📚 Mastering Quadratic Equations for IGCSE Maths | IGCSE 数学二次方程完全攻略
A quadratic equation is one of the most frequently assessed topics in IGCSE Mathematics. It appears in pure algebra questions, in coordinate geometry, and in real-life modelling problems. This article explains the key methods you need to master: factorisation, completing the square, the quadratic formula, and the discriminant, with clear examples and exam-style advice.
二次方程是 IGCSE 数学中最常考查的主题之一。它出现在纯代数题、坐标几何以及现实建模问题中。本文讲解你需要掌握的核心方法：因式分解、配方法、求根公式和判别式，并提供清晰示例和考试技巧。
1. Standard Form and Key Vocabulary | 标准形式与核心术语
ere a, b and c are constants and a ≠ 0. The terms a, b and c are called the quadratic coefficient, linear coefficient and constant term.
一元二次方程的标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。a、b、c 分别称为二次项系数、一次项系数和常数项。
so the condition a ≠ 0 guarantees that the highest power of x is exactly 2. You should always rearrange a quadratic into standard form before choosing a solution method.
如果 a = 0，方程就变成线性方程，因此条件 a ≠ 0 保证 x 的最高次数正好为 2。在选择解法之前，你应该始终把二次方程整理成标准形式。
ne side, to obtain x² - 3x + 2 = 0. Now a = 1, b = -3 and c = 2 can be identified clearly.
例如，从 3x - 4 = x² - 2 开始。将所有项移到一边，得到 x² - 3x + 2 = 0。现在可以清楚地确定 a = 1、b = -3、c = 2。
2. Solving by Factorisation | 因式分解法
. If the product of two factors equals zero, at least one factor must equal zero. This is called the zero product property.
当左边可以写成两个一次因式的乘积 (px + q)(rx + s) = 0 时，可以使用因式分解法。如果两个因式的乘积等于零，那么至少有一个因式必须等于零。这称为零乘积性质。
Step-by-step method: first make the right-hand side equal to 0, then factorise the left-hand side, and finally set each factor equal to 0.
逐步方法：首先使右边等于 0，然后对左边进行因式分解，最后令每个因式等于 0。
- Write the equation in standard form. | 将方程写成标准形式。
- Find two numbers that multiply to ac and add to b. | 找到两个数，使它们相乘等于 ac，相加等于 b。
- Split the middle term if needed, then factor by grouping. | 如需要，拆分中间项，然后分组分解。
- Set each bracket equal to 0 and solve for x. | 令每个括号等于 0 并解出 x。
Example: Solve x² - 5x + 6 = 0. Factorise as (x - 2)(x - 3) = 0, so x = 2 or x = 3.
例：解 x² - 5x + 6 = 0。因式分解为 (x - 2)(x - 3) = 0，所以 x = 2 或 x = 3。
mple: Solve 2x² + 7x + 3 = 0. Factorise as (2x + 1)(x + 3) = 0, giving x = -1/2 or x = -3.
例：解 2x² + 7x + 3 = 0。因式分解为 (2x + 1)(x + 3) = 0，得到 x = -1/2 或 x = -3。
3. Difference of Two Squares | 平方差公式
o squares: a² - b² = (a - b)(a + b). This is useful when the quadratic has no linear term.
因式分解的一个特殊情况是平方差公式：a² - b² = (a - b)(a + b)。当二次方程没有一次项时，这个公式非常有用。
Example: Solve x² - 25 = 0. Write (x - 5)(x + 5) = 0, hence x = 5 or x = -5.
例：解 x² - 25 = 0。写成 (x - 5)(x + 5) = 0，因此 x = 5 或 x = -5。
9 = 0. Here 4x² is (2x)² and 9 is 3², so (2x - 3)(2x + 3) = 0, giving x = 3/2 or x = -3/2.
例：解 4x² - 9 = 0。这里 4x² 是 (2x)²，9 是 3²，所以 (2x - 3)(2x + 3) = 0，得到 x = 3/2 或 x = -3/2。
Recognising perfect squares such as 4, 9, 16, 25, 49, 64 and 100 saves time in exams. Watch for expressions like 9x² - 4, which factorises as (3x - 2)(3x + 2).
识别 4、9、16、25、49、64、100 等完全平方数在考试中可以节省时间。注意像 9x² - 4 这样的表达式，它可以分解为 (3x - 2)(3x + 2)。
4. Completing the Square | 配方法
Completing the square transforms x² + bx into (x + b/2)² - (b/2)². This method is especially useful when the quadratic does not factorise easily, and it also reveals the turning point of a graph.
配方法将 x² + bx 转化为 (x + b/2)² - (b/2)²。当二次方程不容易因式分解时，这种方法特别有用，它还能揭示图像的顶点。
² = 9 to both sides: x² + 6x + 9 = 8, so (x + 3)² = 8. Taking square roots gives x + 3 = ±√8, hence x = -3 ± 2√2.
例：解 x² + 6x + 1 = 0。先写成 x² + 6x = -1。两边加上 (6/2)² = 9：x² + 6x + 9 = 8，所以 (x + 3)² = 8。开平方得到 x + 3 = ±√8，因此 x = -3 ± 2√2。
16 + 5 = (x - 4)² - 11. The turning point is (4, -11), and the line of symmetry is x = 4.
例：将 y = x² - 8x + 5 改写为顶点式。配方：y = (x² - 8x + 16) - 16 + 5 = (x - 4)² - 11。顶点为 (4, -11)，对称轴为 x = 4。
When the coefficient of x² is not 1, first factor it out of the x² and x terms. For example, 2x² + 8x + 1 becomes 2(x² + 4x) + 1, then complete the square inside the bracket.
当 x² 的系数不为 1 时，先将其从 x² 项和 x 项中提出来。例如，2x² + 8x + 1 写成 2(x² + 4x) + 1，然后在括号内完成配方。
5. The Quadratic Formula | 求根公式
x = (-b ± √b² - 4ac) / (2a)
tic equation ax² + bx + c = 0, regardless of whether the roots are rational or irrational.
当因式分解困难或不可能时，使用求根公式。它能解任何二次方程 ax² + bx + c = 0，无论根是有理数还是无理数。
Always substitute a, b and c carefully, including their signs. A common error is to forget the minus sign in -b or to divide only part of the numerator.
代入 a、b、c 时一定要小心，包括它们的正负号。常见错误是漏掉 -b 中的负号，或者只对分子的一部分除以分母。
Example: Solve 2x² - 3x - 2 = 0. Here a = 2, b = -3, c = -2, so x = (3 ± √9 + 16) / 4 = (3 ± 5) / 4. This gives x = 2 or x = -1/2.
例：解 2x² - 3x - 2 = 0。这里 a = 2，b = -3，c = -2，所以 x = (3 ± √9 + 16) / 4 = (3 ± 5) / 4。得到 x = 2 或 x = -1/2。
Example: Solve x² + 4x + 1 = 0. Then x = (-4 ± √16 - 4) / 2 = (-4 ± √12) / 2 = (-4 ± 2√3) / 2 = -2 ± √3. Always simplify the surd fully.
例：解 x² + 4x + 1 = 0。则 x = (-4 ± √16 - 4) / 2 = (-4 ± √12) / 2 = (-4 ± 2√3) / 2 = -2 ± √3。始终要将根式完全化简。
6. The Discriminant and Nature of Roots | 判别式与根的性质
Δ = b² - 4ac
The discriminant tells you how many real roots a quadratic equation has without actually solving it. It is the expression under the square root in the quadratic formula.
判别式告诉你二次方程有多少个实数根，而无需真正求解。它是求根公式中平方根下面的表达式。
