📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are among the most heavily tested topics in IGCSE Mathematics. You will meet them in pure algebra, coordinate geometry, and many practical problems. This revision guide explains four standard solution methods: factorisation, completing the square, the quadratic formula, and graphical methods. We will also examine the discriminant, which tells us the nature of the roots. Every method is demonstrated with fully worked examples in the style of IGCSE exam questions.
二次方程是 IGCSE 数学中考查频率最高的内容之一。无论是纯代数、坐标几何还是实际应用题，都会遇到二次方程。本复习指南讲解四种标准解法：因式分解、配方法、求根公式和图象法，并讨论如何利用判别式判断根的性质。每个方法均配有以 IGCSE 考试风格编写的完整例题。
1. What Is a Quadratic Equation? | 什么是二次方程
A quadratic equation is a polynomial equation in which the highest power of the unknown is 2. Its general form is
二次方程是未知数最高次数为 2 的多项式方程，其一般形式为
ax² + bx + c = 0，其中 a ≠ 0
Here a, b and c are real numbers, and a cannot be zero. If a = 0, the equation becomes linear. Examples of quadratic equations include x² − 4x + 3 = 0, 2x² + 7 = 0 and x² = 9. Note that x² = 9 can be rewritten as x² − 9 = 0, which is still quadratic because it contains the term x².
其中 a、b、c 为实数，且 a 不能为零。若 a = 0，方程就退化为一次方程。例如 x² − 4x + 3 = 0、2x² + 7 = 0 以及 x² = 9 都是二次方程。注意 x² = 9 可改写为 x² − 9 = 0，因为含有 x² 项，所以它仍然是二次方程。
2. Standard Form and Identifying Coefficients | 标准形式与识别系数
Before solving, always rearrange the equation into the standard form ax² + bx + c = 0. Only then can you correctly identify the coefficients a, b and c for use in the quadratic formula or discriminant.
求解之前，必须先将方程整理成标准形式 ax² + bx + c = 0。只有这样，才能正确识别系数 a、b、c，以便使用求根公式或判别式。
Example: solve x² = 6x − 8. Subtract 6x and add 8 to both sides, giving x² − 6x + 8 = 0. Hence a = 1, b = −6, c = 8. A common error is to attempt factorisation before moving all terms to one side.
例如：解方程 x² = 6x − 8。两边同时减去 6x、加上 8，得到 x² − 6x + 8 = 0，因此 a = 1，b = −6，c = 8。常见错误是未把所有项移到等号一侧就开始因式分解。
| Equation 方程 | a | b | c |
|---|---|---|---|
| 2x² + 5x − 3 = 0 | 2 | 5 | −3 |
| x² − 4x = 0 | 1 | −4 | 0 |
| −x² + 9 = 0 | −1 | 0 | 9 |
3. Solving by Factorisation | 因式分解法
Factorisation is quick when the quadratic can be written as the product of two linear brackets. Follow these steps:
当二次式能写成两个一次括号相乘的形式时，因式分解法最为快捷。步骤如下：
- Rearrange the equation so the right-hand side is 0.
将方程整理为右边等于 0 的形式。 - Factorise the left-hand side into two brackets.
将左边因式分解为两个括号的乘积。 - Use the zero product rule: if AB = 0, then A = 0 or B = 0.
利用零积规则：若 AB = 0，则 A = 0 或 B = 0。 - Solve the two resulting linear equations.
分别解两个一次方程。
Worked example: solve x² − 5x + 6 = 0. We need two numbers whose product is 6 and whose sum is −5: (−2) × (−3) = 6 and (−2) + (−3) = −5. Therefore (x − 2)(x − 3) = 0, so x = 2 or x = 3.
例题：解 x² − 5x + 6 = 0。要找两个数，乘积为 6 且和为 −5：(−2) × (−3) = 6，(−2) + (−3) = −5。因此 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。
When the coefficient of x² is not 1, work carefully with the factors. Example: solve 2x² + 5x − 3 = 0. This factorises as (2x − 1)(x + 3) = 0, giving x = ½ or x = −3. You can check by expanding: (2x − 1)(x + 3) = 2x² + 6x − x − 3 = 2x² + 5x − 3.
当 x² 的系数不为 1 时，处理因式要格外小心。例如解 2x² + 5x − 3 = 0，可分解为 (2x − 1)(x + 3) = 0，得到 x = ½ 或 x = −3。可通过展开验算：(2x − 1)(x + 3) = 2x² + 6x − x − 3 = 2x² + 5x − 3。
Special case: the difference of two squares. For example, x² − 25 = 0 gives (x − 5)(x + 5) = 0, so x = 5 or x = −5. Write the answers as x = ±5.
特殊情况：平方差公式。例如 x² − 25 = 0 可化为 (x − 5)(x + 5) = 0，所以 x = 5 或 x = −5，写作 x = ±5。
4. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. For the expression x² + bx, we add (b/2)² to make a perfect square:
配方法将二次式改写成 (x + p)² + q 的形式。对于 x² + bx 这样的表达式，加上 (b/2)² 即可构成完全平方：
x² + bx + (b/2)² = (x + b/2)²
Worked example: solve x² + 6x + 2 = 0. Here b = 6, so (b/2)² = 9. Write x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7. Then (x + 3)² = 7, so x + 3 = ±√7, hence x = −3 ± √7.
例题：解 x² + 6x + 2 = 0。因为 b = 6，所以 (b/2)² = 9。则 x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。于是 (x + 3)² = 7，x + 3 = ±√7，所以 x = −3 ± √7。
The two solutions are x = −3 + √7 ≈ −0.354 and x = −3 − √7 ≈ −5.646. Completing the square is especially useful when the equation does not factorise easily, and it also directly gives the turning point of the parabola: for y = (x + p)² + q, the vertex is at (−p, q).
两个解为 x = −3 + √7 ≈ −0.354 和 x = −3 − √7 ≈ −5.646。当二次方程不易因式分解时，配方法尤为有效；同时，它还能直接给出抛物线的顶点坐标：对于 y = (x + p)² + q，顶点为 (−p, q)。
5. The Quadratic Formula | 求根公式
For any quadratic equation ax² + bx + c = 0, the solutions are given by the quadratic formula:
对于任意二次方程 ax² + bx + c = 0，其解由求根公式给出：
x = (−b ± √(b² − 4ac)) / 2a
Worked example: solve 2x² − 4x − 6 = 0. Here a = 2, b = −4 and c = −6. Substituting into the formula:
例题：解 2x² − 4x − 6 = 0。其中 a = 2，b = −4，c = −6。代入公式：
x = (−(−4) ± √((−4)² − 4 × 2 × (−6))) / (2 × 2)
x = (4 ± √(16 + 48)) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4
So x = (4 + 8)/4 = 3 or
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