📚 Solving Real-World Problems with Quadratic Equations | 用二次方程解决实际问题
Quadratic equations are more than abstract algebraic exercises; they model many real-world situations in physics, geometry, business, and engineering. This article focuses on how to translate word problems into quadratic equations and solve them effectively for IB Mathematics.
二次方程不仅仅是抽象的代数练习，它们在物理、几何、商业和工程等领域中广泛用于建模实际问题。本文重点讲解如何将文字题转化为二次方程，并在 IB 数学考试中高效求解。
1. The General Form and Roots | 二次方程的一般形式与根
A quadratic equation can be written in the general form ax² + bx + c = 0, where a ≠ 0. Its roots are the values of x that satisfy the equation.
二次方程的一般形式为 ax² + bx + c = 0，其中 a ≠ 0。方程的根就是满足该等式的 x 的值。
The roots can be found by factorisation, completing the square, or using the quadratic formula:
求根可以使用因式分解、配方法或二次公式：
x = (-b ± √(b² – 4ac)) / (2a)
When applying this formula to a real-world problem, remember that only roots that make sense in the context are acceptable. For example, a length or time cannot be negative.
在应用题中使用该公式时，只有符合实际情境的根才是可接受的。例如，长度或时间不能为负数。
2. Forming a Quadratic Equation from a Word Problem | 从文字题建立二次方程
The most important skill is translating a word problem into algebra. A reliable process is: define a variable, write an equation from the given relationships, solve it, and interpret the solutions.
最关键的能力是将文字题转化为代数式。可靠的步骤是：设未知数、根据已知关系列出方程、求解，并解释解的含义。
For example: “The product of two consecutive positive integers is 42. Find the integers.” Let the smaller integer be n. Then n(n + 1) = 42, giving n² + n – 42 = 0.
例如：“两个连续正整数的乘积为 42，求这两个整数。”设较小的整数为 n，则 n(n + 1) = 42，得到 n² + n – 42 = 0。
Solving gives n = 6 or n = -7. Since the problem asks for positive integers, the valid answer is 6 and 7.
解得 n = 6 或 n = -7。由于题目要求正整数，因此有效答案为 6 和 7。
3. The Discriminant and the Nature of Roots | 判别式与根的性质
The discriminant is Δ = b² – 4ac. It tells us the nature of the roots without solving the equation.
判别式为 Δ = b² – 4ac。它无需解方程即可判断根的性质。
If Δ > 0, the equation has two distinct real roots. If Δ = 0, there is exactly one real root (a repeated root). If Δ < 0, there are no real roots.
当 Δ > 0 时，方程有两个不相等的实根；当 Δ = 0 时，有一个实根（重根）；当 Δ < 0 时，没有实根。
In applications, a negative discriminant often means the situation described cannot happen. For instance, a ball cannot reach a certain height if the corresponding equation has no real roots.
在实际问题中，判别式为负通常意味着所述情形不可能发生。例如，若对应的方程无实根，则小球无法达到某高度。
4. Projectile Motion | 抛体运动
A classic application is projectile motion. The height h of an object after time t can be modelled by h(t) = -gt² + v₀t + h₀, where g is gravitational acceleration, v₀ is initial velocity, and h₀ is initial height.
抛体运动是经典应用。物体在时间 t 的高度可建模为 h(t) = -gt² + v₀t + h₀，其中 g 是重力加速度，v₀ 是初速度，h₀ 是初始高度。
In IB problems, heights are often given in metres and time in seconds. On Earth, g ≈ 9.8 m/s², so the coefficient is often -4.9t². For example, if a ball is thrown upward from the ground with initial velocity 19.6 m/s, its height is h(t) = -4.9t² + 19.6t.
在 IB 题目中，高度通常以米、时间以秒为单位。地球上 g ≈ 9.8 m/s²，因此二次项系数常为 -4.9t²。例如，若一小球以 19.6 m/s 的初速度从地面竖直上抛，其高度为 h(t) = -4.9t² + 19.6t。
To find when the ball hits the ground, set h(t) = 0 and solve. Factoring gives t(-4.9t + 19.6) = 0, so t = 0 or t = 4 seconds. The positive time after launch is the answer.
求小球何时落地，令 h(t) = 0 并求解。因式分解得 t(-4.9t + 19.6) = 0，所以 t = 0 或 t = 4 秒。发射后大于零的时间即为答案。
The maximum height occurs at the vertex. The time at the vertex is t = -v₀/(2a), where a = -4.9. Here t = 2 s, and h(2) = -4.9(2)² + 19.6(2) = 19.6 m.
最大高度出现在顶点处。到达顶点的时间为 t = -v₀/(2a)，其中 a = -4.9。此处 t = 2 秒，h(2) = -4.9(2)² + 19.6(2) = 19.6 米。
5. Geometry and Area Problems | 几何与面积问题
Many geometric problems lead to quadratic equations. For example, a rectangle has a perimeter of 20 m and an area of 24 m². Let its width be x. The length is 10 – x, so x(10 – x) = 24.
许多几何问题会化为二次方程。例如，一个矩形的周长为 20 米，面积为 24 平方米。设宽为 x，则长为 10 – x，所以 x(10 – x) = 24。
Expanding and rearranging gives x² – 10x + 24 = 0, which factors as (x – 4)(x – 6) = 0. Thus the dimensions are 4 m by 6 m.
展开并整理得 x² – 10x + 24 = 0，因式分解为 (x – 4)(x – 6) = 0。因此矩形的尺寸为 4 米 × 6 米。
When solving area problems, always check that the side lengths are positive and that they satisfy the original conditions.
在求解面积问题时，务必检查边长为正数，并满足原始条件。
6. Business and Economics Applications | 商业与经济中的应用
Quadratic equations appear in profit and revenue problems. Suppose the revenue R from selling x items is given by R(x) = -x² + 100x. To find the number of items that gives a revenue of 2400, solve -x² + 100x = 2400.
二次方程也出现在利润和收入问题中。假设卖出 x 件商品的收入为 R(x) = -x² + 100x。若要求收入为 2400，解 -x² + 100x = 2400。
Rearranging gives x² – 100x + 2400 = 0, so (x – 40)(x – 60) = 0. Either 40 or 60 items can produce this revenue.
整理得 x² – 100x + 2400 = 0，所以 (x – 40)(x – 60) = 0。销售 40 件或 60 件都能获得该收入。
In business problems, the domain of x is usually non-negative, and sometimes it is restricted to integers. Always interpret the result within the given context.
在商业问题中，x 的定义域通常为非负数，有时还要求是整数。务必结合具体情境解释结果。
7. Consecutive Number and Digit Problems | 连续整数与数字问题
Problems involving consecutive integers are a standard source of quadratic equations. For three consecutive odd integers, if the sum of the squares of the first two is equal to the square of the third, we can set up an equation.
连续整数问题是二次方程的常见来源。若三个连续奇数的前两个的平方和等于第三个的平方，我们可以据此建立方程。
Let the integers be n, n + 2, n + 4. Then n² + (n + 2)² = (n + 4)². Expanding and simplifying gives n² – 4n – 12 = 0, so n = 6 or n = -2.
设三个连续奇数为 n、n + 2、n + 4。则 n² + (n + 2)² = (n + 4)²。展开化简得 n² – 4n – 12 = 0，所以 n = 6 或 n = -2。
The odd numbers are therefore 6, 8, 10 (even, so not valid for “odd”) or -2, 0, 2 (also even). Wait, this reveals that the original wording “odd” is impossible. This is a useful lesson: not every equation has a solution that fits the description.
因此得到的数可能是 6、8、10（偶数，不符合“奇数”）或 -2、0、2（也是偶数）。这提醒我们：并非每个方程都有符合描述的解。
A better example: “Find two consecutive positive integers whose product is 72.” Let n(n + 1) = 72, so n² + n – 72 = 0, giving n = 8 or n = -9. The positive integers are 8 and 9.
更好的例子：“求两个连续正整数，使它们的乘积为 72。”设 n(n + 1) = 72，则 n² + n – 72 = 0，得 n = 8 或 n = -9。正整数为 8 和 9。
8. Pythagorean Theorem Applications | 勾股定理应用
Right-triangle problems frequently lead to quadratics. If the hypotenuse of a right triangle is 13 cm and the sum of the legs is 17 cm, what are the legs?
直角三角形问题经常导出二次方程。若直角三角形的斜边为 13 厘米，两直角边之和为 17 厘米，求两直角边的长度。
Let one leg be x, so the other leg is 17 – x. By Pythagoras: x² + (17 – x)² = 169.
设一条直角边为 x，则另一条为 17 – x。由勾股定理：x² + (17 – x)² = 169。
Expanding gives 2x² – 34x + 289 = 169, so x² – 17x + 60 = 0. Thus (x – 5)(x – 12) = 0, so the legs are 5 cm and 12 cm.
展开得 2x² – 34x + 289 = 169，即 x² – 17x + 60 = 0。因此 (x – 5)(x – 12) = 0，两直角边为 5 厘米和 12 厘米。
Always verify that the lengths satisfy the Pythagorean theorem. Here 5² + 12² = 25 + 144 = 169 = 13².
务必验证长度满足勾股定理。这里 5² + 12² = 25 + 144 = 169 = 13²。
9. Optimisation Using Quadratic Functions | 利用二次函数求最优化
Although optimisation problems usually involve finding a maximum or minimum, they are often solved by locating the vertex of a quadratic function rather than solving an equation. The vertex occurs at x = -b/(2a).
虽然最优化问题通常要求最大值或最小值，但它往往通过求二次函数的顶点来解决，而不是解方程。顶点横坐标为 x = -b/(2a)。
For example, a rectangular field is to be fenced on three sides using 80 m of fencing. Let the side perpendicular to the wall be x. Then the area is A(x) = x(80 – 2x) = -2x² + 80x.
例如，一个矩形场地要用 80 米长的围栏围三面。设垂直于墙的边为 x，则面积为 A(x) = x(80 – 2x) = -2x² + 80x。
The maximum area occurs at x = -80/(2 × -2) = 20 m. The corresponding area is A(20) = 20(80 – 40) = 800 m².
最大面积出现在 x = -80/(2 × -2) = 20 米处，此时面积为 A(20) = 20(80 – 40) = 800 平方米。
In IB exams, when a maximum or minimum is required, check whether the vertex lies within the domain given in the problem. If not, the optimum may occur at an endpoint.
在 IB 考试中，当需要最大值或最小值时，请检查顶点是否在题目给定的定义域内。如果不在，最值可能出现在端点处。
10. Common Pitfalls and Exam Tips | 常见错误与考试技巧
One common error is discarding a negative root without checking the context. Sometimes a negative root has no meaning, but in certain problems, such as finding a time before launch, it may be interpreted differently. Always explain why you reject a root.
常见错误之一是不加检查就舍弃负根。有些情况下负根没有实际意义，但在某些问题中，如发射前的时间，可能有其他解释。请务必说明你舍弃某个根的理由。
Another mistake is failing to set the equation to zero before factorising or using the quadratic formula. The formula requires the form ax² + bx + c = 0.
另一个错误是在因式分解或使用二次公式前，没有将方程化为零的形式。二次公式要求方程形如 ax² + bx + c = 0。
Always check units and whether the answer is reasonable. For example, a time of -3 seconds usually means “3 seconds before the experiment began,” which is often not a valid answer unless stated.
始终检查单位以及答案是否合理。例如，时间为 -3 秒通常意味着“实验开始前 3 秒”，除非题目说明，否则往往不是有效答案。
Finally, when a problem asks for a maximum or minimum, use the symmetric property of the parabola or the vertex formula. Do not confuse the vertex with the roots.
最后，当题目要求最大值或最小值时，使用抛物线的对称性或顶点公式。不要将顶点与根混淆。
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