📚 Quadratics | 二次函数与方程
A quadratic is a fundamental algebraic concept that appears throughout the Edexcel A-Level Mathematics course. Mastering quadratic functions, equations, and their graphs forms the backbone for many advanced topics including calculus, coordinate geometry, and modelling. This article provides a comprehensive revision guide covering all essential aspects of quadratics, from solving equations to interpreting discriminants and applying Vieta's formulas.
二次函数是 Edexcel A-Level 数学课程中贯穿始终的基础代数概念。掌握二次函数、二次方程及其图像，是学习微积分、坐标几何和数学建模等更高级主题的基石。本文提供一份全面的复习指南，涵盖二次函数的各个重要方面，从解方程到判别式的解读，以及韦达定理的应用。
1. What is a Quadratic? | 什么是二次函数？
, b and c are constants and a ≠ 0. If a = 0, the expression becomes linear, not quadratic.
二次表达式是一个次数为 2 的多项式，即变量的最高次幂为 2。其标准形式为 ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。若 a = 0，则表达式退化为一次式而非二次式。
f y = ax² + bx + c is a parabola, which opens upwards when a > 0 and downwards when a < 0.
常见的例子有 3x² - 5x + 2、x² - 9 和 -2x² + 4x。y = ax² + bx + c 的图像是一条抛物线，当 a > 0 时开口向上，当 a < 0 时开口向下。
2. Solving Quadratic Equations by Factorising | 因式分解法解二次方程
ion ax² + bx + c = 0 factorises to (px + q)(rx + s) = 0, giving solutions x = -q/p and x = -s/r.
因式分解是解二次方程最简单的办法，前提是表达式能够写成两个一次因式的乘积。方程 ax² + bx + c = 0 可因式分解为 (px + q)(rx + s) = 0，从而得出解 x = -q/p 和 x = -s/r。
. Always check your factors by expanding them to ensure they match the original quadratic.
例如，x² - 7x + 12 = 0 可化为 (x - 3)(x - 4) = 0，因此 x = 3 或 x = 4。始终通过展开来检验因式是否正确。
fficient of x² is not 1, you can apply methods like splitting the middle term or using the 'ac' uation 2x² - x - 6 = 0 factorises as (2x + 3)(x - 2) = 0, giving roots x = -3/2 and x = 2.
即便 x² 的系数不是 1，也可以运用拆中项法或“ac 法”。方程 2x² - x - 6 = 0 因式分解为 (2x + 3)(x - 2) = 0，解得 x = -3/2 和 x = 2。
3. Completing the Square | 配方法
Completing the square rewrites a quadratic as a(x + p)² + q, which is extremely useful for finding the vertex of a parabola and for solving equations that do not factorise easily. The process involves manipulating the expression to create a perfect square trinomial.
配方法将二次式改写为 a(x + p)² + q 的形式，这对于求抛物线的顶点以及解难以因式分解的方程极为有用。其过程是通过代数操作构造出完全平方三项式。
9 + 5 = (x + 3)² - 4. The minimum value of the expression is -4 and it occurs when x = -3.
以 x² + 6x + 5 为例。取 x 系数的一半（即 3），平方得 9，然后写为 (x + 3)² - 9 + 5 = (x + 3)² - 4。该式的最小值为 -4，且在 x = -3 时取得。
If the coefficient of x² is not 1, factor it out first: for 2x² + 8x + 3, take 2 common from the x terms to get 2(x² + 4x) + 3, then complete the square inside the bracket to obtain 2(x + 2)² - 5.
若 x² 的系数不为 1，则先将其提出：对于 2x² + 8x + 3，把含 x 的项提取公因子 2 得 2(x² + 4x) + 3，再对括号内配平方，得到 2(x + 2)² - 5。
4. The Quadratic Formula and the Discriminant | 二次公式与判别式
ula provides a direct method to find the roots of any quadratic equation ax² + bx + c = 0:
x = (-b ± √b² - 4ac) / (2a)
二次公式为任意二次方程 ax² + bx + c = 0 提供了直接的求根方法：
x = (-b ± √b² - 4ac) / (2a)
lled the discriminant. It determines the nature of the roots without solving the equation.
根号内的表达式 Δ = b² - 4ac 称为判别式。它无需解方程即可判断根的性质。
| Discriminant | Nature of roots |
| Δ > 0 | Two distinct real roots |
| Δ = 0 | One repeated real root (tangent situation) |
| Δ < 0 | No real roots (complex roots) |
| 判别式 | 根的性质 |
| Δ > 0 | 两个相异实根 |
| Δ = 0 | 一个重实根（相切情况） |
| Δ < 0 | 无实根（仅复数根） |
5. Sketching Quadratic Graphs | 二次函数图像的绘制
(roots, if real), and the vertex. The shape depends on the sign of a; remembering that a > 0 gives a ∪-shape and a < 0 gives an ∩-shape.
绘制二次函数图像需要确认关键特征：y 截距（令 x = 0）、x 截距（即实的根）以及顶点。图像的形状取决于 a 的符号：a > 0 呈 ∪ 形，a < 0 呈 ∩ 形。
symmetry is halfway between the roots at x = 2, and substituting gives the vertex (2, -1).
对于 y = x² - 4x + 3，y 截距为 (0,3)。解 x² - 4x + 3 = 0 得 (x - 1)(x - 3) = 0，故 x 截距为 (1,0) 和 (3,0)。对称轴位于两根正中间 x = 2，代入得顶点 (2, -1)。
A well-labelled sketch must show the intercepts, vertex coordinates, and the coordinate axes. You do not need a precise scale but must indicate the correct relative positions.
清晰的草图必须标出截距、顶点坐标和坐标轴。无需精确的比例尺，但必须显示正确的相对位置。
6. Vertex and Line of Symmetry | 顶点与对称轴
completed square form a(x - h)² + k, the vertex is (h, k). The line of symmetry is x = h.
每条二次函数图像都有一个顶点（最大值或最小值点）以及过该点的垂直对称轴。利用配方式 a(x - h)² + k，顶点即为 (h, k)，对称轴为 x = h。
nd by substituting this x-value into the function. This formula is derived directly from the quadratic formula or by differentiation.
若二次函数以标准形式 ax² + bx + c 给出，则对称轴为 x = -b/(2a)，代入该 x 值即可求得顶点。该公式可由二次公式或求导直接推导得出。
) = 1, so the vertex is (1, f(1)) = (1, 5). Since a is negative, this vertex is a maximum.
对于 f(x) = -3x² + 6x + 2，对称轴为 x = -6/(2 × -3) = 1，因此顶点为 (1, f(1)) = (1, 5)。由于 a 为负，该顶点为最大值点。
7. Quadratic Inequalities | 二次不等式
es x = 2 and x = 3. Then sketch the parabola or use a sign diagram to determine the intervals where the inequality holds.
解形如 x² - 5x + 6 > 0 的二次不等式，需先求出临界值，即解二次方程 x² - 5x + 6 = 0，得 x = 2 和 x = 3。然后画出抛物线草图或使用符号表，确定不等式成立的区间。
wards, the expression is positive outside the interval between the roots: x < 2 or x > 3. For a 'less than' inequality x² - 5x + 6 < 0, the solution is the interval 2 < x < 3.
由于 y = x² - 5x + 6 的图像开口向上，表达式在两根之间的外侧为正：x < 2 或 x > 3。对于“小于”型不等式 x² - 5x + 6 < 0，解集为区间 2 < x < 3。
Always be careful with the inequality sign, and remember that multiplying or dividing by a negative number reverses the inequality. Write the solution using set notation or interval notation as required by the exam.
务必注意不等号的方向，并记住乘以或除以负数会颠倒不等号。按考试要求使用集合符号或区间符号写出解集。
8. Sum and Product of Roots (Vieta's Formulas) | 根的和与积（韦达定理）
0, then the sum α + β = -b/a and the product αβ = c/a. These relationships, known as Vieta' m given roots or for solving symmetric root problems without finding the individual roots.
若 α 与 β 是二次方程 ax² + bx + c = 0 的根，则 α + β = -b/a，αβ = c/a。这组关系称为韦达定理，在已知根构造方程或求解对称根问题时极为有用，且无需具体求出根的值。
an equation has roots α and β, expressions like α² + β² can be found using (α + β)² - 2αβ.
例如，若根为 3 和 -2，则方程为 x² - (3 + (-2))x + (3)(-2) = 0，即 x² - x - 6 = 0。反之，若方程有根 α 和 β，则 α² + β² 等表达式可通过 (α + β)² - 2αβ 求解。
This topic frequently appears in Edexcel exam questions requiring the formation of a new quadratic whose roots are related to those of a given equation, e.g. roots 2α + 1 and 2β + 1.
这一知识点常出现在 Edexcel 考题中，要求根据已知方程的根构造新的二次方程，例如根为 2α + 1 和 2β + 1 的情形。
9. Hidden Quadratics – Using Substitution | 隐藏二次方程 – 换元法
be turned into one by a suitable substitution. Equations such as x⁴ - 13x² + 36 = 0 or e²ˣ - 5eˣ + 6 = 0 are examples of hidden quadratics.
有时方程形式上看并非关于 x 的二次方程，但通过适当换元便可化为二次方程。形如 x⁴ - 13x² + 36 = 0 或 e²ˣ - 5eˣ + 6 = 0 的方程即为隐藏二次方程。
13u + 36 = 0, factorising to (u - 4)(u - 9) = 0, so u = 4 or u = 9. Reverting gives x² = 4 ⇒ x = ±2 and x² = 9 ⇒ x = ±3.
对于 x⁴ - 13x² + 36 = 0，令 u = x²，则方程化为 u² - 13u + 36 = 0，因式分解得 (u - 4)(u - 9) = 0，故 u = 4 或 u = 9。回代得 x² = 4 ⇒ x = ±2 以及 x² = 9 ⇒ x = ±3。
Similarly, for e²ˣ - 5eˣ + 6 = 0, set t = eˣ to obtain t² - 5t + 6 = 0, giving t = 2 or t = 3. Therefore eˣ = 2 ⇒ x = ln 2 and eˣ = 3 ⇒ x = ln 3. Always check for extraneous solutions when substituting back.
类似地，对于 e²ˣ - 5eˣ + 6 = 0，设 t = eˣ 得 t² - 5t + 6 = 0，解得 t = 2 或 t = 3。因此 eˣ = 2 ⇒ x = ln 2，eˣ = 3 ⇒ x = ln 3。回代时务必检查是否有增根。
10. Quadratic Simultaneous Equations | 二次联立方程组
Solving a system where one equation is linear and the other is quadratic often involves substituting the linear equation into the quadratic. The result is a quadratic equation in one variable, which can be solved to find the points of intersection of a line and a parabola.
求解一个为一次式、另一个为二次式的联立方程组，通常需要将一次方程代入二次方程。最终得到一个单变量的二次方程，解之即可求出一条直线与一条抛物线的交点。
yields x = 1 or x = 4. Substituting back into the line gives the points (1, 3) and (4, 9).
考虑 y = 2x + 1 与 y = x² - 3x + 5。令两者相等得 x² - 5x + 4 = 0，解得 x = 1 或 x = 4。代回直线得交点 (1, 3) 和 (4, 9)。
arabola, yielding exactly one point of contact. This links to the discriminant and tangency conditions often tested.
若所得二次方程 Δ < 0，则直线与抛物线不相交。若 Δ = 0，则直线与抛物线相切，恰好有一个切点。这与判别式及相切条件相关联，常为考试重点。
11. Modelling with Quadratics | 二次函数建模应用
Quadratics are widely used to model real-life situations involving area, revenue, projectile motion, and optimisation. Typically a quadratic function is formed from given information, and you must find its maximum or minimum value using completing the square or differentiation.
二次函数广泛用于涉及面积、收益、抛体运动及最优化的实际生活建模。通常要根据给定信息构造一个二次函数，然后利用配方法或求导找出其最大值或最小值。
-2x² + 100x. Completing the square gives the maximum area and the corresponding dimension.
例如，固定周长的矩形围栏可用一条边长 x 表示，得到面积函数 A = x(100 - 2x) = -2x² + 100x。通过配平方可求得最大面积及相应的边长。
in time t. Solving h(t) = 0 gives the time of flight, and the vertex gives maximum height.
在力学中，上抛物体的高度常表示为 h(t) = -4.9t² + vt + h₀，这是关于时间 t 的二次函数。解 h(t) = 0 可得飞行时间，而顶点则给出最大高度。
12. Key Points for Edexcel Exams | Edexcel 考试关键要点
When tackling quadratics questions in the Edexcel A-Level exam, always read the problem carefully to decide which method is most efficient. Show clear steps when factorising, completing the square, or applying the formula. Correctly simplify surd answers and leave them in exact form unless stated otherwise.
在 Edexcel A-Level 考试中处理二次函数题目时，务必仔细审题，选择最有效的方法。进行因式分解、配平方或套用公式时，要写出清晰的步骤。正确化简根式答案，除非另有说明，否则保留精确值。
Remember that the discriminant is your friend—use it to determine the number of solutions or to prove that a line does or does not intersect a curve. For inequalities, a quick sketch saves time and prevents sign errors. Master Vieta's formulas because they often appear in 'new quadratic' problems.
请记住，判别式是你的好帮手——用它来判断解的个数，或证明一条直线是否与曲线相交。解不等式时，快速画一个草图既能节省时间，又能避免符号错误。熟练掌握韦达定理，因为它常常出现在“新二次方程”类问题中。
Finally, practise a wide range of modelling problems. Being able to translate a real-world scenario into a quadratic expression and interpret the results in context is a high-level skill that examiners love to test.
最后，要多练习多样的建模问题。能够将实际情境转化为二次表达式，并结合实际背景解读结果，这是一项高阶技能，也是考官喜欢考查的内容。
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