Problem Solving with Quadratics | 二次方程解题策略

📚 Problem Solving with Quadratics | 二次方程解题策略

Quadratic equations appear everywhere — from projectile motion in physics to profit optimisation in business. In IB Mathematics, the ability to translate a real-world scenario into a quadratic model, solve it efficiently, and interpret the results in context is an essential skill. This article walks you through a structured approach to problem solving with quadratics, covering the four core solution methods, the discriminant, graphical interpretation, and applications such as area, motion, and economics.

二次方程无处不在——从物理中的抛射体运动到商业中的利润最大化。在IB数学中，将现实情境转化为二次模型、高效求解并合理解读结果是一项关键能力。本文将系统梳理二次方程解题策略，涵盖四个核心解法、判别式、图形解读以及面积、运动和经济等应用问题。

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1. Translating Words into Quadratic Models | 将文字转化为二次模型

Many problems begin with a description of a relationship. Identify the unknown quantity (usually denoted x), then express other quantities in terms of x. Look for keywords such as “product”, “sum”, “area”, “maximum height”, or “break-even” to set up a quadratic equation of the form ax² + bx + c = 0.

许多问题从描述某一关系开始。确定未知量（通常记为x），然后用x表示其他量。留意“乘积”、“和”、“面积”、“最大高度”或“盈亏平衡”等关键词，以建立形如ax² + bx + c = 0的二次方程。

• Example: The product of two consecutive positive integers is 156. Let the smaller be x; then the larger is x+1. The equation is x(x+1) = 156 → x² + x − 156 = 0.
• 示例：两个连续正整数的乘积为156。设较小者为x，则较大者为x+1。方程即x(x+1)=156 → x² + x − 156 = 0。

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2. Choosing the Right Solving Strategy | 选择合适的解法

You have four primary tools: factorisation, completing the square, the quadratic formula, and graphing. The choice depends on the structure of the quadratic and the context. Factorisation is fastest when coefficients are small integers, while the quadratic formula works for any equation. Completing the square is crucial for finding vertex coordinates and for integration techniques, and graphical methods provide visual insight into solutions.

你有四个主要工具：因式分解、配方法、求根公式和图像法。选择取决于方程的结构和问题背景。当系数为较小的整数时，因式分解最快；求根公式则适用于任何方程。配方法对于求顶点坐标和积分技巧至关重要；图像方法则能直观呈现解的情况。

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3. Solving by Factorisation | 因式分解法求解

If the quadratic expression can be written as a product of two linear factors, set each factor to zero. For ax² + bx + c = 0, find two numbers that multiply to ac and add to b. Always verify your factorisation by expanding.

若二次式可写成两个一次因式的乘积，则令每个因式为零。对于ax² + bx + c = 0，找出两个数，其乘积为ac，和为b。始终通过展开来验证分解结果。

Example: Solve x² − 5x + 6 = 0. The numbers are −2 and −3 because (−2)×(−3)=6 and (−2)+(−3)=−5. So (x−2)(x−3)=0 → x=2 or x=3.

示例：解x² − 5x + 6 = 0。所需两数为−2和−3，因为(−2)×(−3)=6且(−2)+(−3)=−5。于是(x−2)(x−3)=0 → x=2 或 x=3。

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4. Applying the Quadratic Formula | 使用求根公式

When factorisation is not straightforward, use the quadratic formula: x = [−b ± √(b² − 4ac)] / (2a). The expression under the square root, Δ = b² − 4ac, is the discriminant. Substitute carefully, especially with negative coefficients.

当不易分解时，使用求根公式：x = [−b ± √(b² − 4ac)] / (2a)。根号下的表达式Δ = b² − 4ac即为判别式。代入时需仔细，尤其当系数为负时。

x = (−b ± √Δ) / (2a) where Δ = b² − 4ac

x = (−b ± √Δ) / (2a)，其中 Δ = b² − 4ac

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5. Using the Discriminant to Analyse Solutions | 用判别式分析解的情况

The discriminant tells you how many real solutions exist without solving the equation. If Δ > 0, there are two distinct real roots. If Δ = 0, there is one repeated real root. If Δ < 0, there are no real roots (two complex roots). In application problems, a negative discriminant often means the scenario is impossible (e.g., a projectile never reaches a given height).

判别式无需解方程即可判断实根个数。若Δ > 0，有两个不等实根；若Δ = 0，有一个重根；若Δ < 0，无实根（两个复根）。在应用题中，负判别式通常意味着该情境不可能发生（如抛射体永远达不到某高度）。

Δ > 0	Two distinct real roots / 两个不等实根
Δ = 0	One repeated real root / 一个重根
Δ < 0	No real roots / 无实根

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6. Completing the Square for Vertex Form | 配方法化为顶点式

Completing the square rewrites a quadratic in the form a(x − h)² + k, where (h, k) is the vertex. This is essential for optimisation problems (maximum/minimum values) and for solving equations where factorisation is messy. To complete the square for x² + bx, add and subtract (b/2)².

配方法将二次式写为a(x − h)² + k的形式，其中(h, k)为顶点。这对于最优化问题（最大值/最小值）以及难以因式分解的方程求解至关重要。对x² + bx配方，则加上并减去(b/2)²。

Example: x² + 6x + 5 = (x² + 6x + 9) − 9 + 5 = (x + 3)² − 4.

示例：x² + 6x + 5 = (x² + 6x + 9) − 9 + 5 = (x + 3)² − 4.

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7. Interpreting Graphs of Quadratics | 解读二次函数图像

The graph of y = ax² + bx + c is a parabola. The sign of a determines whether it opens upwards (a > 0) or downwards (a < 0). The x-intercepts correspond to the real roots of ax² + bx + c = 0; the y-intercept is c. The vertex gives the maximum or minimum value of the function.

y = ax² + bx + c的图像是一条抛物线。a的符号决定开口向上（a > 0）还是向下（a < 0）。x轴截距对应方程ax² + bx + c = 0的实根；y轴截距为c。顶点给出函数的最大值或最小值。

When solving problems, sketching a quick graph can help check whether solutions are reasonable. For example, a projectile’s height equation h(t) = −4.9t² + vt + h₀ opens downward, so the maximum height occurs at the vertex.

在解题时，快速画出草图有助于检查解是否合理。例如，抛射体高度方程h(t) = −4.9t² + vt + h₀开口向下，因此最大高度出现在顶点处。

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8. Area and Geometry Problems | 面积与几何问题

Many quadratic problems involve rectangles, borders, or composite shapes. Define the unknown dimension, express area in terms of that variable, and set it equal to the given area. Often one dimension is expressed as a linear function of the other.

许多二次方程问题涉及矩形、边框或组合形状。定义未知尺寸，用该变量表示面积，并令其等于所给面积。通常一个维度用另一个维度的线性函数来表示。

Example: A rectangular garden is 3 m longer than it is wide. Its area is 40 m². Let width = x, length = x+3 → x(x+3) = 40 → x² + 3x − 40 = 0. Solve to get x = 5 (rejecting the negative root).

示例：一个矩形花园的长比宽多3米，面积为40平方米。设宽为x，长为x+3 → x(x+3)=40 → x²+3x−40=0。解得x=5（舍去负根）。

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9. Projectile Motion and Quadratic Models | 抛射体运动与二次模型

Vertical motion under gravity is modelled by a quadratic function of time: h(t) = −½gt² + v₀t + h₀, where g is the acceleration due to gravity (9.8 m/s² or 32 ft/s² depending on units). Questions often ask for the time when the object hits the ground (h=0), maximum height, or the time to reach a specific height.

重力作用下的竖直运动可用时间的二次函数建模：h(t) = −½gt² + v₀t + h₀，其中g为重力加速度（取决于单位，9.8 m/s²或32 ft/s²）。问题常要求计算物体落地时间（h=0）、最大高度或达到某一高度的时间。

Set h(t) equal to the required value and solve the resulting quadratic. Remember to reject negative times or times beyond the flight’s duration.

令h(t)等于所需数值，求解所得二次方程。记得舍去负时间或超出飞行时段的时间。

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10. Profit, Revenue, and Optimisation | 利润、收入与最优化

Business problems frequently lead to quadratics when revenue or profit depends on a price-demand relationship. For example, if lowering the price increases the number of units sold, revenue R(x) = (price per unit) × (quantity) may become a quadratic. The maximum revenue or profit occurs at the vertex.

当收入或利润依赖价格–需求关系时，商业问题常归结为二次方程。例如，若降价可增加销售量，收入R(x) =（单价）×（数量）可能成为二次函数。最大收入或利润出现在顶点处。

Example: A company finds that when it sells x units, profit P(x) = −5x² + 400x − 3000. The maximum profit is at x = −b/(2a) = −400/(2×(−5)) = 40, giving P(40) = 5000.

示例：某公司销售x件产品时，利润P(x)=−5x²+400x−3000。最大利润出现在x=−b/(2a)=−400/(2×(−5))=40处，得P(40)=5000。

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11. Checking and Interpreting Solutions | 检验与解读解的意义

Always substitute your solutions back into the original context, not just the equation. A mathematically correct root may be extraneous in the real-world situation — for instance, a negative length, a time before launch, or a fractional number of items when only whole items make sense. State your final answer with correct units and rounding as required.

始终将解代回原始情境中，而不仅仅是方程中。数学上正确的根在现实中可能无意义——例如，负长度、发射前的时间，或非整数的物品数量（当只有整数有意义时）。最终答案需写明单位并按要求舍入。

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12. Common Pitfalls and How to Avoid Them | 常见误区及对策

Students often forget to set the equation to zero before factorising, misapply the quadratic formula (especially with minus signs), or ignore the meaning of the discriminant. Another common error is using the formula for the axis of symmetry x = −b/(2a) with the wrong sign. To build confidence, practise problems from varied contexts and always verify your factorisation by expanding.

学生常犯的错误包括：因式分解前忘记将方程设为零，求根公式代入时搞错符号（尤其是负号），或忽略判别式的含义。另一个常见错误是使用对称轴公式x=−b/(2a)时弄错符号。为增强信心，需练习不同情境的题目，并始终通过展开来验证因式分解的正确性。

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