Mastering Quadratic Equations for IGCSE | 掌握 IGCSE 二次方程

📚 Mastering Quadratic Equations for IGCSE | 掌握 IGCSE 二次方程

Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. They appear in algebra, graphs, word problems and even in geometry. Being able to solve a quadratic equation confidently, by factorising, completing the square or using the quadratic formula, is essential for high marks.

二次方程是 IGCSE 数学课程中最重要的主题之一。它出现在代数、图像、应用题甚至几何中。能够熟练地通过因式分解、配方法或求根公式来求解二次方程，是取得高分的关键。

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

A quadratic equation is a polynomial equation of degree 2. The highest power of the variable x is 2, and it can be written in the general form:

二次方程是一个二次多项式方程。未知数 x 的最高次数为 2，它可以写成一般形式：

ax² + bx + c = 0

This means that a quadratic equation always contains an x² term. It may also contain an x term and a constant term, but the coefficient of x² must not be zero.

这意味着二次方程一定含有 x² 项。它也可能含有 x 项和常数项，但 x² 的系数不能为零。

ratic equations. However, x³ + 2x − 1 = 0 is not quadratic because the highest power is 3.

例如，x² − 5x + 6 = 0 和 2x² + 3x − 2 = 0 是二次方程。而 x³ + 2x − 1 = 0 不是二次方程，因为最高次数是 3。

In IGCSE exams, you will usually be asked to solve quadratic equations, sketch their graphs, or use them to model real-life situations.

在 IGCSE 考试中，你通常会被要求解二次方程、绘制其图像，或用它们建立实际问题的模型。

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2. Standard Form and Coefficients | 标准形式与系数

orm of a quadratic equation is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.

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。

c is the constant term. For example, in 3x² − 7x + 2 = 0, we have a = 3, b = −7 and c = 2.

这里，a 称为二次项系数，b 称为一次项系数，c 称为常数项。例如，在 3x² − 7x + 2 = 0 中，a = 3，b = −7，c = 2。

Before solving a quadratic equation, always rearrange it so that one side is equal to 0. This is important for factorising and for using the quadratic formula.

在求解二次方程之前，一定要先移项使其中一边等于 0。这对因式分解和使用求根公式都很重要。

If the equation is given as x² + 4x = 5, move all terms to one side: x² + 4x − 5 = 0.

如果方程写成 x² + 4x = 5，就把所有项移到一边：x² + 4x − 5 = 0。

Equation / 方程	a	b	c
x² − 3x + 2 = 0	1	−3	2
2x² + x − 4 = 0	2	1	−4
−x² + 5 = 0	−1	0	5

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3. Solving by Factorising | 因式分解法求解

Factorising is often the fastest method, but it only works when the quadratic expression can be written as the product of two linear brackets.

因式分解通常是最快的方法，但只有当二次式可以写成两个一次括号的乘积时才适用。

b. If the equation is x² − 5x + 6 = 0, the numbers −2 and −3 multiply to 6 and add to −5.

对于 x² + bx + c，要寻找两个数，使它们相乘得 c，相加得 b。如果方程是 x² − 5x + 6 = 0，那么 −2 和 −3 相乘得 6，相加得 −5。

So x² − 5x + 6 = (x − 2)(x − 3). Setting each bracket to zero gives x = 2 or x = 3.

因此 x² − 5x + 6 = (x − 2)(x − 3)。令每个括号分别等于零，得到 x = 2 或 x = 3。

(x − 2)(x − 3) = 0 ⇒ x = 2 or x = 3

When a ≠ 1, use trial and error or the 'ac method' 3, find factors of 2 × 3 = 6 that add to 7: 6 and 1. Then rewrite and factorise in pairs.

当 a ≠ 1 时，可以使用试错法或 'ac 方法'。对于 2x² + 7x + 3，找出 2 × 3 = 6 中相加得 7 的因数：6 和 1。然后重新拆项并分组分解。

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Therefore x = −1/2 or x = −3.

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。因此 x = −1/2 或 x = −3。

(2x + 1)(x + 3) = 0 ⇒ x = −1/2 or x = −3

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4. Completing the Square | 配方法

Completing the square is a method that rewrites a quadratic expression in the form (x + p)² + q. This is useful for solving equations and for finding the turning point of a graph.

配方法是一种把二次式改写成 (x + p)² + q 形式的方法。它对于解方程和求图像的顶点非常有用。

half of the coefficient of x: 6 ÷ 2 = 3. Then write (x + 3)² and subtract the square of 3.

对于 x² + 6x + 1 = 0，先取 x 项系数的一半：6 ÷ 2 = 3。然后写成 (x + 3)²，再减去 3 的平方。

x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8.

x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。

add 8 to both sides, take the square root, and remember both positive and negative roots.

要解 (x + 3)² − 8 = 0，把 8 移到等号右边，开平方，并记住要取正负两个根。

(x + 3)² = 8 ⇒ x + 3 = ±√8 ⇒ x = −3 ± √8

When the coefficient of x² is not 1, first factor it out from the x² and x terms. For 2x² + 8x + 5, write 2[x² + 4x] + 5, then complete the square inside the brackets.

当 x² 的系数不是 1 时，要先把 x² 和 x 项的公因数提出来。对于 2x² + 8x + 5，写成 2[x² + 4x] + 5，然后在括号内完成配方。

2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.

2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。

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