G-2 Teacher's Book 1152: Solving Quadratic Equations | G-2教师用书1152：二次方程的解法

📚 G-2 Teacher’s Book 1152: Solving Quadratic Equations | G-2教师用书1152：二次方程的解法

This article provides a comprehensive revision guide for solving quadratic equations, a core topic in the IGCSE Mathematics syllabus. It corresponds to the teaching notes in the Teacher’s Book section G-2, page 1152, and is designed to help both teachers and students master the key methods, common pitfalls, and exam-style strategies.

本文为 IGCSE 数学考纲中的核心内容——二次方程解法，提供了全面的复习指导。内容对应教师用书 G-2 部分第 1152 页的教学笔记，旨在帮助师生掌握关键方法、常见陷阱及应试策略。

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

A quadratic equation is a polynomial equation of degree 2, usually written in the standard form:

二次方程是次数为 2 的多项式方程，通常写成标准形式：

ax² + bx + c = 0, where a ≠ 0

Here, \(a\) (avoid LaTeX) — here, a is the coefficient of x², b is the coefficient of x, and c is a constant. The solutions to the equation are called the roots or zeros.

其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。方程的解称为根或零点。

For example, x² – 5x + 6 = 0 has roots x = 2 and x = 3, because substituting either value makes the left side equal zero.

例如，x² – 5x + 6 = 0 的根为 x = 2 和 x = 3，因为代入任意一个值都能使左边等于零。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic expression can be written as a product of two linear factors. The rule is: if (px + q)(rx + s) = 0, then either px + q = 0 or rx + s = 0.

当二次表达式可以写成两个一次因式的乘积时，因式分解法往往是最快捷的方法。规则是：若 (px + q)(rx + s) = 0，则 px + q = 0 或 rx + s = 0。

Example: Solve x² – 7x + 10 = 0. We need two numbers that multiply to 10 and add to –7. These are –2 and –5. So (x – 2)(x – 5) = 0, giving x = 2 or x = 5.

例：解 x² – 7x + 10 = 0。需要找到两个数，乘积为 10，和为 –7。这两个数是 –2 和 –5。因此 (x – 2)(x – 5) = 0，得 x = 2 或 x = 5。

For non-monic quadratics (where a ≠ 1), you may need to factor by grouping or use the “cross method”.

对于非首一二次方程（a ≠ 1），可能需要用分组分解法或“十字相乘法”。

3. Completing the Square | 配方法

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

配方法将二次方程改写为 a(x + p)² + q 的形式。这对于求解方程以及寻找抛物线顶点特别有用。

Example: Solve x² + 6x – 7 = 0 by completing the square. First, x² + 6x = (x + 3)² – 9. Then the equation becomes (x + 3)² – 9 – 7 = 0, so (x + 3)² = 16. Hence x + 3 = ±4, giving x = 1 or x = –7.

例：用配方法解 x² + 6x – 7 = 0。首先，x² + 6x = (x + 3)² – 9。原方程变为 (x + 3)² – 9 – 7 = 0，即 (x + 3)² = 16。于是 x + 3 = ±4，得 x = 1 或 x = –7。

When the equation is in the form ax² + bx + c = 0 with a ≠ 1, factor out a first before completing the square.

当方程为 ax² + bx + c = 0 且 a ≠ 1 时，先提出 a，再进行配方。

4. The Quadratic Formula | 求根公式

The quadratic formula works for any quadratic equation, including those that cannot be factored easily. It is derived from completing the square:

求根公式适用于任何二次方程，包括那些不易因式分解的方程。它由配方法推导而来：

x = [–b ± √(b² – 4ac)] / (2a)

To use it, substitute the values of a, b and c into the formula, then simplify. Be careful with signs.

使用时，将 a、b、c 的值代入公式，然后化简。注意符号问题。

Example: Solve 2x² – 3x – 2 = 0. Here a = 2, b = –3, c = –2. Then the discriminant b² – 4ac = ( –3)² – 4×2×( –2) = 9 + 16 = 25. So x = [3 ± √25] / 4 = [3 ± 5] / 4, giving x = 2 or x = –0.5.

例：解 2x² – 3x – 2 = 0。这里 a = 2，b = –3，c = –2。判别式 b² – 4ac = ( –3)² – 4×2×( –2) = 9 + 16 = 25。所以 x = [3 ± √25] / 4 = [3 ± 5] / 4，得 x = 2 或 x = –0.5。

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

The discriminant Δ = b² – 4ac tells us how many real roots a quadratic equation has, without fully solving it.

判别式 Δ = b² – 4ac 告诉我们二次方程有多少个实数根，而不必完整求解。

Value of Δ	Nature of roots	图形表现
Δ > 0	Two distinct real roots	抛物线与 x 轴相交于两点
Δ = 0	One repeated real root	抛物线与 x 轴相切于一点
Δ < 0	No real roots	抛物线与 x 轴无交点

If Δ is a perfect square, the equation can be factored over the integers; otherwise, the roots are irrational.

若 Δ 是完全平方数，则方程可以在整数范围内因式分解；否则根为无理数。

6. Graphs of Quadratic Functions | 二次函数的图象

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upwards; if a < 0, it opens downwards. The roots of the equation are the x-intercepts of the graph.

y = ax² + bx + c 的图象是一条抛物线。若 a > 0，开口向上；若 a < 0，开口向下。方程的解就是图象与 x 轴交点的横坐标。

The vertex (turning point) can be found by completing the square: for y = a(x – h)² + k, the vertex is (h, k). Alternatively, use the formula x = –b / (2a) to find the axis of symmetry.

顶点（转向点）可通过配方求得：对于 y = a(x – h)² + k，顶点为 (h, k)。另一种方法是利用公式 x = –b / (2a) 求出对称轴。

Example: For y = x² – 4x + 3, a = 1, b = –4, so the axis of symmetry is x = 2. Substituting x = 2 gives y = –1, so the vertex is (2, –1). The roots are x = 1 and x = 3.

例：对于 y = x² – 4x + 3，a = 1，b = –4，因此对称轴为 x = 2。代入 x = 2 得 y = –1，所以顶点为 (2, –1)。根为 x = 1 和 x = 3。

7. Solving Quadratic Inequalities | 二次不等式的解法

Quadratic inequalities, such as x² – x – 6 < 0, can be solved by first finding the roots of the corresponding equation, then testing intervals on a number line.

二次不等式，如 x² – x – 6 < 0，可以先求出对应方程的根，然后在数轴上检验各区间。

Factorise: (x – 3)(x + 2) < 0. The roots are x = 3 and x = –2. The sign of the expression alternates across the intervals (–∞, –2), (–2, 3), and (3, ∞). Since the coefficient of x² is positive, the expression is negative between the roots. Thus the solution is –2 < x < 3.

因式分解得 (x – 3)(x + 2) < 0。根为 x = 3 和 x = –2。表达式的符号在区间 (–∞, –2)、(–2, 3) 和 (3, ∞) 上交替变化。由于 x² 系数为正，表达式在两根之间为负。因此解集为 –2 < x < 3。

Always remember to flip the inequality sign when multiplying or dividing by a negative number.

务必记住：当乘以或除以负数时，要改变不等号方向。

8. Common Pitfalls and Exam Tips | 常见陷阱与应试技巧

Do not forget to set the equation to zero before factorising or using the formula. 在因式分解或使用公式前，务必先将方程化为标准形式并令其等于零。
Check signs carefully, especially when substituting negative values into the quadratic formula. 代入负数到求根公式时，要特别小心符号。
If the equation has no real roots, do not force factorisation; state that there are no real solutions. 如果方程无实数根，不要强行因式分解，应直接说明无实数解。
For word problems, define the variable clearly and reject any root that does not fit the context (e.g., negative length). 对于应用题，要清晰地定义变量，并舍弃不符合实际的根（如负长度）。
Always check your answers by substituting back into the original equation. 务必通过代入原方程来检验答案。

9. Worked Example: Mixed Problem | 综合例题

Problem: A rectangle has length (x + 3) cm and width (x – 1) cm. Its area is 45 cm². Find the value of x and hence the dimensions of the rectangle.

题目：一个矩形的长为 (x + 3) 厘米，宽为 (x – 1) 厘米，面积为 45 平方厘米。求 x 的值及矩形的尺寸。

Solution: The area is (x + 3)(x – 1) = 45. Expanding gives x² + 2x – 3 = 45. Rearranging gives x² + 2x – 48 = 0. Factorising: (x + 8)(x – 6) = 0, so x = –8 or x = 6. Since length cannot be negative, x = 6. The length is 6 + 3 = 9 cm, and the width is 6 – 1 = 5 cm.

解：面积为 (x + 3)(x – 1) = 45。展开得 x² + 2x – 3 = 45。移项得 x² + 2x – 48 = 0。因式分解：(x + 8)(x – 6) = 0，所以 x = –8 或 x = 6。由于长度不能为负，取 x = 6。长为 6 + 3 = 9 厘米，宽为 6 – 1 = 5 厘米。

10. Practice Questions | 练习题目

Try solving these to reinforce your understanding:

请尝试解答以下题目，以巩固理解：

Solve x² – 9x + 14 = 0 by factorisation. 用因式分解法解 x² – 9x + 14 = 0。
Solve 2x² + 5x – 3 = 0 using the quadratic formula. 用求根公式解 2x² + 5x – 3 = 0。
Find the roots of x² + 4x + 4 = 0, and write the equation in completed square form. 求 x² + 4x + 4 = 0 的根，并将其写成配方法形式。
Determine the nature of roots for 3x² + 2x + 5 = 0. 判断 3x² + 2x + 5 = 0 的根的性质。
Solve x² – 5x – 6 > 0. 解不等式 x² – 5x – 6 > 0。

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G-2 Teacher’s Book 1152: Solving Quadratic Equations | G-2教师用书1152：二次方程的解法