📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in almost every exam paper, either as standalone questions or as part of problem-solving exercises. Understanding how to solve them efficiently is essential for achieving a high grade.
二次方程是 IGCSE 数学中最重要的主题之一。几乎每份试卷都会出现二次方程，无论是作为独立题目还是作为解题练习的一部分。掌握如何高效地解二次方程，是取得高分的关键。
1. What Is a Quadratic Equation | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2. The general form is written as ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The variable x is the unknown we aim to find.
二次方程是次数为 2 的多项式方程。其一般形式写作 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。变量 x 是我们要求解的目标。
There are three main methods for solving quadratic equations: (1) factorisation (also called factoring), (2) completing the square, and (3) using the quadratic formula. Each method has its own strengths and is suitable for different situations.
解二次方程主要有三种方法： （1）因式分解法、 （2）配方法 以及 （3）求根公式法。 每种方法各有优势，适用于不同情况。
2. Solving by Factorisation | 因式分解法
Factorisation is often the quickest method when the quadratic expression can be written as the product of two linear factors. The principle is simple: if the product of two expressions equals zero, then at least one of them must be zero.
当二次表达式可以写成两个一次因式的乘积时，因式分解通常是最快捷的方法。其原理很简单：如果两个表达式的乘积等于零，那么其中至少有一个必须为零。
Example: Solve x² + 5x + 6 = 0.
例如：解 x² + 5x + 6 = 0。
We look for two numbers that multiply to give 6 and add to give 5. These numbers are 2 and 3.
我们寻找两个数，它们的乘积为 6，和为 5。这两个数是 2 和 3。
(x + 2)(x + 3) = 0
Therefore, either x + 2 = 0 or x + 3 = 0, giving:
因此，要么 x + 2 = 0，要么 x + 3 = 0，得到：
x = -2 or x = -3
Check by substituting each value back into the original equation. Both values satisfy the equation, so the solution set is { -2, -3 }.
将每个值代回原方程进行检验。两个值均满足方程，因此解集为 { -2, -3 }。
3. Difference of Two Squares | 平方差公式
A special case of factorisation is the difference of two squares. The identity states:
因式分解的一个特殊情形是平方差公式。该恒等式为：
a² - b² = (a + b)(a - b)
This method is extremely useful when the quadratic has no x term. For instance, solve x² - 9 = 0:
当二次方程没有一次项时，此方法非常有用。例如，解 x² - 9 = 0：
x² - 9 = (x + 3)(x - 3) = 0
Therefore, x = 3 or x = -3.
因此，x = 3 或 x = -3。
This works for non-integer squares as well. Solve x² - 5 = 0. Recognise that 5 = (√5)², so:
该方法同样适用于非整数平方。解 x² - 5 = 0。注意 5 = (√5)²，所以：
x² - 5 = (x + √5)(x - √5) = 0
Hence x = √5 or x = -√5.
因此 x = √5 或 x = -√5。
4. Solving by Completing the Square | 配方法
Completing the square is a powerful technique that works for any quadratic equation, including those that cannot be factorised. The idea is to rewrite the quadratic in the form (x + p)² + q.
配方法是一种强大的技巧，适用于任何二次方程，包括那些无法因式分解的方程。其核心思想是将二次式改写为 (x + p)² + q 的形式。
Example: Solve x² + 6x + 1 = 0 by completing the square.
例：用配方法解 x² + 6x + 1 = 0。
Step 1: Move the constant term to the right side.
第一步：将常数项移到右边。
x² + 6x = -1
Step 2: Add (b/2)² to both sides. Here b = 6, so (6/2)² = 9.
第二步：等式两边加上 (b/2)²。此处 b = 6，因此 (6/2)² = 9。
x² + 6x + 9 = -1 + 9
Step 3: Simplify both sides.
第三步：化简两边。
(x + 3)² = 8
Step 4: Take the square root of both sides.
第四步：两边同时开平方。
x + 3 = ±√8
Therefore:
因此：
x = -3 ± 2√2
ng with equations such as x² + 2ax + c = 0, or when deriving the quadratic formula itself.
这种方法在处理形如 x² + 2ax + c = 0 的方程，或推导求根公式本身时尤为有用。
5. Worked Example: Coefficient of x² is Not 1 | 示例：二次项系数不为 1
When a ≠ 1, the process becomes slightly different. Consider 2x² + 8x + 3 = 0.
当 a ≠ 1 时，过程略有不同。考虑 2x² + 8x + 3 = 0。
Step 1: Divide through by the coefficient of x² (in this case, 2):
第一步：除以 x² 的系数（本例中为 2）：
x² + 4x + ∕ = 0
Step 2: Move the constant to the right:
第二步：将常数项移到右边：
x² + 4x = -3∕₂
Step 3: Complete the square by adding (4/2)² = 4:
第三步：两边加上 (4/2)² = 4 进行配方：
x² + 4x + 4 = -3∕₂ + 4
(x + 2)² = 5∕₂
Step 4: Solve:
第四步：求解：
x + 2 = ±√5∕₂ → x = -2 ± √5∕₂
This gives two real roots. Note that we could also factorise the original equation, but if you struggle to spot the factors, completing the square is a reliable alternative.
这样得到两个实数根。注意我们也可以对原方程进行因式分解，但如果难以快速找出因式，配方法是一个可靠的选择。
6. The Quadratic Formula | 求根公式
l quadratic ax² + bx + c = 0. It provides a universal solution for any quadratic equation.
求根公式通过对一般二次方程 ax² + bx + c = 0 配方推导而来。它为任何二次方程提供了一种通用解法。
x = (-b ± √b² - 4ac) / 2a
This formula is used when factorisation is difficult or impossible. It also introduces the concept of the discriminant, given by Δ = b² - 4ac.
该公式在难以或无法因式分解时使用。它还引入了判别式的概念，即 Δ = b² - 4ac。
The discriminant tells us about the nature of the roots:
判别式告诉我们根的性质：
- If Δ > 0, the equation has two distinct real roots.
- 如果 Δ > 0，方程有两个不相等的实数根。
- If Δ = 0, the equation has exactly one repeated real root.
- 如果 Δ = 0，方程有一个重根（两个相等的实数根）。
- If Δ < 0, the equation has no real roots (two complex roots).
- 如果 Δ < 0，方程没有实数根（有两个复数根）。
7. Using the Formula: Example | 使用求根公式：示例
Solve 3x² - 5x - 2 = 0 using the quadratic formula.
用求根公式解 3x² - 5x - 2 = 0。
Here, a = 3, b = -5, c = -2. Substitute into the formula:
这里 a = 3，b = -5，c = -2。代入公式：
x = (5 ± √((-5)² - 4 × 3 × (-2))) / (2 × 3)
x = (5 ± √25 + 24) / 6
x = (5 ± √49) / 6 = (5 ± 7) / 6
Therefore:
因此：
x = (5 + 7)/6 = 2 or x = (5 - 7)/6 = -1∕₃
The solution set is {2, -1∕₃}. You can verify by substituting both values into the original equation.
解集为 {2, -1∕₃}。你可以将两个值代回原方程进行验证。
8. Choosing the Right Method | 如何选择合适的方法
In the exam, you need to decide which method to use. The table below summarises the key considerations.
在考试中，你需要判断使用哪种方法。下表总结了关键考虑因素。
| Method 方法 | Best When 适用情况 | Notes 备注 |
| Factorisation 因式分解 | The quadratic is easy to factor. | Fastest method; requires integer factors. |
| Completing the square 配方法 | a = 1 and coefficients are simple. | Useful for finding turning points of graphs. |
| Quadratic formula 求根公式 | Any quadratic, especially when factorisation is not obvious. | Works universally; use a calculator carefully. |
In general, try factorisation first. If you cannot spot the factors quickly, switch to the quadratic formula or completing the square.
一般情况下，优先尝试因式分解。如果无法快速看出因式排列，则改用求根公式或配方法。
9. Word Problems Involving Quadratics | 二次方程应用题
Quadratic equations often appear in word problems. The key steps are: define the unknown, form the equation, solve it, and then check the answer in the context of the problem.
二次方程经常出现在应用题中。关键步骤是：设未知数、列出方程、求解，然后在问题情境中检验答案。
Example: A rectangle has length (x + 3) cm and width x cm. Its area is 28 cm². Find x.
例：一个长方形的长为 (x + 3) 厘米，宽为 x 厘米，面积为 28 平方厘米。求 x。
The area gives the equation:
面积给出方程：
x(x + 3) = 28
x² + 3x - 28 = 0
Factorise:
因式分解：
(x + 7)(x - 4) = 0
e a length cannot be negative, we reject x = -7. The width is 4 cm and the length is 7 cm.
因此 x = -7 或 x = 4。由于长度不能为负数，我们舍去 x = -7。宽为 4 厘米，长为 7 厘米。
Always remember to state the final answer in the context of the problem.
始终记得在问题情境中写出最终答案。
10. Exam Tips and Common Mistakes | 考试技巧与常见错误
Here are some tips that will help you avoid common pitfalls and score full marks on quadratic equation questions.
以下是一些帮助你在二次方程题目上避免常见陷阱、拿到满分的小贴士。
- Rearrange to the form ax² + bx + c = 0 before attempting to solve. Many students forget to move all terms to one side first.
- 先将方程整理为 ax² + bx + c = 0 的形式再求解。许多学生忘记先把所有项移到一边。
- Check your factorisation by expanding. A quick check can save you from losing several marks.
- 通过展开来检验因式分解是否正确。快速检验能避免丢分。
- When using the formula, write down the values of a, b, c first. This reduces careless substitution errors.
- 使用求根公式时，先写出 a、b、c 的值。这能减少代入时的粗心错误。
- Be careful with signs: b² - 4ac requires correct handling of negative values of b and c.
- 注意符号：b² - 4ac 要求正确处理 b 和 c 的负值。
- Always state both solutions, unless the problem context eliminates one.
- 始终写出两个解，除非题目情境排除了其中一个。
11. Practice Questions | 练习题
Try these questions on your own. The answers are given at the end of this section.
请独立尝试以下题目。本节末尾提供答案。
Question 1: Solve x² - 7x + 10 = 0 by factorisation.
第 1 题：用因式分解法解 x² - 7x + 10 = 0。
Question 2: Solve 2x² + 5x - 3 = 0 using the quadratic formula.
第 2 题：用求根公式解 2x² + 5x - 3 = 0。
uestion 3: Solve x² + 2x - 2 = 0 by completing the square. Leave your answer in surd form.
第 3 题：用配方法解 x² + 2x - 2 = 0。答案保留根号形式。
Question 4: A right-angled triangle has a hypotenuse of 13 cm and two legs of lengths (x + 1) cm and (x + 7) cm. Find x.
第 4 题：一个直角三角形斜边长为 13 厘米，两条直角边长分别为 (x + 1) 厘米和 (x + 7) 厘米。求 x。
12. Answers and Solutions | 答案与解析
Check your answers and review the solutions carefully.
核对你的答案，并仔细复习解题过程。
Answer 1: Factorise as (x - 5)(x - 2) = 0, so x = 5 or x = 2.
答案 1：因式分解为 (x - 5)(x - 2) = 0，所以 x = 5 或 x = 2。
Answer 2: a = 2, b = 5, c = -3. Substitute:
答案 2：a = 2，b = 5，c = -3。代入：
x = (-5 ± √25 + 24) / 4 = (-5 ± 7) / 4
So x = 1∕₂ or x = -3.
因此 x = 1∕₂ 或 x = -3。
Answer 3: x² + 2x = 2. Add (2/2)² = 1 to both sides:
答案 3：x² + 2x = 2。两边加上 (2/2)² = 1：
(x + 1)² = 3
Therefore x = -1 ± √3.
因此 x = -1 ± √3。
Answer 4: By Pythagoras' theorem:
答案 4：由勾股定理：
(x + 1)² + (x + 7)² = 13²
Expanding and simplifying gives x² + 8x + 40 = 169, so x² + 8x - 129 = 0.
展开化简得 x² + 8x + 40 = 169，即 x² + 8x - 129 = 0。
ives (x + 17)(x - 9) = 0. Since lengths are positive, x = 9. The legs are 10 cm and 16 cm.
因式分解得 (x + 17)(x - 9) = 0。由于长度为正，x = 9。两条直角边分别为 10 厘米和 16 厘米。
By mastering these three solving methods and understanding the discriminant, you will be well prepared for any quadratic equation question in your IGCSE exam. Keep practising until the techniques become automatic.
通过掌握这三种求解方法并理解判别式，你将能够从容应对 IGCSE 考试中任何二次方程题目。坚持练习，直到这些技巧变得熟练自如。
Published by TutorHao | 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