📚 Solving Quadratic Equations: Factorising, Formula and Graphs | 解二次方程：因式分解、公式与图像
n be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. Mastering this topic means knowing how to solve quadratics by factorising, by completing the square, by using the quadratic formula, and by interpreting graphs. In this revision guide we break down each method step by step, highlight the common mistakes examiners see every year, and show how exam questions are actually structured.
二次方程贯穿 IGCSE 与 A-Level 数学课程，从基础代数运算到实际情境建模都会出现。二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。掌握这一主题意味着要会通过因式分解、配方法、二次公式以及图像解读来求解。在本复习指南中，我们将逐步拆解每一种方法，指出考官每年都会见到的常见错误，并说明考试题的真实结构。
1. What is a Quadratic Equation? | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2: the highest power of the variable x is 2. Depending on the value of the discriminant, the equation may have two distinct real solutions, one repeated real solution, or no real solutions at all. The general form is shown below.
二次方程是一个次数为 2 的多项式方程：变量 x 的最高次数为 2。根据判别式的不同取值，方程可能有两个不同的实数解、一个重复的实数解，或者没有实数解。一般形式如下所示。
ax² + bx + c = 0, a ≠ 0
Here a is the coefficient of x², b is the coefficient of x, and c is the constant term. If a were allowed to be zero, the x² term would vanish and the equation would collapse into a linear equation, which is why the condition a ≠ 0 is essential. Every quadratic must be rearranged into this standard form before any method is applied.
其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。如果允许 a 等于 0，x² 项就会消失，方程便退化为一次方程，因此 a ≠ 0 这个条件是必要的。在套用任何方法之前，都必须先把方程整理成这种标准形式。
- Examples of quadratics: x² − 5x + 6 = 0, 2x² + 7x + 3 = 0, x² − 9 = 0, 3x² − 12x = 0.
- Examples that are not quadratics: 3x + 4 = 0 (linear), x³ − x = 0 (cubic), 1 ÷ x + x = 2 (not a polynomial).
- 中文要点：二次方程可以缺少 b 项或 c 项（例如 x² − 9 = 0），但绝不能缺少 a 项，否则就不是二次方程。
2. Key Vocabulary and Standard Forms | 关键术语与标准形式
Exam questions use several words that all point to the same idea. A root, a solution, a zero and an x-intercept of the corresponding graph all describe the value of x that makes the expression equal to zero. Recognising this vocabulary is often the difference between understanding a question and misreading it.
考题中会使用几个不同的词，但它们指向同一个概念：根（root）、解（solution）、零点（zero）以及对应图像与 x 轴交点的横坐标，指的都是使表达式等于 0 的 x 的取值。能否识别这些术语，往往就是"看懂题目"和"读错题目"之间的差别。
- Coefficient: the number multiplying a variable, such as 2 in 2x².
- Monic quadratic: a quadratic with a = 1, for example x² + 4x − 5.
- Parabola: the U-shaped curve that a quadratic function produces.
- Vertex: the turning point (minimum or maximum) of the parabola.
- Axis of symmetry: the vertical line through the vertex, x = −b ÷ 2a.
- Discriminant: the quantity b² − 4ac, which tells you how many real roots exist.
- 中文要点：二次方程的标准形式是 ax² + bx + c = 0，右边必须为 0；二次函数的标准写法是 y = ax² + bx + c，两者只差一个等号，但一个求根、一个画图。
3. Solving by Factorising | 因式分解法
Factorising is the fastest method when the quadratic factorises with integer values. The logic rests on the zero product property: if the product of two factors equals zero, then at least one of the factors must be zero. The standard steps are shown below.
当二次式能用整数因式分解时，因式分解法是最快的方法。其逻辑建立在零乘积性质上：如果两个因式的乘积为 0，那么至少有一个因式等于 0。标准步骤如下。
- Step 1: Rearrange the equation into the form ax² + bx + c = 0.
- Step 2: Factorise the left-hand side completely.
- Step 3: Set each factor equal to zero.
- Step 4: Solve the resulting linear equations and check both answers.
- 中文要点：第一步必须先移项使右边为 0，第二步把左边彻底分解，第三步令每个因式等于 0，第四步解出每个一次方程并代回检验。
umbers are −2 and −3, so the factorised form is (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
对于首项系数为 1 的二次式，例如 x² − 5x + 6 = 0，需要找两个数：它们相乘得到 +6，相加得到 −5。这两个数是 −2 和 −3，所以分解结果为 (x − 2)(x − 3) = 0，从而 x = 2 或 x = 3。
s are 6 and 1. Splitting the middle term gives (2x + 1)(x + 3) = 0, therefore x = −1 ÷ 2 or x = −3.
当 a ≠ 1 时，先计算 a 与 c 的乘积，再找两个数，使它们的乘积等于 ac、和等于 b。对于 2x² + 7x + 3 = 0，ac = 6，和为 7，所以这两个数是 6 和 1。拆开中间项后得到 (2x + 1)(x + 3) = 0，因此 x = −1 ÷ 2 或 x = −3。
r case, ax² + bx = x(ax + b). For example, x² − 9 = 0 gives (x − 3)(x + 3) = 0 so x = ±3, while 3x² − 12x = 0 gives 3x(x − 4) = 0 so x = 0 or x = 4.
有两种特殊模式值得牢记，因为它们出现频率极高：平方差 x² − k² = (x − k)(x + k)，以及提取公因式 ax² + bx = x(ax + b)。例如 x² − 9 = 0 得到 (x − 3)(x + 3) = 0，所以 x = ±3；而 3x² − 12x = 0 得到 3x(x − 4) = 0，所以 x = 0 或 x = 4。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q, which exposes the vertex of the parabola immediately and is the standard route to proving the quadratic formula. It also works when a quadratic does not factorise with integers.
配方法把二次式改写成 a(x + p)² + q 的形式，这能立刻看出抛物线的顶点，也是推导二次公式的标准路径。当二次式无法用整数分解时，配方法同样适用。
x² + bx + c = (x + b ÷ 2)² − (b ÷ 2)² + c
r x² + 6x + 1 = 0, half of 6 is 3, so we write (x + 3)² − 9 + 1 = 0, that is (x + 3)² = 8. Taking square roots gives x + 3 = ±√8, so x = −3 ± 2√2.
规则很简单：取 x 的系数的一半，放进括号里，再减去这个数的平方，以保证表达式的值不变。对于 x² + 6x + 1 = 0，6 的一半是 3，于是写成 (x + 3)² − 9 + 1 = 0，即 (x + 3)² = 8。两边开平方得 x + 3 = ±√8，所以 x = −3 ± 2√2。
− 2)² − 3. The vertex is therefore (2, −3) and the minimum value of the expression is −3.
当 a ≠ 1 时，先把 a 从前两项中提出来。对于 2x² − 8x + 5，写成 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3。因此顶点是 (2, −3)，该表达式的最小值为 −3。
- Advantage: gives exact surd answers and reveals the vertex directly.
- Disadvantage: fractions appear when b is odd, so arithmetic errors are more likely.
- 中文要点：配方法的最大价值在于"求最值"与"写顶点式"，也是解答"证明二次公式"这类题的必用工具。
5. The Quadratic Formula | 二次公式
ersal method: it solves every quadratic that has real roots, whether or not it factorises.
二次公式是通过对一般式 ax² + bx + c = 0 配方推导出来的。它是万能方法：只要二次方程有实根，无论能否分解，它都能求解。
x = (−b ± √b² − 4ac) ÷ 2a
− 3 = 0, we have a = 2, b = −4 and c = −3, so b² − 4ac = 16 + 24 = 40. Therefore x = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = (2 ± √10) ÷ 2, which is approximately 2.58 or −0.58.
代入时必须小心，因为 b 为负数时极易出现符号错误。对于 2x² − 4x − 3 = 0，有 a = 2，b = −4，c = −3，所以 b² − 4ac = 16 + 24 = 40。因此 x = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = (2 ± √10) ÷ 2，约等于 2.58 或 −0.58。
- Always write down a, b and c explicitly before substituting.
- Keep the answer in exact surd form unless the question asks for a given number of decimal places.
- Simplify the surd fully: √40 = 2√10.
- 中文要点：先把 a、b、c 明确写出再代入；除非题目要求保留小数位，否则应保留精确的根式形式；最后一定要把根式化简到最简。
6. The Discriminant | 判别式
Its sign tells you how many real roots the equation has, without solving it. This is a very common short exam question.
判别式就是根号下的表达式，Δ = b² − 4ac。它的正负号能在不解方程的情况下告诉你该方程有多少个实根。这是非常常见的短小题型。
Δ = b² − 4ac
| Value of Δ | Nature of roots | Graph behaviour |
| Δ > 0 (not a perfect square) | Two distinct irrational real roots | Crosses the x-axis twice |
| Δ > 0 and a perfect square | Two distinct rational roots (factorises) | Crosses the x-axis twice |
| Δ = 0 | One repeated real root | Touches the x-axis at the vertex |
| Δ < 0 | No real roots | Does not meet the x-axis |
it has no real roots and its parabola sits entirely above the x-axis. Questions often say "show that the equation has no real roots" or "find the value of k for which the equation has equal roots", which simply means setting Δ = 0 and solving for k.
例如 x² + 4x + 5 = 0 的 Δ = 16 − 20 = −4，因此它没有实根，其抛物线完全位于 x 轴上方。题目常问"证明该方程没有实根"，或"求使方程有相等实根的 k 值"，后者只需令 Δ = 0 再解出 k 即可。
7. Graphs of Quadratic Functions | 二次函数图像
Solving the equation ax² + bx + c = 0 is exactly the same as finding where the parabola cuts the x-axis.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，曲线开口向上并有最低点；若 a < 0，曲线开口向下并有最高点。解方程 ax² + bx + c = 0 与求抛物线与 x 轴的交点是同一件事。
- y-intercept: substitute x = 0, giving (0, c).
- Axis of symmetry: the vertical line x = −b ÷ 2a.
- Vertex: substitute the axis value back into the equation to get the y-coordinate.
- Roots: the x-values where y = 0, found by factorising, completing the square or the formula.
- 中文要点：与 y 轴交点令 x = 0 得 (0, c)；对称轴为 x = −b ÷ 2a；顶点横坐标就是对称轴，纵坐标代回原式求得；与 x 轴交点即方程的根。
he roots are x = 1 and x = 3. A complete sketch needs all four of these features labelled.
对于 y = x² − 4x + 3，与 y 轴交点为 (0, 3)，对称轴为 x = 2，顶点为 (2, −1)，因式分解得 (x − 1)(x − 3) = 0，所以根为 x = 1 和 x = 3。一幅完整的草图需要把这四个特征全部标注出来。
Graphs are also used to solve simultaneous problems: the number of intersections between a line and a parabola equals the number of solutions of the corresponding equation, which links straight back to the discriminant.
图像还常用来解决联立问题：直线与抛物线的交点个数，等于对应方程的解的个数，这与判别式直接呼应。
8. Choosing the Best Method | 如何选择最佳方法
Different questions reward different methods. Choosing efficiently saves time and reduces the chance of arithmetic slips, especially in a timed paper.
不同题目适合不同方法。高效选择方法既能节省时间，也能降低算错的风险，在限时考试中尤其重要。
| Situation | Best method | Reason |
| Small integer roots, a = 1 | Factorising | Fastest and clearest |
| Difference of two squares | Factorising | Pattern recognised instantly |
| Question asks for the vertex or maximum/minimum | Completing the square | Vertex form appears directly |
| Awkward coefficients or surd answers | Quadratic formula | Always works |
| Question asks how many roots | Discriminant | No solving needed |
| Graph or intersection question | Sketch plus algebra | Sketch guides the algebra |
In longer papers, the wording of the question is a strong hint. "Factorise and hence solve" means factorising is expected, while "give your answers in surd form" signals the formula. Never ignore an instruction that names the method.
在长题中，题干措辞本身就是强烈提示。"因式分解并由此求解"意味着要用因式分解；"以根式形式给出答案"则暗示要用二次公式。凡是明确点名方法的指令，绝不可忽略。
9. Common Mistakes and Exam Technique | 常见错误与应试技巧
Most lost marks in quadratic questions are not caused by difficulty but by carelessness. The list below collects the errors that appear most often in marked scripts.
二次方程题目丢分大多不是由于难度，而是由于粗心。下面列出的是阅卷时出现频率最高的错误。
- Forgetting to rearrange: applying a method before the right-hand side equals zero.
- Dividing by x: dividing 3x² = 12x by x loses the solution x = 0. Always factorise instead.
- Sign errors with negative b: −(−4) becomes −4 in the formula.
- Losing a root: only one of the two brackets is solved.
- Rounding too early: rounding √40 before finishing changes the final answer.
- Missing a method instruction such as "by completing the square".
- 中文要点：忘记移项、两边同除以 x 导致丢根、b 为负时符号出错、只解出一个因式、过早四舍五入、无视题目指定的方法，都是高频失分点。
of numbers, and use your graphic display calculator to check roots quickly when the paper permits it. Checking a root by substitution takes ten seconds and regularly rescues an entire question.
技巧方面：把每一步代入过程写清楚；最终答案要写成 x = ... 或 x = ...，而不是只丢两个数字；若考试允许使用图形计算器，可用它快速验算根。把根代回原方程检验只需十秒，却常常能挽救整道大题。
10. Worked Exam-Style Examples | 考试风格例题精讲
rs with product 12 and sum −7, namely −3 and −4, so (x − 3)(x − 4) = 0 and x = 3 or x = 4.
例题 1：用因式分解解 x² − 7x + 12 = 0。需要两个数，乘积为 12、和为 −7，即 −3 与 −4，因此 (x − 3)(x − 4) = 0，得 x = 3 或 x = 4。
+ 6x) + 7 = 2[(x + 3)² − 9] + 7 = 2(x + 3)² − 11, so the minimum value is −11 when x = −3.
例题 2：把 2x² + 12x + 7 写成 a(x + p)² + q 的形式，并由此求最小值。提取 2 得 2(x² + 6x) + 7 = 2[(
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