Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, and many real-world problems. Mastering the different methods of solution is essential for exam success.

二次方程是 IGCSE 数学中最重要的内容之一。它们出现在代数、坐标几何以及许多现实生活问题中。掌握不同的求解方法对于考试成功至关重要。

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1. Standard Form | 标准形式

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. Its general form is:

二次方程是最高次数为 2 的多项式方程，即变量的最高次幂为 2。其一般形式为：

ax² + bx + c = 0, where a ≠ 0

Here a, b and c are constants, and a cannot be zero. If a = 0, the equation becomes linear, not quadratic. Examples of quadratic equations include x² − 5x + 6 = 0 and 2x² + 3x − 1 = 0.

其中 a、b、c 为常数，且 a 不能为零。如果 a = 0，方程就变成一次方程，而不是二次方程。二次方程的例子包括 x² − 5x + 6 = 0 和 2x² + 3x − 1 = 0。

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2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method. If the left-hand side can be written as a product of two linear factors, we set each factor equal to zero.

因式分解通常是最快捷的方法。如果等号左边可以写成两个一次因式的乘积，我们就令每个因式等于零。

For example, solve x² − 5x + 6 = 0:

例如，解方程 x² − 5x + 6 = 0：

(x − 2)(x − 3) = 0

Therefore, x − 2 = 0 or x − 3 = 0, so x = 2 or x = 3. Always check that the product of the constant terms equals c and their sum equals b.

因此，x − 2 = 0 或 x − 3 = 0，所以 x = 2 或 x = 3。始终检查常数项的乘积是否等于 c，以及它们的和是否等于 b。

Another special case is the difference of two squares: x² − 9 = (x + 3)(x − 3) = 0, giving x = −3 or x = 3.

另一个特殊情况是平方差公式：x² − 9 = (x + 3)(x − 3) = 0，得到 x = −3 或 x = 3。

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3. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This method works for any quadratic, even when factorisation is difficult, and it also reveals the vertex of the parabola.

配方法将二次式改写为 (x + p)² + q 的形式。这种方法适用于任何二次方程，即使因式分解困难时也有效，并且还能揭示抛物线的顶点。

For a monic quadratic x² + bx + c, we use the identity:

对于首项系数为 1 的二次式 x² + bx + c，我们使用恒等式：

x² + bx + c = (x + b/2)² − (b/2)² + c

Example: solve x² + 6x + 5 = 0.

例：解 x² + 6x + 5 = 0。

(x + 3)² − 9 + 5 = 0 → (x + 3)² = 4

Taking square roots gives x + 3 = ±2, so x = −1 or x = −5. Remember to include both the positive and negative roots.

开平方得 x + 3 = ±2，所以 x = −1 或 x = −5。记住要同时取正根和负根。

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4. The Quadratic Formula | 求根公式

The quadratic formula can solve any quadratic equation. It is especially useful when the factors are not obvious.

求根公式可以求解任何二次方程。当因式不明显时，它尤其有用。

For ax² + bx + c = 0, the solutions are given by:

对于 ax² + bx + c = 0，解由下式给出：

x = (−b ± √(b² − 4ac)) / 2a

Worked example: solve 2x² − 4x − 3 = 0 using the formula. Here a = 2, b = −4, c = −3.

例题：用公式法解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。

First compute the discriminant: b² − 4ac = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40.

先计算判别式：b² − 4ac = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40。

x = (4 ± √40) / 4 = (4 ± 2√10) / 4 =

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