📚 Solving Quadratic Equations for IGCSE | IGCSE 二次方程求解
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from pure algebraic manipulation to modelling real-life situations. This revision guide covers the three main solution methods - factorisation, completing the square and the quadratic formula - and shows you how to choose the most efficient one under exam conditions. It also explains the discriminant, the link between equations and graphs, word-problem technique and the mistakes that cost marks most often.
二次方程在 IGCSE 数学大纲中随处可见，从纯代数变形到现实情境建模都离不开它。本复习指南涵盖三种主要解法——因式分解法、配方法和二次公式法，并说明如何在考试条件下选择最有效的方法。文中还会讲解判别式、方程与图像的关联、应用题技巧，以及最常见、最容易丢分的错误。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation in which the highest power of the variable is 2. It can contain an x² term, an x term and a constant term, but it cannot contain any higher power such as x³, a negative power such as x⁻¹, or a fractional power such as x½.
二次方程是一个多项式方程，其中变量的最高次数为 2。它可以包含 x² 项、x 项和常数项，但不能包含更高次幂（如 x³）、负次幂（如 x⁻¹）或分数次幂（如 x½）。
The general form, also called standard form, is written below, where a, b and c are constants and a ≠ 0. If a were equal to 0, the equation would become linear rather than quadratic.
二次方程的一般形式也叫标准形式，写法如下，其中 a、b、c 是常数且 a ≠ 0。如果 a 等于 0，方程就变成一次方程而不是二次方程。
ax² + bx + c = 0, a ≠ 0
The number a is called the quadratic coefficient, b the linear coefficient and c the constant term. Each has a specific job: a controls whether the parabola opens upwards or downwards, b affects the position of the axis of symmetry, and c gives the y-intercept of the graph.
a 称为二次项系数，b 称为一次项系数，c 称为常数项。它们各有分工：a 决定抛物线开口向上还是向下，b 影响对称轴的位置，c 则直接给出图像的 y 轴截距。
The solutions of a quadratic equation are also called roots, zeros or x-intercepts. In IGCSE questions you may be asked for exact roots in surd form, approximate roots rounded to a given number of decimal places or significant figures, or the condition under which real roots exist at all.
二次方程的解也叫做根、零点或 x 轴截距。在 IGCSE 考题中，你可能会被要求给出根式形式的精确根、按要求保留小数位或有效数字的近似根，或者判断实数根存在的条件。
2. Standard Form and Rearranging | 标准形式与整理方程
erm on one side of the equals sign and leaving 0 on the other side. Skipping this step is the single most common reason why factorisation appears to fail.
在求解任何二次方程之前，都应先把它改写成标准形式 ax² + bx + c = 0。这意味着把所有的项移到等号一侧，让等号另一侧剩下 0。跳过这一步，是因式分解"看起来分解不出来"的最常见原因。
ting 6x from both sides. The coefficients are then easy to read off: a = 1, b = −6, c = 5.
例如，x² + 5 = 6x 在两边同时减去 6x 后变成 x² − 6x + 5 = 0。此时系数一目了然：a = 1，b = −6，c = 5。
d form is x² + x − 20 = 0. Notice that you must expand and move the 14 across - you must not simply set each bracket equal to 14.
如果题目中含有括号，要先展开再整理。例如 (x + 3)(x − 2) = 14 展开得 x² + x − 6 = 14，两边减去 14 后标准形式为 x² + x − 20 = 0。注意必须先展开并把 14 移项，绝不能直接把每个括号等于 14。
If fractions appear, multiply every term by the common denominator to clear them. If a is negative, many students multiply the whole equation by −1 to make a positive, which usually makes factorisation much easier; the roots are unchanged by this step.
如果出现分数，就用所有分母的公倍数乘以每一项来去分母。如果 a 为负数，很多同学会把整个方程乘以 −1 使 a 变为正数，这通常会让因式分解容易得多；这一步不会改变方程的根。
x² + 5 = 6x → x² − 6x + 5 = 0
3. Method 1: Factorisation | 方法一：因式分解
, then at least one of them must be zero. So if (x + p)(x + q) = 0, then x = −p or x = −q.
因式分解是展开括号的逆过程，其依据是零乘积性质：如果两个式子的乘积为 0，那么其中至少有一个为 0。因此若 (x + p)(x + q) = 0，则 x = −p 或 x = −q。
1, find two numbers that multiply to give c and add to give b. Those two numbers become p and q in the factorised form (x + p)(x + q).
对于 a = 1 的简单情形 x² + bx + c = 0，只需找出两个数，使它们的乘积为 c、和为 b。这两个数就是因式形式 (x + p)(x + q) 中的 p 和 q。
to +5 and adding to −6 are −1 and −5, so (x − 1)(x − 5) = 0 and therefore x = 1 or x = 5.
例题：x² − 6x + 5 = 0。乘积为 +5、和为 −6 的两个数是 −1 和 −5，所以 (x − 1)(x − 5) = 0，由此得到 x = 1 或 x = 5。
+ 1)(x + 3) = 0, giving x = −½ or x = −3. Always expand mentally to check the middle term.
当 a ≠ 1 时，寻找形如 (mx + p)(nx + q) 的一对括号，其中 mn = a、pq = c。例题：2x² + 7x + 3 = 0 分解为 (2x + 1)(x + 3) = 0，得到 x = −½ 或 x = −3。一定要在脑中展开检验一次中项。
| Pattern | 类型 | Factorised form | 因式形式 | Roots | 根 |
| x² − 16 = 0 | (x − 4)(x + 4) = 0 | x = ±4 |
| x² + 6x + 9 = 0 | (x + 3)² = 0 | x = −3 (repeated | 重根) |
| x² − 5x = 0 | x(x − 5) = 0 | x = 0 or x = 5 |
quare gives a repeated root, which is exactly the case where the discriminant equals zero.
有两个特例值得记牢：平方差公式 x² − k² = (x − k)(x + k)，以及完全平方公式 x² + 2kx + k² = (x + k)²。完全平方会产生重根，而重根正对应判别式等于 0 的情形。
4. Method 2: Completing the Square | 方法二：配方法
Completing the square rewrites a quadratic so that the variable appears only once, inside a squared bracket. It works for every quadratic, even when factorisation is impossible, and it also gives the vertex of the graph directly.
配方法把二次式改写成变量只出现一次、且位于一个平方括号内的形式。它对任何二次方程都适用，即使无法因式分解也能用，而且还能直接给出图像的顶点。
÷ 2, so half of the coefficient of x goes inside the bracket and its square is subtracted.
关键恒等式来自展开 (x + p)² = x² + 2px + p²。把它与 x² + bx 对比可知 p = b ÷ 2，因此 x 的系数的一半放进括号里，而它的平方要在外面减掉。
x² + bx + c = (x + b ÷ 2)² − (b ÷ 2)² + c
)² − 9 + 1 = 0, that is (x + 3)² = 8. Taking the square root of both sides gives x + 3 = ±√8, so x = −3 ± 2√2, which is about −0.17 or −5.83.
例题：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。两边开平方得 x + 3 = ±√8，所以 x = −3 ± 2√2，约为 −0.17 或 −5.83。
his form is called completed-square form or vertex form, and the vertex of the parabola is at (−2, −5).
当 a ≠ 1 时，先从前两项中提出公因数 a。例如 2x² + 8x + 3 = 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 5。这种形式叫做配方法形式或顶点式，抛物线的顶点就在 (−2, −5)。
Exam questions often phrase this as "express x² + 8x + 11 in the form (x + p)² + q", or ask for the minimum value of a quadratic expression, which is simply the constant term when the squared bracket is at its smallest. Completing the square is also the standard way to prove that a quadratic is always positive.
考题常把它表述为"将 x² + 8x + 11 写成 (x + p)² + q 的形式"，或询问二次式的最小值——当平方括号取最小值时，最值就是那个常数项。配方法也是证明某个二次式恒为正的标准手段。
5. Method 3: The Quadratic Formula | 方法三：二次公式
The quadratic formula solves any quadratic equation in standard form, and it is the method to use when factorisation is not obvious or when the coefficients are decimals. You should memorise it exactly.
二次公式可以求解任何标准形式的二次方程，当因式分解不明显或系数是小数时，它就是首选方法。这个公式必须一字不差地背下来。
x = (−b ± √b² − 4ac) ÷ 2a
4x − 3 = 0. Here a = 2, b = −4 and c = −3, so b² − 4ac = 16 + 24 = 40. Therefore x = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = 1 ± √10 ÷ 2, which is about 2.58 or −0.58 to 2 decimal places.
例题：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3，所以 b² − 4ac = 16 + 24 = 40。因此 x = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = 1 ± √10 ÷ 2，约为 2.58 或 −0.58（保留两位小数）。
gative value of b, always put brackets around it, as in −(−4) = +4. Write down the values of a, b and c before substituting - this earns method marks even if the arithmetic later goes wrong.
整个公式要当作一个整体来处理：除以 2a 的是整个分子，而不只是根号部分。代入负的 b 时一定要加括号，例如 −(−4) = +4。代入前先把 a、b、c 的值写出来——即使后面运算出错，这一步也能拿到方法分。
IGCSE questions usually state the required accuracy, such as "give your answers correct to 3 significant figures". Surd answers should be simplified, for example √40 = 2√10, and fractions should be cancelled to their lowest terms.
IGCSE 题目通常会给出精度要求，例如"答案保留三位有效数字"。根式答案要化简，比如 √40 = 2√10；分数要约到最简。
6. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression under the square root in the quadratic formula is called the discriminant, usually written as D or Δ. Its value alone tells you how many real roots the equation has, without solving it.
二次公式中根号下的表达式称为判别式，通常记作 D 或 Δ。仅凭它的值就能判断方程有多少个实根，无需真正求解。
Δ = b² − 4ac
| Value of Δ | 判别式的值 | Nature of roots | 根的性质 | Graph | 图像 |
| Δ > 0 | Two distinct real roots | 两个不同实根 | Crosses the x-axis twice |
| Δ = 0 | One repeated real root | 一个重实根 | Touches the x-axis once |
| Δ < 0 | No real roots | 没有实根 | Does not meet the x-axis |
= 0 has equal roots. Equal roots mean Δ = 0, so k² − 36 = 0 and therefore k = 6 or k = −6.
典型考题是求某个未知系数的值。例如，若 x² + kx + 9 = 0 有两个相等的根，求 k。相等的根意味着 Δ = 0，因此 k² − 36 = 0，得到 k = 6 或 k = −6。
Note the difference between "no real roots" and "no solutions" quation still has solutions, they are simply not real numbers, so at IGCSE level you write "no real roots" tion mentions tangency to a curve or a line touching a curve, that is also the Δ = 0 case.
要注意"没有实根"和"没有解"的区别：当 Δ < 0 时方程仍有解，只是不是实数解，所以在 IGCSE 阶段应写"没有实根"。如果题目提到直线与曲线相切、或曲线与 x 轴相切，那同样属于 Δ = 0 的情形。
7. Choosing the Best Method | 如何选择最优解法
All three methods give the same answers, so the choice is about speed and accuracy. Under exam pressure, a few seconds spent deciding the method saves minutes of calculation.
三种方法得到的答案完全相同，所以选择的关键在于速度与准确率。在考试压力下，花几秒钟判断用哪种方法，能省下几分钟的计算时间。
- Use factorisation when a, b and c are small integers and the roots look like simple fractions or whole numbers. | 当 a、b、c 都是较小的整数、根看起来是简单的整数或分数时，用因式分解法。
- Use the quadratic formula when factorisation fails, when the coefficients are decimals, or when the question says "give your answers to 3 significant figures". | 当因式分解不成功、系数为小数，或题目要求"保留三位有效数字"时，用二次公式法。
- Use completing the square when the question asks for the vertex, the maximum or minimum value, or requires the form (x + p)² + q. | 当题目要求顶点、最大值或最小值，或要求写成 (x + p)² + q 的形式时，用配方法。
- Use the discriminant when the question only asks how many roots there are, or asks for a condition on an unknown coefficient. | 当题目只问根的个数，或要求未知系数满足的条件时，用判别式。
A quick mental test helps: if you cannot see a factor pair within about ten seconds, switch to the formula. Do not spend three minutes hunting for a factorisation that may not exist.
一个快速的心理检验很有用：如果十秒内看不出因数对，就改用公式。不要花三分钟去寻找一个可能根本不存在的因式分解。
Also check your exam board's calculator rules. Some papers are non-calculator, in which case surd manipulation and simple factorisation matter much more. On calculator papers, you can still earn accuracy marks by writing the substituted values before pressing the buttons.
同时要确认你所在考试局的計算器规定。有些试卷不允许使用计算器，此时根式化简和简单因式分解就格外重要。在允许使用计算器的试卷上，先写出
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