Quadratic Functions and Equations | 二次函数与方程

📚 Quadratic Functions and Equations | 二次函数与方程

Quadratic functions and equations appear throughout the IGCSE Mathematics syllabus. They are used to model projectile motion, area problems, profit functions and many other real-life situations. In this revision guide, you will learn how to expand, factorise, solve and graph quadratic expressions, and how to interpret the discriminant and the turning point.

二次函数与方程贯穿 IGCSE 数学大纲。它们用于描述抛体运动、面积问题、利润函数以及许多其他实际情境。在本复习指南中，你将学习如何展开、因式分解、求解和绘制二次表达式图像，以及如何解释判别式和顶点。

1. What is a Quadratic? | 什么是二次表达式

A quadratic expression is a polynomial of degree 2. Its standard form is written as ax² + bx + c, where a, b and c are constants and a ≠ 0. If a = 0, the expression becomes linear, not quadratic.

二次表达式是次数为 2 的多项式。其标准形式写作 ax² + bx + c，其中 a、b 和 c 是常数，且 a ≠ 0。如果 a = 0，表达式就变成一次式，而不是二次式。

For example, 2x² + 3x − 5 is quadratic because the highest power of x is 2. The term 2x² is the quadratic term, 3x is the linear term and −5 is the constant term.

例如，2x² + 3x − 5 是二次式，因为 x 的最高次数是 2。其中 2x² 是二次项，3x 是一次项，−5 是常数项。

In IGCSE questions, you may be asked to identify whether an expression is quadratic, write it in standard form, or use its coefficients in later calculations. Always check that the coefficient of x² is not zero before applying quadratic methods.

在 IGCSE 题目中，你可能会被要求判断一个表达式是否为二次式、将其写成标准形式，或在后续计算中使用各项系数。在应用二次方法之前，务必先确认 x² 的系数不为零。

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2. Expanding Double Brackets | 展开双括号

To expand two binomials such as (x + p)(x + q), use the distributive property or the FOIL method: First, Outer, Inner, Last. Multiply each term in the first bracket by each term in the second bracket, then collect like terms.

要展开两个二项式，例如 (x + p)(x + q)，可以使用分配律或 FOIL 方法：首项、外项、内项、末项。将第一个括号中的每一项分别乘以第二个括号中的每一项，然后合并同类项。

(x + p)(x + q) = x² + (p + q)x + pq

For example, expanding (x + 3)(x − 2) gives x² − 2x + 3x − 6, which simplifies to x² + x − 6.

例如，展开 (x + 3)(x − 2) 得到 x² − 2x + 3x − 6，化简后为 x² + x − 6。

When a coefficient appears before the bracket, such as 3(x + 1)(x − 4), expand the brackets first, then multiply every term by 3. For example, (x + 1)(x − 4) = x² − 3x − 4, so 3(x + 1)(x − 4) = 3x² − 9x − 12.

当括号前有系数时，例如 3(x + 1)(x − 4)，先展开括号，再将每一项乘以 3。例如，(x + 1)(x − 4) = x² − 3x − 4，所以 3(x + 1)(x − 4) = 3x² − 9x − 12。

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3. Factorising Quadratics | 二次式因式分解

Factorising is the reverse of expanding. For a monic quadratic x² + bx + c, we look for two numbers that multiply to give c and add to give b. If those numbers are p and q, then x² + bx + c = (x + p)(x + q).

因式分解是展开的逆运算。对于首项系数为 1 的二次式 x² + bx + c，我们需要找到两个数，它们的乘积等于 c，和等于 b。如果这两个数是 p 和 q，那么 x² + bx + c = (x + p)(x + q)。

Example: Factorise x² + 5x + 6. We need two numbers that multiply to 6 and add to 5. The numbers 2 and 3 work, so x² + 5x + 6 = (x + 2)(x + 3).

例题：因式分解 x² + 5x + 6。我们需要两个数，乘积为 6，和为 5。数字 2 和 3 满足条件，所以 x² + 5x + 6 = (x + 2)(x + 3)。

When the coefficient of x² is not 1, use the ac method or splitting the middle term. For example, 2x² + 7x + 3 becomes 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

当 x² 的系数不是 1 时，可以使用 ac 法或拆分中间项。例如，2x² + 7x + 3 可写成 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

Always check your factorisation by expanding the brackets back. This will catch sign errors and missing terms quickly.

务必通过重新展开括号来检验因式分解是否正确。这样可以快速发现符号错误和漏项。

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4. Solving by Factorising | 因式分解法解方程

If a quadratic equation ax² + bx + c = 0 can be factorised, we can set each factor equal to zero. This works because if a product equals zero, at least one factor must equal zero.

如果一个二次方程 ax² + bx + c = 0 可以进行因式分解，我们可以令每个因式等于零。这是因为如果乘积等于零，那么至少有一个因式必须等于零。

Example: Solve x² − 3x − 10 = 0. Factorise as (x − 5)(x + 2) = 0. Setting each factor to zero gives x − 5 = 0 or x + 2 = 0, so x = 5 or x = −2.

例题：解方程 x² − 3x − 10 = 0。因式分解为 (x − 5)(x + 2) = 0。令每个因式为零，得 x − 5 = 0 或 x + 2 = 0，所以 x = 5 或 x = −2。

Always check your solutions by substituting them back into the original equation. If both sides are equal, the solution is correct.

始终将解代回原方程进行检验。如果两边相等，则解是正确的。

Some equations have a repeated root, such as x² − 4x + 4 = 0, which factorises to (x − 2)² = 0, giving only x = 2. In IGCSE, you should still list this as one solution or explain that it is a repeated root.

有些方程有重根，例如 x² − 4x + 4 = 0，因式分解为 (x − 2)² = 0，只得到 x = 2。在 IGCSE 中，你仍应将其列为一个解，或说明它是重根。

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5. Quadratic Formula | 二次公式

When a quadratic cannot easily be factorised, use the quadratic formula. For ax² + bx + c = 0, the solutions are:

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