📚 Chapter Review 3: Equations and Inequalities | 第3章复习：方程与不等式
Chapter Review 3 for Edexcel A-Level Mathematics consolidates the core techniques needed to solve linear equations, quadratic equations, simultaneous equations and inequalities. The chapter builds on GCSE algebra and introduces the discriminant, completing the square and graphical interpretation of solutions. A strong grasp of this chapter is essential for later topics such as coordinate geometry, differentiation and modelling.
Edexcel A-Level 数学第3章复习巩固求解一元一次方程、二次方程、联立方程与不等式所需的核心技巧。本章在 GCSE 代数基础上引入判别式、配方法以及解的图像解释。扎实掌握本章对后续的坐标几何、微分和建模等主题至关重要。
In the exam, questions often combine several skills: setting up an equation from a worded problem, solving it exactly, using the discriminant to classify roots, and presenting an inequality solution using set notation or a sketch. This review walks through each skill group with worked examples and common pitfalls.
在考试中，题目往往综合考查多种技能：从文字题建立方程、精确求解、利用判别式判断根的类型，并用集合符号或草图表示不等式解。本复习将通过例题和常见错误逐一梳理各技能点。
1. Solving Linear Equations | 解一元一次方程
Linear equations are the simplest equation type in Chapter 3, but they must be handled accurately under time pressure. The usual method is to remove brackets, clear fractions, collect like terms, and isolate the unknown variable on one side of the equation.
一元一次方程是第3章中最简单的方程类型，但在时间压力下必须准确处理。常规方法是去括号、清除分数、合并同类项，并将未知数单独移到方程的一边。
When an equation contains fractions, multiply every term by the lowest common denominator. For example, solve (x + 2)/3 = (2x − 1)/5. Cross-multiply to get 5(x + 2) = 3(2x − 1), so 5x + 10 = 6x − 3, giving x = 13.
当方程含有分数时，应将每一项乘以最小公分母。例如，解 (x + 2)/3 = (2x − 1)/5。交叉相乘得 5(x + 2) = 3(2x − 1)，因此 5x + 10 = 6x − 3，解得 x = 13。
Always check your final answer by substituting it back into the original equation. This is quick and catches sign errors, especially when expanding a negative bracket such as −2(x − 3) = −2x + 6.
始终将最终答案代回原方程检验。这个方法快速且能发现符号错误，尤其是在展开负括号时，例如 −2(x − 3) = −2x + 6。
- Remove brackets first and watch negative signs.
- 先去括号并注意负号。
- Clear denominators by multiplying by the lowest common multiple.
- 方程两边乘以最小公倍数以清除分母。
- Collect variable terms on one side and constants on the other.
- 将含未知数的项移到一边，常数项移到另一边。
2. Solving Quadratic Equations by Factorising | 用因式分解法解二次方程
A quadratic equation has the standard form ax² + bx + c = 0, where a ≠ 0. If the quadratic expression factorises, you can use the zero product property: if pq = 0, then p = 0 or q = 0.
二次方程的标准形式为 ax² + bx + c = 0，其中 a ≠ 0。如果二次式可以因式分解，就可使用零乘积性质：若 pq = 0，则 p = 0 或 q = 0。
For a monic quadratic such as x² − 5x + 6 = 0, find two numbers that multiply to 6 and add to −5. They are −2 and −3, so the equation becomes (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
对于首项系数为1的二次方程，如 x² − 5x + 6 = 0，找出两个数，它们的乘积为 6、和为 −5。这两个数是 −2 和 −3，因此方程化为 (x − 2)(x − 3) = 0，得到 x = 2 或 x = 3。
For non-monic quadratics, use the ac method. For example, solve 6x² + 5x − 6 = 0. Here ac = −36, so find two numbers that multiply to −36 and add to 5: they are 9 and −4. Split the middle term to get 6x² + 9x − 4x − 6 = 0, then factor by grouping as (3x − 2)(2x + 3) = 0. The solutions are x = 2/3 and x = −3/2.
对于首项系数不为1的二次方程，使用 ac 法。例如，解 6x² + 5x − 6 = 0。这里 ac = −36，找出两个数，它们的乘积为 −36、和为 5：它们是 9 和 −4。拆解中间项得 6x² + 9x − 4x − 6 = 0，再分组因式分解为 (3x − 2)(2x + 3) = 0。解为 x = 2/3 和 x = −3/2。
Do not divide both sides of a quadratic equation by x or by any expression involving x, as this can remove a valid solution. Instead, bring all terms to one side and factorise.
不要将二次方程两边同时除以 x 或任何含 x 的式子，因为这会丢失有效解。正确做法是将所有项移到一边并因式分解。
3. The Quadratic Formula and Exact Solutions | 二次公式与精确解
When a quadratic does not factorise neatly, use the quadratic formula. For ax² + bx + c = 0, the solutions are given by:
当二次方程不易因式分解时，使用二次公式。对于 ax² + bx + c = 0，解由下式给出：
x = [ −b ± √(b² − 4ac) ] / (2a)
Example: solve 2x² + 3x − 4 = 0. Here a = 2, b = 3 and c = −4. The formula gives x = [ −3 ± √(9 + 32) ] / 4 = ( −3 ± √41 ) / 4. These are the exact solutions, and √41 should not be rounded unless the question asks for a decimal approximation.
示例：解 2x² + 3x − 4 = 0。这里 a = 2，b = 3，c = −4。代入公式得 x = [ −3 ± √(9 + 32) ] / 4 = ( −3 ± √41 ) / 4。这些是精确解，除非题目要求小数近似，否则不要对 √41 取近似值。
You should also be able to simplify surds correctly. For instance, √72 = 6√2, so an answer such as ( −6 ± √72 ) / 6 simplifies to ( −6 ± 6√2 ) / 6 = −1 ± √2.
你还应该能够正确化简二次根式。例如，√72 = 6√2，因此诸如 ( −6 ± √72 ) / 6 的答案可化简为 ( −6 ± 6√2 ) / 6 = −1 ± √2。
The quadratic formula is especially useful when the discriminant is not a perfect square, or when the coefficients are large. In exam questions that ask for exact form, leave your answer in surd form.
当判别式不是完全平方数或系数较大时，二次公式尤其有用。在要求精确形式的考题中，答案应保留根号形式。
4. Completing the Square | 配方法
Completing the square transforms a quadratic expression into the form a(x + p)² + q. This form reveals the turning point of the graph and is useful for solving equations, proving results and finding maximum or minimum values.
配方法将二次式转化为 a(x + p)² + q 的形式。该形式能显示图像的转折点，并可用于解方程、证明结论以及求最大值或最小值。
x² + 6x + 10 = (x + 3)² + 1
Because (x + 3)² ≥ 0 for all real x, the whole expression is always at least 1. Therefore the minimum value of x² + 6x + 10 is 1, and it occurs when x = −3. The vertex of the parabola y = x² + 6x + 10 is (−3, 1).
由于对所有实数 x，(x + 3)² ≥ 0，因此整个式子始终至少为 1。所以 x² + 6x + 10 的最小值为 1，且在 x = −3 时取得。抛物线 y = x² + 6x + 10 的顶点为 (−3, 1)。
For a non-monic quadratic, factor out the leading coefficient first. Example: 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3. The minimum value of the original expression is −3.
对于首项系数不为1的二次式，先提出首项系数。示例：2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3。原式的最小值为 −3。
To solve an equation by completing the square, isolate the squared term and take the square root of both sides, remembering to include ±. For instance, x² + 6x + 10 = 0 leads to (x + 3)² = −1, which has no real solutions because a squared real number cannot be negative.
要用配方法解方程，先分离平方项，再对两边开平方，并记得加上 ± 号。例如，x² + 6x + 10 = 0 化为 (x + 3)² = −1，该方程无实数解，因为实数的平方不可能为负。
5. The Discriminant | 判别式
The discriminant of ax² + bx + c = 0 is defined as Δ = b² − 4ac. It tells you the number and nature of the real roots without solving the equation.
方程 ax² + bx + c = 0 的判别式定义为 Δ = b² − 4ac。它无需解方程即可告诉你实根的个数与类型。
| Condition | Number of real roots | Graph behaviour |
|---|---|---|
| Δ > 0 | Two distinct real roots | Curve crosses the x-axis twice |
更多咨询请联系16621398022（同微信）
CommentsMore posts |
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导