Solving Quadratic Equations by Factorisation | IGCSE 数学：因式分解法求解二次方程

📚 Solving Quadratic Equations by Factorisation | IGCSE 数学：因式分解法求解二次方程

Quadratic equations appear frequently in IGCSE Mathematics, and factorisation is one of the most direct methods for solving them. Mastering this technique allows you to solve many equations quickly and accurately without needing the quadratic formula.

二次方程在 IGCSE 数学中频繁出现，而因式分解是求解二次方程最直接的方法之一。掌握这一技巧，你无需使用求根公式，就能快速而准确地解决许多方程。

1. Understanding Quadratic Equations | 理解二次方程

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a ≠ 0, and a, b, c are constants. The highest power of the variable x is 2, which gives the equation its name ‘quadratic’.

二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a ≠ 0，a、b、c 为常数。变量 x 的最高次数为 2，因此得名“二次”方程。

2. Why Factorisation Works | 因式分解法的原理

The foundation of solving quadratic equations by factorisation lies in the zero product property: if the product of two factors is zero, then at least one of the factors must be zero. In symbols, if p × q = 0, then p = 0 or q = 0.

用因式分解法求解二次方程的基础是零积性质：如果两个因数的乘积为零，那么至少有一个因数必须为零。用符号表示，若 p × q = 0，则 p = 0 或 q = 0。

If (x − m)(x − n) = 0, then x = m or x = n.

若 (x − m)(x − n) = 0，则 x = m 或 x = n。

3. Solving Equations of the Form x² = k | 求解 x² = k 形式的方程

Before dealing with general quadratics, consider the simple case x² = k, where k > 0. Taking square roots on both sides gives two solutions: x = √k or x = −√k. For example, x² = 25 leads to x = 5 or x = −5.

在应对一般的二次方程之前，先考虑简单情形 x² = k，其中 k > 0。两边同时开平方，得到两个解：x = √k 或 x = −√k。例如，x² = 25 可解得 x = 5 或 x = −5。

x² = k ⇒ x = ±√k

4. Factorising ax² + bx + c with a = 1 | 当 a = 1 时的因式分解

When the coefficient of x² is 1, we look for two numbers whose sum is b and whose product is c. For example, to factorise x² + 7x + 12, we need two numbers that add to 7 and multiply to 12, namely 3 and 4. Thus x² + 7x + 12 = (x + 3)(x + 4).

当 x² 的系数为 1 时，我们需要找到两个数，使它们的和为 b、积为 c。例如，要将 x² + 7x + 12 因式分解，需要找到两个数，它们的和为 7、积为 12，即 3 和 4。因此 x² + 7x + 12 = (x + 3)(x + 4)。

5. Solving Quadratics by Factorisation | 用因式分解法求解二次方程

To solve a quadratic equation using factorisation, first ensure the equation is in the form ax² + bx + c = 0. Factorise the left-hand side, then apply the zero product property to find the values of x.

用因式分解法求解二次方程时，首先确保方程的形式为 ax² + bx + c = 0。然后对左边进行因式分解，再利用零积性质求出 x 的值。

Example 1: Solve x² − 5x + 6 = 0.
例 1：解方程 x² − 5x + 6 = 0。

Two numbers with sum 5 and product 6 are 2 and 3, hence (x − 2)(x − 3) = 0, so x = 2 or x = 3.

和为 5、积为 6 的两个数是 2 和 3，因此 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。

Example 2: Solve x² − 4x = 0.
例 2：解方程 x² − 4x = 0。

Factorise out x to get x(x − 4) = 0, so x = 0 or x = 4.

提取公因式 x，得到 x(x − 4) = 0，所以 x = 0 或 x = 4。

6. Dealing with Negative and Zero Constants | 处理负常数项与零常数项

When c is negative, the two numbers have opposite signs. For example, x² − x − 6 = 0 can be factorised as (x − 3)(x + 2) = 0, giving x = 3 or x = −2.

当 c 为负数时，两个数的符号相反。例如，x² − x − 6 = 0 可以分解为 (x − 3)(x + 2) = 0，从而得到 x = 3 或 x = −2。

When b = 0, the equation reduces to x² = −c/a, and the solutions are x = ±√(−c/a) if −c/a is positive.

当 b = 0 时，方程简化为 x² = −c/a，若 −c/a 为正数，则解为 x = ±√(−c/a)。

7. Difference of Two Squares | 平方差公式

An important special case is the difference of two squares: a² − b² = (a − b)(a + b). For example, x² − 9 = (x − 3)(x + 3), so the equation x² − 9 = 0 has solutions x = 3 and x = −3.

一个重要的特殊情形是平方差公式：a² − b² = (a − b)(a + b)。例如，x² − 9 = (x − 3)(x + 3)，因此方程 x² − 9 = 0 的解为 x = 3 和 x = −3。

a² − b² = (a − b)(a + b)

8. Perfect Square Trinomials | 完全平方三项式

A perfect square trinomial is of the form a² ± 2ab + b² = (a ± b)². For instance, x² + 6x + 9 = (x + 3)², so the equation x² + 6x + 9 = 0 has a repeated root x = −3.

完全平方三项式的形式为 a² ± 2ab + b² = (a ± b)²。例如，x² + 6x + 9 = (x + 3)²，因此方程 x² + 6x + 9 = 0 有重根 x = −3。

9. Factorising ax² + bx + c with a ≠ 1 | 当 a ≠ 1 时的因式分解

When a ≠ 1, factorisation requires more care. One method is to find two numbers whose product is ac and whose sum is b, then split the middle term and factor by grouping.

当 a ≠ 1 时，因式分解需要更加小心。一种方法是找到两个数，使它们的积为 ac、和为 b，然后拆分中间项并分组因式分解。

Example: Solve 2x² + 5x − 3 = 0.

例：解方程 2x² + 5x − 3 = 0。

Here a = 2, b = 5, c = −3, so ac = −6. Two numbers with product −6 and sum 5 are 6 and −1. Rewriting the middle term: 2x² + 6x − x − 3 = 0. Grouping gives 2x(x + 3) − 1(x + 3) = 0, so (2x − 1)(x + 3) = 0. Hence x = ½ or x = −3.

这里 a = 2，b = 5，c = −3，所以 ac = −6。积为 −6、和为 5 的两个数是 6 和 −1。改写中间项：2x² + 6x − x − 3 = 0。分组因式分解得到 2x(x + 3) − 1(x + 3) = 0，即 (2x − 1)(x + 3) = 0。所以 x = ½ 或 x = −3。

10. Equations Requiring Rearrangement | 需要变形整理的情形

Some equations are not given in standard quadratic form. For example, the equation x(x − 3) = 10 must first be expanded and rearranged: x² − 3x − 10 = 0, then factorised to (x − 5)(x + 2) = 0, giving x = 5 or x = −2.

有些方程并非以标准二次形式给出。例如，方程 x(x − 3) = 10 必须先展开并整理：x² − 3x − 10 = 0，然后分解为 (x − 5)(x + 2) = 0，得到 x = 5 或 x = −2。

Similarly, when fractions or brackets appear, multiply out and collect like terms before factorising.

同样，当出现分数或括号时，应先去括号、合并同类项，然后再进行因式分解。

11. Fractional and Negative Roots | 分数根与负根

Solutions to quadratic equations may be fractions or negative numbers. For instance, solving (3x + 2)(x − 5) = 0 gives x = −⅔ or x = 5. Always check the factorisation by expanding your factors back to the original quadratic to ensure correctness.

二次方程的解可能是分数或负数。例如，解方程 (3x + 2)(x − 5) = 0 可得 x = −⅔ 或 x = 5。始终通过将因式展开回原二次式来检查因式分解是否正确。

12. Common Mistakes and Exam Tips | 常见错误与考试提示

Never forget to rearrange the equation to zero on one side before factorising.
在因式分解之前，务必把方程整理为一侧等于 0 的形式。
When taking square roots, remember to include both the positive and negative roots.
开平方时，要同时考虑正根和负根。
Check your solutions by substituting them back into the original equation.
将解代回原方程进行检验。
If a quadratic cannot be factorised, another method such as the quadratic formula may be needed, but factorisation is often the fastest when it works.
如果二次式无法因式分解，可能需要使用其他方法如公式法，但在可行时因式分解通常是最快的。

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Solving Quadratic Equations by Factorisation | IGCSE 数学：因式分解法求解二次方程