Solving Quadratic Equations and Understanding Graphs | 解二次方程与理解函数图像

📚 Solving Quadratic Equations and Understanding Graphs | 解二次方程与理解函数图像

Quadratic equations and their graphs form one of the most important topics in the Edexcel IGCSE Mathematics syllabus. Whether you are factorising, using the quadratic formula, or interpreting a parabola, a clear understanding of this topic is essential for exam success. This article explains the key methods step by step, with worked examples and common pitfalls highlighted.

二次方程及其图像是 Edexcel IGCSE 数学大纲中最重要的主题之一。无论是因式分解、使用求根公式，还是解读抛物线，清晰理解这一主题都是考试成功的关键。本文将逐步讲解核心方法，并附有例题和常见易错点提示。

1. The Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The highest power of the variable x is 2, which is why it is called “quadratic”. If a = 0, the equation becomes linear, not quadratic.

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。变量 x 的最高次数为 2，因此称为”二次”。若 a = 0，方程变为一次方程，而非二次方程。

ax² + bx + c = 0, a ≠ 0

For example, x² – 5x + 6 = 0 is a quadratic equation with a = 1, b = -5, and c = 6. In contrast, 3x + 2 = 0 is linear because the highest power of x is 1.

例如，x² – 5x + 6 = 0 是一个二次方程，其中 a = 1，b = -5，c = 6。相比之下，3x + 2 = 0 是一次方程，因为 x 的最高次数为 1。

2. Solving by Factorisation | 因式分解法求解

Factorisation is often the quickest method when the quadratic has simple integer roots. The idea is to write ax² + bx + c as a product of two brackets, then set each bracket equal to zero.

当二次方程具有简单的整数根时，因式分解通常是最快的方法。其核心思想是将 ax² + bx + c 写成两个括号的乘积，然后令每个括号等于零。

For a monic quadratic (where a = 1), look for two numbers that multiply to give c and add to give b. For example, solve x² – 5x + 6 = 0. We need two numbers whose product is 6 and whose sum is -5. These numbers are -2 and -3. Therefore,

对于首项系数为 1 的二次方程（a = 1），需找到两个数，其乘积为 c，其和为 b。例如，解 x² – 5x + 6 = 0。我们需要两个数，乘积为 6，和为 -5。这两个数是 -2 和 -3。因此，

(x – 2)(x – 3) = 0

Setting each bracket to zero gives x – 2 = 0 or x – 3 = 0, so x = 2 or x = 3. Always check your answers by substituting back into the original equation.

令每个括号为零，可得 x – 2 = 0 或 x – 3 = 0，因此 x = 2 或 x = 3。务必通过代入原方程来检验答案。

3. Factorising Non-Monic Quadratics | 非首项系数为 1 的二次方程因式分解

When a ≠ 1, factorisation requires more care. Consider 2x² + 7x + 3 = 0. We look for two numbers that multiply to a × c = 2 × 3 = 6 and add to b = 7. These numbers are 6 and 1.

当 a ≠ 1 时，因式分解需要更加小心。考虑 2x² + 7x + 3 = 0。我们需要找到两个数，其乘积为 a × c = 2 × 3 = 6，其和为 b = 7。这两个数是 6 和 1。

Rewrite the middle term 7x as 6x + x, then factor by grouping:

将中间项 7x 改写为 6x + x，然后分组因式分解：

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

Hence (2x + 1)(x + 3) = 0, giving x = -½ or x = -3. This method, sometimes called the “ac method” or “split the middle term”, works for any quadratic that factorises over the integers.

因此 (2x + 1)(x + 3) = 0，得到 x = -½ 或 x = -3。这种方法有时称为”ac 法”或”拆中项法”，适用于任何可在整数范围内因式分解的二次方程。

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

Completing the square transforms a quadratic into the form p(x + q)² + r = 0. This method is useful for solving equations that do not factorise and for finding the vertex of a parabola.

配方法将二次方程转化为 p(x + q)² + r = 0 的形式。该方法适用于求解无法因式分解的方程，并可用于求抛物线的顶点。

For x² + 6x + 2 = 0, take half of 6, which is 3, and write (x + 3)² = x² + 6x + 9. Since we only have +2, we adjust:

对于 x² + 6x + 2 = 0，取 6 的一半得 3，写成 (x + 3)² = x² + 6x + 9。由于原方程只有 +2，我们需要调整：

(x + 3)² – 9 + 2 = 0

(x + 3)² = 7

Taking square roots, x + 3 = ±√7, so x = -3 ± √7. This gives two exact solutions: x = -3 + √7 or x = -3 – √7.

对方程两边开平方，x + 3 = ±√7，因此 x = -3 ± √7。这给出两个精确解：x = -3 + √7 或 x = -3 – √7。

5. The Quadratic Formula | 求根公式

The quadratic formula solves any quadratic equation ax² + bx + c = 0. You are given this formula in the Edexcel IGCSE formula booklet, but you must know how to apply it correctly.

求根公式可以求解任何二次方程 ax² + bx + c = 0。Edexcel IGCSE 公式手册中会给出此公式，但你必须知道如何正确使用它。

x = [-b ± √(b² – 4ac)] / 2a

The expression b² – 4ac is called the discriminant. It determines the number of real roots of the equation. If b² – 4ac > 0, there are two distinct real roots; if b² – 4ac = 0, there is exactly one repeated root; and if b² – 4ac < 0, there are no real roots.

表达式 b² – 4ac 称为判别式。它决定方程实数根的个数。若 b² – 4ac > 0，则有两个不相等的实数根；若 b² – 4ac = 0，则恰有一个重根；若 b² – 4ac < 0，则没有实数根。

6. Worked Example Using the Formula | 使用公式的完整例题

Solve 3x² – 5x – 2 = 0 using the quadratic formula. Here, a = 3, b = -5, and c = -2.

使用求根公式解 3x² – 5x – 2 = 0。这里 a = 3，b = -5，c = -2。

x = [5 ± √((-5)² – 4 × 3 × (-2))] / (2 × 3)

x = [5 ± √(25 + 24)] / 6 = [5 ± √49] / 6 = [5 ± 7] / 6

Thus x = (5 + 7)/6 = 12/6 = 2, or x = (5 – 7)/6 = -2/6 = -⅓. The solutions are x = 2 and x = -⅓.

因此 x = (5 + 7)/6 = 12/6 = 2，或 x = (5 – 7)/6 = -2/6 = -⅓。解为 x = 2 和 x = -⅓。

Always simplify surds where possible and leave answers in exact form unless the question asks for decimals.

务必尽可能化简根式，除非题目要求保留小数，否则以精确形式给出答案。

7. The Graph of y = ax² + bx + c | 二次函数图像 y = ax² + bx + c

The graph of a quadratic function is a curve called a parabola. When a > 0, the parabola opens upwards and has a minimum point; when a < 0, it opens downwards and has a maximum point.

二次函数的图像是抛物线。当 a > 0 时，抛物线开口向上，有最小值点；当 a < 0 时，抛物线开口向下，有最大值点。

The roots of the equation ax² + bx + c = 0 are the x-coordinates where the parabola crosses the x-axis. The y-intercept is the point (0, c), where the graph crosses the y-axis.

方程 ax² + bx + c = 0 的根是抛物线与 x 轴交点的 x 坐标。y 截距是图像与 y 轴的交点 (0, c)。

If b² – 4ac > 0, the parabola crosses the x-axis at two points.
若 b² – 4ac > 0，抛物线在 x 轴上有两个交点。
If b² – 4ac = 0, the parabola touches the x-axis at one point (the vertex lies on the axis).
若 b² – 4ac = 0，抛物线在 x 轴上有一个切点（顶点位于轴上）。
If b² – 4ac < 0, the parabola does not intersect the x-axis at all.
若 b² – 4ac < 0，抛物线与 x 轴没有任何交点。

8. Finding the Vertex by Completing the Square | 通过配方法求顶点

Completing the square is the most reliable way to find the coordinates of the vertex. For y = x² – 4x + 1, complete the square:

配方法是求顶点坐标最可靠的方法。对于 y = x² – 4x + 1，进行配方：

y = (x – 2)² – 4 + 1 = (x – 2)² – 3

In this form, the vertex is at (2, -3), and the axis of symmetry is the vertical line x = 2. Using the general form y = (x – h)² + k, the vertex is (h, k).

在这种形式下，顶点为 (2, -3)，对称轴是垂直线 x = 2。利用一般形式 y = (x – h)² + k，顶点为 (h, k)。

9. Sketching Quadratic Graphs | 画二次函数图像草图

To sketch a quadratic graph accurately, you need four key features: the y-intercept, the roots (if any), the vertex, and the direction of opening.

要准确地画出二次函数图像的草图，你需要四个关键特征：y 截距、根（如果有）、顶点和开口方向。

For y = x² – 2x – 3:

对于 y = x² – 2x – 3：

Y-intercept: set x = 0, giving y = -3. So the point is (0, -3).
y 截距：令 x = 0，得 y = -3。因此点为 (0, -3)。
Roots: factorise x² – 2x – 3 = (x – 3)(x + 1), so roots are x = 3 and x = -1.
根：因式分解 x² – 2x – 3 = (x – 3)(x + 1)，所以根为 x = 3 和 x = -1。
Vertex: complete the square to get y = (x – 1)² – 4, so the vertex is (1, -4).
顶点：配方得到 y = (x – 1)² – 4，因此顶点为 (1, -4)。
Direction: a = 1 > 0, so the parabola opens upwards.
方向：a = 1 > 0，所以抛物线开口向上。

Plot these points and draw a smooth symmetric curve through them. Mark the axis of symmetry x = 1.

标出这些点，并画一条平滑的对称曲线穿过它们。标出对称轴 x = 1。

10. Solving Quadratic Equations by Graphing | 通过图像解二次方程

You can solve a quadratic equation by drawing its graph and reading off the x-coordinates of the points where the curve crosses the x-axis. This method gives approximate solutions suitable for questions that ask for one decimal place.

你可以通过绘制二次函数的图像并读取曲线与 x 轴交点的 x 坐标来解二次方程。这种方法给出的近似解适用于要求保留一位小数的题目。

For example, to solve x² – 2x – 2 = 0 graphically, plot y = x² – 2x – 2 for x from -2 to 4. The curve crosses the x-axis at approximately x = -0.7 and x = 2.7.

例如，用图像法解 x² – 2x – 2 = 0，在 x 从 -2 到 4 的范围内绘制 y = x² – 2x – 2。曲线与 x 轴的交点约为 x = -0.7 和 x = 2.7。

If the equation does not factorise, you may need to rearrange it into a suitable form before plotting. Sometimes you are given a graph and asked to solve a related equation by adding a line. The intersection points of the curve and the line give the solutions.

如果方程无法因式分解，你可能需要在作图前将其整理为合适的形式。有时题目会给出图像，并要求通过添加一条直线来解相关方程。曲线与直线的交点即为方程的解。

11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students frequently lose marks in quadratic questions due to avoidable errors. Here are some common pitfalls and how to avoid them.

学生常常因为可避免的错误而在二次方程题目中丢分。以下是一些常见的易错点及避免方法。

Forgetting to rearrange the equation into the form ax² + bx + c = 0 before applying the formula. Always bring all terms to one side first.
在使用求根公式之前忘记将方程整理为 ax² + bx + c = 0 的形式。务必先将所有项移到等号一侧。
Sign errors when substituting negative values into the quadratic formula. Use brackets carefully in every substitution.
在求根公式中代入负值时出现符号错误。每次代入时都要仔细使用括号。
Misidentifying the y-intercept. It is always the value of c, not the constant in the completed-square form.
错误识别 y 截距。它始终是 c 的值，而不是配方形式中的常数项。
Drawing the parabola with the wrong direction. Check the sign of a first.
抛物线开口方向画反。先检查 a 的符号。

Always check your solutions by substituting them back into the original equation. This simple step catches most arithmetic errors.

务必通过将解代回原方程来检验。这个简单步骤能发现大多数计算错误。

12. Exam-Style Practice Question | 考试风格练习题

Try this question in preparation for your exam: Solve 2x² – 3x – 5 = 0, giving your answers correct to two decimal places.

请在备考时尝试此题：解 2x² – 3x – 5 = 0，答案精确到小数点后两位。

Using the quadratic formula with a = 2, b = -3, and c = -5:

使用求根公式，a = 2，b = -3，c = -5：

x = [3 ± √(9 + 40)] / 4 = [3 ± √49] / 4

x = (3 + 7) / 4 = 2.50 or x = (3 – 7) / 4 = -1.00

So the solutions are x = 2.50 and x = -1.00. As check, (2.50)² × 2 – 3(2.50) – 5 = 12.50 – 7.50 – 5 = 0, which confirms the solution.

因此解为 x = 2.50 和 x = -1.00。验证：2 × (2.50)² – 3(2.50) – 5 = 12.50 – 7.50 – 5 = 0，确认解正确。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Quadratic Equations and Understanding Graphs | 解二次方程与理解函数图像