Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式

📚 Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式

Quadratic equations appear throughout IGCSE Mathematics, from algebraic manipulation to real-world projectile problems. A quadratic equation is any equation that can be written in the form ax² + bx + c = 0 where a ≠ 0. In this revision guide, we explore three standard solution methods: factorisation, completing the square, and the quadratic formula. We also examine the discriminant, graphical interpretation, and common exam pitfalls.

二次方程贯穿 IGCSE 数学，从代数运算到现实世界的抛体运动问题。二次方程是可以写成 ax² + bx + c = 0 形式且 a ≠ 0 的方程。本复习指南将讲解三种标准解法：因式分解法、配方法和求根公式法，同时分析判别式、图像意义以及常见考试易错点。

1. Recognising a Quadratic Equation | 识别二次方程

A quadratic equation contains a variable raised to the second power as its highest power. The general form is ax² + bx + c = 0, where a, b and c are constants and a cannot be zero. If a = 0, the equation becomes linear. For example, 3x² – 5x + 2 = 0 is quadratic, while 2x + 7 = 0 is linear.

二次方程的最高次数是 2，标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a 不能为 0。若 a = 0，方程就退化为一次方程。例如 3x² – 5x + 2 = 0 是二次方程，而 2x + 7 = 0 是一次方程。

Some equations need rearranging before they look quadratic. For instance, x(x + 3) = 10 can be expanded to x² + 3x – 10 = 0. Always simplify and set one side to zero first.

有些方程需要先整理才能看出是二次方程。例如 x(x + 3) = 10 展开后可写成 x² + 3x – 10 = 0。解答前务必先化简并将一边设为零。

2. Standard Form ax² + bx + c = 0 | 标准形式 ax² + bx + c = 0

Writing an equation in standard form is essential because all solution methods rely on identifying a, b and c correctly. Consider 5 – 2x = x². Rearranging gives -x² – 2x + 5 = 0, or equivalently x² + 2x – 5 = 0 after multiplying by -1. Here a = 1, b = 2, and c = -5.

将方程写成标准形式至关重要，因为所有解法都依赖于正确识别 a、b、c。例如 5 – 2x = x²，移项得 -x² – 2x + 5 = 0，乘以 -1 后等价于 x² + 2x – 5 = 0。此时 a = 1、b = 2、c = -5。

Pay close attention to signs. In 2x² – 7 = 0, there is no x term, so b = 0. In 4x² + 3x = 0, there is no constant term, so c = 0. Missing terms are often tested in IGCSE papers.

要特别注意符号。在 2x² – 7 = 0 中没有 x 项，因此 b = 0。在 4x² + 3x = 0 中没有常数项，因此 c = 0。缺项情况在 IGCSE 试卷中经常出现。

3. Solving by Factorisation | 因式分解法

Factorisation works when the quadratic can be written as a product of two linear factors equal to zero. If pq = 0, then p = 0 or q = 0. For example, solve x² – 5x + 6 = 0. We look for two numbers that multiply to 6 and add to -5: they are -2 and -3. Therefore (x – 2)(x – 3) = 0, giving x = 2 or x = 3.

当二次方程可以分解为两个一次因式的乘积并等于零时，就能用因式分解法。若 pq = 0，则 p = 0 或 q = 0。例如解 x² – 5x + 6 = 0。我们寻找两个数，乘积为 6，和为 -5：这两个数是 -2 和 -3。因此 (x – 2)(x – 3) = 0，得到 x = 2 或 x = 3。

When the coefficient of x² is not 1, use trial and error or grouping. For 2x² + 7x + 3, factors are (2x + 1)(x + 3). Check by expanding: 2x² + 6x + x + 3 = 2x² + 7x + 3. Setting to zero gives x = -1/2 or x = -3.

当 x² 的系数不是 1 时，可采用试错法或分组法。例如 2x² + 7x + 3 可分解为 (2x + 1)(x + 3)。展开检验：2x² + 6x + x + 3 = 2x² + 7x + 3。令其等于零，得 x = -1/2 或 x = -3。

4. Completing the Square | 配方法

Completing the square transforms the quadratic into the form (x + p)² = q. This method is especially useful when the quadratic does not factorise neatly. Start with x² + 6x + 2 = 0. Move the constant: x² + 6x = -2. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 7. Write as (x + 3)² = 7. Then x + 3 = ±√7, so x = -3 ± √7.

配方法将二次方程变形为 (x + p)² = q 的形式。当方程无法整洁地因式分解时，这种方法尤其有用。以 x² + 6x + 2 = 0 为例，先将常数移项：x² + 6x = -2。两边加上 (6/2)² = 9：x² + 6x + 9 = 7。写成 (x + 3)² = 7。于是 x + 3 = ±√7，所以 x = -3 ± √7。

If the coefficient a is not 1, factor it out first. For 2x² + 8x + 5 = 0, write 2(x² + 4x) = -5. Add (4/2)² = 4 inside the bracket: 2(x² + 4x + 4) = -5 + 8. Thus 2(x + 2)² = 3, so (x + 2)² = 3/2. Then x = -2 ± √(3/2).

如果系数 a 不是 1，需要先将其提出来。例如 2x² + 8x + 5 = 0，写成 2(x² + 4x) = -5。在括号内加上 (4/2)² = 4：2(x² + 4x + 4) = -5 + 8。于是 2(x + 2)² = 3，所以 (x + 2)² = 3/2。最终 x = -2 ± √(3/2)。

5. The Quadratic Formula | 求根公式

The quadratic formula solves any quadratic equation ax² + bx + c = 0. The formula is:

求根公式可以解任意二次方程 ax² + bx + c = 0。公式为：

x = (-b ± √(b² – 4ac)) / (2a)

This formula is derived by completing the square on the general form. It always works, even when factorisation is difficult or impossible. Substitute the values of a, b and c carefully, and pay attention to negative signs.

该公式通过对一般形式配方推导而来。即使因式分解困难或无法进行时，它也一定能使用。代入 a、b、c 时要仔细，尤其注意负号。

Example: Solve 3x² – 4x – 2 = 0. Here a = 3, b = -4, c = -2. Compute b² – 4ac = (-4)² – 4(3)(-2) = 16 + 24 = 40. Thus x = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3.

示例：解 3x² – 4x – 2 =

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