📚 Mastering Quadratic Equations for IGCSE | IGCSE 二次方程求解精讲
Quadratic equations are one of the most important algebra topics in IGCSE Mathematics. They appear in pure algebra questions, word problems, graphical problems and even in some geometry and trigonometry contexts. A strong understanding of how to recognise, rearrange and solve quadratic equations by different methods is essential for a high grade.
二次方程是 IGCSE 数学代数部分最重要的主题之一。它们出现在纯代数题、文字题、图像题，甚至一些几何和三角学问题中。扎实掌握如何识别、整理并用不同方法求解二次方程，是取得高分的关键。
1. Recognising Quadratic Equations | 识别二次方程
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 condition a ≠ 0 is essential because if a = 0, the equation becomes linear rather than quadratic. In IGCSE exam papers, quadratic equations often appear hidden inside word problems or after rearranging an expression.
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 为常数且 a ≠ 0。a ≠ 0 这一条件至关重要，因为如果 a = 0，方程就退化为一次方程。在 IGCSE 考试中，二次方程经常隐藏在文字题中，或需要移项整理后才会显现。
For example, 2x² + 3x − 5 = 0 and x² − 4x = 0 are quadratic equations. However, 0x² + 2x + 1 = 0 is just 2x + 1 = 0, so it is not quadratic.
例如，2x² + 3x − 5 = 0 和 x² − 4x = 0 都是二次方程。然而，0x² + 2x + 1 = 0 实际上就是 2x + 1 = 0，因此它不是二次方程。
2. Standard Form and Coefficients | 标准形式与系数
Before choosing a method, you should always rewrite the equation in standard form: ax² + bx + c = 0. This means one side must be zero, and the terms should usually be arranged in descending powers of x. Once in standard form, identify the coefficients a, b and c carefully, including their signs.
在选择解法之前，你应当先把方程写成标准形式：ax² + bx + c = 0。这意味着等式一边必须为零，而且各项通常按照 x 的降幂排列。写成标准形式后，要仔细确定系数 a、b、c，包括它们的正负号。
For example, rearrange x² = 5x − 6 into x² − 5x + 6 = 0. Here a = 1, b = −5 and c = 6. Note that b is negative, which is a common source of sign errors.
例如，将 x² = 5x − 6 移项为 x² − 5x + 6 = 0。这里 a = 1，b = −5，c = 6。注意 b 是负数，这是常见的符号错误来源。
3. Solving by Factorisation | 因式分解法求解
Factorisation is often the fastest method when the quadratic has simple integer roots. The idea is to write ax² + bx + c as a product of two linear factors. For a monic quadratic, x² + (p+q)x + pq = (x + p)(x + q), so we need two numbers whose sum is b and whose product is c.
当二次方程具有简单整数根时，因式分解通常是最快的方法。其思路是将 ax² + bx + c 写成两个一次因式的乘积。对于首项系数为 1 的二次式，x² + (p+q)x + pq = (x + p)(x + q)，因此我们需要找到两个数，使它们的和等于 b，乘积等于 c。
x² + (p + q)x + pq = (x + p)(x + q)
Solve x² + 5x + 6 = 0. We need two numbers with sum 5 and product 6: 2 and 3. So (x + 2)(x + 3) = 0, giving x = −2 or x = −3.
求解 x² + 5x + 6 = 0。我们需要两个数，和为 5，积为 6：它们是 2 和 3。因此 (x + 2)(x + 3) = 0，得到 x = −2 或 x = −3。
When a ≠ 1, use splitting the middle term or trial and error. For 2x² + 7x + 3, split 7x as 6x + x, then factor in pairs: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 a ≠ 1 时，可以使用裂项法或尝试法。例如 2x² + 7x + 3，把 7x 拆成 6x + x，然后分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
4. Solving by Completing the Square | 配方法求解
Completing the square is a powerful method because it works for any quadratic equation and also helps with finding the vertex of a parabola. The key identity is (x + p)² = x² + 2px + p², so half the coefficient of x becomes p.
配方法是一种重要的方法，因为它适用于任何二次方程，也有助于求抛物线的顶点。关键恒等式是 (x + p)² = x² + 2px + p²，因此 x 项系数的一半就是 p。
(x + p)² = x² + 2px + p²
Solve x² + 6x + 2 = 0. First write x² + 6x as (x + 3)² − 9, so the equation becomes (x + 3)² − 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7, giving x = −3 ± √7.
求解 x² + 6x + 2 = 0。首先把 x² + 6x 写成 (x + 3)² − 9，因此方程变为 (x + 3)² − 7 = 0。然后 (x + 3)² = 7，所以 x + 3 = ±√7，得到 x = −3 ± √7。
5. Solving by the Quadratic Formula | 二次公式求解
The quadratic formula gives the solutions of ax² + bx + c = 0 directly. It is especially useful when factorisation is difficult or when the coefficients are not integers.
二次公式可以直接给出 ax² + bx + c = 0 的解。当因式分解困难或系数不是整数时，它尤其有用。
x = (−b ± √(b² − 4ac)) ÷ (2a)
Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substitute: x = (−3 ± √(3² − 4×2×(−2))) ÷ (2×2) = (−3 ± √(9 + 16)) ÷ 4 = (−3 ± 5) ÷ 4. Therefore x = 0.5 or x = −2.
求解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入公式：x = (−3 ± √(3² − 4×2×(−2))) ÷ (2×2) = (−3 ± √(9 + 16)) ÷ 4 = (−3 ± 5) ÷ 4。因此 x = 0.5 或 x = −2。
6. The Discriminant and Nature of Roots | 判别式与根的性质
The expression b² − 4ac is called the discriminant, often written as Δ = b² − 4ac. It tells us about the number and type of roots without solving the equation.
表达式 b² − 4ac 称为判别式，通常写作 Δ = b² − 4ac。它可以在不求解方程的情况下告诉我们根的数量和类型。
| Discriminant | Nature of roots | 根的性质 |
| Δ > 0 | Two distinct real roots | 更多咨询请联系16621398022（同微信）
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