📚 Problems Leading to Quadratic Equations | 二次方程应用题
In the Edexcel IGCSE Mathematics syllabus, quadratic equations rarely appear in isolation. More often, you are given a word problem that must be translated into a quadratic equation before solving. This article explores the classic problem types, a reliable step-by-step strategy, and the most common pitfalls, so that you can confidently tackle any question that “leads to a quadratic equation.”
在 Edexcel IGCSE 数学大纲中，二次方程很少单独出现。更常见的情况是，你拿到一道应用题，必须先将其翻译成二次方程，然后再求解。本文将探讨经典题型、可靠的解题步骤策略以及最常见的易错点，帮助你自信应对任何”可转化为二次方程”的题目。
1. The Standard Form and Solving Methods | 标准形式与解法
Every quadratic equation can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. In word problems, you first obtain this form by expanding and rearranging an equation derived from the situation described in the question.
任何一个二次方程都可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。在应用题中，你首先需要根据题目描述的情境列式，再通过展开和移项得到这一标准形式。
Three solution methods are commonly tested: factorisation, the quadratic formula x = (−b ± √(b² − 4ac)) / 2a, and completing the square. In IGCSE word problems, factorisation is often the fastest when the roots are integers; the quadratic formula is the reliable fallback for non-integer roots.
三种解法是考试常见考点：因式分解法、求根公式 x = (−b ± √(b² − 4ac)) / 2a 以及配方法。在 IGCSE 应用题中，当根为整数时，因式分解通常最快；当根不是整数时，求根公式是最可靠的备用方法。
ax² + bx + c = 0, with a ≠ 0
x = (−b ± √(b² − 4ac)) / 2a
2. Number Problems | 数字问题
Number problems ask you to find an unknown number from a relationship involving squares or products. For example: “The square of a number plus three times the number equals 40. Find the number.” Let x be the unknown number. The equation is x² + 3x = 40.
数字问题要求你根据涉及平方或乘积的关系求出未知数。例如：”一个数的平方加上这个数的三倍等于 40，求这个数。”设未知数为 x，则方程为 x² + 3x = 40。
Rearranging gives x² + 3x − 40 = 0. Factorising, we obtain (x + 8)(x − 5) = 0, so x = −8 or x = 5. If the problem does not specify “positive”, both solutions are valid; if it asks for a positive number, we select x = 5.
移项得 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。如果题目没有特别说明”正数”，则两个解都成立；如果要求正数，则取 x = 5。
- Read carefully: does the problem allow negative answers? 仔细审题：题目是否允许负数答案？
- Always define your variable at the start. 解题之初务必设未知数。
- Check each solution in the original wording. 将每个解代回原题文字中验证。
3. Consecutive Integer Problems | 连续整数问题
Consecutive integer problems are a special family of number problems. A classic example: “The product of two consecutive positive integers is 132. Find the integers.” Let the smaller integer be n; the next integer is then n + 1.
连续整数问题是数字问题中的一个特殊类型。经典例题：”两个连续正整数的乘积为 132，求这两个整数。”设较小整数为 n，则下一个整数为 n + 1。
We write n(n + 1) = 132, which expands to n² + n = 132. Rearranging gives n² + n − 132 = 0. Factorise: (n + 12)(n − 11) = 0, hence n = −12 or n = 11. Since the integers are positive, we take n = 11 and n + 1 = 12.
我们列式 n(n + 1) = 132，展开得 n² + n = 132。移项得 n² + n − 132 = 0。因式分解：(n + 12)(n − 11) = 0，因此 n = −12 或 n = 11。因为要求正整数，所以取 n = 11，n + 1 = 12。
n(n + 1) = 132 → n² + n − 132 = 0 → (n + 12)(n − 11) = 0
Consecutive odd or even integers differ by 2, so they are represented as n and n + 2. Consecutive integers differ by 1, so they are n and n + 1. Always check the wording of the question.
连续奇数或连续偶数相差 2，因此表示为 n 和 n + 2。连续整数相差 1，因此表示为 n 和 n + 1。务必仔细查看题目措辞。
4. Rectangle and Area Problems | 矩形与面积问题
Area problems are among the most common quadratic applications. Example: “A rectangle is 5 cm longer than it is wide. Its area is 84 cm². Find its dimensions.” Let the width be x cm. Then the length is (x + 5) cm.
面积问题是最常见的二次方程应用之一。例如：”一个矩形的长比宽长 5 cm，面积为 84 cm²，求其尺寸。”设宽为 x cm，则长为 (x + 5) cm。
The area equation is x(x + 5) = 84. Expanding and rearranging gives x² + 5x − 84 = 0. Factorising: (x + 12)(x − 7) = 0, so x = −12 or x = 7. A width cannot be negative, so x = 7. The rectangle is 7 cm by 12 cm.
面积方程为 x(x + 5) = 84。展开并移项得 x² + 5x − 84 = 0。因式分解得 (x + 12)(x − 7) = 0，因此 x = −12 或 x = 7。宽度不能为负，所以 x = 7。该矩形为 7 cm × 12 cm。
- Width = x, length = x + 5, area = width × length. 宽 = x，长 = x + 5，面积 = 宽 × 长。
- Reject negative lengths immediately. 直接舍去负的长度值。
- State the final answer with units. 最终答案务必带单位。
5. Right-Angled Triangle Problems | 直角三角形问题
Pythagoras’ theorem often produces quadratic equations when the two legs are related. Example: “The hypotenuse of a right-angled triangle is 13 cm. One leg is 7 cm longer than the other. Find the lengths of the legs.”
当两条直角边之间存在关系时，勾股定理常常会产生二次方程。例如：”直角三角形的斜边为 13 cm，一条直角边比另一条长 7 cm，求两条直角边的长度。”
Let the shorter leg be x cm. Then the other leg is (x + 7) cm. By Pythagoras: x² + (x + 7)² = 13². Expand: x² + x² + 14x + 49 = 169, so 2x² + 14x − 120 = 0. Divide by 2: x² + 7x − 60 = 0.
设较短直角边为 x cm，则另一条直角边为 (x + 7) cm。由勾股定理：x² + (x + 7)² = 13²。展开：x² + x² + 14x + 49 = 169，所以 2x² + 14x − 120 = 0。两边除以 2：x² + 7x − 60 = 0。
x² + (x + 7)² = 169 → 2x² + 14x − 120 = 0 → x² + 7x − 60 = 0
Factorising (x + 12)(x − 5) = 0 gives x = −12 or x = 5. Since a length cannot be negative, the legs are 5 cm and 12 cm. This is the famous 5-12-13 Pythagorean triple.
因式分解 (x + 12)(x − 5) = 0 得 x = −12 或 x = 5。由于长度不能为负，两条直角边分别为 5 cm 和 12 cm。这正是著名的 5-12-13 勾股数。
6. Projectile and Motion Problems | 抛体与运动问题
Motion problems use the formula s = ut + ½at², where s is displacement, u is initial velocity, t is time and a is acceleration. Under gravity on Earth, a ≈ −10 m/s², so a quadratic in t frequently appears.
运动问题使用公式 s = ut + ½at²，其中 s 为位移，u 为初速度，t 为时间，a 为加速度。在地球重力作用下，a ≈ −10 m/s²，因此经常出现关于 t 的二次方程。
Example: “A ball is thrown upward with initial velocity 20 m/s. At what times is the ball 15 m above the ground? Use a = −10 m/s².” Substitute: 15 = 20t − 5t².
例：”小球以 20 m/s 的初速度竖直上抛，在哪些时间小球离地 15 m？取 a = −10 m/s²。”代入得：15 = 20t − 5t²。
Rearrange: 5t² − 20t + 15 = 0. Divide by 5: t² − 4t + 3 = 0. Factorise: (t − 1)(t − 3) = 0, so t = 1 or t = 3. The ball reaches 15 m on the way up at 1 s and on the way down at 3 s.
移项得 5t² − 20t + 15 = 0。除以 5：t² − 4t + 3 = 0。因式分解得 (t − 1)(t − 3) = 0，所以 t = 1 或 t = 3。小球在上升途中于 1 s 到达 15 m，在下落途中于 3 s 再次到达 15 m。
- Two positive times are often both valid for an upward throw. 上抛问题中两个正时间通常都有效。
- Write s, u, t, a in a column first to avoid substitution errors. 先列好 s、u、t、a 再代入，避免代错。
- The negative root often represents a time before the throw; reject it. 负根通常表示抛出之前的时间，应舍去。
7. Cost and Profit Problems | 成本与利润问题
Business problems sometimes give profit as a product of price and quantity, leading to a quadratic. Example: “A shop sells x items per day. The daily profit is given by P = x(20 − x) pounds. Find x so that the daily profit is £96.”
商业问题中，利润有时表示为价格与数量的乘积，从而产生二次方程。例如：”某商店每天售出 x 件商品，日利润为 P = x(20 − x) 英镑。求使日利润为 96 英镑的 x 值。”
Set x(20 − x) = 96. Expand: 20x − x² = 96. Rearrange: x² − 20x + 96 = 0. Factorise: (x − 8)(x − 12) = 0, so x = 8 or x = 12. Both answers make sense: selling 8 items or 12 items both yield £96 profit.
令 x(20 − x) = 96。展开得 20x − x² = 96。移项得 x² − 20x + 96 = 0。因式分解得 (x − 8)(x − 12) = 0，因此 x = 8 或 x = 12。两个答案都合理：售出 8 件或 12 件都能获得 96 英镑利润。
x(20 − x) = 96 → x² − 20x + 96 = 0 → (x − 8)(x − 12) = 0
Always check whether both roots are acceptable in the context. If the question demands “maximum profit”, you would instead complete the square: P = −(x − 10)² + 100, giving a maximum of £100 at x = 10.
务必检查两个根在具体情境中是否都可行。如果题目要求”最大利润”，则应改用配方法：P = −(x − 10)² + 100，最大利润为 100 英镑，此时 x = 10。
8. Border and Frame Problems | 边框与画框问题
Another frequent geometry application involves adding a uniform border around a rectangle. Example: “A rectangular picture is 10 cm by 8 cm. A border of uniform width x cm is added. The total area becomes 120 cm². Find x.”
另一个常见的几何应用是在矩形周围加上等宽的边框。例如：”一幅矩形画为 10 cm × 8 cm，四周加上宽度为 x cm 的等宽边框后，总面积为 120 cm²。求 x。”
The outer dimensions become (10 + 2x) and (8 + 2x). Hence (10 + 2x)(8 + 2x) = 120. Expand: 80 + 36x + 4x² = 120, so 4x² + 36x − 40 = 0. Divide by 4: x² + 9x − 10 = 0.
外框尺寸变为 (10 + 2x) 和 (8 + 2x)。因此 (10 + 2x)(8 + 2x) = 120。展开得 80 + 36x + 4x² = 120，所以 4x² + 36x − 40 = 0。除以 4 得 x² + 9x − 10 = 0。
Factorise (x + 10)(x − 1) = 0, giving x = −10 or x = 1. The border width must be positive, so x = 1 cm. Notice that x = −10 would make the outer dimensions negative, which is impossible.
因式分解得 (x + 10)(x − 1) = 0，得到 x = −10 或 x = 1。边框宽度必须为正数，所以 x = 1 cm。注意 x = −10 会使外框尺寸为负，这是不可能的。
9. A Complete Worked Example | 完整例题演示
Let us combine everything into one careful, exam-style solution. Question: “A rectangular garden is 3 m longer than it is wide. The area of the garden is 70 m². Find the perimeter of the garden.”
让我们将所学内容整合到一道完整的考试风格例题中。题目：”一个矩形花园的长比宽多 3 m，面积为 70 m²，求该花园的周长。”
Step 1: Define variables. Let the width be w m. Then the length is (w + 3) m. Step 2: Write the area equation: w(w + 3) = 70. Step 3: Expand and rearrange: w² + 3w − 70 = 0.
第一步：设未知数。设宽为 w m，则长为 (w + 3) m。第二步：写出面积方程：w(w + 3) = 70。第三步：展开并移项：w² + 3w − 70 = 0。
Step 4: Factorise: (w + 10)(w − 7) = 0, so w = −10 or w = 7. Reject w = −10. Therefore width = 7 m and length = 10 m. Step 5: Perimeter = 2(7 + 10) = 34 m.
第四步：因式分解：(w + 10)(w − 7) = 0，因此 w = −10 或 w = 7。舍去 w = −10。因此宽 = 7 m，长 = 10 m。第五步：周长 = 2(7 + 10) = 34 m。
Perimeter = 2(7 + 10) = 34 m
Always check your final answer against the question. A 7 m by 10 m rectangle has area 70 m² and its length is 3 m more than its width, so the answer is consistent.
始终将最终答案与原题对照检查。7 m × 10 m 的矩形面积为 70 m²，长比宽多 3 m，答案完全一致。
10. A General Strategy | 通用解题策略
Success in these problems depends on a consistent routine. The following five-step strategy works for every quadratic word problem you will meet in the IGCSE examination.
解这类题目的关键在于一套一致的解题流程。下面五步策略适用于 IGCSE 考试中遇到的所有二次方程应用题。
| Step 步骤 | Action 操作 |
|---|---|
| 1 | Read the question and underline the unknown. 读题并标出未知量。 |
| 2 | Let x (or another letter) represent the unknown. 设 x（或其他字母）为未知数。 |
| 3 | Form an equation from the given conditions. 根据已知条件列方程。 |
| 4 | Rearrange to ax² + bx + c = 0 and solve. 化为 ax² + bx + c = 0 并求解。 |
| 5 | Reject invalid roots and answer the original question. 舍去不合理根并回答原问题。 |
In step 4, choose factorisation if you spot factors quickly. Otherwise, use the quadratic formula. In step 5, reject roots that are negative lengths, negative times, or numbers that contradict the wording of the problem.
在第四步中，如果能快速找到因式就优先因式分解；否则使用求根公式。在第五步中，舍去负长度、负时间以及与原题文字矛盾的根。
11. Common Mistakes to Avoid | 常见易错点
Many students lose marks not in solving the equation but in interpreting the problem. Here are the five most common errors seen in Edexcel IGCSE examination scripts.
许多学生丢分并非因为解方程，而是因为对题意的理解。以下是 Edexcel IGCSE 答卷中最常见的五类错误。
- Forgetting to reject a negative root when the context demands a positive value. 当题目情境要求正数时，忘记舍去负根。
- Setting up the equation incorrectly, e.g. using x + 2 instead of x + 7 for “7 longer”. 方程列错，例如将”长 7″误写成 x + 2。
- Expanding brackets incorrectly, especially with (x + 7)² = x² + 14x + 49. 展开括号出错，尤其 (x + 7)² = x² + 14x + 49。
- Answering with x but not the quantity asked for, such as giving x = 7 when the question asks for the length = 12. 只答 x 而未回答题目所问的量，例如题目问长 = 12 时只写 x = 7。
- Omitting units in the final answer. 最终答案漏写单位。
After solving, always substitute your answers back into the original word problem, not just into the equation. This catches most interpretation errors.
求解后，务必将答案代回原题文字中进行检验，而不仅仅代回方程。这样可以发现大多数理解性错误。
12. Quick Practice Problems | 快速练习题
Test yourself with these four questions in exactly the style of the Edexcel IGCSE examination. Solve them fully before checking the answers below.
请完成以下四道 Edexcel IGCSE 风格的自测题。先完整求解，再核对下方答案。
| Question 题目 | Answer 答案 |
|---|---|
| The product of two consecutive odd integers is 143. Find the integers. 两个连续奇数的乘积为 143，求这两个数。 | 11 and 13 11 和 13 |
| A right-angled triangle has legs x and x + 2, with hypotenuse √34. Find x. 直角三角形两直角边为 x 与 x + 2，斜边为 √34，求 x。 | x = 3 (legs 3 and 5) x = 3（直角边为 3 和 5） |
| A ball is thrown so that its height is h = 14t − 5t². Find t when h = 8. 球的高度为 h = 14t − 5t²，求 h = 8 时的 t。 | t = 0.5 or t = 2.3 (3 s.f.) t = 0.5 或 t = 2.3（3 位有效数字） |
| A rectangle has area 40 m² and length 3 m more than twice its width. Find the width. 矩形面积为 40 m²，长比宽的 2 倍多 3 m，求宽。 | Width = 4 m 宽 = 4 m |
For question 4, let width = w, length = 2w + 3. Then w(2w + 3) = 40, so 2w² + 3w − 40 = 0. Factorise: (2w − 5)(w + 8) = 0? No, correct factorisation is (2w + 8) is wrong. Let us solve by formula: w = (−3 + √(9 + 320)) / 4 = (−3 + √329) / 4 ≈ 4.0 m; check: 4 × 11 = 44 ≠ 40. Correct derivation: w(2w + 3) = 40 → 2w² + 3w − 40 = 0 → (2w + 8)(w − 5) = 0 is incorrect because 2w × w = 2w², 8w − 5w = 3w, 8 × (−5) = −40. Wait: (2w + 8)(w − 5) = 2w² − 10w + 8w − 40 = 2w² − 2w − 40, not correct. Correct factorisation: (2w − 5)(w + 8) = 2w² + 16w − 5w − 40 = 2w² + 11w − 40, no. Let me solve properly: 2w² + 3w − 40 = 0. Factors of 2w²: 2w and w. Factors of −40: need pair summing to +3 with cross terms: 2w × 5 = 10w and w × (−8) = −8w, sum 2w, no. 2w × −5 = −10w and w × 8 = 8w, sum −2w, no. So use quadratic formula: w = (−3 ± √(9 + 320)) / 4 = (−3 ± √329) / 4. √329 ≈ 18.14, so w ≈ (−3 + 18.14)/4 ≈ 3.785 =
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