📚 Mixed Exercise 2: Quadratics | 二次函数综合练习
Mixed Exercise 2 serves as a comprehensive recap of everything covered in the Quadratics chapter of the Edexcel A-Level Mathematics syllabus. In this article, we revisit key techniques such as completing the square, the quadratic formula, discriminant analysis, sketching graphs, and solving inequalities. By working through a wide range of example problem types, students can consolidate their understanding and build the fluency required for AS and A-Level examinations.
《混合练习2》是对 Edexcel A-Level 数学课程中“二次函数”章节所有内容的全面回顾。本文将重温配方法、求根公式、判别式分析、图像绘制以及不等式求解等核心技巧。通过梳理各类典型例题，学生可以巩固理解，并培养应对 AS 和 A-Level 考试所需的熟练度。
1. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form a(x + p)² + q, revealing the turning point and making it easier to solve equations. For example, x² + 6x + 5 becomes (x + 3)² − 4. The general steps are: halve the coefficient of x, square it, and adjust the constant term accordingly. This method is essential when the quadratic does not factorise neatly.
配方法将二次式重写为 a(x + p)² + q 的形式，从而显示出转折点并使方程更容易求解。例如，x² + 6x + 5 可化为 (x + 3)² − 4。一般步骤为：将 x 的系数除以2，平方，然后相应地调整常数项。当二次式无法整洁地因式分解时，这一方法至关重要。
f(x) = ax² + bx + c → f(x) = a(x + b/(2a))² − (b² − 4ac)/(4a)
Remember that when a ≠ 1, factor out the coefficient of x² first before completing the square. Mastery of completing the square also helps in deriving the quadratic formula and in integration later in the course.
请记住，当 a ≠ 1 时，要先提取 x² 的系数，然后再配方。熟练掌握配方法也有助于推导求根公式以及后续课程中的积分计算。
2. Solving Quadratic Equations | 解二次方程
A quadratic equation can be solved by factorising, completing the square, or using the quadratic formula. Factorising relies on finding two numbers that multiply to ac and sum to b. The equation is set to zero, and the zero-product property gives the solutions directly. Practice with a mix of positive and negative coefficients builds confidence.
二次方程可以通过因式分解、配方法或使用求根公式求解。因式分解依赖于找到两个数，其乘积为 ac、和为 b。将方程设为零，利用零乘积性质直接得出解。通过练习正负系数的混合题目可以增强信心。
- Step 1: Write in standard form ax² + bx + c = 0.
- Step 1: 写成标准形式 ax² + bx + c = 0。
- Step 2: Choose the most efficient method.
- Step 2: 选择最高效的方法。
- Step 3: Check solutions in the original equation.
- Step 3: 将解代入原方程检验。
Always consider whether the equation can be simplified first, for example by dividing through by a common factor.
始终先考虑方程是否可以化简，例如除以公因数。
3. Quadratic Formula | 求根公式
The quadratic formula, x = [−b ± √b² − 4ac] / (2a), is a universal tool that works for any quadratic equation, even when factorisation is impossible. It is derived directly from completing the square. Students must memorise the formula and know how to substitute values carefully, especially with negative coefficients.
求根公式 x = [−b ± √b² − 4ac] / (2a) 是一个万能工具，适用于任何二次方程，即使无法进行因式分解时亦可使用。它直接由配方法推导而来。学生必须熟记公式，并懂得如何仔细代入数值，尤其是在处理负系数的时候。
If ax² + bx + c = 0, then x = (-b ± √Δ) / (2a), where Δ = b² − 4ac
The term under the square root, b² − 4ac, is called the discriminant. Pay attention to the order of operations: calculate b² and 4ac separately before subtracting. Also, be mindful that the formula gives both roots in one line; you may need to evaluate the '+' and '−' versions separately for exact or decimal answers.
平方根号下的式子 b² − 4ac 被称为判别式。注意运算顺序：先分别计算 b² 和 4ac，然后再相减。此外，注意该公式一行即给出两个根；你可能需要分别计算“+”和“−”的情形，以得到精确解或小数解。
4. Discriminant Analysis | 判别式分析
e is exactly one real repeated root (the graph touches the x-axis). If Δ < 0, there are no real roots, and the graph does not intersect the x-axis. This analysis is frequently tested in problems involving tangency and intersection conditions.
判别式 Δ = b² − 4ac 决定了二次方程根的性质和数量。若 Δ > 0，则有两个不同的实根。若 Δ = 0，则恰有一个重实根（图像与 x 轴相切）。若 Δ < 0，则无实根，图像不与 x 轴相交。这种分析经常出现在涉及相切和相交条件的题目中。
| Condition | Number of real roots | Graph behaviour |
| Δ > 0 | 2 | Crosses x-axis twice |
| Δ = 0 | 1 (repeated) | Touches x-axis |
| Δ < 0 | 0 | Entirely above/below x-axis |
For problems like 'find the set of values of k for which the equation has two distinct real roots', set up Δ > 0 and solve the resulting inequality. Remember to check the original context: sometimes k must be real or positive.
对于“求使方程有两个相异实根的 k 的取值范围”这类问题，应建立 Δ > 0 并求解得到的不等式。记得检查原题背景：有时 k 必须为实数或正数。
5. Factorising Quadratics | 因式分解
uch as splitting the middle term or using trial and error become necessary. For instance, 2x² + 7x + 3 can be factorised to (2x + 1)(x + 3) by finding numbers that multiply to ac = 6 and sum to 7.
对于 a = 1 的二次式，因式分解需要找到两个数，其乘积为 c、和为 b。例如，x² − 5x + 6 = (x − 2)(x − 3)。当 a > 1 时，就需要使用拆分中项或试错等方法。例如，2x² + 7x + 3 可通过找到两数乘积为 ac = 6、和为 7 来分解为 (2x + 1)(x + 3)。
Always check your factorisation by expanding the brackets. If a common factor exists in all terms, factor it out first to simplify the problem. Not all quadratics factorise nicely, but practising this skill sharpens algebraic manipulation for harder topics.
始终通过展开括号来检验因式分解。若所有项都有公因数，可先将其提取出来以简化问题。并非所有二次式都能整洁地分解，但练习这一技能可以锻炼代数操作能力，为后续更难的课题打下基础。
6. Sketching Quadratic Graphs | 绘制二次函数图像
Sketching a quadratic graph involves identifying the shape (∪ or ∩ ards; if a < 0, it opens downward. The vertex can be found by completing the square or using x = −b/(2a).
绘制二次图像需要确定形状（∪ 或 ∩）、y 轴截距、x 轴截距（根）以及顶点（转折点）。y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上；若 a < 0，则开口向下。顶点可通过配方法或使用 x = −b/(2a) 求得。
Label all key points including the coordinates of the vertex and roots. Make sure the sketch is consistent with the discriminant: if Δ < 0, the graph does not cross the x-axis. Use a smooth curve, never straight lines, and show symmetry about the vertical line through the vertex.
标注所有关键点，包括顶点坐标和根。确保绘制的图像与判别式一致：若 Δ < 0，则图像不与 x 轴相交。使用平滑曲线，切忌使用直线，并表现出关于过顶点的铅垂线的对称性。
7. Quadratic Inequalities | 二次不等式
3. Then sketch the graph to determine where the function is below zero. Because the parabola opens upwards, the inequality holds for 1 < x < 3. Express the solution using set notation or interval notation as required.
要解如 x² − 4x + 3 < 0 这样的二次不等式，首先通过求解 x² − 4x + 3 = 0 得到根 x = 1 和 x = 3。然后绘制图像以确定函数值小于零的区间。因为抛物线开口向上，所以在 1 < x < 3 时不等式成立。按要求使用集合表示法或区间表示法写出解。
For inequalities with a negative leading coefficient, multiply through by −1 and reverse the inequality sign. Always check critical values and test points in intervals to confirm. Quadratic inequalities often appear in conjunction with domain restrictions or optimisation problems.
对于首项系数为负的不等式，可通过乘以 −1 并反转不等号来处理。务必检查临界值，并在各区间内取测试点进行验证。二次不等式常与定义域限制或优化问题一起出现。
8. Modelling with Quadratics | 二次函数建模
xpress the quantity to be maximised or minimised as a quadratic function of one variable, then find the vertex. For example, a rectangle with fixed perimeter P has area A = x(P/2 − x), which is a quadratic opening downward; the maximum area occurs at the vertex.
现实问题经常涉及二次模型，如抛射体运动、面积优化和利润最大化等。关键是将需要最大化或最小化的量表示为一个变量的二次函数，然后求顶点。例如，周长为定值 P 的矩形，其面积 A = x(P/2 − x) 是一个开口向下的二次函数；最大面积出现在顶点处。
Remember to define your variables clearly and consider the practical constraints (e.g., x must be positive). Once a quadratic model is set up, use completing the square or the formula for the vertex to find the optimum value. Interpret the mathematical solution in the context of the original problem, and include appropriate units.
记住要清晰定义变量，并考虑实际约束（例如 x 必须为正）。一旦建立了二次模型，就使用配方法或顶点公式求出最优值。将数学解放在原题背景中解读，并附上恰当的单位。
9. Hidden Quadratics | 隐藏的二次方程
ns like x⁴ − 5x² + 4 = 0 (let u = x² → u² − 5u + 4 = 0) or exponential equations such as e2x − 5ex + 6 = 0 (let t = ex). After solving the quadratic, back-substitute to find the original variable and check for extraneous solutions.
有些方程乍看之下并非二次，但可通过代换化作二次方程。例如 x⁴ − 5x² + 4 = 0（令 u = x² → u² − 5u + 4 = 0），或指数方程如 e2x − 5ex + 6 = 0（令 t = ex）。求解二次方程后，再回代求出原变量，并注意检查增根。
equations. Always state the range of the substituted variable to avoid invalid solutions.
这种技巧在三角二次方程（例如通过令 s = sinθ 来求解 2sin²θ − sinθ − 1 = 0）和对数方程中尤为有用。始终注明代换变量的取值范围，以避免无效解。
10. Simultaneous Equations (One Linear, One Quadratic) | 联立方程（一次与二次）
Solving a system where one equation is linear and the other is quadratic involves substitution. From the linear equation, express one variable in terms of the other, then substitute into the quadratic. This yields a quadratic in one variable, which can be solved by factorising or using the formula. Each solution gives a corresponding value for the other variable, leading to pairs of coordinates of intersection points.
求解一个方程是一次、另一个是二次的方程组时，使用代入法。从一次方程中用一个变量表示另一个变量，然后代入二次方程。这会得到一个单变量的二次方程，可通过因式分解或求根公式求解。每个解会给出另一个变量的对应值，从而得到交点的坐标对。
Be systematic: after obtaining the quadratic in one variable, solve it fully before back-substituting. Sometimes the linear equation might need rearrangement. The discriminant of the resulting quadratic can indicate whether the line and the parabola intersect, touch, or miss each other.
要有条理：在得到一个变量的二次方程后，先将其完全解出，再进行回代。有时一次方程可能需要重新整理。所得二次方程的判别式可以表明直线与抛物线是相交、相切还是相离。
11. Vertex and Axis of Symmetry | 顶点与对称轴
rm y = ax² + bx + c, the x-coordinate of the vertex is −b/(2a), and the y-coordinate is found by substituting this x-value. The vertex is the minimum point when a > 0 and the maximum point when a < 0.
对于形如 y = a(x − h)² + k 的二次函数，顶点为 (h, k)，对称轴为铅垂线 x = h。若二次函数以标准形式 y = ax² + bx + c 给出，顶点 x 坐标为 −b/(2a)，y 坐标则通过代入该 x 值求得。当 a > 0 时，顶点为极小点；当 a < 0 时，顶点为极大点。
th vertex at (−2, −3), the range is y ≥ −3. This is heavily used in optimisation contexts.
快速找到顶点有助于绘制图像并确定函数的值域。例如，若二次函数开口向上、顶点在 (−2, −3)，则其值域为 y ≥ −3。这在优化问题中应用广泛。
12. Summary Tips | 总结技巧
When tackling Mixed Exercise 2, develop a checklist: identify the method required, check the discriminant for root nature, and always verify solutions by substitution. Keep an eye out for hidden quadratics and opportunities to factor out common terms. Practise switching between standard form, factorised form, and completed square form, as each gives different information quickly.
在处理《混合练习2》时，建立一份检查清单：确定所需方法，检查判别式以了解根的性质，并始终通过代入来验证解。留意隐藏的二次方程以及提取公因项的机会。练习在标准式、因式分解式和顶点式之间切换，因为每种形式都能快速提供不同的信息。
Drawing a sketch, even if not asked, often clarifies the problem and reduces careless errors. Make a habit of labelling all key points on graphs. Mixed exercises are best approached by first scanning all questions and identifying which techniques are being tested. This builds strategic thinking for exam success.
即使题目未要求，画个草图也常能理清题意并减少粗心错误。养成标注图像上所有关键点的习惯。做综合练习的最佳方法是先浏览所有题目，识别出所考查的技巧。这将为考试成功培养策略性思维。
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