Mastering Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 掌握二次方程：因式分解、配方法与求根公式

📚 Mastering Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 掌握二次方程：因式分解、配方法与求根公式

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, geometry and real-life modelling, so a reliable solving strategy is essential for exam success.

二次方程是 IGCSE 数学中最重要的主题之一。它出现在代数、图像、几何和实际建模中，因此掌握可靠的求解策略对考试成功至关重要。

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

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the unknown x is 2, which is why it is called quadratic.

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 为常数且 a ≠ 0。未知数 x 的最高次数是 2，因此被称为二次方程。

The graph of y = ax² + bx + c is a parabola. Solving the equation means finding the x-values where the parabola crosses the x-axis, called roots or solutions.

y = ax² + bx + c 的图像是一条抛物线。解方程就是找出抛物线与 x 轴交点的 x 值，这些值称为根或解。

For example, x² − 5x + 6 = 0 is quadratic, but 3x + 2 = 0 is linear. In IGCSE questions, you may need to recognise a quadratic even when the terms are not yet arranged neatly.

例如，x² − 5x + 6 = 0 是二次方程，而 3x + 2 = 0 是一次方程。在 IGCSE 题目中，即使各项尚未整齐排列，你也要能识别出二次方程。

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2. Standard Form and Key Vocabulary | 标准形式与关键术语

Before solving, you should rearrange the equation into standard form ax² + bx + c = 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.

在求解之前，应先将方程整理成标准形式 ax² + bx + c = 0。系数 a 是 x² 前面的数，b 是 x 前面的数，c 是常数项。

Key terms include: roots, solutions, zeros, factorisation, perfect square, discriminant and turning point. Examiners often use these words in questions, so you must know what they mean.

关键术语包括：根、解、零点、因式分解、完全平方、判别式和顶点。考官经常在题目中使用这些词，因此你必须了解它们的含义。

The turning point is the maximum or minimum point of the parabola. If a > 0, the parabola opens upwards and the turning point is a minimum; if a < 0, it opens downwards and the turning point is a maximum.

顶点是抛物线的最大值点或最小值点。若 a > 0，抛物线开口向上，顶点为最小值；若 a < 0，抛物线开口向下，顶点为最大值。

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3. Solving by Factorising: The Basic Idea | 因式分解法的基本思路

Factorising relies on the zero product property: if the product of two brackets is zero, then at least one bracket must be zero. This gives two linear equations that are easy to solve.

因式分解法依赖零乘积性质：如果两个括号的乘积为零，那么至少有一个括号为零。这样就可以得到两个容易求解的一元一次方程。

For example, if (x + 2)(x − 5) = 0, then either x + 2 = 0 or x − 5 = 0. The solutions are x = −2 and x = 5. Always check that the equation equals zero before factorising.

例如，若 (x + 2)(x − 5) = 0，则 x + 2 = 0 或 x − 5 = 0。解为 x = −2 和 x = 5。因式分解前务必确认方程等于零。

The main steps are: rearrange to standard form, factorise into two brackets, set each bracket equal to zero, and solve the two linear equations. This method is fastest when the roots are simple integers or fractions.

主要步骤是：整理成标准形式，因式分解为两个括号，令每个括号等于零，然后解两个一元一次方程。当根为简单整数或分数时，这种方法最快。

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4. Worked Example: Factorising x² + 5x + 6 = 0 | 例题：因式分解 x² + 5x + 6 = 0

We need two numbers that multiply to c = 6 and add to b = 5. The numbers 2 and 3 satisfy both: 2 × 3 = 6 and 2 + 3 = 5. Therefore, the factorised form is (x + 2)(x + 3) = 0.

我们需要两个数，它们相乘等于 c = 6，相加等于 b = 5。数字 2 和 3 同时满足：2 × 3 = 6，2 + 3 = 5。因此，因式分解形式为 (x + 2)(x + 3) = 0。

Setting each bracket to zero gives x + 2 = 0 or x + 3 = 0, so x = −2 or x = −3. You can check by substituting: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0.

令每个括号为零，得 x + 2 = 0 或 x + 3 = 0，所以 x = −2 或 x = −3。你可以代入检验：(−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0。

If both numbers are negative, the signs in the brackets will both be negative. If the constant term is negative, the two numbers will have opposite signs. Always check your expansion mentally to avoid sign errors.

如果两个数都为负，括号内的符号将都为负。如果常数项为负，这两个数将异号。务必在心里展开检查，避免符号错误。

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5. When a ≠ 1: Splitting the Middle Term | 当 a ≠ 1 时：拆分中项

When the coefficient of x² is not 1, factorising is slightly harder. One reliable method is to find two numbers that multiply to a × c and add to b, then split the middle term and factorise by grouping.

当 x² 的系数不是 1 时，因式分解稍难一些。一个可靠的方法是找到两个数，它们相乘等于 a × c，相加等于 b，然后拆分中项并按组因式分解。

For 2x² + 7x + 3, we need two numbers that multiply to 2 × 3 = 6 and add to 7. The numbers 6 and 1 work, so rewrite as 2x² + 6x + x + 3, then group to get (2x + 1)(x + 3).

对于 2x² + 7x + 3，我们需要两个数，相乘等于 2 × 3 = 6，相加等于 7。数字 6 和 1 符合要求，因此改写为 2x² + 6x + x + 3，然后分组得到 (2x + 1)(x + 3)。

Setting each bracket to zero gives x = −1/2 or x = −3. Practice this grouping method because it is frequently tested in IGCSE non-calculator papers.

令每个括号为零，得 x = −1/2 或 x = −3。请多加练习这种分组方法，因为它在 IGCSE 非计算器试卷中经常考查。

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6. Completing the Square | 配方法

Completing the square transforms a quadratic into the form a(x + p)² + q = 0. This is useful when the quadratic does not factorise easily and also helps find the vertex of a parabola.

配方法将二次方程转化为 a(x + p)² + q = 0 的形式。当二次方程不容易因式分解时，这种方法非常有用，同时也有助于求抛物线的顶点。

For a monic quadratic x² + bx + c,

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