📚 Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程精讲
Quadratic equations are a central topic in the IGCSE Mathematics syllabus. They appear in algebra, coordinate geometry, and real-life modelling questions. Understanding how to solve them by factorising, completing the square, and using the quadratic formula is essential for success in both Core and Extended papers. This article explains the key concepts, methods, and exam techniques you need.
二次方程是 IGCSE 数学大纲中的核心主题。它们出现在代数、坐标几何和现实情境建模题中。掌握因式分解法、配方法和求根公式对于 Core 和 Extended 试卷都至关重要。本文讲解关键概念、解题方法和考试技巧。
1. What is a Quadratic Equation? | 什么是二次方程
d a is not equal to zero. The highest power of the unknown x is 2. If a were zero, the equation would become linear.
二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数且 a 不等于零。未知数 x 的最高次数是 2。如果 a 为零，方程就变成了一次方程。
= 0 and 4x² - 9 = 0. Some equations may need rearranging before they are in standard form.
例如 x² - 5x + 6 = 0、2x² + 3x - 2 = 0 和 4x² - 9 = 0。有些方程在化为标准形式之前需要先整理。
2. Standard Form and Coefficients | 标准形式与系数
coefficient a is the number multiplying x², b is the number multiplying x, and c is the constant term. Identifying a, b and c correctly is essential before using the quadratic formula or the discriminant.
标准形式是 ax² + bx + c = 0。系数 a 是 x² 前的数，b 是 x 前的数，c 是常数项。正确识别 a、b 和 c 对使用求根公式或判别式至关重要。
2 = 0, we have a = 3, b = -5 and c = 2. In -x² + 4x = 0, we have a = -1, b = 4 and c = 0.
例如，在 3x² - 5x + 2 = 0 中，a = 3，b = -5，c = 2。在 -x² + 4x = 0 中，a = -1，b = 4，c = 0。
3. Solving by Factorising | 因式分解法
Factorising is often the quickest method when the quadratic can be written as a product of two linear brackets. The key idea is the zero-product property: if two factors multiply to give zero, then at least one factor must be zero.
当二次式可以写成两个一次括号的乘积时，因式分解通常是最快的方法。核心思想是零乘积性质：如果两个因式的乘积为零，那么至少有一个因式必须为零。
factorise it as (x - 2)(x - 3) = 0. This gives x - 2 = 0 or x - 3 = 0, so x = 2 or x = 3.
解 x² - 5x + 6 = 0，我们把它因式分解为 (x - 2)(x - 3) = 0。由此得到 x - 2 = 0 或 x - 3 = 0，所以 x = 2 或 x = 3。
x + 3 = 0, which factorises to (2x + 1)(x + 3) = 0. The solutions are x = -1/2 and x = -3.
当 a 不等于 1 时，例如 2x² + 7x + 3 = 0，要找两个数，乘积为 ac = 2 × 3 = 6，和为 b = 7。这两个数是 6 和 1，因此 2x² + 6x + x + 3 = 0，分解为 (2x + 1)(x + 3) = 0。解为 x = -1/2 和 x = -3。
4. Solving by Completing the Square | 配方法
s method is useful when factorising is not straightforward and is also the basis for the quadratic formula.
配方法把二次式转化为 (x + p)² = q 的形式。当因式分解不那么直接时，这个方法很有用，它也是求根公式的基础。
9 = 4. This gives (x + 3)² = 4. Taking square roots gives x + 3 = ±2, so x = -1 or x = -5.
例如：解 x² + 6x + 5 = 0。先把常数项移项：x² + 6x = -5。两边加上 (6/2)² = 9：x² + 6x + 9 = 4。得到 (x + 3)² = 4。开平方得 x + 3 = ±2，所以 x = -1 或 x = -5。
If the coefficient of x² is not 1, divide the whole equation by a before completing the square.
如果 x² 的系数不是 1，在配方前应先把整个方程除以 a。
5. The Quadratic Formula | 求根公式
The quadratic formula gives the solutions of ax² + bx + c = 0 directly. The formula is:
求根公式直接给出 ax² + bx + c = 0 的解。公式为：
x = (-b ± √b² - 4ac) / (2a)
solve 2x² - 4x - 6 = 0. Here a = 2, b = -4, c = -6. Substitute into the formula: x = (4 ± √16 + 48) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4. This gives x = 3 or x = -1.
例如：解 2x² - 4x - 6 = 0。这里 a = 2，b = -4，c = -6。代入公式：x = (4 ± √16 + 48) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4。得到 x = 3 或 x = -1。
Always write the formula with brackets and use a calculator carefully when the discriminant is not a perfect square.
书写公式时始终使用括号，当判别式不是完全平方数时要仔细使用计算器。
6. The Discriminant | 判别式
= b² - 4ac. It tells us how many real solutions the equation has without solving it fully.
判别式是求根公式中平方根号内的表达式：Δ = b² - 4ac。它可以在不解方程的情况下告诉我们方程有多少个实数解。
ted real root. If Δ < 0, there are no real roots, and the graph does not cross the x-axis.
如果 Δ > 0，有两个不同的实根。如果 Δ = 0，有一个重复的实根。如果 Δ < 0，没有实根，图像不与 x 轴相交。
or example, x² - 4x + 4 = 0 has b² - 4ac = 16 - 16 = 0, so it has one repeated root x = 2.
例如，x² - 4x + 4 = 0 的 b² - 4ac = 16 -
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