IGCSE Mathematics T-5-1099: Solving Quadratic Equations | IGCSE数学T-5-1099：解二次方程

📚 IGCSE Mathematics T-5-1099: Solving Quadratic Equations | IGCSE数学T-5-1099：解二次方程

A quadratic equation is a fundamental topic in IGCSE Mathematics. This guide explains the key methods for solving quadratic equations, including factorisation, the quadratic formula, and completing the square, along with graphical insights and common pitfalls.

二次方程是IGCSE数学中的基础主题。本指南讲解解二次方程的主要方法，包括因式分解法、求根公式和配方法，并涵盖图像解读与常见错误。

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

A quadratic equation is an equation that can be written in the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The term ax² is the quadratic term, bx is the linear term, and c is the constant term.

二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数，且 a ≠ 0。ax² 是二次项，bx 是一次项，c 是常数项。

For example, 2x² – 3x + 1 = 0 is a quadratic equation with a = 2, b = -3, and c = 1. The highest power of x is 2, which distinguishes it from linear equations.

例如，2x² – 3x + 1 = 0 是一个二次方程，其中 a = 2，b = -3，c = 1。x 的最高次数是2，这使它区别于一次方程。

2. Solving by Factorisation | 因式分解法

One of the simplest methods is factorisation. We look for two numbers that multiply to give ac and add to give b. Then we rewrite the middle term and factor by grouping.

最简单的方法之一是因式分解。我们要找到两个数，它们相乘等于 ac，相加等于 b。然后重写中间项并分组因式分解。

For example, solve x² + 5x + 6 = 0. We need two numbers with product 6 and sum 5: these are 2 and 3. Hence (x + 2)(x + 3) = 0. Therefore x = -2 or x = -3.

例如，解 x² + 5x + 6 = 0。我们需要两个数乘积为6、和为5：即2和3。因此 (x + 2)(x + 3) = 0。所以 x = -2 或 x = -3。

When the coefficient a is not 1, we multiply a by c and then split the middle term. For example, 2x² + 7x + 3 = 0: ac = 6, and the required numbers are 6 and 1 (since 6 × 1 = 6 and 6 + 1 = 7).

当 a ≠ 1 时，我们将 a 乘以 c，然后拆分中间项。例如，2x² + 7x + 3 = 0：ac = 6，所需的两个数是6和1（因为 6 × 1 = 6 且 6 + 1 = 7）。

Then rewrite: 2x² + 6x + x + 3 = 0, factor: 2x(x + 3) + 1(x + 3) = 0, so (2x + 1)(x + 3) = 0. Thus x = -1/2 or x = -3.

于是改写：2x² + 6x + x + 3 = 0，因式分解：2x(x + 3) + 1(x + 3) = 0，所以 (2x + 1)(x + 3) = 0。因此 x = -1/2 或 x = -3。

3. Checking Factorisability | 判断是否可因式分解

Not every quadratic expression can be factorised over the integers. We can use the discriminant Δ = b² – 4ac to decide. If Δ is a perfect square, the quadratic can be factorised over the rational numbers.

并非所有二次表达式都能在整数范围内因式分解。我们可以用判别式 Δ = b² – 4ac 来判断。如果 Δ 是一个完全平方数，则二次式可以在有理数范围内因式分解。

For example, x² + 3x + 1 = 0 has Δ = 9 – 4 = 5, which is not a perfect square, so the roots are irrational and the quadratic cannot be factorised into rational factors.

例如，x² + 3x + 1 = 0 的 Δ = 9 – 4 = 5，不是完全平方数，因此根是无理数，该二次式不能分解为有理因式。

If Δ < 0, there are no real roots, so the expression cannot be factorised over the real numbers.

如果 Δ < 0，则没有实数根，因此该二次表达式不能在实数范围内因式分解。

4. The Quadratic Formula | 求根公式

For any quadratic equation ax² + bx + c = 0, the solutions are given by:

对于任意二次方程 ax² + bx + c = 0，解由以下公式给出：

x = (-b ± √(b² – 4ac)) / (2a)

This formula works for all types of roots: real distinct, real equal, or complex.

此公式适用于所有类型的根：两个不等的实根、两个相等的实根或复数根。

For example, solve 2x² – 4x – 3 = 0. Here a = 2, b = -4, c = -3. Then b² – 4ac = 16 + 24 = 40. So x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √10/2.

例如，解 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。则 b² – 4ac = 16 + 24 = 40。所以 x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √

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