Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学：掌握二次方程

📚 Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学：掌握二次方程

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebraic manipulation to graphs and real-world problems. This revision guide covers the core methods you need to solve any quadratic equation with confidence and accuracy.

二次方程贯穿 IGCSE 数学课程的各个部分，从代数运算到图像和实际应用问题。本复习指南涵盖你需要掌握的核心方法，帮助你自信、准确地解答任何二次方程。

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

A quadratic equation is an equation of degree 2. This means the highest power of the variable is 2. For example, x² – 4x + 3 = 0 and 2x² + 7x – 5 = 0 are both quadratic equations because they contain an x² term.

二次方程是次数为 2 的方程。这意味着变量的最高次幂是 2。例如，x² – 4x + 3 = 0 和 2x² + 7x – 5 = 0 都是二次方程，因为它们含有 x² 项。

In IGCSE Mathematics, a quadratic equation may be given in different arrangements. Some will already be factorised, some will have terms on both sides, and some may include fractions or brackets. Your first job is usually to rearrange the equation into a standard form.

在 IGCSE 数学中，二次方程可能以不同形式出现。有些已经因式分解，有些两边都有项，有些可能含有分数或括号。你的第一步通常是把方程整理成标准形式。

For example, 3x(x – 2) = 5 can be expanded and rearranged to 3x² – 6x – 5 = 0. This standard form makes the solution methods easier to apply.

例如，3x(x – 2) = 5 可以展开并整理为 3x² – 6x – 5 = 0。这种标准形式使解题方法更容易应用。

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

The standard form of a quadratic equation is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The condition a ≠ 0 is essential because if a were zero, the equation would not be quadratic.

二次方程的标准形式是 ax² + bx + c = 0，其中 a、b 和 c 是常数，且 a ≠ 0。a ≠ 0 这个条件很重要，因为如果 a 为 0，方程就不是二次方程了。

The letters a, b and c are called coefficients. In the equation 3x² – 2x + 5 = 0, we have a = 3, b = -2 and c = 5. Notice that the sign belongs to the coefficient, so the value of b is negative when the x term is subtracted.

字母 a、b 和 c 称为系数。在方程 3x² – 2x + 5 = 0 中，a = 3，b = -2，c = 5。注意符号属于系数，所以当 x 项是减法时，b 的值是负数。

To rearrange into standard form, move every term to one side so the other side equals zero. Then write the terms in descending powers of x: the x² term first, the x term second, and the constant term last.

要整理成标准形式，需要把所有项移到一边，使另一边等于 0。然后按 x 的降幂排列：先写 x² 项，再写 x 项，最后写常数项。

Example: 5 – 3x + 2x² = 0 becomes 2x² – 3x + 5 = 0

For this equation, a = 2, b = -3 and c = 5. Getting the standard form right is the foundation for factorising, completing the square and using the quadratic formula.

对于这个方程，a = 2，b = -3，c = 5。正确写出标准形式是因式分解、配方法和使用二次公式的基础。

3. Solving by Factorising | 因式分解法

Factorising is usually the fastest method when it works. First rearrange the equation to the form ax² + bx + c = 0. Then write the quadratic expression as a product of two brackets. Finally, set each bracket equal to zero and solve the resulting linear equations.

因式分解法通常是适用时最快的方法。首先把方程整理成 ax² + bx + c = 0 的形式。然后把二次式写成两个括号的乘积。最后令每个括号等于 0，并解所得的一次方程。

For x² – 7x + 12 = 0, we need two numbers that multiply to 12 and add to -7. These numbers are -3 and -4, so the factorisation is (x – 3)(x – 4) = 0.

对于 x² – 7x + 12 = 0，我们需要两个数，它们相乘得 12，相加得 -7。这两个数是 -3 和 -4，因此因式分解为 (x – 3)(x – 4) = 0。

(x – 3)(x – 4) = 0 ⇒ x = 3 or x = 4

When the coefficient of x² is not 1, factorising takes more care. For 2x² + 5x – 3 = 0, we look for two numbers that multiply to a × c = -6 and add to b = 5. These are 6 and -1. We split the middle term: 2x² + 6x – x – 3 = 0, then group and factorise to (2x – 1)(x + 3) = 0.

当 x² 的系数不是 1 时，因式分解需要更仔细。对于 2x² + 5x – 3 = 0，我们需要两个数，它们相乘得 a × c = -6，相加得 b = 5。这两个数是 6 和 -1。我们拆分中间项：2x² + 6x – x – 3 = 0，然后分组并因式分解为 (2x – 1)(x + 3) = 0。

From this we get x = 1/2 or x = -3. Always expand your brackets mentally to check that the product gives the original quadratic.

由此得到 x = 1/2 或 x = -3。务必在脑中展开括号，检查乘积是否等于原来的二次式。

4. Solving by Completing the Square | 配方法

Completing the square rewrites x² + bx + c in the form (x + p)² + q. The number p is half the coefficient of x, and q is chosen to keep the expression equivalent. This method is especially useful when the quadratic does not factorise easily.

配方法将 x² + bx + c 写成 (x + p)² + q 的形式。p 是 x 系数的一半，q 用来保持表达式等价。当二次式不容易因式分解时，这种方法特别有用。

For x² + 8x + 5, half of 8 is 4, so we write (x + 4)². Expanding (x + 4)² gives x² + 8x + 16, but we only need +5, so we subtract 11. Therefore x² + 8x + 5 = (x + 4)² – 11.

对于 x² + 8x + 5，8 的一半是 4，所以我们写 (x + 4)²。展开 (x + 4)² 得到 x² + 8x + 16，但我们只需要 +5，所以要减去 11。因此 x² + 8x +

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

更多咨询请联系16621398022（同微信）