📚 IGCSE Mathematics: Solving Quadratic Equations | IGCSE 数学：解一元二次方程
tic equation is any equation that can be written in the form ax² + bx + c = 0 with a ≠ 0. This article covers the standard form, the key vocabulary, the main methods of solution (factorisation, completing the square and the quadratic formula), the discriminant, and the link between algebraic roots and the x-intercepts of a graph.
一元二次方程是 IGCSE 数学中最重要的代数主题之一，在核心课程（Core）和扩展课程（Extended）试卷中都会出现。凡是可以写成 ax² + bx + c = 0（a ≠ 0）形式的方程，都是一元二次方程。本文将介绍二次方程的标准形式、关键术语、主要求解方法（因式分解、配方法、求根公式）、判别式，以及代数根与图像上 x 轴截距之间的联系。
By the end you should be able to choose the fastest method for a given equation, show clear working for method marks, and explain what your answers mean on a graph.
学完本文，你应当能够为给定的方程选择最快的方法、写出清晰的过程以拿到方法分，并解释所求得的答案在图像上意味着什么。
1. What Is a Quadratic Equation? | 什么是一元二次方程？
expressed as ax² + bx + c = 0, where a, b and c are constants and a is not zero. The word 'quadratic' If the highest power were 1 the equation would be linear; if it were 3 it would be cubic.
一元二次方程是含有一个未知数、且可以表示为 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a 不等于零。“二次”这个名称来源于平方，因为 x 的最高次数是 2。如果最高次数是 1，方程就是一次方程；如果是 3，就是三次方程。
understand how the solutions connect to the x-intercepts of the graph of y = ax² + bx + c.
在 IGCSE 阶段，你需要会通过因式分解、配方法和求根公式解一元二次方程，还必须理解方程的解与函数 y = ax² + bx + c 图像上的 x 轴截距之间的联系。
An equation always has the '= 0' ou simplify, factorise or complete the square of an expression, but you solve an equation.
方程一定有等号（通常是“= 0”或其他数），而像 ax² + bx + c 这样的式子没有等号。式子是“化简、因式分解、配方”的对象，方程才是“求解”的对象。
2. Standard Form and Key Terms | 标准形式与关键术语
The standard form of a quadratic equation is shown below. Every method in this article assumes the equation has already been rearranged into this form.
一元二次方程的标准形式如下。本文中的所有方法都假定方程已经整理成这个形式。
ax² + bx + c = 0, a ≠ 0
2, b = −3 and c = 4. Notice that the sign in front of a term travels with the coefficient, so b is negative here.
在标准形式中，a 是 x² 的系数，b 是 x 的系数，c 是常数项。例如，在 2x² − 3x + 4 = 0 中，系数为 a = 2，b = −3，c = 4。注意项前面的符号要跟着系数一起走，所以这里的 b 是负数。
- ax² is the quadratic term — 二次项。
- bx is the linear term — 一次项。
- c is the constant term — 常数项。
- a is the leading coefficient — 首项系数，且 a ≠ 0。
which is a linear equation, not a quadratic one. There is no such restriction on b or c: either or both of them may be zero.
a 的值不能为零，因为如果 a = 0，方程就变成 bx + c = 0，这是一次方程，而不是二次方程。b 和 c 没有这个限制：它们中的一个或两个都可以为零。
² + 4x − 7 = 0. Writing the equation in standard form first avoids almost all sign errors.
使用任何方法之前，都要先整理。例如 3x² = 5x − 1 要写成 3x² − 5x + 1 = 0，x(x + 4) = 7 要写成 x² + 4x − 7 = 0。先化为标准形式，几乎可以避免所有符号错误。
| Equation 方程 | a | b | c |
| x² − 7x + 12 = 0 | 1 | −7 | 12 |
| 3x² + 5x = 0 | 3 | 5 | 0 |
| 2x² − 8 = 0 | 2 | 0 | −8 |
| −x² + 4x − 3 = 0 | −1 | 4 | −3 |
3. Solving by Factorisation | 因式分解法求解
Factorising is usually the fastest method when the quadratic has simple integer roots. First set the equation equal to zero, then write the quadratic expression as a product of two linear factors. The key idea is the zero product rule: if two numbers multiply to give zero, then at least one of them must be zero.
当二次方程有较简单的整数根时，因式分解通常是最快的方法。首先把方程整理成等于零的形式，然后把二次式写成两个一次因式的乘积。核心思想是“零积法则”：如果两个数相乘等于零，那么其中至少有一个数为零。
If (x − p)(x − q) = 0, then x = p or x = q.
ly to +6 and add to −5; these are −2 and −3. So (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
例：解 x² − 5x + 6 = 0。我们需要两个数，它们的乘积是 +6、和是 −5，即 −2 和 −3。于是 (x − 2)(x − 3) = 0，得到 x = 2 或 x = 3。
get 6, and find two numbers with product 6 and sum 7: these are 6 and 1. Split the middle term, then factorise in pairs.
当首项系数不是 1 时，因式分解会稍难一些。对于 2x² + 7x + 3 = 0，先算 a × c = 6，再找两个乘积为 6、和为 7 的数，即 6 和 1。把中间项拆开，再分组提取公因式。
2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)
−3. Always expand your factors mentally to check that you recover the original expression.
所以 x = −1 ÷ 2 = −0.5 或 x = −3。一定要在脑中把因式乘开检验，看是否回到原来的式子。
If the quadratic does not factorise with nice integers, do not force it. Use the quadratic formula instead — that is what it is for.
如果二次式无法用漂亮的整数因式分解，不要硬凑，改用求根公式——公式正是为此而存在的。
4. Common Factors and the Difference of Two Squares | 公因式与平方差
Two special cases appear constantly in IGCSE papers and can be solved in seconds. The first is a common factor: if every term shares a factor, take it out first, and remember that one root is then zero.
有两种特殊情况在 IGCSE 试卷中频繁出现，几秒钟就能解出。第一种是提取公因式：如果每一项都有公共因式，先提取出来，并记住此时有一个根是零。
x² − 4x = 0 → x(x − 4) = 0 → x = 0 or x = 4
s a² − b² = (a + b)(a − b). This pattern works for numbers as well as for algebraic terms.
第二种是平方差，它总可以分解为 a² − b² = (a + b)(a − b)。这个模式对数字和代数式都成立。
x² − 9 = 0 → (x + 3)(x − 3) = 0 → x = 3 or x = −3
= ±2.5. Exams often reward recognising this pattern rather than reaching for the formula.
像 4x² − 25 = 0 这样的方程给出 (2x + 5)(2x − 5) = 0，所以 x = ±2.5。考试中，能识别出这个模式往往比直接套公式更受青睐（也更省时间）。
Watch out for a very common exam trap: the sum of two squares, such as x² + 9, does not factorise over the real numbers. Only a difference of squares splits into two real linear factors.
要注意一个常见的考试陷阱：平方和，例如 x² + 9，在实数范围内不能分解。只有“平方差”才能分解成两个实系数一次因式。
5. Solving by Completing the Square | 配方法求解
you then solve by taking square roots. It works for every quadratic, including ones that do not factorise, and it also reveals the turning point of the graph.
配方法把二次式改写成 (x + p)² = q 的形式，然后通过开平方求解。它对所有二次方程都适用，包括无法因式分解的，同时还能直接看出图像的顶点。
To complete the square for x² + bx, halve the coefficient of x and write (x + b ÷ 2)², then subtract (b ÷ 2)² to keep the expression balanced.
对 x² + bx 配方的方法是：把 x 的系数除以 2，写成 (x + b ÷ 2)²，再减去 (b ÷ 2)²，以保持式子的值不变。
x² + bx + c = (x + b ÷ 2)² − (b ÷ 2)² + c
, so 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, which is approximately x = −0.172 or x = −5.828.
例：解 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。开平方得 x + 3 = ±√8，所以 x = −3 ± 2√2，约等于 x = −0.172 或 x = −5.828。
x² − 12x + 7 = 0, divide by 2 to get x² − 6x + 3.5 = 0, then complete the square as usual.
当 a 不是 1 时，先把整个方程除以 a。对于 2x² − 12x + 7 = 0，两边除以 2 得 x² − 6x + 3.5 = 0，再照常配方。
metry x = h. For y = x² + 6x + 1, the vertex is (−3, −8) and the minimum value of y is −8.
配方形式还能给出顶点：y = a(x − h)² + k 的顶点是 (h, k)，对称轴是 x = h。对于 y = x² + 6x + 1，顶点为 (−3, −8)，y 的最小值为 −8。
6. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation, whether or not it factorises. It is printed on most IGCSE formula sheets, but you must still substitute correctly and show your working.
求根公式能解任何一元二次方程，无论它能否因式分解。大多数 IGCSE 公式表上都会给出这个公式，但你仍需正确代入并写出过程。
x = (−b ± √b² − 4ac) ÷ (2a)
e x² + 4x − 1 = 0. Here a = 1, b = 4, c = −1, so b² − 4ac = 16 − (−4) = 20. Then x = (−4 ± √20) ÷ 2 = (−4 ± 2√5) ÷ 2 = −2 ± √5, giving x ≈ 0.236 or x ≈ −4.236.
例：解 x² + 4x − 1 = 0。此处 a = 1，b = 4，c = −1，所以 b² − 4ac = 16 − (−4) = 20。于是 x = (−4 ± √20) ÷ 2 = (−4 ± 2√5) ÷ 2 = −2 ± √5，得到 x ≈ 0.236 或 x ≈ −4.236。
ulation under the square root in brackets, and never drop the ± sign. If the question says 'give your answers to 3 significant figures', round only at the very end.
代入时的规范性比速度更重要。先明确写出 a、b、c，再把根号下的整个式子用括号括起来，且绝不能丢掉 ± 号。如果题目要求“答案保留三位有效数字”，只在最后一步取近似。
| Situation 情形 | Best method 最佳方法 |
| Simple integer roots 整数根很“好看” | Factorisation 因式分解 |
| No common factor, a = 1, awkward c 不好分解 | Formula 求根公式 |
| Question asks for vertex or turning point 求顶点 | Completing the square 配方法 |
| Missing linear term, e.g. x² − 16 = 0 缺一次项 | Square root or difference of squares 开平方/平方差 |
7. The Discriminant: How Many Real Roots? | 判别式：有几个实根？
The expression b² − 4ac is called the discriminant, often written Δ. It tells you how many real solutions the equation has before you do any solving, and it is the quickest way to answer questions about tangency and intersection.
式子 b² − 4ac 称为判别式，通常记作 Δ。它能在你真正求解之前就告诉你方程有几个实数解，是判断相切与相交问题最快的方法。
Δ = b² − 4ac
- If Δ > 0, there are two distinct real roots — 有两个不相等实根，图像与 x 轴有两个交点。
- If Δ = 0, there is one repeated real root — 有一个重根（两个相等实根），图像与 x 轴相切。
- If Δ < 0, there are no real roots — 没有实根，图像与 x 轴不相交。
no real roots. For x² − 4x + 4 = 0, Δ = 16 − 16 = 0, so there is a repeated root at x = 2.
例：对于 x² − 4x + 5 = 0，Δ = 16 − 20 = −4，所以没有实根。对于 x² − 4x + 4 = 0，Δ = 16 − 16 = 0，所以在 x = 2 处有一个重根。
The discriminant is also used in 'find the value of k' the two equations equal and requiring Δ = 0 produces an equation in k that you can solve.
判别式也常用于“求 k 的值”这类题。如果一条直线与曲线相切，把两个方程联立后令 Δ = 0，就得到一个关于 k 的方程，解出即可。
Note that the discriminant only tells you about real roots. Some Extended questions mention complex roots; at IGCSE you simply state that there are no real solutions.
注意：判别式只说明实根的情况。某些扩展题会提到复根，但在 IGCSE 阶段，你只需说明“没有实数解”即可。
8. Roots, x-Intercepts and Graphs | 根、x 轴截距与图像
+ bx + c. This is why solving an equation and sketching a graph are two views of exactly the same mathematics.
ax² + bx + c = 0 的每一个实数解，都是抛物线 y = ax² + bx + c 上的一个 x 轴截距。这正是“解方程”和“画图像”其实是同一件事的两种视角的原因。
The shape of the parabola depends on a: if a > 0 the curve opens upwards and has a minimum point, while if a < 0 it opens downwards and has a maximum point. The larger the value of |a|, the narrower the curve.
抛物线的形状取决于 a：若 a > 0，曲线开口向上，有最小值点；若 a < 0，开口向下，有最大值点。|a| 越大，曲线越“窄”。
Axis of symmetry: x = −b ÷ (2a) Vertex: (−b ÷ 2a, c − b² ÷ 4a)
ays (0, c). Combining the y-intercept, the x-intercepts and the vertex gives you everything needed for an accurate sketch.
y = ax² + bx + c 的 y 轴截距总是 (0, c)。把 y 轴截距、x 轴截距和顶点结合起来，就足以画出一幅准确的草图。
crosses the x-axis at 1.4 and −2.6, then the equation has solutions x ≈ 1.4 and x ≈ −2.6.
图像题常要求你借助草图解不等式，或读出近似根。如果曲线在 1.4 和 −2.6 处与 x 轴相交，那么方程的解约为 x ≈ 1.4 和 x ≈ −2.6。
9. Forming Equations from Word Problems | 由应用题列方程
Many IGCSE questions hide a quadratic inside a word problem about areas, consecutive numbers, speeds or ages. Your job is to define a variable, translate the information into an equation, then solve it.
许多 IGCSE 题目把二次方程隐藏在面积、连续整数、速度或年龄的应用题里。你的任务是设未知数、把条件翻译成方程，然后求解。
3x − 40 = 0. Factorising gives (x + 8)(x − 5) = 0, so x = 5 (rejecting x = −8 because a length cannot be negative). The rectangle is 5 cm by 8 cm.
例：一个长方形比宽长 3 cm，面积为 40 cm²。设宽为 x cm，则长为 (x + 3) cm，于是 x(x + 3) = 40，即 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，所以 x = 5（舍去 x = −8，因为长度不能为负）。这个长方形为 5 cm × 8 cm。
Always check whether both roots are sensible in context. Negative lengths, negative ages and negative times must be rejected, and examiners give credit for that written reasoning.
一定要检验两个根在实际情境中是否合理。负的长度、负的年龄、负的时间都必须舍去，阅卷老师会给这部分文字推理加分。
For consecutive-integer problems, write the integers as n and n + 1 (or n and n + 2 for consecutive even or odd integers) and build the equation from the given sum or product.
对于连续整数问题，把整数写成 n 和 n + 1（连续偶数或奇数则写成 n 和 n + 2），再根据题目给出的和或积列方程。
10. Quadratics in Simultaneous Equations | 联立方程中的二次方程
When a line meets a curve, you solve one linear and one quadratic equation simultaneously. The standard method is substitution: rearrange the linear equation for y (or x), substitute into the quadratic, and solve the resulting quadratic equation.
当直线与曲线相交时，需要联立一个一次方程和一个二次方程。标准方法是代入法：先把一次方程整理成 y（或 x）的表达式，代入二次方程，解出得到的二次方程。
= 2 with y = 3, or x = −1 with y = 0. The two intersection points are (2, 3) and (−1, 0).
例：联立求解 y = x + 1 与 y = x² − 1。代入得 x² − 1 = x + 1，即 x² − x − 2 = 0，分解为 (x − 2)(x + 1) = 0。因此 x = 2 时 y = 3，或 x = −1 时 y = 0。两个交点分别是 (2,
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