Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. They appear in almost every examination paper, whether as direct solving questions, word problems, or graph interpretation tasks. A solid command of quadratic equations is essential for securing top grades.

一元二次方程是 IGCSE 数学大纲中最核心的考点之一，几乎每份试卷都会出现，无论是直接求解、应用题，还是图像分析题。熟练掌握二次方程，是冲击高分的必备技能。

---

1. What Is a Quadratic Equation | 什么是一元二次方程

A quadratic equation is a polynomial equation in which the highest power of the variable is 2. The general (standard) form is:

一元二次方程是变量最高次数为 2 的多项式方程，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, because if a = 0 the equation becomes linear, not quadratic. The values of x that make the equation true are called the roots or solutions of the equation.

其中 a、b、c 为常数，且 a ≠ 0。如果 a = 0，方程就变成了一次方程（线性方程），不再是二次方程。使方程成立的 x 值称为方程的根或解，也叫零点。

---

2. The Three Forms of a Quadratic | 二次方程的三种基本形式

Depending on the purpose, a quadratic can be written in three main forms. Each form makes a different piece of information easy to read.

根据不同的使用目的，二次方程（或二次函数）有三种主要形式，每种形式都能让你快速读取不同的信息。

• Standard form: y = ax² + bx + c. This form directly shows the y-intercept (0, c) when it is a function.
• 标准形式：y = ax² + bx + c，直接给出函数图像与 y 轴的交点 (0, c)。
• Factored form: y = a(x − p)(x − q). This shows the roots p and q, which are the x-intercepts.
• 因式分解形式：y = a(x − p)(x − q)，直接给出两个根 p 和 q，即图像与 x 轴的交点。
• Vertex (completed square) form: y = a(x − h)² + k. This shows the turning point of the graph at (h, k).
• 顶点式（配方式）：y = a(x − h)² + k，直接给出图像顶点（转向点）的坐标 (h, k)。

In IGCSE exams, you must be able to convert between these forms quickly and accurately.

在 IGCSE 考试中，你需要快速、准确地在三种形式之间进行转换。

---

3. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method, provided the quadratic factorises neatly with integer coefficients.

因式分解法通常是最快的求解方法，但前提是二次式能用整数系数简便地分解。

Step 1: Arrange the equation in standard form ax² + bx + c = 0.
Step 2: Factorise the quadratic expression into two brackets.
Step 3: Use the zero product rule: if A × B = 0, then A = 0 or B = 0.
Step 4: Solve the two resulting linear equations.

第一步：将方程整理为标准形式 ax² + bx + c = 0；第二步：把二次式分解为两个括号的乘积；第三步：利用”零乘积法则”——若 A × B = 0，则 A = 0 或 B = 0；第四步：解这两个一次方程。

Example: Solve x² − 5x + 6 = 0.

例：解方程 x² − 5x + 6 = 0。

We need two numbers whose product is 6 and whose sum is −5. These are −2 and −3. Therefore:

我们需要找到乘积为 6、和为 −5 的两个数，它们是 −2 和 −3。因此：

(x − 2)(x − 3) = 0

So x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3. Always check your answers by substituting them back into the original equation.

所以 x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。记得把答案代回原方程检验。

---

4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x − h)² + k. It is especially useful when the equation cannot be factorised, and it is also the standard technique for finding the turning point of a quadratic graph.

配方法把二次式改写为 a(x − h)² + k 的形式。当方程无法因式分解时，配方法尤为有用，同时它也是求二次函数图像顶点坐标的重要工具。

For the simple case x² + bx + c, take half of b, square it, then add and subtract this value:

对于简单的 x² + bx + c，取 b 的一半并平方，然后加上再减去这个数：

x² + bx + c = (x + b/2)² − (b/2)² + c

Example: Solve x² + 6x + 4 = 0 by completing the square.

例：用配方法解 x² + 6x + 4 = 0。

Half of 6 is 3, and 3² = 9.

6 的一半是 3，且 3² = 9。

(x + 3)² − 9 + 4 = 0 ⇒ (x + 3)² − 5 = 0

(x + 3)² = 5 ⇒ x + 3 = ±√5 ⇒ x = −3 ± √5

When the coefficient a is not 1, factor it out first before completing the square. For example, 2x² + 8x + 3 = 2(x² + 4x) + 3, then complete the square inside the bracket.

当二次项系数 a 不为 1 时，需要先把 a 提取出来，再对括号内部配方。例如 2x² + 8x + 3 = 2(x² + 4x) + 3，然后在括号内完成配方。

---

5. The Quadratic Formula | 求根公式

The quadratic formula works for every quadratic equation, regardless of whether the expression can be factorised. If your quadratic is in the form ax² + bx + c = 0, the formula is:

求根公式适用于所有一元二次方程，无论能否因式分解。若方程为 ax² + bx + c = 0，公式为：

x = (−b ± √(b² − 4ac)) / (2a)

Example: Solve 2x² − 4x − 3 = 0 using the formula.

例：用求根公式解 2x² − 4x − 3 = 0。

Here a = 2, b = −4, c = −3. Substitute:

这里 a = 2，b = −4，c = −3。代入公式：

x = (4 ± √((−4)² − 4 × 2 × (−3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4

x = (4 ± 2√10) / 4 = 1 ± √10/2

The formula may be given in the formula booklet in some versions of the IGCSE syllabus, but you must still know precisely when and how to apply it. Do not confuse the signs: −b changes sign, and the entire numerator is divided by 2a.

某些版本的 IGCSE

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）