📚 Mastering Quadratic Equations: Factorising, Completing the Square, and the Quadratic Formula | 掌握二次方程：因式分解、配方法与求根公式
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebra and graphs to problem solving in geometry and physics. Being able to solve them confidently using factorising, completing the square, and the quadratic formula is a core skill that examiners test every year. This article breaks down the methods step by step, highlights common mistakes, and provides revision-style examples so that you can choose the fastest method in the exam hall.
二次方程贯穿 IGCSE 数学大纲，从代数、图像到几何与物理中的问题求解都有出现。能够熟练运用因式分解、配方法和求根公式解二次方程，是每年考官都会考查的核心技能。本文逐步拆解这些方法，指出常见错误，并提供复习型例题，帮助你在考场上迅速选择最快的解法。
1. Recognising a Quadratic Equation | 认识二次方程
re real numbers and a ≠ 0. The condition a ≠ 0 is essential; if a were 0, the equation would collapse into a linear equation and would no longer be quadratic.
一元二次方程是任何可以整理为标准形式 a x² + b x + c = 0 的方程，其中 a、b、c 为实数且 a ≠ 0。条件 a ≠ 0 至关重要；如果 a = 0，方程就退化为一次方程，不再属于二次方程。
quadratic because it can be rearranged to x² − 5x + 6 = 0, giving a = 1, b = −5 and c = 6.
例如，x² − 4x + 3 = 0 是二次方程，其中 a = 1、b = −4、c = 3。而 x² = 5x − 6 同样是二次方程，因为它可以整理为 x² − 5x + 6 = 0，此时 a = 1、b = −5、c = 6。
efore choosing a solution method, because factorising, completing the square and the formula all assume the equation is written as (expression) = 0. The solutions are also called the roots or the zeros of the quadratic expression.
在 IGCSE 题目中，二次方程可能隐藏在文字题、分式或公式之中。务必先将其化为标准形式，再选择解法，因为因式分解、配方法和求根公式都以"某表达式 = 0"为前提。方程的解也称为该二次表达式的根或零点。
2. Solving by Factorising | 因式分解法
wo numbers whose product is c and whose sum is b. Write them into two brackets and read off the roots with the sign reversed.
当二次式能够干净地拆成两个一次因式时，因式分解法最快。对于 a = 1 的简单二次式，需要找到两个数，它们的乘积为 c、和为 b。把这两个数写进两个括号，取相反数即可读出方程的根。
d sum 5; they are 2 and 3. Thus x² + 5x + 6 = (x + 2)(x + 3) = 0, giving x = −2 or x = −3.
例题：解 x² + 5x + 6 = 0。需要两个数，乘积为 6、和为 5；它们是 2 和 3。因此 x² + 5x + 6 = (x + 2)(x + 3) = 0，得到 x = −2 或 x = −3。
When a ≠ 1, use the 'ac method' 7. The numbers are 6 and 1, so 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 a ≠ 1 时，可使用 ac 法：找出乘积为 a c、和为 b 的两个数，拆开中间项，再分组提取公因式。例题：对于 2x² + 7x + 3，a c = 6，b = 7。所需两个数为 6 和 1，因此 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
- Difference of two squares: x² − 9 = (x − 3)(x + 3); 4x² − 25 = (2x − 5)(2x + 5).
- Perfect square: x² − 10x + 25 = (x − 5)², which repeats one root twice.
- Always check by expanding: (x + 2)(x + 3) = x² + 5x + 6 ✓.
- 平方差：x² − 9 = (x − 3)(x + 3)；4x² − 25 = (2x − 5)(2x + 5)。
- 完全平方：x² − 10x + 25 = (x − 5)²，此时一根重复出现两次。
- 务必展开检验：(x + 2)(x + 3) = x² + 5x + 6 ✓。
+ 5x + 6 = 6, do not set each bracket to 6; rearrange to x² + 5x = 0 and factorise first.
常见警告：只有当乘积等于零时，因式分解才能直接给出根。如果方程是 x² + 5x + 6 = 6，不要把每个括号分别等于 6；应先整理为 x² + 5x = 0 再因式分解。
3. The Zero Product Property | 零乘积性质
A × B = 0 then A = 0 or B = 0. This property applies only to zero, which is why standard form is compulsory.
因式分解之所以成立，依据的是零乘积性质：若两个数的乘积为零，则其中至少有一个为零。用符号表示，若 A × B = 0，则 A = 0 或 B = 0。该性质只适用于零，这正是必须化为标准形式的原因。
n factorise as x(x − 5) = 0, so x = 0 or x = 5. Both roots are genuine and must be listed.
例题：解 x² = 5x。整理为 x² − 5x = 0，再因式分解为 x(x − 5) = 0，因此 x = 0 或 x = 5。两个根都是真实解，必须全部写出。
r three marks. Never divide an equation by a variable unless you have separately proved that the variable cannot be zero.
这里的经典陷阱是两边同时除以 x。这一步会悄悄删掉根 x = 0，通常要丢掉两到三分。除非你已经单独证明该变量不可能为零，否则永远不要用变量去除方程。
Each bracket gives exactly one root, and an equation of degree n has at most n real roots.
同样的思路可推广到三个因式：若 (x − 1)(x + 2)(x − 4) = 0，则 x = 1、x = −2 或 x = 4。每个括号恰好给出一个根，n 次方程最多有 n 个实根。
4. Completing the Square | 配方法
s involving surds rather than rounded decimals. The key identity for a = 1 is shown below.
配方法把二次式改写为 (x + p)² + q 的形式。它总是可行，即使因式分解失败也能用，而且给出的是含根号的精确答案，而不是四舍五入的小数。当 a = 1 时的关键恒等式如下。
x² + b x + c = (x + b/2)² − (b/2)² + c
ant. Example: x² + 6x + 1 = 0 becomes (x + 3)² − 9 + 1 = 0, so (x + 3)² = 8 and x = −3 ± 2√2.
用文字表述：取 x 系数的一半放进括号，再减去它的平方，最后保留原来的常数项。例题：x² + 6x + 1 = 0 变为 (x + 3)² − 9 + 1 = 0，所以 (x + 3)² = 8，得 x = −3 ± 2√2。
ply back. Example: 2x² − 8x + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3.
当 a ≠ 1 时，先把 a 从含 x 的项中提出（常数项不要提出去），在括号内配方，再乘回去。例题：2x² − 8x + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3。
- Always halve the coefficient of x, not the constant term.
- Subtract (b/2)² inside the bracket; forgetting the subtraction is the most frequent error.
- Finish by taking the square root of both sides and remembering the ± sign.
- 永远是对 x 的系数取一半，而不是对常数项取一半。
- 要在括号内减去 (b/2)²；忘记减掉是最常见的错误。
- 最后两边开平方，并记得写出 ± 号。
5. Deriving the Quadratic Formula | 推导求根公式
ing the derivation once makes the formula far easier to remember and to trust in the exam.
求根公式并非凭空而来，它就是把配方法用在一般式 a x² + b x + c = 0 上。跟着推导一遍，公式会变得容易记忆，用起来也更有信心。
+ b ÷ 2a)² − (b ÷ 2a)² + c/a = 0. Step 3: move the constant terms to the right-hand side.
第一步：每项除以 a，得 x² + (b/a)x + c/a = 0。第二步：把 x 的系数取一半，写成 (x + b ÷ 2a)² − (b ÷ 2a)² + c/a = 0。第三步：把常数项移到等号右边。
(x + b ÷ 2a)² = (b² − 4ac) ÷ 4a²
sitive number has two square roots. Step 5: subtract b ÷ 2a from both sides and combine the two fractions over the common denominator 2a.
第四步：两边开平方，并记住 ± 号，因为正数有两个平方根。第五步：两边同时减去 b ÷ 2a，并把两个分式通分到公分母 2a 上。
x = (−b ± √b² − 4ac) ÷ 2a
Notice that the denominator is 2a, the whole expression b² − 4ac sits under the square root, and the minus sign in front of b applies before anything else. Reading the formula aloud in that order prevents most sign slips.
请注意：分母是 2a，整个 b² − 4ac 都在根号内，而 b 前面的负号要最先处理。按这个顺序把公式读出来，可以避免大部分符号错误。
6. Using the Quadratic Formula | 使用求根公式
The formula works for every quadratic, so it is the reliable fallback when factorising fails and when the question asks for answers to three significant figures. Write down a, b and c with their signs first, then substitute carefully.
求根公式适用于任何二次方程，因此当因式分解失效、或题目要求答案保留三位有效数字时，它是最可靠的兜底方法。先写出带符号的 a、b、c，再代入计算。
5) ÷ 4 = −3. Factorising the same expression gives (2x + 1)(x + 3), confirming both roots.
例题：解 2x² + 7x + 3 = 0。此处 a = 2、b = 7、c = 3，所以 b² − 4ac = 49 − 24 = 25。于是 x = (−7 ± 5) ÷ 4，得 x = (−7 + 5) ÷ 4 = −½ 或 x = (−7 − 5) ÷ 4 = −3。对同一式子因式分解得 (2x + 1)(x + 3)，验证了两个根。
solve x² − 4x + 1 = 0. Here a = 1, b = −4, c = 1, so b² − 4ac = 16 − 4 = 12. Then x = (4 ± √12) ÷ 2 = (4 ± 2√3) ÷ 2 = 2 ± √3, so x ≈ 3.73 or x ≈ 0.27 to three significant figures.
含无理根的例题：解 x² − 4x + 1 = 0。此处 a = 1、b = −4、c = 1，所以 b² − 4ac = 16 − 4 = 12。于是 x = (4 ± √12) ÷ 2 = (4 ± 2√3) ÷ 2 = 2 ± √3，即 x ≈ 3.73 或 x ≈ 0.27（保留三位有效数字）。
- Substitute with brackets: for b = −4, compute −(−4), not −4.
- Divide the whole numerator by 2a, not just the square root part.
- Give the exact surd form first, then the decimal if the question asks for it.
- 代入时加括号：若 b = −4，要算 −(−4)，而不是 −4。
- 整个分子都要除以 2a，而不是只除根号那一部分。
- 先写精确的根式形式，题目要求时再写小数。
7. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression under the square root, b² − 4ac, is called the discriminant and is usually written as Δ. It tells you how many real roots the equation has before you do any further work, and it links algebra directly to the graph.
根号下的表达式 b² − 4ac 称为判别式，通常记作 Δ。它能在继续计算之前告诉你方程有几个实根，并把代数与图像直接联系起来。
| Discriminant | Nature of roots | Graph of y = a x² + b x + c |
| Δ > 0 | Two distinct real roots | Curve crosses the x-axis twice |
| Δ = 0 | One repeated real root | Curve touches the x-axis |
| Δ < 0 | No real roots | Curve never meets the x-axis |
| 判别式 | 根的性质 | y = a x² + b x + c 的图像 |
| Δ > 0 | 两个不同实根 | 抛物线与 x 轴相交两次 |
| Δ = 0 | 一个重根 | 抛物线与 x 轴相切 |
| Δ < 0 | 无实根 | 抛物线与 x 轴不相交 |
he question specifies two distinct roots, write Δ > 0 and solve the inequality instead.
例题：要使 x² + kx + 9 = 0 有相等的根，需要 Δ = 0，即 k² − 36 = 0，得 k = 6 或 k = −6。若题目要求两个不同实根，则应写 Δ > 0 并解不等式。
One extra detail: if a, b, c are all integers and Δ is a perfect square, the roots are rational, which also means the expression factorises with integer brackets. If Δ is not a perfect square, the roots are irrational and factorising over the integers is impossible.
补充一点：若 a、b、c 都是整数且 Δ 是完全平方数，则根为有理数，也意味着该式可以用整数系数因式分解。若 Δ 不是完全平方数，则根为无理数，无法在整数范围内因式分解。
8. Completing the Square: Vertex, Range and Graph | 配方法的应用：顶点、值域与图像
ue of the expression is q and it occurs when x = −p. For a < 0 the same bracket gives a maximum value of q.
把二次式改写为 a(x + p)² + q，可立刻看出图像的顶点。当 a > 0 时，括号内的平方永不为负，因此表达式的最小值为 q，且在 x = −p 处取得。当 a < 0 时，同样的括号结构给出最大值 q。
o the vertex is (3, 2) and the range of y is y ≥ 2. The line x = 3 is the axis of symmetry
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图