Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

this revision guide, we will learn how to solve quadratic equations by factorisation, completing the square, and the quadratic formula, and how to interpret the discriminant and graphs.

二次方程是 IGCSE 数学中最重要的主题之一，在核心课程和扩展课程中都会出现。它是形如 ax² + bx + c = 0 的方程，其中 a ≠ 0。在本复习指南中，我们将学习如何用因式分解法、配方法和二次公式解二次方程，以及如何理解判别式和函数图像。

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

he standard form is ax² + bx + c = 0, where a, b and c are constants and a cannot be zero.

二次方程是一个二次多项式方程。变量 x 的最高次数是 2。标准形式为 ax² + bx + c = 0，其中 a、b 和 c 是常数，且 a 不能为零。

ax² + bx + c = 0, a ≠ 0

ar, not quadratic. For example, 2x² − 3x + 1 = 0 is quadratic, while 3x + 5 = 0 is linear.

如果 a = 0，方程变为 bx + c = 0，这是一次方程，不是二次方程。例如，2x² − 3x + 1 = 0 是二次方程，而 3x + 5 = 0 是一次方程。

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

c in standard form ax² + bx + c = 0. Identify a, b and c carefully, including their signs.

要使用任何标准方法，首先要把二次方程写成标准形式 ax² + bx + c = 0。仔细识别 a、b 和 c，包括它们的符号。

1 = 0, we have a = 2, b = −3 and c = 1. For −x² + 5x = 0, we have a = −1, b = 5 and c = 0.

对于 2x² − 3x + 1 = 0，我们有 a = 2，b = −3，c = 1。对于 −x² + 5x = 0，我们有 a = −1，b = 5，c = 0。

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

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3. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic has integer roots. We look for two numbers that multiply to give ac and add to give b, then split the middle term.

当二次方程有整数根时，因式分解通常是最快的方法。我们找到两个数，它们的乘积为 ac，和为 b，然后拆分中间项。

ct is 6 and sum is 5: 2 and 3. So x² + 5x + 6 = (x + 2)(x + 3) = 0. Thus 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 5：2 和 3。因此 x² + 5x + 6 = (x + 2)(x + 3) = 0。所以 x = −2 或 x = −3。

Always set each bracket equal to zero because if a product is zero, at least one factor must be zero.

一定要让每个括号等于零，因为如果乘积为零，那么至少有一个因数为零。

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4. Solving by Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This method is useful for deriving the quadratic formula and for finding the vertex of a parabola.

配方法把 ax² + bx + c 写成 a(x + p)² + q 的形式。这种方法对于推导二次公式和求抛物线的顶点很有用。

o get 9. Write x² + 6x + 9 = 4, so (x + 3)² = 4. Then x + 3 = ±2, giving x = −1 or x = −5.

对于 x² + 6x + 5 = 0，取 x 系数的一半，即 3，平方得 9。写成 x² + 6x + 9 = 4，所以 (x + 3)² = 4。然后 x + 3 = ±2，得到 x = −1 或 x = −5。

When a ≠ 1, factor out a from the first two terms before completing the square inside the bracket.

当 a ≠ 1 时，先从前两项中提取 a，再在括号内完成配方。

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5. The Quadratic Formula | 二次公式

y quadratic equation ax² + bx + c = 0, even when factorisation is difficult or impossible.

二次公式可以解任何二次方程 ax² + bx + c = 0，即使因式分解很难或不可能。

x = [−b ± √b² − 4ac] ÷ 2a

root. The term b² − 4ac is called the discriminant and determines the nature of the roots.

这里 √ 表示平方根。b² − 4ac 被称为判别式，它决定了根的性质。

0, a = 2, b = −4, c = −6. First b² − 4ac = (−4)² − 4×2×(−6) = 16 + 48 = 64. Then x = [4 ± √64] ÷ 4 = [4 ± 8] ÷ 4, giving x = 3 or x = −1.

对于 2x² − 4x − 6 = 0，a = 2，b = −4，c = −6。首先 b² − 4ac = (−4)² − 4×2×(−6) = 16 + 48 = 64。然后 x = [4 ± √64] ÷ 4 = [4 ± 8] ÷ 4，得到 x = 3 或 x = −1。

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6. Discriminant and Nature of Roots | 判别式与根的性质

ant D = b² − 4ac tells us how many real roots a quadratic equation has without solving it.

判别式 D = b² − 4ac 可以告诉我们在不解方程的情况下，二次方程有多少个实数根。

there is one repeated real root. If D < 0, there are no real roots, but two complex roots.

如果 D > 0，有两个不同的实数根。如果 D = 0，有一个重复的实数根。如果 D < 0，没有实数根，但有两个复数根。

x + 5 = 0, D = 2² − 4×1×5 = 4 − 20 = −16. Since D < 0, the equation has no real solutions.

对于 x² + 2x + 5 = 0，D = 2² − 4×1×5 = 4 − 20 = −16。由于 D < 0，该方程没有实数解。

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7. Graphs of Quadratic Functions | 二次函数图像

. If a > 0, the parabola opens upward and has a minimum point. If a < 0, it opens downward and has a maximum point.

y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有一个

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