📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear in almost every IGCSE Mathematics paper, and they form the foundation for much of the later algebra in A-Level study. This revision guide covers every standard solution method, the discriminant, graphical interpretation and common exam-style problems.
二次方程几乎出现在每份 IGCSE 数学试卷中，它也是后续 A-Level 代数学习的重要基础。本复习指南涵盖所有标准解法、判别式、图像意义以及常见的考试题型。
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 is 2. The word "quadratic" comes from the Latin word "quadratus", meaning square.
二次方程是未知数的最高次数为 2 的方程。“二次”一词源自拉丁语 "quadratus"，意为“平方”。
The general form is:
一般形式为：
ax² + bx + c = 0, where a ≠ 0
Here a, b and c are real constants. The coefficient a must not be zero, because then the degree falls to 1 and the equation becomes linear.
其中 a、b、c 为实数常数。系数 a 一定不能为零，否则方程次数降为 1，就变成了一次方程。
In IGCSE, you will be expected to solve quadratics by factorisation, by completing the square, by using the quadratic formula and sometimes by reading roots from a graph.
在 IGCSE 中，你应当掌握因式分解法、配方法、求根公式法，并且有时还需要从图像中直接读出方程的根。
2. Standard Form and Key Terms | 标准形式与关键术语
e standard form ax² + bx + c = 0. The terms are usually written in descending powers of x.
在解任何二次方程之前，都必须先把它整理成标准形式 ax² + bx + c = 0，各项通常按 x 的降幂排列。
equation 2x² = 3x + 5 must be rewritten as 2x² − 3x − 5 = 0 before identifying a, b and c.
例如，方程 2x² = 3x + 5 必须先改写为 2x² − 3x − 5 = 0，才能确定 a、b、c 的值。
ax² is called the quadratic term, and a is the leading coefficient. | ax² 称为二次项，a 是首项系数。
bx is called the linear term. | bx 称为一次项。
c is called the constant term, because it has no variable. | c 称为常数项，因为它不含变量。
we read a = 3, b = −5 and c = 2. Take special care with negative signs when substituting.
在 3x² − 5x + 2 = 0 中，可读出 a = 3，b = −5，c = 2。代入时要特别注意负号。
3. Solving by Factorisation | 因式分解法
Factorisation is often the quickest method. If the quadratic can be written as a product of two linear factors, then the equation is solved by setting each factor equal to zero.
因式分解通常是最快捷的方法。如果二次式能写成两个一次因式的乘积，那么只需令每个因式等于零即可求解。
Example 1: solve x² + 5x + 6 = 0.
例 1：解方程 x² + 5x + 6 = 0。
x² + 5x + 6 = (x + 2)(x + 3) = 0
So x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.
因此 x + 2 = 0 或 x + 3 = 0，得到 x = −2 或 x = −3。
Example 2: solve 2x² + 5x − 3 = 0.
例 2：解方程 2x² + 5x − 3 = 0。
2x² + 5x − 3 = (2x − 1)(x + 3) = 0
Hence x = 1/2 or x = −3.
从而 x = 1/2 或 x = −3。
Step 1: Write the equation in standard form. | 第一步：写成标准形式。
Step 2: Factorise the quadratic expression. | 第二步：对二次式进行因式分解。
Step 3: Set each factor to zero. | 第三步：令每个因式等于零。
Step 4: Solve the resulting linear equations. | 第四步：解所得的两个一次方程。
ultiply to give a × c and add to give b; this helps split the middle term before grouping.
当系数 a 大于 1 时，先找两个数，使它们相乘等于 a × c、相加等于 b，然后用拆项分组的方法完成分解。
4. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form p(x + q)² + r. This method is especially useful when the quadratic does not factorise neatly, and it is also used to find turning points of parabolas.
配方法把二次式改写为 p(x + q)² + r 的形式。当二次式不容易因式分解时，这种方法尤其有用，它也常用于求抛物线的顶点。
For x² + bx, add and subtract (b/2)² to create a perfect square:
对 x² + bx，需要加上并减去 (b/2)²，从而构造完全平方：
x² + bx + c = (x + b/2)² − (b/2)² + c
Example: solve x² + 6x + 8 = 0 by completing the square.
例：用配方法解方程 x² + 6x + 8 = 0。
(x + 3)² − 1 = 0 → (x + 3)² = 1 → x + 3 = ±1
So x = −3 + 1 = −2, or x = −3 − 1 = −4.
所以 x = −3 + 1 = −2，或 x = −3 − 1 = −4。
If a ≠ 1, divide the whole equation by a first, then complete the square on the remaining x² term.
如果 a ≠ 1，应先把整个方程除以 a，再对剩下的 x² 项配方。
5. The Quadratic Formula | 求根公式
The quadratic formula can solve any quadratic equation, whether it factorises or not. It is derived from completing the square and is a reliable final resort in the exam.
求根公式可以求解任何二次方程，无论它能否因式分解。它由配方法推导而来，是考试中值得信赖的“万能方法”。
x = (−b ± √b² − 4ac) / 2a
To use it, substitute the values of a, b and c carefully, then simplify the expression step by step.
使用时，把 a、b、c 的值仔细代入公式，再逐步化简表达式。
Example: solve 2x² + 3x − 2 = 0.
例：解方程 2x² + 3x − 2 = 0。
x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2)
x = (−3 ± √9 + 16) / 4 = (−3 ± 5) / 4
Therefore x = (−3 + 5)/4 = 1/2, or x = (−3 − 5)/4 = −2.
因此 x = (−3 + 5)/4 = 1/2，或 x = (−3 − 5)/4 = −2。
Always simplify surds where possible; in IGCSE, you may be asked to give answers in exact form rather than rounded decimals.
要尽量化简根式；在 IGCSE 中，题目有时会要求以精确形式给出答案，而不是保留多位小数。
6. The Discriminant Δ = b² − 4ac | 判别式 Δ = b² − 4ac
The expression inside the square root in the quadratic formula is called the discriminant. It tells us the number and type of roots before we actually solve the equation.
求根公式中被开方数部分称为判别式。它能在实际求解之前告诉我们根的个数和类型。
If b² − 4ac > 0, the equation has two distinct real roots. | 若 b² − 4ac > 0，方程有两个不相等的实数根。
If b² − 4ac = 0, the equation has exactly one repeated real root. | 若 b² − 4ac = 0，方程有一个二重实数根。
If b² − 4ac < 0, the equation has no real roots. | 若 b² − 4ac < 0，方程没有实数根。
= 0, Δ = (−4)² − 4 × 1 × 4 = 0, so the graph touches the x-axis at the single point x = 2.
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