📚 Problem Solving with Quadratics | 二次方程问题求解
Quadratic functions and equations form the backbone of algebraic problem solving in IB Mathematics. From optimising areas and maximising profits to analysing projectile motion, mastery of quadratic models allows you to connect abstract algebra with real-world situations. This article explores essential techniques—completing the square, the discriminant, Vieta’s formulas, and more—highlighting how each supports the analytical approach demanded by IB exam questions.
二次函数与方程是 IB 数学中代数问题求解的核心。从优化面积、最大化利润到分析抛体运动，掌握二次模型能让你将抽象代数与现实情境联系起来。本文探索核心技巧——配方法、判别式、韦达定理等——并强调每种技巧如何支撑 IB 考试所要求的分析性思维。
1. Quadratic Basics: Forms and Key Features | 二次基础：形式与关键特征
A quadratic can be expressed in standard form y = ax² + bx + c, vertex form y = a(x – h)² + k, or factored form y = a(x – p)(x – q). Each form instantly reveals different information: the standard form shows the y-intercept, the vertex form gives the turning point (h,k), and the factored form exposes the x-intercepts (roots) p and q.
二次函数可表示为一般式 y = ax² + bx + c、顶点式 y = a(x – h)² + k 或因式分解式 y = a(x – p)(x – q)。每种形式能立即揭示不同信息：一般式显示 y 轴截距，顶点式给出转折点 (h,k)，因式分解式则暴露 x 轴截距（根）p 和 q。
In problem solving, choosing the right form is often the first strategic decision. If a question asks for a maximum value, vertex form is ideal; if it describes roots, factored form may lead to a quicker solution.
在问题求解中，选择合适的表达式形式往往是第一个策略决策。如果题目要求最大值，顶点式最理想；如果描述了根，因式分解式可能更快得出答案。
- Standard: y = 2x² – 4x + 1 → y-intercept (0,1)
- Vertex: y = 2(x – 1)² – 1 → minimum at (1, –1)
- Factored: y = (x – 2)(x + 3) → roots 2 and –3
2. Modelling with Quadratic Functions | 二次函数建模
Many IB problems present a real-life scenario—such as the path of a ball, the area of a rectangular enclosure, or the revenue from selling items—and require you to construct a quadratic model. You typically identify two independent variables, express one in terms of the other, and then write a quadratic expression for the quantity to be optimised or solved.
许多 IB 问题会给出一个现实场景——比如球的轨迹、矩形围栏面积或销售商品收入——并要求你构建一个二次模型。通常需要确定两个自变量，将一个用另一个表示，然后为待优化或求解的量写出二次表达式。
with 40 m of fencing: if the side perpendicular to the wall is x, the parallel side is (40 – 2x). The area A(x) = x(40 – 2x) = –2x² + 40x, a quadratic model.
例如，一个农民靠墙用 40 米围栏建矩形围场：若垂直于墙的边长为 x，则平行边长为 (40 – 2x)。面积 A(x) = x(40 – 2x) = –2x² + 40x，这是一个二次模型。
Once the model is established, you can apply algebraic techniques to find maximum area, dimensions for a given area, or conditions that make no solution possible.
一旦建立模型，你就可以运用代数技巧求解最大面积、给定面积下的尺寸，或判断无解的条件。
3. Completing the Square and the Vertex | 配方法与顶点
Completing the square transforms ax² + bx + c into a(x – h)² + k, immediately revealing the vertex (h,k). This is essential for optimisation problems where the maximum or minimum value of a quadratic function is required.
配方法将 ax² + bx + c 转化为 a(x – h)² + k，立即揭示顶点 (h,k)。这在需要求二次函数最大值或最小值的优化问题中至关重要。
, add and subtract it inside the bracket, then rewrite as a perfect square. For f(x) = 2x² – 8x + 5, you get f(x) = 2(x – 2)² – 3. The vertex is (2, –3) and since a > 0, this is a minimum.
过程：从前两项中提取 a，取 b/a 的一半并平方，在括号内加减该值，然后写成完全平方形式。对于 f(x) = 2x² – 8x + 5，可得 f(x) = 2(x – 2)² – 3。顶点为 (2, –3)，由于 a > 0，这是最小值。
of symmetry (x = h) and to state the range of the function directly from the vertex form.
在 IB 考试中，你可能还需要写出对称轴 (x = h) 并根据顶点式直接写出函数的值域。
x = –b/(2a) gives the x-coordinate of the vertex
4. The Discriminant and Nature of Roots | 判别式与根的性质
The discriminant Δ = b² – 4ac blems that ask whether a line intersects a curve, or for finding conditions on parameters.
判别式 Δ = b² – 4ac 决定二次方程 ax² + bx + c = 0 的实根数量和类型。对于判断直线是否与曲线相交，或求参数条件，它是一个强有力的工具。
| Δ > 0 | Δ = 0 | Δ < 0 |
|---|---|---|
| Two distinct real roots | One repeated real root (tangent) | No real roots |
| 两个不同的实根 | 一个重根（相切） | 没有实根 |
t to the parabola y = x² + 3x + 1. Set equations equal: x² + 3x + 1 = 2x + k → x² + x + (1 – k) = 0. For tangency, Δ = 0 → 1 – 4(1)(1 – k) = 0 → 1 – 4 + 4k = 0 → k = 3/4.
经典的 IB 问题：求 k 值使得直线 y = 2x + k 与抛物线 y = x² + 3x + 1 相切。令方程相等：x² + 3x + 1 = 2x + k → x² + x + (1 – k) = 0。相切要求 Δ = 0 → 1 – 4(1)(1 – k) = 0 → 1 – 4 + 4k = 0 → k = 3/4。
Always set the equation to zero before applying Δ. This technique extends to intersections of any two curves.
在应用 Δ 之前一定要将方程化为等于零的形式。该技巧可推广到任意两条曲线的交点问题。
5. Vieta's Formulas: Sum and Product of Roots | 韦达定理：根的和与积
a quadratic equation ax² + bx + c = 0 with roots α and β, Vieta’s formulas state: α + β = – bypass the need to find the roots explicitly, offering elegant shortcuts in many problems.
对于有根 α 和 β 的二次方程 ax² + bx + c = 0，韦达定理给出：α + β = –b/a 且 αβ = c/a。这些关系绕过了显式求根的需要，为许多问题提供了优雅的捷径。
Use Vieta when the question involves symmetric expressions like α² + β², (α – β)², or 1/α + 1/β. For instance, α² + β² = (α + β)² – 2αβ, easily computed without solving the equation.
当问题涉及对称表达式如 α² + β²、(α – β)² 或 1/α + 1/β 时，可运用韦达定理。例如 α² + β² = (α + β)² – 2αβ，无需解方程即可轻松计算。
Example: If roots of 2x² – ind α²β + αβ². Factor as αβ(α + β). Here α+β = 3, αβ = 3/2, so the value is (3/2)·3 = 9/2.
示例：若 2x² – 6x + 3 = 0 的根为 α 和 β，求 α²β + αβ²。因式分解为 αβ(α + β)。这里 α+β = 3, αβ = 3/2，所以值为 (3/2)·3 = 9/2。
6. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities such as x² – 5x + 6 < 0 are solved by first finding the roots, then testing intervals on the number line. A sign chart or a rough sketch of the parabola helps visualise where the function is positive or negative.
解二次不等式如 x² – 5x + 6 < 0，首先求出根，然后在数轴上检验区间。符号表格或抛物线的简图有助于直观看到函数在何处为正或为负。
Factorise: (x – 2)(x – he parabola opens upward, so the expression is negative between the roots. The solution is 2 < x < 3. If the inequality was > 0, the solution would be x < 2 or x > 3.
因式分解：(x – 2)(x – 3) < 0。根为 2 和 3。由于 a = 1 > 0，抛物线开口向上，因此表达式在两根之间为负。解为 2 < x < 3。如果不等式是 > 0，解则为 x < 2 或 x > 3。
Critical tip: always rearrange the inequality to have zero on one side. Do not multiply or divide by x unless you are certain of its sign, as this can reverse the inequality.
关键提示：始终将不等式整理成一边为零的形式。除非确定 x 的符号，否则不要乘以或除以 x，因为这可能改变不等号方向。
7. Optimization Problems: Maximizing Area or Profit | 优化问题：面积或利润最大化
Optimisation is a direct application of vertex form. Given a quadratic model representing area, profit, or any physical quantity, the maximum or minimum occurs at the vertex. If the coefficient of x² is negative, the vertex gives a maximum; if positive, a minimum.
优化是顶点式的直接应用。给定一个代表面积、利润或任意物理量的二次模型，最大值或最小值出现在顶点处。若 x² 的系数为负，顶点给出最大值；若为正，则为最小值。
Consider a profit function P(x) = –5x² + 200x – 1500, where x is the number of items sold. The vertex x-coordinate is –b/(2a) = –200/(–10) = 20. The maximum profit is P(20) = –5(400) + 4000 – 1500 = 500. So selling 20 items maximises profit at $500.
考虑利润函数 P(x) = –5x² + 200x – 1500，其中 x 为销售数量。顶点 x 坐标为 –b/(2a) = –200/(–10) = 20。最大利润为 P(20) = –5(400) + 4000 – 1500 = 500。因此销售 20 件商品可获得最大利润 $500。
Remember to interpret the results in context, and check whether x must be an integer or whether endpoints of a domain need testing.
记住要在题目情境中解释结果，并检查 x 是否必须为整数，或者是否需要检验定义域的端点。
8. Kinematics and Projectile Motion | 运动学与抛体运动
In physics-related problems, the quadratic often describes the height of a projectile over time: h(t) = –½gt² + v₀t + h₀ /s², but IB questions may use simplified values like 10 m/s² or ignore units for algebraic practice.
在与物理相关的问题中，二次函数通常描述抛射体高度随时间的变化：h(t) = –½gt² + v₀t + h₀，其中 g 为重力加速度，v₀ 为初速度的竖直分量，h₀ 为初始高度。在地球上 g ≈ 9.8 m/s²，但 IB 题目可能使用简化值如 10 m/s²，或忽略单位进行代数练习。
eight is attained. Setting h(t) equal to a target value gives a quadratic to solve, and you must select the value that makes physical sense.
常见问题包括求最大高度（在顶点处）、到达地面的时间（h = 0），或达到某一高度的时间。将 h(t) 设为目标值会产生一个需解的二次方程，你必须选取符合物理意义的那个值。
Example: h(t) = –5t² + 10t + 20. Maximum height at t = –10/(–10) = 1 s, h(1) = 25 m. To find when it hits the ground, solve –5t² + 10t + 20 = 0 → t² – 2t – 4 = 0 → t = 1 ± √5, take positive t ≈ 3.24 s.
示例：h(t) = –5t² + 10t + 20。最大值在 t = –10/(–10) = 1 s 时，h(1) = 25 m。要求落地时间，解 –5t² + 10t + 20 = 0 → t² – 2t – 4 = 0 → t = 1 ± √5，取正值 t ≈ 3.24 s。
9. Real-World Contexts and Interpreting Solutions | 实际问题情境与解的解释
IB emphasises authentic applications. You might encounter a quadratic modelling the spread of a rumour, the braking distance of a car, or the concentration of a chemical over time. In such contexts, not all algebraic solutions will be valid—negative times or lengths are discarded.
IB 强调真实应用。你可能会遇到模拟谣言传播、汽车制动距离或化学浓度随时间变化的二次模型。在此类情境下，并非所有代数解都有效——负的时间或长度将被舍去。
g sentence that links the mathematics back to the scenario. For instance, after solving x² – ate what these numbers mean (e.g., the object is at ground level at 3 seconds and 7 seconds).
始终写出将数学与情景联系起来的总结句。例如，解 x² – 10x + 21 = 0 得到 x = 3 或 7，你应该陈述这些数字表示什么（比如，物体在 3 秒和 7 秒时处于地面）。
Be mindful of domain constraints. If x represents the number of people, only non-negative integers are acceptable. If it's a length, consider possible range based on geometry.
注意定义域的限制。如果 x 代表人数，则只能接受非负整数。如果是长度，要考虑几何中可能的范围。
10. Using Technology: GDC Tips for Quadratics | 使用技术：GDC 求解二次方程技巧
For IB papers where a graphic display calculator is allowed, efficient GDC use can save time and reduce errors. You can graph the quadratic, find roots or maxima directly, or use the equation solver. However, you must still show analytical working unless a question specifically allows technology-only solutions.
在允许使用图形计算器的 IB 试卷中，高效利用 GDC 可以节省时间并减少错误。你可以绘制二次函数图像，直接求根或最大值，或使用方程求解器。但是，你仍须展示解析步骤，除非题目明确允许仅用技术求解。
Key GDC functions:
- Graph and use 'zero' or 'root' to find x-intercepts.
- 'Maximum'/'Minimum' to find vertex coordinates.
- Polynomial solver (degree 2) for quick root finding.
- Table of values to check integer solutions.
关键 GDC 功能：
- 绘图并使用“零点”或“根”功能求 x 轴截距。
- “最大值”/“最小值”功能求顶点坐标。
- 多项式求解器（二次）快速求根。
- 数值表检查整数解。
Always check the reasonableness of GDC outputs; a calculator may give a root at x = –2.5, but if x is a length, you must reject it and explain why.
始终检查 GDC 输出的合理性；计算器可能给出 x = –2.5 的根，但若 x 是长度，你必须拒绝它并解释原因。
11. Common Errors and How to Avoid Them | 常见错误及避免方法
Misinterpreting the sign of the quadratic coefficient is a frequent mistake. If a < 0, the parabola opens downward and the vertex is a maximum—but students often still call it a minimum out of habit. Always visualise the graph.
错误理解二次项系数的符号是一个常见错误。如果 a < 0，抛物线开口向下，顶点为最大值——但学生常常出于习惯仍称其为最小值。始终要想象图像。
Another pitfall is forgetting to set the equation to zero before applying the discriminant or factorising. Also, when solving inequalities, multiplying by a negative quantity or an x without considering its sign leads to incorrect intervals.
另一个陷阱是在应用判别式或因式分解前忘记将方程设为零。同样，解不等式时，乘以负数或不考虑符号的 x 会导致错误的区间。
he quadratic to standard form can cause errors in identifying a, b, c. For example, in (2x – 1)² = 3x + 4, expand to 4x² – 4x + 1 = 3x + 4 → 4x² – 7x – 3 = 0 before proceeding.
最后，未将二次式化简为标准形式可能导致识别 a、b、c 出错。例如，在 (2x – 1)² = 3x + 4 中，先展开为 4x² – 4x + 1 = 3x + 4 → 4x² – 7x – 3 = 0 再进行后续操作。
Practice varied problems to build confidence in selecting the most efficient method. With quadratics, there is often more than one route to the solution; choose the one that minimises algebraic manipulation.
通过练习多样题目来增强信心，学会选择最有效的方法。对于二次问题，通常存在多种求解途径；选择代数操作最少的那一条。
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