📚 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, geometry, and even statistics questions, and mastering them is essential for exam success.
一元二次方程是 IGCSE 数学中最重要的考点之一。它出现在代数、几何乃至统计题目中，掌握它是考试成功的关键。
1. What is a Quadratic Equation? | 什么是一元二次方程
wer of the variable is 2. The standard form is ax² + bx + c = 0, where a must not be zero.
一元二次方程是未知数最高次数为 2 的整式方程，标准形式为 ax² + bx + c = 0，其中 a 不能为零。
. If a = 0, the equation becomes linear (bx + c = 0), so the condition a ≠ 0 is essential.
例如，2x² + 3x - 5 = 0 是一元二次方程，因为 x 的最高指数是 2。如果 a = 0，方程就变成一次方程 (bx + c = 0)，因此 a ≠ 0 这一条件至关重要。
In the IGCSE syllabus, you need to know four methods of solving quadratic equations: factorisation, completing the square, the quadratic formula, and graphical methods. Each has its own advantages, and the exam may ask you to use a specific one.
在 IGCSE 考纲中，你需要掌握四种解一元二次方程的方法：因式分解法、配方法、求根公式法以及图像法。每种方法各有优势，考试有时会指定使用某一种方法。
2. The Standard Form and Key Terms | 标准形式与关键术语
into the standard form ax² + bx + c = 0 before solving. This applies to all four methods.
每个一元二次方程在求解前都应整理成标准形式 ax² + bx + c = 0，这对四种方法都适用。
- a is the coefficient of x² and must be non-zero.
- a 是 x² 的系数，必须不为零。
- b is the coefficient of x. It can be zero, as in x² - 4 = 0.
- b 是 x 的系数，可以为零，例如 x² - 4 = 0。
- c is the constant term. It can also be zero, as in x² + 2x = 0.
- c 是常数项，也可以为零，例如 x² + 2x = 0。
The solutions of a quadratic equation are called its roots. A typical IGCSE question provides an equation in expanded form and asks you to solve it; always check that your answer is in the required form before moving on.
一元二次方程的解称为它的根 (roots)。典型的 IGCSE 题目会给出展开形式的方程并要求求解；动笔之前务必确认题目要求的答案形式。
3. Solving by Factorisation | 因式分解法
Factorisation is the fastest method when the quadratic can be written as a product of two linear factors. You look for two numbers that multiply to give ac and add to give b.
当二次式能写成两个一次因式的乘积时，因式分解是最快捷的方法。你需要找两个数，相乘得 ac、相加得 b。
Example: Solve x² + 5x + 6 = 0.
例：解方程 x² + 5x + 6 = 0。
We need two numbers that multiply to 6 and add to 5. These numbers are 2 and 3.
我们需要两个数，相乘得 6、相加得 5。这两个数是 2 和 3。
x² + 5x + 6 = (x + 2)(x + 3) = 0
Using the zero product property, either x + 2 = 0 or x + 3 = 0, so x = -2 or x = -3.
根据零积性质，x + 2 = 0 或 x + 3 = 0，所以 x = -2 或 x = -3。
by grouping: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0. So x = -1/2 or x = -3.
当 x² 的系数不是 1 时，过程稍长。例如，解 2x² + 7x + 3 = 0。将 a 和 c 相乘：2 × 3 = 6。找两个数相乘得 6、相加得 7，即 6 和 1。改写中间项：2x² + 6x + x + 3 = 0，再分组因式分解：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0，所以 x = -1/2 或 x = -3。
4. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. This method always works, even when factorisation is difficult.
配方法将二次式改写为 (x + p)² + q 的形式。即使因式分解困难，这种方法也总是有效。
Example: Solve x² + 6x - 7 = 0 by completing the square.
例：用配方法解 x² + 6x - 7 = 0。
Step 1: Move the constant term to the right side.
第一步：把常数项移到右侧。
x² + 6x = 7
p 2: Halve the coefficient of x (6 ÷ 2 = 3), square it (3² = 9), and add it to both sides.
第二步：将 x 的系数除以 2 (6 ÷ 2 = 3)，再取平方 (3² = 9)，然后两边同时加上 9。
x² + 6x + 9 = 7 + 9
Step 3: Write the left side as a perfect square.
第三步：把左边写成完全平方式。
(x + 3)² = 16
Step 4: Take the square root of both sides, remembering the ± sign.
第四步：两边同时开平方，注意 ± 号。
x + 3 = ±4
Therefore x = -3 + 4 = 1 or x = -3 - 4 = -7.
因此 x = -3 + 4 = 1 或 x = -3 - 4 = -7。
le, to solve 2x² + 8x - 10 = 0, divide by 2 to get x² + 4x - 5 = 0, then proceed as above.
如果 x² 的系数不是 1，应先将整个方程除以该系数。例如，解 2x² + 8x - 10 = 0 时，先除以 2 得 x² + 4x - 5 = 0，再按上述步骤进行。
5. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation directly from its coefficients. It is especially useful when factorisation is not easy and completing the square is too time-consuming.
求根公式可以从系数直接解出任何一元二次方程，特别适用于因式分解困难、配方法又太耗时的情况。
