📚 Solving Equations: Complete A-Level Methods and Exam Guide | 解方程：A-Level 完整方法与考点指南
Equations are at the heart of A-level Mathematics. In the AQA specification, "solving equations" brings together exact algebraic techniques, simultaneous equations, factor theorem methods, and numerical methods such as iteration and Newton-Raphson. This article explains each method with worked examples, the key conditions you must remember in the exam, and the errors that cost most marks.
方程是 A-Level 数学的核心内容。在 AQA 考试大纲中，"解方程"涵盖精确代数技巧、联立方程、因式定理方法以及迭代和牛顿-拉弗森等数值方法。本文结合例题讲解每种方法，并给出考试中必须记住的关键条件，以及最容易丢分的常见错误。
The guide follows the order in which the topics normally appear across Year 12 and Year 13: exact methods first, then non-linear systems, then numerical methods, and finally exam technique. Every section ends with the conditions that examiners expect you to state explicitly.
本文按照 Year 12 到 Year 13 的常规教学顺序展开：先讲精确方法，再讲非线性方程组，然后是数值方法，最后是应试技巧。每一节都会给出考官希望你明确写出的条件。
1. The idea of solving an equation | 解方程的核心思想
An equation states that two algebraic expressions are equal. Solving the equation means finding every value of the unknown, usually x, that makes the equality true. In A-level work you will meet linear, quadratic, simultaneous, cubic and numerically solved equations. A value that satisfies the equation is called a root or a solution, and the set of all such values is the solution set.
方程表示两个代数表达式相等。解方程意味着找到使等式成立的所有未知数（通常是 x）的值。在 A-Level 学习中你会遇到线性方程、二次方程、联立方程、三次方程以及需要数值求解的方程。使方程成立的值称为根或解，所有这样的值构成的集合称为解集。
Always remember that an equation may have zero, one, several or infinitely many solutions, depending on its type and domain. An equation that is true for every value in the domain is called an identity, while one that is never true has no solution.
始终记住，一个方程可能有零个、一个、多个或无限多个解，具体取决于方程类型和定义域。对定义域内所有值都成立的方程称为恒等式，而永远不成立的方程则无解。
2x + 3 = 2x + 7 ⇒ 3 = 7, no solution (无解)
2(x + 1) = 2x + 2, true for all x (恒等式，无穷多解)
Two equations are equivalent if they have exactly the same solution set. Adding the same quantity to both sides, multiplying both sides by a non-zero quantity, and applying the same one-to-one function to both sides all preserve equivalence. Squaring both sides does not, which is why checking is compulsory whenever you square.
如果两个方程的解集完全相同，就称它们为同解方程。两边同时加上同一个量、两边同时乘以一个非零量、以及对两边同时施加同一个一一对应的函数，都能保持同解性。但两边同时平方不能保持同解性，这正是凡平方必检验的原因。
2. Linear equations | 线性方程
nt of x. If a = c and b = d the equation is an identity; if a = c but b ≠ d there is no solution.
线性方程的形式为 ax + b = cx + d。求解时，把所有含 x 的项移到一边，常数项移到另一边，然后除以 x 的系数。若 a = c 且 b = d，方程为恒等式；若 a = c 但 b ≠ d，则方程无解。
3(x − 2) = 2x + 5 ⇒ 3x − 6 = 2x + 5 ⇒ x = 11
o the solution is correct. Substitution is a fast and reliable check that also guards against sign slips.
检验：3(11 − 2) = 27，2(11) + 5 = 27，所以解是正确的。代入检验既快速又可靠，还能防止符号错误。
When fractions appear, multiply both sides by the lowest common multiple of the denominators. This clears all fractions in one step and reduces the chance of arithmetic mistakes.
当方程中出现分数时，两边同乘所有分母的最小公倍数。这样可以一步消去全部分数，减少计算错误。
(2x + 1) ÷ 3 = (x − 4) ÷ 2 ⇒ 2(2x + 1) = 3(x − 4) ⇒ 4x + 2 = 3x − 12 ⇒ x = −14
If the unknown appears in a denominator, the domain excludes the values that make any denominator zero. State this restriction before you start, then compare your final answers with it.
如果未知数出现在分母中，定义域要排除使任一分母为零的值。开始求解前先写出这一限制条件，最后再与所得答案对照。
3. Quadratic equations | 二次方程
ndard methods: factorising, completing the square, and the quadratic formula. In an exam you should choose factorising when the factors are obvious, the formula when the coefficients are awkward, and completing the square when the question asks for a turning point or an exact surd form.
二次方程的形式为 ax² + bx + c = 0，其中 a ≠ 0。三种标准方法是：因式分解、配方、求根公式。考试中，因式明显时用因式分解，系数复杂时用求根公式，题目要求顶点或精确根式形式时用配方。
x = (−b ± √b² − 4ac) ÷ (2a)
ompleting the square gives (x − 2.5)² = 0.25, so x = 2.5 ± 0.5, which again gives 2 and 3.
例题：解 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。配方得 (x − 2.5)² = 0.25，所以 x = 2.5 ± 0.5，同样得到 2 和 3。
< 0 there are no real roots. When Δ is a perfect square and a, b, c are integers, the quadratic factorises over the rationals.
判别式 Δ = b² − 4ac 决定实数根的个数。若 Δ > 0，有两个不同实根；若 Δ = 0，有一个重根；若 Δ < 0，无实根。当 a、b、c 为整数且 Δ 是完全平方数时，二次式可在有理数范围内因式分解。
| Discriminant | 判别式 | Number of real roots | 实根个数 | Graph | 图像 |
| b² − 4ac > 0 | Two distinct roots | 两个不同根 | Crosses the x-axis twice | 与 x 轴交于两点 |
| b² − 4ac = 0 | One repeated root | 一个重根 | Touches the x-axis | 与 x 轴相切 |
| b² − 4ac < 0 | No real roots | 无实根 | Does not meet the x-axis | 与 x 轴无交点 |
4. Simultaneous equations | 联立方程
Two equations in two unknowns are solved either by elimination or by substitution. Use elimination when both equations are linear, and substitution when one equation is linear and the other is quadratic or circular.
两个未知数、两个方程组成的方程组，可用消元法或代入法求解。两个方程都是线性时用消元法；一个线性、另一个是二次或圆的方程时用代入法。
x − 3 = 0. Factorising gives (x + 3)(x − 1) = 0, so x = 1 and y = 3, or x = −3 and y = −1.
例题：联立求解 y = x + 2 与 x² + y² = 10。代入得 x² + (x + 2)² = 10，即 2x² + 4x − 6 = 0，也就是 x² + 2x − 3 = 0。因式分解得 (x + 3)(x − 1) = 0，所以 x = 1、y = 3，或 x = −3、y = −1。
Each solution pair must be written together. A common error is to mix an x value from one solution with a y value from the other, which produces a pair that satisfies neither equation.
每一组解必须成对写出。常见错误是把一组解的 x 与另一组解的 y 拼在一起，得到的数对连一个方程都不满足。
= 0, the line is a tangent. This link between algebra and geometry is frequently examined.
交点个数等于实数解的个数。若消元后的二次方程 Δ < 0，直线与曲线不相交；若 Δ = 0，直线为切线。这种代数与几何之间的联系是高频考点。
5. Polynomials and the factor theorem | 多项式与因式定理
er theorem states that when f(x) is divided by (x − a) the remainder is f(a). Together they allow you to factorise cubics quickly.
因式定理指出：对于多项式 f(x)，(x − a) 是 f(x) 的因式，当且仅当 f(a) = 0。余数定理指出：f(x) 除以 (x − a) 的余数为 f(a)。两者结合可以快速分解三次多项式。
ich factorises as (x − 2)(x − 3). Therefore f(x) = (x − 1)(x − 2)(x − 3) and the roots are x = 1, 2, 3.
例题：将 f(x) = x³ − 6x² + 11x − 6 完全分解。先试小因数：f(1) = 1 − 6 + 11 − 6 = 0，所以 (x − 1) 是因式。做除法得 x² − 5x + 6，即 (x − 2)(x − 3)。因此 f(x) = (x − 1)(x − 2)(x − 3)，方程的根为 x = 1、2、3。
For a cubic, candidates for integer roots are the factors of the constant term divided by the factors of the leading coefficient. Always test in a systematic order and record each value of f(a) so the examiner can see your method.
对三次多项式，整数根的候选值是常数项的因数除以首项系数的因数。要按固定顺序逐一尝试，并把每个 f(a) 的值写出来，让考官看到你的方法。
f(x) = (x − a)q(x) + f(a)
6. Hidden quadratics and algebraic fractions | 隐蔽二次方程与分式方程
A hidden quadratic is an equation that becomes quadratic after a substitution. Recognising the repeated pattern x² inside an equation is the key skill here.
隐蔽二次方程是指经过换元后变成二次方程的方程。识别方程中反复出现的 x² 结构是这一部分的关键技能。
= 1 or u = 4. Therefore x² = 1 or x² = 4, so x = ±1 or x = ±2. There are four real roots.
例一：解 x⁴ − 5x² + 4 = 0。设 u = x²，得 u² − 5u + 4 = 0，所以 u = 1 或 u = 4。因此 x² = 1 或 x² = 4，即 x = ±1 或 x = ±2，共有四个实根。
Example two: solve 2²ˣ − 5 × 2ˣ + 4 = 0. Let u = 2ˣ, so u² − 5u + 4 = 0, giving u = 1 or u = 4. Since 2ˣ is always positive, both are allowed, so x = 0 or x = 2.
例二：解 2²ˣ − 5 × 2ˣ + 4 = 0。设 u = 2ˣ，得 u² − 5u + 4 = 0，所以 u = 1 或 u = 4。由于 2ˣ 恒为正，两者均合法，故 x = 0 或 x = 2。
For algebraic fractions, multiply through by the common denominator, solve the resulting polynomial, then reject any value that makes a denominator zero. Those rejected values are extraneous solutions.
对于分式方程，两边同乘公分母，解出多项式方程，然后舍去使分母为零的值。被舍去的这些值称为增根。
7. Exponential and logarithmic equations | 指数与对数方程
To solve aᶠ⁽ˣ⁾ = aᵍ⁽ˣ that the exponential function is one-to-one, so f(x) = g(x). To solve equations such as 3ˣ here the bases cannot be matched, take logarithms of both sides and use log(aⁿ) = n log a.
解 aᶠ⁽ˣ⁾ = aᵍ⁽ˣ⁾（a > 0 且 a ≠ 1）时，利用指数函数是一一对应的性质，得 f(x) = g(x)。对于 3ˣ = 20 这类无法统一底数的方程，两边取对数，并使用 log(aⁿ) = n log a。
3ˣ⁺¹ = 81 ⇒ 3ˣ⁺¹ = 3⁴ ⇒ x + 1 = 4 ⇒ x = 3
Logarithmic equations require the domain condition that every logarithmic argument must be positive. Solve the equation first, then check each answer against the domain.
对数方程要求定义域条件：每个对数的真数必须为正。先解方程，再逐个检验答案是否满足定义域。
−2. Since x must be greater than 2 for both logs to be defined, x = −2 is rejected and the only solution is x = 4.
例题：解 log₂ x + log₂(x − 2) = 3。合并对数得 log₂[x(x − 2)] = 3，所以 x(x − 2) = 8，即 x² − 2x − 8 = 0，也就是 (x − 4)(x + 2) = 0，故 x = 4 或 x = −2。由于两个对数都有意义要求 x > 2，故舍去 x = −2，唯一解为 x = 4。
8. Trigonometric equations | 三角方程
Trigonometric equations have infinitely many solutions unless an interval is given. Solve for the principal value, then use the symmetry of the graph or the CAST diagram to find every solution in the required interval.
除非给出区间，三角方程有无限多解。先求出主值，再利用图像的对称性或 CAST 图找出指定区间内的全部解。
e in the second quadrant, the other solution is 180° − 30° = 150°. So θ = 30° or θ = 150°.
例题：在 0° ≤ θ ≤ 360° 内解 sin θ = 0.5。主值为 30°；由于正弦在第二象限也为正，另一解为 180° − 30° = 150°。所以 θ = 30° 或 150°。
nterval, then divide every solution by 2 to return to θ. Forgetting to widen the interval is one of the most common errors in this topic.
对于 sin 2θ = 0.5 这类方程，先令 u = 2θ 把区间相应扩大，在扩大后的区间内解 u，再把每个解除以 2 还原为 θ。忘记扩大区间是本专题最常见的错误之一。
rctan k + 180°n, where n is any integer. In radians the same rule reads θ = arctan k + πn.
对于 tan θ = k，解以 180° 为周期重复，通解为 θ = arctan k + 180°n，其中 n 为任意整数。用弧度表示时，同样的规则写作 θ = arctan k + πn。
9. Numerical methods: sign change and iteration | 数值方法：符号变化与迭代
Some equations cannot be solved exactly. Numerical methods find a root to a required degree of accuracy. The first step is always to locate the root by finding an interval where f(x) changes sign.
有些方程无法精确求解。数值方法可以把根求到指定的精确度。第一步始终是找到一个使 f(x) 变号的区间，从而定位根的位置。
If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then by the change of sign rule there is at least one root in (a, b). You must state continuity to earn the mark.
若 f 在 [a, b] 上连续，且 f(a) 与 f(b) 异号，则由符号变化法则可知 (a, b) 内至少存在一个根。要拿到这一分，必须写明"连续"。
f(x) = x³ − x − 1, f(1) = −1 < 0, f(2) = 5 > 0 ⇒ root in (1, 2)
ₙ₊₁ = g(xₙ). The iteration converges to a root when the starting value is close enough and |g'(x)| < 1 near that root.
迭代法需要把方程整理成 x = g(x) 的形式，并使用 xₙ₊₁ = g(xₙ)。当初值足够接近根、且在根附近 |g'(x)| < 1 时，迭代收敛到该根。
Worked example: from x³ − x − 1 = 0 we get x = ∛(x + 1), so g(x) = ∛(x + 1) and xₙ₊₁ = ∛(xₙ + 1). Starting with x₀ = 1: x₁ = ∛ , x₂ ≈ 1.3123, x₃ ≈ 1.3224, x₄ ≈ 1.3243. The values converge towards approximately 1.3247.
例题：由 x³ − x − 1 = 0 得 x = ∛(x + 1)，故 g(x) = ∛(x + 1)，迭代式 xₙ₊₁ = ∛(xₙ + 1)。取 x₀ = 1：x₁ = ∛2 ≈ 1.2599，x₂ ≈ 1.3123，x₃ ≈ 1.3224，x₄ ≈ 1.3243。数值逐渐收敛到约 1.3247。
10. The Newton-Raphson method | 牛顿-拉弗森法
Newton-Raphson is a fast iterative method based on the tangent line to the curve. It usually converges much more quickly than simple iteration, but it can fail if the starting value is poor or if f'(xₙ) is close to zero.
牛顿-拉弗森法是基于曲线切线的快速迭代法。它通常比简单迭代收敛快得多，但初值取得不好或 f'(xₙ) 接近零时会失效。
xₙ₊₁ = xₙ − f(xₙ) ÷ f'(xₙ)
Worked example: for f(x) = x³ − x − 1, we have f'(x) = 3x² − 1. Taking x₀ = 1: f(1) = −1 and f'(1) = 2, so x₁ = 1 − (−1) ÷ 2 = 1.5. Then f(1.5) = 0.875
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