📚 Quadratic Equations and Graphs | 二次方程与图像
Quadratic equations lie at the heart of IGCSE algebra, connecting factorisation, graphs, and problem-solving. In this revision guide, you will learn how to recognise, solve, and sketch quadratic equations confidently, with clear methods and exam-focused examples.
二次方程是 IGCSE 代数的核心内容，它将因式分解、图像和实际问题紧密相连。在本复习指南中，你将系统掌握二次方程的识别、求解与图像绘制方法，并通过清晰的步骤和考点示例建立信心。
1. What is a Quadratic Equation? | 什么是二次方程？
are constants and a ≠ 0. If a were zero, the equation would become linear, not quadratic.
二次方程是最高次数为 2 的多项式方程。其标准形式为 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。如果 a 等于 0，方程就会退化成一元一次方程。
The term ax² is called the quadratic term, bx is the linear term, and c is the constant term. In IGCSE, you will be expected to rearrange equations into this form before solving.
ax² 称为二次项，bx 称为一次项，c 称为常数项。在 IGCSE 考试中，你需要先将方程整理成标准形式再进行求解。
2. Solving by Factorisation | 因式分解法
If a quadratic expression can be written as a product of two linear factors, we can solve the equation by setting each factor to zero. This is often the fastest method when the roots are integers or simple fractions.
如果二次式可以写成两个一次因式的乘积，我们就可以令每个因式等于 0 来求解。当根为整数或简单分数时，这通常是最快速的方法。
Factorising gives (x + 2)(x + 3) = 0, so x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.
例如，解方程 x² + 5x + 6 = 0。因式分解得 (x + 2)(x + 3) = 0，因此 x + 2 = 0 或 x + 3 = 0，解得 x = −2 或 x = −3。
Always check that the equation equals zero first. If it does not, rearrange all terms to one side before factorising.
务必先确认方程等于 0。如果不等于 0，需要先把所有项移到一边，再进行因式分解。
3. The Quadratic Formula | 二次公式
rmula will solve any quadratic equation. For ax² + bx + c = 0, the solutions are given by:
当因式分解比较困难或不可能时，二次公式可以求解任何二次方程。对于 ax² + bx + c = 0，解为：
x = (−b ± √b² − 4ac) ÷ (2a)
This formula comes from completing the square and is provided in many IGCSE papers, but you must be able to substitute a, b and c accurately, including negative signs.
这个公式来源于配方法，在许多 IGCSE 试卷中会给出，但你必须能够准确地代入 a、b 和 c，尤其是注意负号。
For example, to solve 2x² − 3x − 5 = 0, use a = 2, b = −3, c = −5. Then x = (3 ± √9 + 40) ÷ 4 = (3 ± 7) ÷ 4, giving x = 2.5 or x = −1.
例如，解 2x² − 3x − 5 = 0，取 a = 2，b = −3，c = −5。代入公式得 x = (3 ± √9 + 40) ÷ 4 = (3 ± 7) ÷ 4，解得 x = 2.5 或 x = −1。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This is useful for solving equations, finding the turning point, and dealing with problems where the quadratic formula is less convenient.
配方法将二次式写成 a(x + p)² + q 的形式。它对解方程、求顶点坐标以及处理二次公式不方便的问题都很有用。
efficient of x, which is 3, and square it to get 9. Write (x + 3)² − 9 + 5 = (x + 3)² − 4.
对 x² + 6x + 5 进行配方：取 x 项系数的一半 3，平方得 9，然后写成 (x + 3)² − 9 + 5 = (x + 3)² − 4。
Setting (x + 3)² − 4 = 0 gives (x + 3)² = 4, so x + 3 = ±2, hence x = −1 or x = −5.
令 (x + 3)² − 4 = 0，得 (x + 3)² = 4，所以 x + 3 = ±2，最终 x = −1 或 x = −5。
5. The Discriminant | 判别式
sion Δ = b² − 4ac. It tells us the nature of the roots without solving the equation fully.
二次方程的判别式是表达式 Δ = b² − 4ac。它可以在不完整求解方程的情况下告诉我们根的性质。
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