Mastering Quadratic Equations | 掌握二次方程

📚 Mastering Quadratic Equations | 掌握二次方程

Quadratic equations are one of the most fundamental topics in IGCSE Mathematics. They appear in algebra, graphs, geometry, and real-world problems. This guide will take you from expanding brackets to solving equations and sketching parabolas with confidence.

二次方程是IGCSE数学中最基础的主题之一。它们出现在代数、图像、几何和实际问题中。本指南将带你从展开括号到解方程、绘制抛物线，建立充分自信。

1. What is a Quadratic Equation? | 什么是二次方程

A quadratic equation is an equation in which the highest power of the unknown variable is 2. It can be written in the standard form:

二次方程是未知数最高次数为2的方程，其标准形式为：

ax² + bx + c = 0, where a ≠ 0

Here, a, b and c are constants, and a cannot be zero. If a were zero, the equation would become linear. The name ‘quadratic’ comes from the Latin word ‘quadratus’, meaning square. Common examples include x² + 5x + 6 = 0, 2x² − 4x + 1 = 0, and x² − 9 = 0.

其中a、b、c为常数，且a不能为零。若a为零，方程就退化为一次方程。”quadratic”一词源自拉丁语”quadratus”，意为”平方”。常见例子包括 x² + 5x + 6 = 0、2x² − 4x + 1 = 0 和 x² − 9 = 0。

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2. Expanding Double Brackets | 展开双括号

To expand (x + p)(x + q), multiply each term in the first bracket by each term in the second bracket. This is often remembered as FOIL: First, Outer, Inner, Last. The general result is:

展开 (x + p)(x + q) 时，将第一个括号中的每一项分别乘以第二个括号中的每一项。这常记为FOIL法则：首项、外项、内项、末项。一般结果为：

(x + p)(x + q) = x² + (p + q)x + pq

For example, expand (x + 3)(x + 5): x × x = x², x × 5 = 5x, 3 × x = 3x, 3 × 5 = 15. Collecting like terms gives x² + 8x + 15. Notice that 8 = 3 + 5 and 15 = 3 × 5.

例如，展开 (x + 3)(x + 5)：x × x = x²，x × 5 = 5x，3 × x = 3x，3 × 5 = 15。合并同类项得 x² + 8x + 15。注意 8 = 3 + 5，15 = 3 × 5。

Two special products are especially important:

两个特殊乘积尤其重要：

• Difference of two squares | 平方差公式：(x + a)(x − a) = x² − a²
• Perfect square | 完全平方公式：(x + a)² = x² + 2ax + a²

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3. Factorising Quadratics | 因式分解二次式

Factorising is the reverse of expanding. To factorise x² + bx + c, find two integers m and n such that m × n = c and m + n = b. Then the expression can be written as (x + m)(x + n).

因式分解是展开的逆运算。要对 x² + bx + c 进行因式分解，需找到两个整数m和n，使 m × n = c 且 m + n = b，则该式可写成 (x + m)(x + n)。

Factorise x² + 7x + 12. We need two numbers that multiply to 12 and add to 7: these are 3 and 4. So x² + 7x + 12 = (x + 3)(x + 4).

分解 x² + 7x + 12。需要两个数相乘为12、相加为7：这两个数是3和4。所以 x² + 7x + 12 = (x + 3)(x + 4)。

For quadratics with a leading coefficient not equal to 1, such as 2x² + 7x + 3, first multiply a and c: 2 × 3 = 6. Find two numbers whose product is 6 and sum is 7: these are 1 and 6. Split the middle term: 2x² + x + 6x + 3. Then factor by grouping: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).

对于首项系数不为1的二次式，如 2x² + 7x + 3，先将a与c相乘：2 × 3 = 6。找到乘积为6且和为7的两个数：1和6。拆分中间项：2x² + x + 6x + 3。再分组因式分解：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。

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4. Solving by Factorisation | 用因式分解法解方程

If a product of two factors equals zero, then at least one factor must be zero. This is called the zero product property, and it is the key to solving quadratic equations by factorisation.

如果两个因式的乘积为零，则至少有一个因式为零。这就是零积性质，是用因式分解法解二次方程的关键。

Solve x² − 5x + 6 = 0. First factorise: (x − 2)(x − 3) = 0. Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

解方程 x² − 5x + 6 = 0。先因式分解：(x − 2)(x − 3) = 0。因此 x − 2 = 0 或 x − 3 = 0，故 x = 2 或 x = 3。

Always substitute your answers back into the original equation to check. When x = 2, 4 − 10 + 6 = 0 ✓; when x = 3, 9 − 15 + 6 = 0 ✓. Before factorising, make sure the equation is in standard form; rearrange terms if necessary.

务必把解代回原方程检验。当 x = 2 时，4 − 10 + 6 = 0 ✓；当 x = 3 时，9 − 15 + 6 = 0 ✓。因式分解前，确保方程已化为标准形式，必要时先移项整理。

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5. The Quadratic Formula | 二次公式

When factorisation is difficult or impossible, the quadratic formula can always be used. For any quadratic equation ax² + bx + c = 0:

当因式分解困难或无法进行时，二次公式始终可用。对于任意二次方程 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) / (2a)

Solve 2x² − 7x + 3 = 0 using the formula. Here a = 2, b = −7, c = 3. Substituting these values gives:

用公式解 2x² − 7x + 3 = 0。这里 a = 2，b = −7，c = 3。代入这些值得：

x = (7 ± √((−7)² − 4 × 2 × 3)) / (2 × 2) = (7 ± √(49 − 24)) / 4 = (7 ± √25) / 4 = (7 ± 5) / 4

Therefore x = (7 + 5) / 4 = 3, or x = (7 − 5) / 4 = 0.5. The solutions are x = 3 and x = ½. The ± symbol means you work out two separate values, one with addition and

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