📚 Exercise 3B: Solving Quadratic Equations and Graphing Parabolas | 练习3B：解二次方程与绘制抛物线
In IB Mathematics, particularly in the Analysis and Approaches (AA) course, Exercise 3B often builds on the foundational concepts of quadratic functions introduced in Chapter 3. This set of problems typically focuses on solving quadratic equations using factoring, the quadratic formula, and completing the square, as well as interpreting graphs of parabolas. Mastering these skills is essential not only for exam success but also for understanding more advanced topics such as functions, calculus, and optimisation.
在IB数学中，特别是分析与方法(AA)课程中，练习3B通常建立在第3章二次函数基础概念之上。这组题目重点在于利用因式分解、二次公式和配方法求解二次方程，并解读抛物线图像。熟练掌握这些技能不仅对考试成功至关重要，也有助于理解更高级的主题，如函数、微积分和优化问题。
1. Overview of Quadratic Equations | 二次方程概述
rresponding parabolic graph y = ax² + bx + c. The solutions, or roots, correspond to the x‑ aph, and the vertex provides either a maximum or minimum value depending on the sign of a.
二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数且 a ≠ 0。练习3B通常要求你求解此类方程，并将代数解与对应的抛物线图像 y = ax² + bx + c 关联起来。解，即根，对应图像与 x 轴的交点，而顶点则根据 a 的符号提供最大值或最小值。
2. Factoring Quadratic Expressions | 因式分解二次式
Before solving, it is often useful to factor the quadratic expression. Factoring reverses the expansion of two binomials. For a simple monic quadratic x² + bx + c, look for two numbers that multiply to c and add to b. For example, x² + 5x + 6 factors into (x + 2)(x + 3). Always check for a greatest common factor first: in 3x² – 12, factor out 3 to get 3(x² – 4), then recognise the difference of squares to obtain 3(x – 2)(x + 2).
在求解之前，对二次式进行因式分解通常很有帮助。因式分解是将两个二项式乘积展开的逆过程。对于简单的首一二次式 x² + bx + c，寻找乘积为 c、和为 b 的两个数。例如，x² + 5x + 6 分解为 (x + 2)(x + 3)。务必先检查公因式：在 3x² – 12 中，提取公因数 3 得到 3(x² – 4)，再利用平方差公式得 3(x – 2)(x + 2)。
When the leading coefficient a is not 1, use the 'ac method' + 3) = (2x + 1)(x + 3). This technique is a frequent requirement in Exercise 3B problems.
当首项系数 a 不为 1 时，采用“ac 法”。以 2x² + 7x + 3 为例，计算 ac = 6。寻找乘积为 6 且和为 7 的两个数：6 和 1。改写中间项：2x² + 6x + x + 3，然后分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。该技巧是练习3B中的常见要求。
3. Solving Quadratic Equations by Factoring | 用因式分解法解二次方程
roduct of factors equals zero, then at least one factor must be zero. For example, from (x – 4)(x + 2) = 0, set each factor to zero to obtain x = 4 or x = –2. Exercise 3B often presents equations like x² – x – 6 = 0, which factors to (x – 3)(x + 2) = 0, yielding x = 3, x = –2. Always rearrange all terms to one side so the equation equals zero before factoring—neglecting this is a common source of errors.
一旦二次式化为因式乘积形式，就可以应用零乘积性质：如果几个因式的乘积等于零，则至少有一个因式为零。例如，由 (x – 4)(x + 2) = 0，令每个因式等于零，得到 x = 4 或 x = –2。练习3B中常出现如 x² – x – 6 = 0 的方程，分解为 (x – 3)(x + 2) = 0，解得 x = 3、x = –2。务必先将所有项移到等号一侧使方程等于零再进行分解——忽略这一点是常见的错误来源。
4. The Quadratic Formula | 二次公式
When factoring is difficult or the roots are non‑integer, the quadratic formula is your go‑to method. For any quadratic equation ax² + bx + c = 0, the roots are given by:
当因式分解较为困难或根为非整数时，二次公式是你的首选方法。对于任意二次方程 ax² + bx + c = 0，其根由下式给出：
x = [–b ± √b² – 4ac] / (2a)
Let's solve 2x² – 3x – 5 = 0. Here a = 2, b = –3, c = –5. Substitute carefully: x = [–(–3) ± √((–3)² – 4(2)(–5))] / (2 × 2) = [3 ± √9 + 40] / 4 = [3 ± √49] / 4 = [3 ± 7] / 4. So x = 10/4 = 2.5 or x = –4/4 = – nswers in exact simplified surd form unless the problem specifies a decimal approximation.
以 2x² – 3x – 5 = 0 为例。这里 a = 2, b = –3, c = –5。仔细代入：x = [–(–3) ± √((–3)² – 4(2)(–5))] / (2 × 2) = [3 ± √9 + 40] / 4 = [3 ± √49] / 4 = [3 ± 7] / 4。所以 x = 10/4 = 2.5 或 x = –4/4 = –1。在代入负数时使用括号以避免符号错误。在练习3B中，除非题目要求小数近似值，否则请始终给出精确的简化根式形式。
5. Discriminant and Nature of Roots | 判别式与根的性质
The discriminant Δ = b² – 0, there is exactly one real root (a repeated root), and the parabola is tangent to the x‑ o find the range of a parameter k for which an equation has, say, two distinct real roots.
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