📚 Quadratic Equations in Word Problems: From Problem to Equation | 二次方程应用题：从题目到方程的转化
One of the most challenging skills in IGCSE Mathematics is not solving the quadratic equation itself, but rather translating a word problem into the correct equation. Many students can solve \(x^2 – 5x + 6 = 0\) with ease, yet struggle when the same equation is hidden inside a story about rectangles, moving objects, or product prices. In this article, we will break down a reliable strategy for converting word problems into quadratic equations, using Edexcel-style examples throughout.
IGCSE 数学中最具挑战性的技能之一，往往不是解二次方程本身，而是如何将一道应用题准确地转化为方程。许多学生能轻松解出 \(x^2 – 5x + 6 = 0\)，但当同一个方程隐藏在矩形、运动物体或商品价格的故事中时，他们却无从下手。本文将为你拆解一套可靠的策略，帮助你学会将应用题转化为二次方程，并贯穿 Edexcel 风格的例题讲解。
1. Understanding the Key Steps | 理解关键步骤
Every quadratic word problem can be solved through a consistent set of steps. First, read the problem carefully and identify what is being asked. Second, choose a variable \(x\) to represent the unknown quantity. Third, express other quantities in terms of \(x\) using the relationships described in the problem. Fourth, identify the mathematical condition that links these quantities and write it as an equation. Finally, solve the equation and check whether your solutions make sense in the context of the problem.
每一道二次方程应用题都可以通过一套固定的步骤来解决。第一步，仔细阅读题目，明确问题要求什么；第二步，选择一个变量 \(x\) 来表示未知量；第三步，利用题目中的关系，用含 \(x\) 的代数式表达其他量；第四步，找出连接这些量的数学条件，并将其写成方程；最后，解出方程，并检查解是否在题目情境中合理。
Read → Define → Express → Equate → Solve → Check
阅读 → 设未知数 → 表达 → 列方程 → 求解 → 检验
2. The Variable: Choosing What \(x\) Represents | 设未知数：\(x\) 代表什么
In most Edexcel IGCSE quadratic word problems, you are told what \(x\) should represent, for example “Let \(x\) be the width of the rectangle” or “Let \(x\) be the smaller integer.” When not specified, you must choose the variable yourself. The best choice is usually the quantity that is either (a) directly asked for in the question, or (b) the one that other quantities are described in relation to.
在大多数 Edexcel IGCSE 二次方程应用题中，题目会明确告诉你 \(x\) 代表什么，例如“设 \(x\) 为矩形的宽”或“设 \(x\) 为较小的整数”。当题目没有指定时，你需要自己选择变量。最佳选择通常是题目直接要求的量，或者其他量所描述的对象。
For example, if a problem states: “A rectangle is 3 cm longer than it is wide,” then the width is the natural choice for \(x\), while the length becomes \(x + 3\). If a problem asks you to find two consecutive integers, choosing \(x\) as the first integer makes the second integer \(x + 1\). The golden rule is: always define \(x\) in a single sentence before writing anything else.
例如，如果题目说：“一个矩形的长比宽长 3 cm”，那么宽就是 \(x\) 的自然选择，而长则变成 \(x + 3\)。如果题目要求找到两个连续整数，将 \(x\) 设为第一个整数，则第二个整数为 \(x + 1\)。黄金法则是：在写任何别的内容之前，先用一句话明确 \(x\) 的含义。
3. Translating Words into Algebraic Expressions | 将文字翻译成代数表达式
Many students struggle because they do not have a systematic way of converting English phrases into algebra. Some common translations include: “sum” means addition (+), “difference” means subtraction (−), “product” means multiplication (×), “is” or “equals” means (=), and “of” often means multiplication. Expressions such as “the square of a number” become \(x^2\), while “the sum of the number and its square” becomes \(x + x^2\).
许多学生感到困难，是因为他们没有一套系统的方法来将英文短语转换成代数式。常见的翻译包括：“sum”表示加法（+），”difference”表示减法（−），”product”表示乘法（×），”is”或”equals”表示等号（=），而”of”通常表示乘法。例如，“一个数的平方”表示为 \(x^2\)，而“这个数与它的平方之和”则表示为 \(x + x^2\)。
Consider this sentence: “The product of two consecutive integers is 42.” The phrase “two consecutive integers” translates to \(x\) and \(x + 1\). The word “product” tells you to multiply them, and “is 42” tells you the result equals 42. Thus the equation is \(x(x + 1) = 42\). This kind of direct translation is the foundation of all word problem solving.
看这个句子：“两个连续整数的乘积是 42”。“两个连续整数”翻译为 \(x\) 和 \(x + 1\)，“乘积”告诉你将它们相乘，而“是 42”告诉你结果等于 42。因此方程为 \(x(x + 1) = 42\)。这种直接翻译是解决所有应用题的基础。
4. The Classic Rectangle Problem | 经典矩形的面积问题
A very common Edexcel IGCSE question type involves rectangles. Consider this example: “A rectangle has perimeter 30 cm and area 54 cm². Find the dimensions of the rectangle.” Students often try to solve this with a single equation, but the key is to use two pieces of information. Let \(x\) be the length and \(y\) be the width. From the perimeter we get \(2(x + y) = 30\), so \(x + y = 15\), meaning \(y = 15 – x\).
一个非常常见的 Edexcel IGCSE 题型涉及矩形。看这个例子：“一个矩形的周长为 30 cm，面积为 54 cm²。求这个矩形的尺寸。”学生常常试图用一个方程来解，但关键在于使用两条信息。设 \(x\) 为长，\(y\) 为宽。由周长可得 \(2(x + y) = 30\)，因此 \(x + y = 15\)，即 \(y = 15 − x\)。
Now use the area: \(xy = 54\). Substituting \(y = 15 – x\) gives \(x(15 – x) = 54\), which simplifies to \(x^2 – 15x + 54 = 0\). Solving this by factorisation or the quadratic formula gives \(x = 6\) or \(x = 9\). Both values are valid: one represents the length and the other the width. The rectangle is 6 cm by 9 cm.
接着利用面积：\(xy = 54\)。将 \(y = 15 − x\) 代入得 \(x(15 − x) = 54\)，化简为 \(x^2 − 15x + 54 = 0\)。通过因式分解或求根公式，解出 \(x = 6\) 或 \(x = 9\)。两个解都有效：一个代表长，另一个代表宽。矩形的尺寸为 6 cm × 9 cm。
5. Consecutive Numbers and Integer Puzzles | 连续整数问题
Problems involving consecutive numbers are a staple of the Edexcel IGCSE syllabus. A typical question might read: “The sum of the squares of two consecutive positive integers is 61. Find the integers.” Here, the first integer is \(x\) and the second is \(x + 1\). The phrase “the sum of the squares” translates as \(x^2 + (x + 1)^2\), and “is 61” gives the equation.
涉及连续整数的题目是 Edexcel IGCSE 大纲中的常见题型。一道典型题目可能是：“两个连续正整数的平方和是 61。求这两个整数。”这里，第一个整数为 \(x\)，第二个为 \(x + 1\)。“平方和”翻译为 \(x^2 + (x + 1)^2\)，“是 61”则给出方程。
Expanding \((x + 1)^2\) to get \(x^2 + 2x + 1\), the equation becomes \(x^2 + x^2 + 2x + 1 = 61\), which simplifies to \(2x^2 + 2x – 60 = 0\), or \(x^2 + x – 30 = 0\). Factoring gives \((x + 6)(x – 5) = 0\), so \(x = -6\) or \(x = 5\). Since the problem specifies positive integers, \(x = 5\), and the two integers are 5 and 6.
展开 \((x + 1)^2\) 得到 \(x^2 + 2x + 1\)，原方程变为 \(x^2 + x^2 + 2x + 1 = 61\)，化简后为 \(2x^2 + 2x − 60 = 0\)，即 \(x^2 + x − 30 = 0\)。因式分解得 \((x + 6)(x − 5) = 0\)，因此 \(x = −6\) 或 \(x = 5\)。题目指定正整数，故 \(x = 5\)，这两个整数是 5 和 6。
6. Product and Sum Patterns | 积与和模式
Another frequently tested pattern asks you to relate the product and sum of numbers. For instance: “A number squared minus twice the same number equals 15. Find all possible values of the number.” Let \(x\) be the number. The phrase “squared” gives \(x^2\), “minus twice the same number” gives \(−2x\), and “equals 15” gives the equation \(x^2 − 2x = 15\).
另一种常见模式要求你关联数字的积与和。例如：“一个数的平方减去这个数的两倍等于 15。求这个数的所有可能值。”设 \(x\) 为这个数。“平方”给出 \(x^2\)，“减去这个数的两倍”给出 \(−2x\)，“等于 15”给出方程 \(x^2 − 2x = 15\)。
Rearranging to standard form: \(x^2 – 2x – 15 = 0\). This factors to \((x – 5)(x + 3) = 0\), so \(x = 5\) or \(x = -3\). Both values are valid because the problem did not restrict the type of number. This is an important lesson: always check the constraints mentioned in the question before discarding a root.
整理为标准形式：\(x^2 − 2x − 15 = 0\)。分解因式得 \((x − 5)(x + 3) = 0\)，因此 \(x = 5\) 或 \(x = −3\)。在确保两侧开方时为正…… 实际上，这两个值都有效，因为题目没有限制数字类型。这是一个重要的教训：在舍弃某个根之前，务必检查题目中给出的约束条件。
7. The Pythagorean Theorem Approach | 勾股定理的应用题
Geometry questions that produce quadratic equations often come from the Pythagorean theorem. Consider: “The hypotenuse of a right-angled triangle is 10 cm, and one leg is 2 cm longer than the other. Find the length of the longer leg.” Let the shorter leg be \(x\) cm. Then the longer leg is \(x + 2\) cm, and the hypotenuse is 10 cm. The Pythagorean theorem gives \(x^2 + (x + 2)^2 = 10^2\).
能产生二次方程的几何题目通常来自勾股定理。考虑：“一个直角三角形的斜边长为 10 cm，一条直角边比另一条长 2 cm。求较长直角边的长度。”设较短的直角边为 \(x\) cm，则较长的直角边为 \(x + 2\) cm，斜边为 10 cm。由勾股定理得 \(x^2 + (x + 2)^2 = 10^2\)。
Expanding: \(x^2 + x^2 + 4x + 4 = 100\), so \(2x^2 + 4x – 96 = 0\), or \(x^2 + 2x – 48 = 0\). Factoring gives \((x + 8)(x – 6) = 0\), hence \(x = -8\) (rejected, since length cannot be negative) or \(x = 6\). The shorter leg is 6 cm and the longer leg is 8 cm. This example shows how one root is often discarded for practical reasons.
展开：\(x^2 + x^2 + 4x + 4 = 100\)，得到 \(2x^2 + 4x − 96 = 0\)，即 \(x^2 + 2x − 48 = 0\)。因式分解得 \((x + 8)(x − 6) = 0\)，因此 \(x = −8\)（舍去，因为长度不能为负）或 \(x = 6\)。较短直角边为 6 cm，较长直角边为 8 cm。这个例子展示了在实际情境中舍弃一个根的情形。
8. Calculating with Money: Rates and Costs | 金钱计算：价格与费用问题
Exam questions frequently frame quadratic equations in financial contexts. For example: “The cost of buying \(n\) identical books is £12. If each book had cost £1 less, the same £12 could buy 4 more books. Find \(n\).” Here the price per book is \(\frac{12}{n}\). In the hypothetical situation, the price per book would be \(\frac{12}{n} – 1\), and the number of books would be \(n + 4\).
考试题目经常将二次方程置于金融背景中。例如：“购买 \(n\) 本相同的书共花费 12 英镑。如果每本书便宜 1 英镑，同样的 12 英镑可以多买 4 本书。求 \(n\)。”这里每本书的价格为 \(\frac{12}{n}\)。在假设情境中，每本书的价格为 \(\frac{12}{n} − 1\)，而购买数量为 \(n + 4\)。
Since the cost per book is the total cost divided by the number of books, we must have both expressions describing the same quantity: \(\frac{12}{n + 4} = \frac{12}{n} – 1\). Multiplying through by \(n(n+4)\) eliminates fractions: \(12n = 12(n + 4) – n(n + 4)\). Expanding, we get \(12n = 12n + 48 – n^2 – 4n\), or \(0 = 48 – n^2 – 4n\). Rearranging: \(n^2 + 4n – 48 = 0\).
由于每本书的价格等于总价除以数量，两个表达式必须描述同一个量：\(\frac{12}{n + 4} = \frac{12}{n} − 1\)。两边同时乘以 \(n(n + 4)\) 以消去分母：\(12n = 12(n + 4) − n(n + 4)\)。展开得 \(12n = 12n + 48 − n^2 − 4n\)，即 \(0 = 48 − n^2 − 4n\)。整理后：\(n^2 + 4n − 48 = 0\)。
Factoring \((n + 8)(n – 6) = 0\) yields \(n = 6\) (reject \(n = -8\) as negative). So 6 books were originally bought at £2 each. Financial word problems commonly produce quadratic equations because multiplying two quantities — price and quantity — naturally creates a second-degree expression.
因式分解 \((n + 8)(n − 6) = 0\) 得 \(n = 6\)（舍去 \(n = −8\)，因为数量为负）。因此最初买了 6 本书，每本 2 英镑。金融类应用题之所以常产生二次方程，是因为两个量——价格与数量——相乘时自然会产生二次表达式。
9. Rejecting Solutions: Checking Your Roots | 检验根：如何舍去无效解
After solving a quadratic equation, you usually end up with two possible values for \(x\). It is essential to check each one against the original problem’s context. Three common reasons for rejecting a root are: the value is negative when the problem requires a positive quantity (such as a length, a count of items, or a price); the value produces a negative dimension when substituted back into expressions like \(x + 3\); or the value violates an explicit condition stated in the question, such as “positive integers.”
解出二次方程后，通常能得到 \(x\) 的两个可能值。必须将每个值放回原题情境中验证。舍去一个根的三个常见原因是：值为负数而题目要求正数（如长度、物品数量或价格）；将值代回 \(x + 3\) 等表达式时产生负的维度；或者数值违反了题目中明确给出的条件，例如“正整数”。
This step is especially important in Edexcel IGCSE because alternative solutions are not always physically meaningful. Always write a short line in your solution: “\(x = -6\) is rejected because a length cannot be negative.” Not only does this clarify your reasoning for examiners, but it also demonstrates a complete understanding of the modelling process.
这一步骤在 Edexcel IGCSE 中尤为重要，因为额外的解并非总是具有实际意义。在解答中一定要写一行短句：“因为长度不能为负，所以 \(x = −6\) 被舍去。”这不仅能让考官看清你的推理过程，也表明你对建模过程有完整的理解。
10. A Complete Worked Example | 完整例题精讲
Let us now work through a full Edexcel-style problem from start to finish. Question: “A rectangular garden has a length of \((x + 5)\) metres and a width of \((x – 1)\) metres. Its area is 20 m². Find \(x\).” Step one: identify the known relationships. The area of a rectangle is \( \text{length} \times \text{width}\). Thus \((x + 5)(x – 1) = 20\).
现在我们从头到尾完成一道完整的 Edexcel 风格题目。题目：“一个矩形花园的长为 \((x + 5)\) 米，宽为 \((x − 1)\) 米，面积为 20 平方米。求 \(x\)。”第一步：确认已知关系。矩形的面积为长乘宽，因此 \((x + 5)(x − 1) = 20\)。
Expanding the left side: \(x^2 + 4x – 5 = 20\). Subtract 20 from both sides: \(x^2 + 4x – 25 = 0\). This does not factor nicely, so use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\) with \(a = 1\), \(b = 4\), \(c = -25\).
展开左边：\(x^2 + 4x − 5 = 20\)。两边同时减去 20：\(x^2 + 4x − 25 = 0\)。该式不易因式分解，因此使用求根公式 \(x = \frac{−b ± \sqrt{b^2 − 4ac}}{2a}\)，其中 \(a = 1\)、\(b = 4\)、\(c = −25\)。
This gives \(x = \frac{-4 \pm \sqrt{16 + 100}}{2} = \frac{-4 \pm \sqrt{116}}{2} = \frac{-4 \pm 2\sqrt{29}}{2}\). Hence \(x = -2 + \sqrt{29}\) or \(x = -2 – \sqrt{29}\). Since the width \(x – 1\) must be positive, we reject the negative root. So \(x = -2 + \sqrt{29} \approx 3.385\), and the width is approximately 2.385 m, with length approximately 8.385 m.
于是 \(x = \frac{−4 ± \sqrt{16 + 100}}{2} = \frac{−4 ± \sqrt{116}}{2} = \frac{−4 ± 2\sqrt{29}}{2}\)。因此 \(x = −2 + \sqrt{29}\) 或 \(x = −2 − \sqrt{29}\)。由于宽 \(x − 1\) 必须为正，我们舍去负根。所以 \(x = −2 + \sqrt{29} \approx 3.385\)，宽约为 2.385 m，长约为 8.385 m。
11. Common Pitfalls and How to Avoid Them | 常见错误与规避方法
Students make several predictable mistakes when working on quadratic word problems. One common pitfall is forgetting to expand and rearrange the equation into the standard form \(ax^2 + bx + c = 0\) before applying the quadratic formula. Another is solving the equation but never checking whether the roots make sense in the original context: negative lengths or fractional counts of people are immediate red flags.
学生在做二次方程应用题时，常犯若干可预见的错误。一个常见陷阱是忘记展开并整理为标准形式 \(ax^2 + bx + c = 0\) 就直接套用求根公式。另一个错误是解出方程后不检查根是否在原题情境中合理：长度为负或人数为非整数都是明显的警示信号。
Additionally, students often misread the phrase “one number is 3 more than another” and incorrectly write \(x + 3\) as the relationship in both directions. Careful reading and explicit definition of the variable can prevent these issues. Always translate the problem into your own words before translating it into algebra, and always write equations in non-abbreviated sentences first if you find yourself struggling.
此外，学生常常错误解读“一个数比另一个数多 3”这种表述，并在两个方向都错误地写成 \(x + 3\)。仔细阅读题目并明确设变量可避免这些问题。在将文字转换为代数之前，先用自己的话复述题目；如果觉得困难，可以先把文字关系写成完整的句子，再转成方程。
12. Summary and Exam Strategy | 总结与考试策略
In summary, converting word problems into quadratic equations is a skill that improves with practice and a systematic method: read carefully, define one variable, express all other quantities in terms of that variable, identify the linking condition, write the equation, solve it, and check validity. In the Edexcel IGCSE exam, these problems are usually worth 4 to 6 marks: 1 mark for defining the variable, 1-2 marks for forming the equation, 1-2 marks for solving it correctly, and 1 mark for the final valid answer with units.
总结而言，将应用题转化为二次方程是一项可以通过练习和系统方法逐步提高的技能：仔细阅读、设定一个变量、用该变量表达其他所有量、找出连接条件、写出方程、求解并检验有效性。在 Edexcel IGCSE 考试中，这类题目通常为 4 到 6 分：设变量 1 分，列方程 1−2 分，正确求解 1−2 分，最终含单位且合理的答案 1 分。
We also recommend a two-minute verification strategy: after obtaining your roots, substitute them back into the original word problem (not just the equation) rather than merely into the algebra. This confirms that the entire problem — not just the equation — is satisfied. With regular practice, you will find that the translation step becomes faster and more intuitive, and solving these problems will become one of the most reliable parts of your exam technique.
我们还推荐一个两分钟的验证策略：求得根之后，将它们代回原题（而不仅仅是方程）中，确认整个问题——而非只有方程——都得到满足。通过定期练习，你会发现转化步骤会越来越快、越来越直觉化，解决这类题将成为你考试中最稳妥的部分之一。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply