📚 Solving Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | IGCSE 数学二次方程求解
Quadratic equations are a central topic in IGCSE Mathematics, appearing in algebra, graphs, and real-world problems. A quadratic equation has degree 2 and can often be solved by factorising, completing the square, or applying the quadratic formula. Understanding how to choose and use these methods is essential for exam success.
二次方程是 IGCSE 数学的核心内容，广泛出现在代数、图像和实际问题中。二次方程的次数为 2，通常可以用因式分解法、配方法或二次公式求解。掌握如何选择并使用这些方法对考试成功至关重要。
1. What is a Quadratic Equation? | 什么是二次方程？
here a, b and c are constants and a ≠ 0. If a were zero, the equation would become linear.
二次方程是次数为 2 的方程，即变量的最高次数为 2。其标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。若 a 等于 0，方程将退化为一次方程。
e also called roots, and they correspond to x-intercepts on the graph of y = ax² + bx + c.
二次方程可以有 0 个、1 个或 2 个实数解。这些解也称为根，它们在 y = ax² + bx + c 的图像上对应 x 轴截距。
ots of the equation ax² + bx + c = 0 are the points where the parabola crosses the x-axis.
二次函数的图像是一条抛物线。方程 ax² + bx + c = 0 的根就是抛物线与 x 轴交点的横坐标。
2. Standard Form and Key Terms | 标准形式与关键术语
s on one side, with zero on the other. For example, 3x² = 5x - 2 becomes 3x² - 5x + 2 = 0.
求解之前，一定要先把方程整理成标准形式 ax² + bx + c = 0，即所有项移到一边，另一边为 0。例如 3x² = 5x - 2 要写成 3x² - 5x + 2 = 0。
The coefficient a is the number in front of x², b is the number in front of x, and c is the constant term. These values are used directly in the quadratic formula and completing the square.
系数 a 是 x² 前面的数，b 是 x 前面的数，c 是常数项。这些值会直接用于二次公式和配方法。
| Term | 中文术语 | Example in 2x² - 4x - 6 |
|---|---|---|
| Quadratic term | 二次项 | 2x² |
| Linear term | 一次项 | -4x |
| Constant term | 常数项 | -6 |
| Leading coefficient | 首项系数 | a = 2 |
dentifying a, b and c. For example, x(x - 3) = 4 must first be written as x² - 3x - 4 = 0.
在 IGCSE 题目中，你可能需要先展开括号并化简，再确定 a、b、c。例如 x(x - 3) = 4 必须先写成 x² - 3x - 4 = 0。
3. Solving by Factorising | 因式分解法
q)(rx + s) = 0, then either px + q = 0 or rx + s = 0. This uses the zero product property.
当二次式可以分解为两个一次因式的乘积时，可以使用因式分解法。若 (px + q)(rx + s) = 0，则 px + q = 0 或 rx + s = 0。这利用了零乘积性质。
and add to b. For example, x² + 7x + 10 = (x + 2)(x + 5), since 2 × 5 = 10 and 2 + 5 = 7.
对于首项系数为 1 的二次式 x² + bx + c，找到两个数，使它们相乘得 c，相加得 b。例如 x² + 7x + 10 = (x + 2)(x + 5)，因为 2 × 5 = 10，2 + 5 = 7。
+ 6x, then factor by grouping: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1).
如果 a 不等于 1，可以使用尝试法或拆中项法。对于 2x² + 7x + 3，把 7x 拆成 x + 6x，再分组分解：2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)。
4. Worked Example: Factorising | 因式分解法例题
Solve x² - 3x - 18 = 0 by factorising.
用因式分解法解 x² - 3x - 18 = 0。
3 because (-6) × 3 = -18 and (-6) + 3 = -3. So the factorised form is (x - 6)(x + 3) = 0.
我们需要两个数，乘积为 -18，和为 -3。这两个数是 -6 和 3，因为 (-6) × 3 = -18，(-6) + 3 = -3。因此因式分解为 (x - 6)(x + 3) = 0。
ero: x - 6 = 0 gives x = 6, and x + 3 = 0 gives x = -3. The solutions are x = 6 or x = -3.
令每个因式等于 0：x - 6 = 0 得 x = 6，x + 3 = 0 得 x = -3。解为 x = 6 或 x = -3。
x = 6, 6² - 3(6) - 18 = 36 - 18 - 18 = 0. For x = -3, (-3)² - 3(-3) - 18 = 9 + 9 - 18 = 0.
一定要通过代入检验：x = 6 时，6² - 3(6) - 18 = 36 - 18 - 18 = 0。x = -3 时，(-3)² - 3(-3) - 18 = 9 + 9 - 18 = 0。
5. Completing the Square | 配方法
This method is useful when factorising is difficult or when finding the vertex of a parabola.
配方法将二次式转化为 a(x + p)² + q = 0 的形式。当因式分解较困难或需要求抛物线的顶点时，这个方法很有用。
To complete the square for x² + bx, add and subtract half of b, squared: (b/2)². This creates a perfect square trinomial. For example, x² + 6x becomes (x + 3)² - 9.
对 x² + bx 配方时，加上并减去 b 的一半的平方，即 (b/2)²。这会形成一个完全平方三项式。例如 x² + 6x 可写成 (x + 3)² - 9。
