📚 Chapter Review 2: Quadratics | 二次函数章节复习
This chapter review covers the core techniques for working with quadratic functions in the Edexcel A-Level Pure Mathematics 1 course. You will revise completing the square, factorising, the quadratic formula, the discriminant, graphing, inequalities, roots and coefficients, simultaneous equations, and modelling. Mastery of these skills is essential because quadratics appear throughout the AS and A-Level specification.
本章复习涵盖 Edexcel A-Level 纯数学 1 中二次函数的核心技巧。你将复习配方法、因式分解、二次公式、判别式、图像、不等式、根与系数、联立方程以及建模。熟练掌握这些技能至关重要，因为二次函数贯穿 AS 和 A-Level 的整个考试范围。
1. Completing the Square | 配方法
Completing the square rewrites a quadratic expression ax² + bx + c in the form a(x + p)² + q. This form reveals the vertex of the parabola and is often used to solve equations or sketch graphs.
配方法将二次表达式 ax² + bx + c 化为 a(x + p)² + q 的形式。这种形式可以揭示抛物线的顶点，常用于解方程或绘制图像。
For example, x² + 6x + 5 becomes (x + 3)² − 4 because (x + 3)² = x² + 6x + 9, so we subtract 4 to keep the value unchanged.
例如，x² + 6x + 5 可化为 (x + 3)² − 4，因为 (x + 3)² = x² + 6x + 9，因此我们需要减去 4 以保持原值不变。
When the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 3, write 2(x² + 4x) + 3 = 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 5.
当 x² 的系数不为 1 时，需要先将其提出来。对于 2x² + 8x + 3，先写成 2(x² + 4x) + 3 = 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 5。
The vertex of y = a(x + p)² + q is at (−p, q). The line x = −p is the axis of symmetry.
y = a(x + p)² + q 的顶点坐标为 (−p, q)。直线 x = −p 是对称轴。
2. Solving by Factorising | 因式分解法求解
To solve a quadratic equation by factorising, first set the equation equal to zero, then factorise the quadratic expression into two linear brackets.
用因式分解法解二次方程时，首先令方程等于零，然后将二次表达式分解为两个一次因式的乘积。
For x² − 5x + 6 = 0, we look for two numbers that multiply to 6 and add to −5, giving (x − 2)(x − 3) = 0. Therefore x = 2 or x = 3.
对于 x² − 5x + 6 = 0，我们需要找到两个数，它们的乘积为 6，和为 −5，得到 (x − 2)(x − 3) = 0。因此 x = 2 或 x = 3。
If an equation is not equal to zero, rearrange it first. For x² + 4x = 12, rewrite as x² + 4x − 12 = 0, then factorise to (x + 6)(x − 2) = 0, giving x = −6 or x = 2.
如果方程不等于零，需要先移项。对于 x² + 4x = 12，先化为 x² + 4x − 12 = 0，然后因式分解为 (x + 6)(x − 2) = 0，得到 x = −6 或 x = 2。
When the coefficient of x² is greater than 1, use the ac method or trial and error. For 2x² + 7x + 3, split 7x into 6x + x: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 x² 的系数大于 1 时，可使用 ac 法或尝试法。对于 2x² + 7x + 3，将 7x 拆分为 6x + x：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
3. The Quadratic Formula | 二次公式
The quadratic formula solves any equation of the form ax² + bx + c = 0. The solutions are given by:
二次公式可以求解任何形如 ax² + bx + c = 0 的方程。其解为：
x = ( −b ± √( b² − 4ac ) ) ÷ 2a
Use this formula when factorising is difficult or when the roots are irrational. For 2x² − 4x − 3 = 0, a = 2, b = −4, c = −3, so:
当因式分解困难或根为无理数时使用该公式。对于 2x² − 4x − 3 = 0，a = 2，b = −4，c = −3，因此：
x = ( 4 ± √( (−4)² − 4 × 2 × (−3) ) ) ÷ 4 = ( 4 ± √( 16 + 24 ) ) ÷ 4 = ( 4 ± √40 ) ÷ 4
The roots simplify to (4 ± 2√10) ÷ 4, which gives x = (2 ± √10) ÷ 2. Always simplify surds fully.
根可化简为 (4 ± 2√10) ÷ 4，得到 x = (2 ± √10) ÷ 2。始终要将根式化简到最简形式。
Remember to quote the formula accurately in exams. A sign error in b or c is one of the most common mistakes.
考试中务必准确写出公式。b 或 c 的符号错误是最常见的错误之一。
4. The Discriminant | 判别式
The discriminant of ax² + bx + c = 0 is Δ = b² − 4ac. It determines the nature and number of real roots without solving the equation.
ax² + bx + c = 0 的判别式为 Δ = b² − 4ac。它可以在不解方程的情况下判断实根的性质和个数。
- If Δ > 0, the equation has two distinct real roots.
- 如果 Δ > 0，方程有两个不相等的实根。
- If Δ = 0, the equation has one repeated real root.
- 如果 Δ = 0，方程有一个重复的实根。
- If Δ < 0, the equation has no real roots.
- 如果 Δ < 0，方程没有实根。
Edexcel often asks you to find the range of a parameter for which a quadratic has real roots. For kx² + 2x + 1 = 0, the discriminant is 2² − 4 × k × 1 = 4 − 4k. For real roots, 4 − 4k ≥ 0, so k ≤ 1.
Edexcel 常要求找出使二次方程具有实根的参数范围。对于 kx² + 2x + 1 = 0，判别式为 2² − 4 × k × 1 = 4 − 4k。要有实根，需要 4 − 4k ≥ 0，因此 k ≤ 1。
Also, a graph of y = ax² + bx + c touches the x-axis when Δ = 0 and does not cross it when Δ < 0. Link the discriminant to the number of x-axis intersections.
此外，当 Δ = 0 时，y = ax² + bx + c 的图像与 x 轴相切；当 Δ < 0 时，图像不与 x 轴相交。要将判别式与 x 轴交点的数目联系起来。
5. Sketching Quadratic Graphs | 二次函数图像草图
A quadratic graph y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point; if a < 0, it opens downwards and has a maximum point.
二次函数 y = ax² + bx + c 的图像是抛物线。如果 a > 0，抛物线开口向上，有最小值点；如果 a < 0，抛物线开口向下，有最大值点。
To sketch a quadratic, find the y-intercept by setting x = 0, find the x-intercepts by solving ax² + bx + c = 0, and find the vertex by completing the square.
要绘制二次函数草图，可令 x = 0 求 y 轴截距，解 ax² + bx + c = 0 求 x 轴截距，并通过配方法求顶点。
For y = x² − 4x + 3, the y-intercept is (0, 3). Factorising gives (x − 1)(x − 3) = 0, so x-intercepts are at x = 1 and x = 3. Completing the square gives y = (x − 2)² − 1, so the vertex is (2, −1).
对于 y = x² − 4x + 3，y 轴截距为 (0, 3)。因式分解得到 (x − 1)(x − 3) = 0，因此 x 轴截距为 x = 1 和 x = 3。配方法得到 y = (x − 2)² − 1，因此顶点为 (2, −1)。
Label all key points clearly on your sketch. Examiners award marks for shape, intercepts, and the turning point.
在草图中要清晰标注所有关键点。考官会根据图像形状、截距和转折点给分。
6. Quadratic Inequalities | 二次不等式
To solve a quadratic inequality, first rearrange it so one side is zero. Then factorise the quadratic and identify the critical values where the expression equals zero.
解二次不等式时，首先移项使一边为零。然后因式分解二次式，并找出使表达式等于零的临界值。
For x² − 5x + 6 < 0, factorise to (x − 2)(x − 3) < 0. The critical values are x = 2 and x = 3. Test intervals: x < 2, 2 < x < 3, and x > 3. The product is negative only between 2 and 3, so the solution is 2 < x < 3.
对于 x² − 5x + 6 < 0，因式分解为 (x − 2)(x − 3) < 0。临界值为 x = 2 和 x = 3。检验区间：x < 2、2 < x < 3 和 x > 3。乘积仅在 2 与 3 之间为负，因此解集为 2 < x < 3。
If the inequality is x² − 5x + 6 > 0, the solution is the opposite region: x < 2 or x > 3. Use strict or non-strict signs carefully.
如果不等式为 x² − 5x + 6 > 0，则解集为相反的区域：x < 2 或 x > 3。要仔细区分严格与不严格的符号。
Sketching the parabola y = (x − 2)(x − 3) helps you see which intervals lie above or below the x-axis. This visual method reduces errors.
绘制抛物线 y = (x − 2)(x − 3) 有助于判断哪些区间位于 x 轴上方或下方。这种直观方法可以减少错误。
7. Roots and Coefficients | 根与系数关系
If α and β are the roots of ax² + bx + c = 0, then the sum of the roots is α + β = −b ÷ a and the product is αβ = c ÷ a.
如果 α 和 β 是 ax² + bx + c = 0 的根，则根的和为 α + β = −b ÷ a，根的积为 αβ = c ÷ a。
These relationships can be derived from the factorised form a(x − α)(x − β) = 0. Expanding gives ax² − a(α + β)x + aαβ, so comparing coefficients gives the formulas.
这些关系可由因式分解形式 a(x − α)(x − β) = 0 推导出来。展开得到 ax² − a(α + β)x + aαβ，通过比较系数即可得到上述公式。
For example, if α + β = 5 and αβ = 6, a quadratic with these roots is x² − 5x + 6 = 0. This is useful for forming equations from given roots.
例如，如果 α + β = 5 且 αβ = 6，则以这些为根的二次方程为 x² − 5x + 6 = 0。这在由给定根构造方程时非常有用。
Edexcel may ask you to find α² + β² or 1/α + 1/β. Use the identities α² + β² = (α + β)² − 2αβ and 1/α + 1/β = (α + β) ÷ αβ.
Edexcel 可能要求计算 α² + β² 或 1/α + 1/β。可使用恒等式 α² + β² = (α + β)² − 2αβ 和 1/α + 1/β = (α + β) ÷ αβ。
8. Simultaneous Equations with Quadratics | 含二次式的联立方程
To solve one linear and one quadratic equation simultaneously, substitute the linear equation into the quadratic to create a single quadratic in one variable.
要联立求解一个一次方程和一个二次方程，可将一次方程代入二次方程，得到只含一个变量的二次方程。
For y = 2x + 1 and y = x² + x − 1, substitute to get 2x + 1 = x² + x − 1. Rearrange to x² − x − 2 = 0, factorise to (x − 2)(x + 1) = 0, so x = 2 or x = −1. Then find the corresponding y-values.
对于 y = 2x + 1 和 y = x² + x − 1，代入得到 2x + 1 = x² + x − 1。移项得到 x² − x − 2 = 0，因式分解为 (x − 2)(x + 1) = 0，因此 x = 2 或 x = −1。然后求出对应的 y 值。
The solutions are (2, 5) and (−1, −1). Always pair each x-value with its correct y-value from the linear equation.
解为 (2, 5) 和 (−1, −1)。始终要将每个 x 值与一次方程中对应的 y 值配对。
Geometrically, the solutions are the intersection points of a line and a parabola. If the discriminant of the resulting quadratic is negative, the line and parabola do not meet.
从几何上看，解就是直线与抛物线的交点。如果所得二次方程的判别式为负，则直线与抛物线不相交。
9. Modelling with Quadratics | 二次函数建模
Quadratic models describe projectile motion, area problems, revenue and profit, and many other real-life situations. The key is to identify the variable, write a quadratic expression, and then use differentiation or completing the square to find a maximum or minimum.
二次模型可用于描述抛体运动、面积问题、收益与利润以及许多其他实际情境。关键是确定变量、写出二次表达式，然后使用微分或配方法求最大值或最小值。
For a rectangular pen with perimeter 40 m, let the width be x. The length is 20 − x, so the area is A = x(20 − x) = 20x − x². Completing the square gives A = −(x − 10)² + 100, so the maximum area is 100 m² when x = 10.
对于一个周长为 40 m 的矩形围栏，设宽为 x。长为 20 − x，因此面积为 A = x(20 − x) = 20x − x²。配方法得到 A = −(x − 10)² + 100，因此当 x = 10 时最大面积为 100 m²。
In modelling questions, define all variables clearly and check that your solution makes sense in context. For example, negative lengths must be rejected.
在建模题中，要清楚地定义所有变量，并检验解在实际情境中是否合理。例如，负的长度必须舍去。
Edexcel often includes a modelling question in each paper. Practise extracting the quadratic from a written problem before doing any calculation.
Edexcel 每份试卷通常包含一道建模题。在做任何计算之前，要练习从文字题中提取二次关系。
10. Exam Strategy and Common Pitfalls | 考试策略与常见错误
When answering quadratic questions, always write down the method step by step. This earns method marks even if a numerical slip occurs later.
解答二次函数题目时，务必逐步写出解题过程。即使后面的数值计算出现失误，也能获得方法分。
Common errors include forgetting to divide by the coefficient of x² before completing the square, mishandling negative signs in the quadratic formula, and not writing the full solution set for inequalities.
常见错误包括：配方法前忘记除以 x² 的系数，二次公式中负号处理错误，以及解不等式时没有写出完整的解集。
Check your answers by substituting them back into the original equation. For graph sketches, verify that the intercepts and vertex are consistent with the algebra.
通过将答案代回原方程进行检查。对于图像草图，要验证截距和顶点是否与代数运算一致。
| Task | Key Check |
|---|---|
| Factorising | Expanding gives original expression |
| Completing the square | Vertex and intercepts match |
| Quadratic formula | Signs of b and c correct |
| Inequality | Test a value in each interval |
任务
关键检查
| 因式分解 | 展开后得到原式 |
|---|---|
| 配方法 | 顶点和截距一致 |
| 二次公式 | b 和 c 的符号正确 |
| 不等式 | 在每个区间检验一个值 |
Use past paper questions to build speed and accuracy. Pay attention to command words such as “hence”, “show that”, and “given that” because they tell you which method to use.
利用历年真题提高速度和准确性。注意题干中的指令词，如 “hence”、“show that” 和 “given that”，它们会提示你应使用哪种方法。
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