Exercise 1C: Solving Quadratic Equations | 解二次方程

📚 Exercise 1C: Solving Quadratic Equations | 解二次方程

Exercise 1C focuses on one of the most fundamental skills in IB Mathematics: solving quadratic equations. Whether you are following the Analysis and Approaches (AA) or Applications and Interpretation (AI) route, you will frequently encounter quadratics in topics like functions, calculus, optimisation, and modelling. This article provides a thorough walkthrough of the techniques covered in Exercise 1C — factorisation, completing the square, the quadratic formula, and the discriminant — along with tips for recognising which method to use. By the end, you will be well-prepared to tackle any quadratic equation that appears in your practice set.

练习 1C 聚焦于 IB 数学中最基本的技能之一：解二次方程。无论你选择的是分析与方法 (AA) 还是应用与解释 (AI)，你都会在函数、微积分、优化和建模等主题中频繁遇到二次方程。本文详细讲解了练习 1C 所涵盖的技巧——因式分解、配方法、二次公式和判别式——并给出识别使用哪种方法的提示。读完本文，你将能应对练习中出现的任何二次方程。

1. Factoring Monic Quadratics | 因式分解首一二次三项式

A monic quadratic has the form x² + bx + c, where the coefficient of x² is 1. To factorise, look for two numbers that multiply to c and add to b. The factorised form will be (x + p)(x + q), where p + q = b and p q = c.

首一二次三项式的形式为 x² + bx + c，其中 x² 的系数为 1。因式分解时，寻找两个数，它们的乘积等于 c，和等于 b。分解后的形式为 (x + p)(x + q)，其中 p + q = b，p q = c。

Example: Solve x² – 5x + 6 = 0 by factorisation.

例子：用因式分解法解方程 x² – 5x + 6 = 0。

We need two numbers with product 6 and sum –5. The numbers –2 and –3 satisfy this, so (x – 2)(x – 3) = 0. Hence, x = 2 or x = 3.

我们需要两个数，积为 6，和为 –5。–2 和 –3 满足条件，因此 (x – 2)(x – 3) = 0。所以 x = 2 或 x = 3。

Be careful with signs: if c is positive and b is negative, both factors are negative; if c is negative, the factors have opposite signs.

注意符号：若 c 为正且 b 为负，则两因式均为负数；若 c 为负，则两因式符号相反。

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2. Factoring Non-Monic Quadratics | 因式分解非首一二次三项式

When the leading coefficient a is not 1, factorisation becomes more involved. One popular method is the ‘ac method’: multiply a and c, find two numbers that multiply to ac and add to b, then split the middle term and factor by grouping.

当首项系数 a 不等于 1 时，因式分解变得更复杂。一种常用方法是 ‘ac 法’：将 a 与 c 相乘，找到两个数，其积为 ac，和为 b，然后拆中项，分组分解。

Example: Factorise 2x² + 7x + 3.

例子：分解 2x² + 7x + 3。

ac = 2×3 = 6. Find two numbers that multiply to 6 and add to 7: 6 and 1. Rewrite 7x as 6x + x: 2x² + 6x + x + 3. Group: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

ac = 2×3 = 6。找两数乘积为 6，和为 7：6 和 1。将 7x 改写为 6x + x：2x² + 6x + x + 3。分组：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

Then set each factor to zero to solve the equation if needed.

如有需要，令每个因式等于零即可求解方程。

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3. Completing the Square: The Basic Idea | 配方法的基本思想

Completing the square transforms a quadratic expression into a perfect square trinomial plus a constant. For x² + bx, we add (b/2)² to obtain (x + b/2)². This technique is essential for deriving the quadratic formula and for analysing the vertex of a parabola.

配方法将二次式转化为一个完全平方三项式加上一个常数。对于 x² + bx，我们加上 (b/2)²，得到 (x + b/2)²。这一技巧对于推导二次公式和分析抛物线顶点至关重要。

Example: Express x² + 6x + 5 in completed square form.

例子：将 x² + 6x + 5 写成配方式。

Take half of 6, which is 3, square it to get 9. Write x² + 6x = (x + 3)² – 9. Then add the constant: (x + 3)² – 9 + 5 = (x + 3)² – 4.

6 的一半是 3，平方得 9。x² + 6x = (x + 3)² – 9。再加上常数：(x + 3)² – 9 + 5 = (x + 3)² – 4。

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4. Completing the Square When a ≠ 1 | 当 a ≠ 1 时的配方法

If the quadratic is 2x² + 8x + 5, first factor out the leading coefficient from the x² and x terms: 2(x² + 4x) + 5. Complete the square inside the bracket: (x² + 4x) = (x + 2)² – 4, so 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3.

若二次式为 2x² + 8x + 5，首先将 x² 和 x 项提取系数：2(x² + 4x) + 5。在括号内配方：x² + 4x = (x + 2)² – 4，因此 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3。

Remember to multiply the constant term correctly when expanding.

展开时务必正确乘以常数项。

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5. Deriving the Quadratic Formula | 推导二次公式

The quadratic formula is obtained by completing the square on the general quadratic ax² + bx + c = 0. Move c to the right, divide by a, complete the square, and solve for x. The result is:

二次公式是通过对一般二次方程 ax² + bx + c = 0 进行配方得到的。将 c 移项，除以 a，完成配方，解出 x。结果为：

x = [ –b ± √(b² – 4ac) ] / (2a)

This formula is a powerful tool that always works, provided the discriminant is non-negative.

该公式是一个强有力的工具，只要判别式非负，始终有效。

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6. Applying the Quadratic Formula | 应用二次公式

To solve 3x² – 4x – 2 = 0, identify a = 3, b = –4, c = –2. Substitute into the formula:

解方程 3x² – 4x – 2 = 0，确定 a = 3，b = –4，c = –2。代入公式：

x = [4 ± √( (–4)² – 4×3×(–2) )] / (2×3) = [4 ± √(16 + 24)] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = (2 ± √10) / 3

Always simplify the surd and the fraction where possible. Leave answers in exact form unless instructed otherwise.

尽量简化根式和分数。除非另有要求，答案保留精确值。

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7. The Discriminant (Δ) and Root Analysis | 判别式 (Δ) 与根的分析

The discriminant Δ = b² – 4ac reveals the nature of the roots without solving the equation.

判别式 Δ = b² – 4ac 无需解方程即可揭示根的性质。

• If Δ > 0: two distinct real roots.
• If Δ = 0: one repeated real root (or two equal real roots).
• If Δ < 0: no real roots (two complex conjugate roots, which you may encounter in HL).

• 若 Δ > 0：有两个不等实根。
• 若 Δ = 0：有一个重实根（或两个相等实根）。
• 若 Δ < 0：无实根（两共轭复数根，HL 可能涉及）。

Example: For 2x² – 4x + 3 = 0, Δ = (–4)² – 4(2)(3) = 16 – 24 = –8 < 0, so there are no real solutions.

例子：方程 2x² – 4x + 3 = 0，Δ = (–4)² – 4×2×3 = 16 – 24 = –8 < 0，因此无实数解。

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8. Solving Word Problems Involving Quadratics | 解涉及二次方程的应用题

Many real-world problems lead to quadratic equations, such as projectile motion, area optimisation, and number puzzles. The key is to translate the description into an equation, solve it, and then interpret the solution in context (discarding impossible values like negative lengths).

许多实际问题都会导出二次方程，如抛物运动、面积优化和数字谜题。关键是将描述转化为方程，求解，然后结合实际情况解释解（舍弃不可能的值，如负长度）。

Example: A rectangle has a length 3 m longer than its width. If its area is 40 m², find the dimensions. Let width = w, length = w + 3, area = w(w + 3) = 40 → w² + 3w – 40 = 0. Factorise: (w + 8)(w – 5) = 0. Since width > 0, w = 5 m, length = 8 m.

例子：一个矩形，长比宽多 3 米，面积为 40 平方米，求尺寸。设宽为 w，长为 w + 3，面积 w(w + 3) = 40 → w² + 3w – 40 = 0。因式分解得 (w + 8)(w – 5) = 0。因为宽大于 0，w = 5 m，长 = 8 m。

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9. Common Mistakes in Exercise 1C | 练习 1C 中的常见错误

Students often forget to set the equation to zero before factorising, mishandle signs, or misapply the quadratic formula (e.g. misplacing the minus sign for b). Another frequent error is forgetting to consider the discriminant when the question asks about the number of solutions.

学生常常在因式分解前忘记将方程设为零、符号处理不当或因式公式误用（例如漏掉 b 的负号）。另一个常见错误是，当问题询问解的个数时，忽略了判别式。

Always double-check your factorisation by expanding. In the formula, use brackets when squaring negative b: (–b)² is positive.

务必通过展开验证因式分解。在公式中，对负的 b 平方时使用括号：(–b)² 为正。

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10. How to Excel in Exercise 1C | 如何在练习 1C 中脱颖而出

Tackle problems systematically: first try factorising (fastest if possible); if that fails, use the quadratic formula or completing the square. Familiarise yourself with the perfect square trinomials to speed up completing the square. Mix up practice between pure algebra and contextual problems. Finally, create a quick reference sheet with the quadratic formula, discriminant rules, and completing-the-square steps.

系统地处理问题：首先尝试因式分解（可能最快）；若不行，使用二次公式或配方法。熟悉完全平方三项式以加快配方法的速度。将纯代数练习与实际应用题混合练习。最后，制作一张速查表，包括二次公式、判别式规则和配方法步骤。

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