📚 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 – s 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
inear 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。
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
. 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), (– ient 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 = – 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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