📚 Exercise 6B.1: Solving Quadratic Equations | 习题 6B.1：解二次方程
Welcome to your step-by-step guide to Exercise 6B.1, where we focus on solving quadratic equations using multiple methods. Quadratic equations appear throughout the IB Mathematics syllabus, from algebraic manipulation to modelling real-world scenarios. This article will explore factoring, completing the square, the quadratic formula, and graphical approaches, with worked examples and common pitfalls to avoid.
欢迎来到 6B.1 习题的逐步讲解，我们重点关注用多种方法求解二次方程。二次方程贯穿 IB 数学课程，从代数运算到现实情境建模无处不在。本文将探讨因式分解法、配方法、二次公式法和图解方法，并配有例题和常见错误提醒。
1. Understanding the Standard Form | 认识标准形式
nd a ≠ 0. The highest power of the variable x is 2. Recognising this structure is the very first step in Exercise 6B.1, as each solving method depends on correctly identifying the coefficients a, b, and c.
二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。未知数 x 的最高次数是 2。识别这一结构是 6B.1 习题的第一步，因为每种解法都依赖于正确找出系数 a、b 和 c。
For example, 3x² – 12x + 9 = 0 has a = 3, b = – ore you can see the standard form, such as bringing all terms to one side and simplifying.
例如，方程 3x² – 12x + 9 = 0 中 a = 3，b = –12，c = 9。有时需要对方程进行整理才能看到标准形式，比如将所有项移至一边并化简。
2. Solving by Factoring | 因式分解法
ise 6B.1 often starts with simple trinomials where a = 1, then moves to cases where a > 1.
因式分解法依赖于将 ax² + bx + c 表达为两个一次多项式的乘积。当方程等于零时，我们可以应用零乘积性质。6B.1 习题通常从 a = 1 的简单三项式入手，再进阶到 a > 1 的情形。
Example: Solve x² – 5x + 6 = 0.
例题：求解 x² – 5x + 6 = 0。
- Find two numbers that multiply to c = 6 and add to b = –5: –2 and –3.
- 找到两个相乘得 c = 6、相加得 b = –5 的数：–2 和 –3。
- Factor: (x – 2)(x – 3) = 0.
- 因式分解：(x – 2)(x – 3) = 0。
- Set each factor to zero: x – 2 = 0 or x – 3 = 0 → x = 2 or x = 3.
- 令每个因式为零：x – 2 = 0 或 x – 3 = 0 → x = 2 或 x = 3。
3. The Zero Product Property | 零乘积性质
The Zero Product Property states that if the product of two or more factors is zero, then at least one factor must be zero. This is the logical backbone of the factoring method. Without it, we could not justify splitting the binomials into separate equations.
零乘积性质指出，若两个或多个因式的乘积为零，则至少有一个因式为零。这是因式分解法的逻辑支柱。没有它，我们就无法将两个一次多项式拆分为独立的方程。
s property. It is a common mistake in Exercise 6B.1 to factor an expression set equal to a number other than zero, forgetting to bring all terms to one side first.
在应用该性质前，务必检查方程是否为 ( ... ) = 0 的形式。6B.1 习题中常见的错误是，将等于非零数的表达式进行因式分解，而忘记先把所有项移到同一边。
4. Completing the Square | 配方法
Completing the square transforms a quadratic equation into a perfect square trinomial, making it easy to solve by taking square roots. This technique is particularly important in IB Mathematics because it reveals the vertex form of a quadratic function: a(x – h)² + k.
配方法将二次方程转化为完全平方三项式，通过取平方根轻松求解。这一技巧在 IB 数学中尤为重要，因为它能揭示二次函数的顶点式：a(x – h)² + k。
Example: Solve x² + 6x – 7 = 0 by completing the square.
例题：用配方法解 x² + 6x – 7 = 0。
- Move the constant term: x² + 6x = 7.
- 移常数项：x² + 6x = 7。
- Take half of b = 6, square it: (6/2)² = 9. Add 9 to both sides: x² + 6x + 9 = 7 + 9.
- 取一次项系数 b = 6 的一半并平方：(6/2)² = 9。两边加 9：x² + 6x + 9 = 7 + 9。
- Write as a perfect square: (x + 3)² = 16.
- 写成完全平方：(x + 3)² = 16。
- Take square root: x + 3 = ±4 → x = –3 ± 4 → x = 1 or x = –7.
- 开平方：x + 3 = ±4 → x = –3 ± 4 → x = 1 或 x = –7。
5. The Quadratic Formula | 二次公式
g through Exercise 6B.1 must memorise this formula and use it confidently when factoring is difficult or impossible.
二次公式为任意形如 ax² + bx + c = 0 的二次方程提供了通解。学生在完成 6B.1 习题时，必须熟记该公式，并在因式分解困难或不可行时自信地使用它。
x = [ –b ± √b² – 4ac ] / (2a)
Be careful to substitute the values of a, b, and c with their signs. For 2x² – 3x – 5 = 0, we have a = 2, b = –3, c = –5. Plugging into the formula gives:
代入时务必注意系数的正负号。对于方程 2x² – 3x – 5 = 0，有 a = 2，b = –3，c = –5，代入公式得：
x = [ –(–3) ± √((–3)² – 4(2)(–5)) ] / (2·2) = [ 3 ± √9 + 40 ] / 4 = [ 3 ± √49 ] / 4 = [ 3 ± 7 ] / 4
Thus x = 10/4 = 2.5 or x = –4/4 = –1.
因此 x = 10/4 = 2.5 或 x = –4/4 = –1。
6. Understanding the Discriminant | 理解判别式
The discriminant Δ = b² – sked to determine whether the solutions are real and distinct, real and equal, or complex.
判别式 Δ = b² – 4ac 能在不彻底求解方程的情况下，告诉我们根的性质。在 6B.1 习题中，你可能会被问到解是相异实根、相等实根还是复数根。
| Discriminant / 判别式 | Nature of roots / 根的性质 |
| Δ > 0 | Two distinct real roots / 两个相异实根 |
| Δ = 0 | One real repeated root / 一个二重实根 |
| Δ < 0 | Two complex conjugate roots / 一对共轭复根 |
For instance, x² – 4x + 5 = 0 has Δ = (–4)² – 4(1)(5) = 16 – 20 = –4, so the equation has no real solutions, only complex ones.
例如，x² – 4x + 5 = 0 的判别式 Δ = (–4)² – 4(1)(5) = 16 – 20 = –4，因此该方程没有实数解，只有复数解。
7. Graphical Interpretation | 图解方法
n links algebraic solutions to their geometric meaning. If a > 0, the parabola opens upward; if a < 0, it opens downward.
求解 ax² + bx + c = 0 等同于寻找抛物线 y = ax² + bx + c 的 x 截距。6B.1 习题常将代数解与其几何意义联系起来。若 a > 0，抛物线开口向上；若 a < 0，则开口向下。
The vertex of the parabola is at x = – riminant: two intercepts for Δ > 0, one touch point for Δ = 0, and no intercept for Δ < 0.
抛物线的顶点位于 x = –b/(2a)，纵坐标可通过代入求得。x 截距的个数与判别式一致：Δ > 0 有两个截距，Δ = 0 有一个切点，Δ < 0 则无截距。
8. Choosing the Right Method | 选择正确的方法
An important skill tested in Exercise 6B.1 is selecting the most efficient method for a given quadratic. Use factoring when the trinomial factorises easily with integers. Use completing the square when b is even and you need the vertex form. Apply the quadratic formula for decimal or irrational coefficients, or when factoring seems too slow.
6B.1 习题考查的一项重要技能是针对给定的二次方程选择最高效的方法。当三项式容易用整数分解时就使用因式分解法；当 b 是偶数且你需要顶点式时使用配方法；当系数是小数或无理数，或因式分解太慢时使用二次公式。
Sometimes you may start with one method and switch to another. For example, attempting to factor 3x² + 2x – 6 may not yield integer factors, so the quadratic formula becomes the better choice.
有时你可能从一种方法开始，转而使用另一种方法。例如，尝试分解 3x² + 2x – 6 可能无法得到整数因式，此时二次公式便是更佳选择。
9. Word Problems and Modelling | 应用题与建模
Exercise 6B.1 may present real-life contexts such as projectile motion, area problems, or profit functions, all of which lead to quadratic equations. The key is to define the variable, write the equation, solve, and interpret the answer in context, discarding extraneous solutions.
6B.1 习题可能给出实际情境，如抛体运动、面积问题或利润函数，它们都会导出二次方程。关键是定义变量、列出方程、求解，并在情境中解释答案，舍去不合题意的解。
less than its length. If the area is 54 m², find the dimensions. Let length = x, width = x – 3. Area: x(x – 3) = 54 → x² – 3x – 54 = 0 → (x – 9)(x + 6) = 0 → x = 9 (length), width = 6 m. The negative solution x = –6 is rejected.
例题：一个矩形花园的宽比长短 3 米。若面积为 54 平方米，求尺寸。设长为 x，则宽为 x – 3。面积：x(x – 3) = 54 → x² – 3x – 54 = 0 → (x – 9)(x + 6) = 0 → x = 9（长），宽 = 6 米。负解 x = –6 舍去。
10. Common Mistakes to Avoid | 常见错误需避免
Even top students can stumble on careless errors in Exercise 6B.1. One frequent mistake is forgetting to set the equation to zero before factoring. Another is misapplying the quadratic formula by incorrectly substituting negative signs, especially when b is negative.
即便是优等生也会在 6B.1 习题中因粗心而犯错。一个常见错误是在因式分解前忘记令方程等于零。另一个错误是在应用二次公式时错误地代入负号，尤其是在 b 为负数时。
Also, when completing the square, remember to add (b/2)² to both sides, not just the left. Always check your solutions by plugging them back into the original equation. Finally, do not round intermediate steps too early when using the quadratic formula with irrational roots.
此外，在配方时要记住将 (b/2)² 同时加到两边，而非只加在左边。始终要将解代回原方程进行检验。最后，当二次公式产生无理根时，不要在中间步骤过早进行四舍五入。
11. Practice with Variation | 多样化练习
ise 6B.1, practice a wide range of equations: ones with fractional coefficients (e.g., ½x² – x – 1 = 0), equations requiring rearrangement (e.g., 2x(x – with parameters (e.g., kx² + 4x + 1 = 0) where you analyse conditions on k for real roots.
要熟练掌握 6B.1 习题，需练习各类方程：带有分数系数的（如 ½x² – x – 1 = 0）、需要整理的（如 2x(x – 1) = 3），以及含参方程（如 kx² + 4x + 1 = 0），要求分析 k 满足什么条件时方程有实根。
Create your own variations by changing the coefficients and see if factoring, completing the square, or the formula feels more natural. Over time, your fluency will improve and you will handle exam-style questions with confidence.
通过改变系数来自编变式，体验因式分解、配方法或公式法哪种更顺手。久而久之，你的熟练度将提升，处理考试风格的问题时也会信心十足。
12. Connecting to the IB Syllabus | 与 IB 课程大纲衔接
Exercise 6B.1 aligns with the IB Analysis and Approaches and Applications and Interpretation syllabi, which require students to solve quadratic equations analytically and graphically, understand the discriminant, and apply these concepts to modelling. This exercise is foundational for topics like functions, inequalities, and calculus, where finding intercepts and vertex coordinates becomes routine.
6B.1 习题与 IB 分析与方法和应用与解释课程大纲一致，这些大纲要求学生能用解析法和图像法求解二次方程、理解判别式并将这些概念应用于建模。该习题是函数、不等式和微积分等课题的基础，在那里面求解截距和顶点坐标将成为常规操作。
By thoroughly working through Exercise 6B.1, you build the algebraic fluency that supports success in later chapters, from polynomial equations to optimisation problems. Treat each solution not just as an answer, but as a building block for deeper mathematical thinking.
通过扎扎实实地完成 6B.1 习题，你能建立起支持后续章节（从多项式方程到优化问题）成功的代数流利度。要把每一个解不仅视为答案，更视为构建深层数学思维的基石。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图