Quadratic Equations and Functions | 二次方程与函数

📚 Quadratic Equations and Functions | 二次方程与函数

Quadratic equations and functions form one of the most important topics in IGCSE Mathematics. Understanding how to manipulate and solve quadratic expressions enables students to model real-world situations, analyse parabolas, and succeed in higher-level mathematics. This article provides a complete, step-by-step revision guide with worked examples and common exam tips.

二次方程与函数是 IGCSE 数学中最重要的话题之一。掌握二次表达式的化简与求解，不仅能帮助同学们解决实际问题、分析抛物线图形，也为更高级的数学学习奠定坚实基础。本文将提供完整的分步复习指南，附带典型例题与考试技巧。

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

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The term “quadratic” comes from the Latin word “quadratus”, meaning square, because the highest power of the variable is 2.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。 “quadratic” 一词源自拉丁语 “quadratus”，意为“平方”，因为变量的最高次数是 2。

For example, 2x² + 3x − 5 = 0 is a quadratic equation, while x³ + 2x = 0 is not, because the highest exponent is 3.

例如，2x² + 3x − 5 = 0 是二次方程，而 x³ + 2x = 0 不是，因为最高指数为 3。

ax² + bx + c = 0

The coefficient a cannot be zero; if a = 0, the equation becomes linear. The constants b and c may be zero, so x² = 4, 2x² − 8 = 0 and x² + 5x = 0 are all valid quadratic equations.

系数 a 不能为零；若 a = 0，方程就变为一次方程。常数 b 和 c 可以为零，因此 x² = 4、2x² − 8 = 0 和 x² + 5x = 0 都是合法的二次方程。

2. Expanding and Factorising Quadratics | 展开与因式分解二次式

Before solving quadratic equations, you must be comfortable expanding and factorising quadratic expressions. Expanding means removing brackets: (x + 2)(x + 3) = x² + 5x + 6. The most common method is FOIL: First, Outer, Inner, Last.

在解二次方程之前，你必须熟练展开和因式分解二次表达式。展开就是去掉括号：(x + 2)(x + 3) = x² + 5x + 6。最常用的方法是 FOIL：先（First）、外（Outer）、内（Inner）、后（Last）。

Factorising is the reverse process: writing an expression as a product of two brackets. For a simple quadratic x² + bx + c, we look for two numbers that multiply to give c and add to give b.

因式分解是相反的过程：将一个表达式写成两个括号的乘积。对于简单的二次式 x² + bx + c，我们寻找两个数，使其乘积等于 c，和等于 b。

Example: Factorise x² + 7x + 12. We need two numbers that multiply to 12 and add to 7: 3 and 4. So x² + 7x + 12 = (x + 3)(x + 4).

例：将 x² + 7x + 12 因式分解。我们需要两个数，乘积为 12，和为 7：即 3 和 4。所以 x² + 7x + 12 = (x + 3)(x + 4)。

When the coefficient of x² is not 1, use the “ac method” or trial and error. For example, 2x² + 7x + 3 = (2x + 1)(x + 3). It is essential to check by expanding.

当 x² 的系数不为 1 时，可使用“ac 法”或试错法。例如，2x² + 7x + 3 = (2x + 1)(x + 3)。务必通过展开来验证。

3. Solving by Factorising | 用因式分解法求解

The simplest way to solve a quadratic equation is to factorise the expression and then use the zero product property: if AB = 0, then A = 0 or B = 0.

解二次方程最简单的方法是将表达式因式分解，然后利用零积性质：若 AB = 0，则 A = 0 或 B = 0。

Example: Solve x² + 7x + 12 = 0. Factorise to get (x + 3)(x + 4) = 0. Therefore x + 3 = 0 or x + 4 = 0, giving x = −3 or x = −4.

例：解 x² + 7x + 12 = 0。因式分解得 (x + 3)(x + 4) = 0。因此 x + 3 = 0 或 x + 4 = 0，得到 x = −3 或 x = −4。

Always rearrange the equation to the form ax² + bx + c = 0 before factorising. For example, solve x² = 5x + 6. First rewrite as x² − 5x − 6 = 0, then (x − 6)(x + 1) = 0, so x = 6 or x = −1.

因式分解前务必先将方程整理成 ax² + bx + c = 0 的形式。例如，解 x² = 5x + 6。先改写为 x² − 5x − 6 = 0，然后 (x − 6)(x + 1) = 0，所以 x = 6 或 x = −1。

If the quadratic is a difference of squares, use a² − b² = (a − b)(a + b). For instance, x² − 25 = (x − 5)(x + 5) = 0, so x = 5 or x = −5.

若二次式是平方差，则利用 a² − b² = (a − b)(a + b)。例如，x² − 25 = (x − 5)(x + 5) = 0，所以 x = 5 或 x = −5。

4. Solving by Completing the Square | 用配方法求解

Completing the square is a method that rewrites a quadratic expression as a perfect square plus a constant. The general process works for any quadratic, even when factorisation is difficult.

配方法是将二次表达式改写为一个完全平方加上一个常数的过程。该方法适用于任何二次式，即使因式分解困难时也有效。

Starting with x² + bx + c, we add and subtract (b/2)² to create a perfect square. For example, x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.

从 x² + bx + c 出发，我们加上并减去 (b/2)² 以构成完全平方。例如，x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。

To solve an equation such as x² + 6x + 2 = 0, first complete the square: (x + 3)² − 7 = 0, so (x + 3)² = 7. Taking the square root gives x + 3 = ±√7, so x = −3 ± √7.

要解形如 x² + 6x + 2 = 0 的方程，先配方：(x + 3)² − 7 = 0，所以 (x + 3)² = 7。两边开平方得 x + 3 = ±√7，因此 x = −3 ± √7。

This method also reveals the coordinates of the vertex of the parabola y = (x + h)² + k, which is (−h, k).

该方法的另一个优点是可以直接得到抛物线 y = (x + h)² + k 的顶点坐标 (−h, k)。

5. The Quadratic Formula | 求根公式法

The quadratic formula is the most reliable method for solving any quadratic equation. It works for all cases, including those that cannot be factorised easily.

求根公式是解任何二次方程最可靠的方法。它适用于所有情况，包括那些难以因式分解的方程。

x = (−b ± √(b² − 4ac)) / (2a)

Given ax² + bx + c = 0, you substitute the values of a, b and c into the formula. The symbol ± means you must calculate two separate values: one with a plus sign and one with a minus sign.

对于 ax² + bx + c = 0，将 a、b、c 的值代入公式。符号 ± 表示必须计算两个值：一个用加号，一个用减号。

Example: Solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Substitute into the formula:

例：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。代入公式：

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (2 ± √10) / 2

Thus x ≈ 2.58 or x ≈ −0.58. Always remember to simplify surds where possible, and check your answers by substituting back.

因此 x ≈ 2.58 或 x ≈ −0.58。记住要尽量化简根式，并通过代回原方程检验答案。

6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: Δ = b² − 4ac. It tells us how many real roots a quadratic equation has without solving it completely.

判别式是求根公式中根号下的部分：Δ = b² − 4ac。它能在不完整解方程的情况下告诉我们二次方程有多少个实数根。

If b² − 4ac > 0, the equation has two distinct real roots.

若 b² − 4ac > 0，方程有两个不同的实数根。
If b² − 4ac = 0, the equation has exactly one real root (a repeated root).

若 b² − 4ac = 0，方程恰好有一个实数根（重根）。
If b² − 4ac < 0, the equation has no real roots (two complex roots).

若 b² − 4ac < 0，方程没有实数根（有两个复数根）。

Example: For x² − 4x + 4 = 0, the discriminant is 16 − 16 = 0, so there is one repeated root, x = 2. Graphically, this means the parabola touches the x-axis at one point.

例：对于 x² − 4x + 4 = 0，判别式为 16 − 16 = 0，因此有一个重根 x = 2。图形上，这意味着抛物线在一点处与 x 轴相切。

The discriminant is also useful when determining whether a quadratic intersects, touches, or misses the x-axis entirely.

判别式还可用于判断二次函数的图像与 x 轴是相交、相切还是完全不相交。

7. Graphs of Quadratic Functions | 二次函数的图像

The graph of a quadratic function y = ax² + bx + c is always a parabola. If a > 0, the parabola opens upward (U-shaped); if a < 0, it opens downward (∩-shaped).

二次函数 y = ax² + bx + c 的图像总是抛物线。若 a > 0，抛物线开口向上（U 形）；若 a < 0，开口向下（∩ 形）。

The roots of the equation y = 0 correspond to the x-intercepts of the graph. The constant c gives the y-intercept, since when x = 0, y = c.

方程 y = 0 的根对应图像与 x 轴的交点。常数 c 给出 y 轴截距，因为当 x = 0 时，y = c。

To sketch a quadratic graph, find the x-intercepts, the y-intercept, and the vertex. For example, y = x² − 2x − 3 has intercepts at x = 3, x = −1, and y = −3. The vertex lies halfway between the roots, at x = 1.

画二次函数草图时，通常需要找出 x 轴截距、y 轴截距和顶点。例如，y = x² − 2x − 3 的 x 轴截距为 x = 3 和 x = −1，y 轴截距为 y = −3。顶点的 x 坐标位于两个根的正中间，即 x = 1。

8. The Vertex and Axis of Symmetry | 顶点与对称轴

Every parabola has an axis of symmetry, a vertical line that divides the graph into two mirror images. For a quadratic function y = ax² + bx + c, the axis of symmetry is given by the vertical line x = −b/(2a).

每条抛物线都有一条对称轴，即一条竖直直线，使图像分成两个镜像部分。对于二次函数 y = ax² + bx + c，对称轴为竖直线 x = −b/(2a)。

The vertex is the point on the parabola that lies on the axis of symmetry. Its x-coordinate is the same value, −b/(2a). Substitute this into the function to find the y-coordinate.

顶点是抛物线上位于对称轴上的点。其 x 坐标同样是 −b/(2a)。将这一值代入函数即可求出 y 坐标。

Example: For y = x² − 6x + 5, the axis of symmetry is x = 6/2 = 3. Substituting x = 3 gives y = 9 − 18 + 5 = −4. So the vertex is (3, −4).

例：对于 y = x² − 6x + 5，对称轴为 x = 6/2 = 3。代入 x = 3 得 y = 9 − 18 + 5 = −4。因此顶点为 (3, −4)。

Completing the square provides the vertex directly: if y = (x − h)² + k, then the vertex is (h, k) and the axis of symmetry is x = h.

配方法可直接得到顶点：若 y = (x − h)² + k，则顶点为 (h, k)，对称轴为 x = h。

9. Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities involve expressions like x² − 3x − 4 > 0 or x² + 2x ≤ 8. Solving them requires finding the roots first, then determining the sign of the quadratic on different intervals of the x-axis.

二次不等式涉及如 x² − 3x − 4 > 0 或 x² + 2x ≤ 8 的表达式。解这类不等式需要先求根，再判断二次式在 x 轴不同区间内的正负号。

Example: Solve x² − 3x − 4 > 0. Factorise to get (x − 4)(x + 1) > 0. The roots are x = 4 and x = −1. The graph of y = (x − 4)(x + 1) opens upward, so it is positive outside the interval between the roots.

例：解 x² − 3x − 4 > 0。因式分解得 (x − 4)(x + 1) > 0。根为 x = 4 和 x = −1。函数 y = (x − 4)(x + 1) 开口向上，因此在两根之间以外的区间为正。

x < −1 或 x > 4

For inequalities with ≤ or ≥, include the roots. For x² − 3x − 4 ≤ 0, the solution is −1 ≤ x ≤ 4.

对于含有 ≤ 或 ≥ 的不等式，应包括根。对于 x² − 3x − 4 ≤ 0，解为 −1 ≤ x ≤ 4。

Always sketch the parabola or use a sign table to avoid mistakes on the direction of the inequality.

为了避免方向错误，建议画草图或使用符号表来确定解集。

10. Applications of Quadratics | 二次方程的应用

Quadratic equations appear in many real-life contexts, including projectile motion, area problems, economics, and engineering. In examination problems, you often need to form a quadratic equation from the given information, then solve it.

二次方程出现在许多生活场景中，包括抛物运动、面积问题、经济学和工程学。在考试题目中，你常常需要根据已知信息列出二次方程，然后求解。

Example: A rectangle has length 5 cm more than its width. Its area is 36 cm². Let the width be w. Then w(w + 5) = 36, so w² + 5w − 36 = 0. Factorising gives (w + 9)(w − 4) = 0, so w = 4 or w = −9. Since width cannot be negative, w = 4 cm.

例：一个长方形的长比宽多 5 cm，面积为 36 cm²。设宽为 w。则 w(w + 5) = 36，所以 w² + 5w − 36 = 0。因式分解得 (w + 9)(w − 4) = 0，因此 w = 4 或 w = −9。宽度不能为负，所以 w = 4 cm。

In projectile motion, the height h of an object thrown upward can be modelled by h = −gt² + v₀t + h₀, a quadratic in time t. Setting h = 0 gives the time when the object hits the ground.

在抛物运动中，物体被抛起后的高度 h 可用时间 t 的二次函数 h = −gt² + v₀t + h₀ 表示。令 h = 0 即可得到物体落地的时间。

Exam tip: Read the question carefully, define your variable, and check whether any solution must be rejected due to the real-world context.

考试提示：仔细阅读题目，设定变量，并检查是否有解因实际意义需要舍去。

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Quadratic Equations and Functions | 二次方程与函数