing the square and graphical methods – is essential for a top grade.
解一元二次方程是 IGCSE 数学（核心与扩展课程）中最重要的内容之一。它既会作为纯代数题出现，也会融入几何、三角和实际应用题中。掌握因式分解、求根公式、配方和图像法这四大核心方法是取得高分的关键。
1. What Is a Quadratic Equation? | 什么是一元二次方程
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The word ‘quadratic’ comes from the Latin word ‘quadratus’, meaning ‘square’.
一元二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数且 a ≠ 0。’二次’一词源自拉丁语 ‘quadratus’，意为’正方形’。
The highest power of the unknown x is 2, so a quadratic equation has at most two solutions, which are also called roots. These roots can be real and different, real and equal, or not real at all.
未知数 x 的最高次数是 2，所以一元二次方程至多有两个解，也称为根。这些根可能是两个不同的实数、两个相等的实数，或者根本不是实数。
Examples: x² – 5x + 6 = 0; 2x² + 3x – 7 = 0; x² = 4.
例如：x² – 5x + 6 = 0；2x² + 3x – 7 = 0；x² = 4。
2. Solving by Factorisation | 因式分解法
Factorisation is the fastest method when the quadratic expression can be written as a product of two linear brackets. For a quadratic of the form x² + bx + c, find two numbers that multiply to give c and add to give b.
当二次式能够写成两个一次括号相乘的形式时，因式分解法是最快的。对于 x² + bx + c 形式的二次方程，需要找到两个数，使其乘积等于 c，和等于 b。
Example: Solve x² – 5x + 6 = 0. We need two numbers that multiply to 6 and add to -5: these are -2 and -3. Therefore (x – 2)(x – 3) = 0, so x = 2 or x = 3.
例：解方程 x² – 5x + 6 = 0。需要找到两个数，乘积为 6、和为 -5，即 -2 和 -3。因此 (x – 2)(x – 3) = 0，所以 x = 2 或 x = 3。
If the leading coefficient is not 1, for example 2x² + 5x + 3 = 0, use the ‘product-sum’ method: multiply a and c (2 × 3 = 6), then find two numbers that multiply to 6 and add to 5, which are 2 and 3. Rewrite the middle term as 2x + 3x and factor by grouping.
如果首项系数不是 1，例如 2x² + 5x + 3 = 0，可以使用’乘积-和’法：先把 a 与 c 相乘（2 × 3 = 6），再找两个数使其乘积为 6、和为 5，即 2 和 3。将中间项改写成 2x + 3x，然后分组因式分解。
Always use the zero product property: if A × B = 0, then A = 0 or B = 0. This is exactly why factorisation only works when the equation is rearranged to equal zero.
务必使用零乘积性质：若 A × B = 0，则 A = 0 或 B = 0。这正是因式分解仅在方程整理成等于零的形式时才有效的原因。
3. The Quadratic Formula | 求根公式
When factorisation is difficult or impossible, use the quadratic formula:
当因式分解困难或无法进行时，可运用求根公式：
x = (-b ± √(b² – 4ac)) / (2a)
Here a, b and c are the coefficients from ax² + bx + c = 0. The symbol ± means you must calculate both x = (-b + √(b² – 4ac)) / (2a) and x = (-b – √(b² – 4ac)) / (2a).
其中 a、b、c 是方程 ax² + bx + c = 0 的系数。符号 ± 表示需要分别计算 x = (-b + √(b² – 4ac)) / (2a) 和 x = (-b – √(b² – 4ac)) / (2a)。
Example: Solve 2x² – 4x – 3 = 0. Here a = 2, b = -4, c = -3. Substituting into the formula gives x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √10 / 2. Therefore x ≈ 2.58 or x ≈ -1.58 to 2 decimal places.
例：解方程 2x² – 4x – 3 = 0。此时 a = 2，b = -4，c = -3。代入求根公式得 x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √10 / 2。因此 x ≈ 2.58 或 x ≈ -1.58（精确到两位小数）。
Always check that the equation is in standard form ax² + bx + c = 0 before substituting. A common mistake is using c = 0 for the equation x² – 4x = 3; the correct form is x² – 4x – 3 = 0, so c = -3.
代入数值前，务必检查方程是否已化为标准形式 ax² + bx + c = 0。常见错误是在方程 x² – 4x = 3 中误取 c = 0；正确形式应为 x² – 4x – 3 = 0，即 c = -3。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. For a monic quadratic where a = 1, add and subtract (b/2)².
配方法把二次式改写为 a(x + p)² + q 的形式。对于首项系数为 1 的二次式，需要加减 (b/2)²。
Example: x² + 6x – 4 = 0. Take half of 6, which is 3, and square it to get 9. Then x² + 6x = (x + 3)² – 9. The equation becomes (x + 3)² – 9 – 4 = 0, that is (x + 3)² = 13. Taking square roots gives x + 3 = ±√13, so x = -3 ± √13.
例：x² + 6x – 4 = 0。取 6 的一半即 3，平方得 9。于是 x² + 6x = (x + 3)² – 9。方程变为 (x + 3)² – 9 – 4 = 0，即 (x + 3)² = 13。开平方得 x + 3 = ±√13，所以 x = -3 ± √13。
For a non-monic quadratic such as 2x² + 8x + 3 = 0, first factor out the 2: 2(x² + 4x) + 3 = 0. Then complete the square inside the bracket: 2[(x + 2)² – 4] + 3 = 2(x + 2)² – 8 + 3 = 2(x + 2)² – 5 = 0.
对于首项系数不为 1 的二次式，如 2x² + 8x + 3 = 0，先把 2 提取出来：2(x² + 4x) + 3 = 0。再在括号内配方：2[(x + 2)² – 4] + 3 = 2(x + 2)² – 8 + 3 = 2(x + 2)² – 5 = 0。
Completing the square is also the fastest way to find the turning point of a quadratic graph: for y = a(x + p)² + q, the vertex is (-p, q).
配方法也是快速求二次函数图像顶点的方法：对于 y = a(x + p)² + q，顶点坐标为 (-p, q)。
5. The Discriminant | 判别式
The expression b² – 4ac under the square root in the quadratic formula is called the discriminant, often written as Δ (Delta). It tells us the nature of the roots without solving the whole equation.
求根公式根号内的表达式 b² – 4ac 称为判别式，常用 Δ（大写希腊字母 Delta）表示。它让我们不必解方程就能判断根的性质。
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If b
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