📚 IGCSE Maths Topic 2: Solving Quadratic Equations by Factorising, Completing the Square, and Quadratic Formula | IGCSE 数学主题二：解二次方程——因式分解、配方法与求根公式
c 6 (graphs), and the same three techniques keep coming back: factorising, completing the square and the quadratic formula. In this article we work carefully through all three methods for equations written as ax² + bx + c = 0, show how the discriminant predicts how many real roots a quadratic has, and highlight the mistakes that cost marks most often in the exam.
二次方程是 IGCSE 数学的核心内容，出现在主题 2（代数）、主题 3（方程、公式与恒等式）和主题 6（图像）中，而三种方法会反复出现：因式分解、配方法与求根公式。本文将完整讲解如何解形如 ax² + bx + c = 0 的方程，说明判别式如何预判实根个数，并总结最容易丢分的常见错误。
1. What Is a Quadratic Equation? | 什么是二次方程？
er of the unknown x is 2, which is exactly what makes the equation quadratic. If a = 0 the equation collapses into a linear equation, so a must never be zero for a genuine quadratic.
二次方程是指可以整理成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数且 a ≠ 0。未知数 x 的最高次数是 2，这正是它被称为二次方程的原因。如果 a = 0，方程就退化为一次方程，因此真正的二次方程必须满足 a 不为零。
ax² + bx + c = 0, a ≠ 0
rs at all. A root is a value of x that makes the expression equal to zero, and on a graph those roots are the x-intercepts of the parabola y = ax² + bx + c.
每个二次方程都有两个根（解），只不过有时两根相同，有时两根不是实数。所谓根，就是使表达式等于零的 x 值；在图像上，根就是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。
2. Standard Form and Coefficients | 标准形式与系数
t be rewritten as x² - 3x - 2 = 0, giving a = 1, b = -3 and c = -2. Skipping this step is the single most common reason students lose marks.
在求解之前，一定要把方程整理成一边为零、各项按 x 的降幂排列的形式。例如 3x + 2 = x² 必须改写为 x² - 3x - 2 = 0，此时 a = 1，b = -3，c = -2。跳过这一步是学生丢分最常见的原因。
The coefficient a is the number multiplying x²; b is the number multiplying x; c is the constant term (including its sign). Identifying a, b and c correctly is essential before using the quadratic formula, and misreading the sign of b or c will ruin the whole calculation.
系数 a 是 x² 前面的数；b 是 x 前面的数；c 是常数项（连同正负号）。在使用求根公式之前，正确识别 a、b、c 至关重要；看错 b 或 c 的正负号会让整道题的计算全错。
| Equation 方程 | a | b | c |
| x² - 5x + 6 = 0 | 1 | -5 | 6 |
| 2x² + 7x + 3 = 0 | 2 | 7 | 3 |
| x² = 9 | 1 | 0 | -9 |
| 3x² - 4x = 0 | 3 | -4 | 0 |
Note that when a term is missing, its coefficient is zero, not absent. Writing the zero explicitly in your working keeps the arithmetic honest and makes the substitution into the formula much safer.
注意，当某一项缺失时，它的系数是 0，而不是"不存在"。在演算中显式写出这个 0，可以让计算更可靠，代入公式时也不容易出错。
3. Solving by Factorising: The Basic Idea | 因式分解法：基本思想
ates that if A × B = 0 then A = 0 or B = 0. So we set each bracket equal to zero and solve two simple linear equations.
如果二次式能分解为两个一次因式的乘积，例如 (px + q)(rx + s) = 0，那么乘积为零。零乘积性质指出：若 A × B = 0，则 A = 0 或 B = 0。因此我们让每个括号分别等于零，再解两个简单的一次方程。
(x + 2)(x - 5) = 0 ⇒ x = -2 or x = 5
s straight off the brackets without flipping the sign lose the mark almost every time. Factorising is the fastest method when the roots are integers or simple fractions, and examiners often expect it when the numbers are friendly.
注意符号要反过来：括号 (x + 2) 给出的根是 x = -2，而 (x - 5) 给出的根是 x = 5。直接把括号里的数字当作根而不变号的同学，几乎每次都丢分。当根是整数或简单分数时，因式分解是最快的方法，在数字"好看"的题目中，考官通常也期望用这种方法。
4. Factorising When a = 1 | 首项系数为 1 的因式分解
= b. For x² - 5x + 6, the pair is -2 and -3 because (-2) × (-3) = 6 and (-2) + (-3) = -5.
当 a = 1 时，我们要找两个数，它们的乘积等于 c，和等于 b。于是 x² + bx + c = (x + p)(x + q)，其中 pq = c 且 p + q = b。对于 x² - 5x + 6，这对数是 -2 和 -3，因为 (-2) × (-3) = 6 且 (-2) + (-3) = -5。
x² + bx + c = (x + p)(x + q), pq = c, p + q = b
Use the signs as a guide. If c is positive, both numbers have the same sign (both positive if b is positive, both negative if b is negative). If c is negative, the two numbers have opposite signs, and the larger one takes the sign of b. This single rule removes most of the guesswork.
可以用符号规律来快速判断。若 c 为正，则两数同号（b 为正时两数都正，b 为负时两数都负）。若 c 为负，则两数异号，且绝对值较大的那个数取 b 的符号。掌握这条规律，大部分"试数"的过程就可以省掉。
5x + 6, which matches. Ten seconds of checking prevents a whole question from collapsing.
一定要在脑中把括号展开验算。对于 (x - 2)(x - 3)，展开得 x² - 3x - 2x + 6 = x² - 5x + 6，与题目一致。花十秒检查，可以避免整道题崩盘。
5. Factorising When a ≠ 1 | 首项系数不为 1 的因式分解
When a is not 1, the search is a little wider. Multiply a and c to get the target product, then find two numbers that multiply to ac and add to b. Split the middle term with those two numbers and factorise by grouping. This is known as the ac method.
当 a 不为 1 时，试数的范围更大。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。然后把这个中间项拆成两项，用分组分解法完成因式分解。这就是所谓的 ac 法。
2x² + 7x + 3: ac = 6, and 6 × 1 = 6, 6 + 1 = 7
= -½ and x = -3. Both roots should be checked by substitution, especially fractional ones.
因此 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。令每个括号等于零，得 x = -½ 和 x = -3。两个根（尤其是分数根）都应代回原方程检验。
s look for that common factor before starting the ac method, because it makes everything smaller and easier.
有时表达式先有公因式可提。对于 2x² - 8，先提 2 得 2(x² - 4) = 2(x - 2)(x + 2)，所以根是 x = 2 与 x = -2。在用 ac 法之前一定要先看有没有公因式，这会让后续计算更简单。
6. Completing the Square: The Method | 配方法：方法与步骤
in: x² + bx + c = (x + b/2)² - (b/2)² + c. The expression is now in completed-square form.
配方法把一个二次式改写为"完全平方 + 常数"的形式。对于 x² + bx + c，取 b 的一半、平方，加上再减去：x² + bx + c = (x + b/2)² - (b/2)² + c。此时表达式就变成了配方形式。
x² + bx + c = (x + b/2)² + c - b²/4
(x + 3)² - 8. The minimum value of the expression is therefore -8, occurring when x = -3.
例题：x² + 6x + 1。6 的一半是 3，3² = 9。因此 x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8。所以该表达式的最小值是 -8，在 x = -3 时取得。
) + c, complete the square inside the bracket, then expand the a back. The final form is a(x + b/2a)² + c - b²/4a.
如果 a 不为 1，先把 a 从含 x 的项中提出：ax² + bx + c = a(x² + (b/a)x) + c，在括号内配方，再把 a 乘回去。最终形式为 a(x + b/2a)² + c - b²/4a。
7. Solving Equations by Completing the Square | 用配方法解方程
f both sides, remembering the ± sign. If (x + 3)² - 8 = 0, then (x + 3)² = 8, so x + 3 = ±√8 and x = -3 ± √8.
用配方法解方程时，先把完全平方项单独放好，然后两边开平方，切记保留 ± 号。若 (x + 3)² - 8 = 0，则 (x + 3)² = 8，所以 x + 3 = ±√8，即 x = -3 ± √8。
(x + 3)² = 8 ⇒ x = -3 ± √8 ⇒ x = -3 ± 2√2
√8 simplifies to 2√2, so the exact roots are x = -3 + 2√2 and x = -3 - 2√2, which are approximately 0.17 and -6.17 (2 decimal places). If the question says "leave your answer in surd form", write the exact surd version; if it says "correct to 3 significant figures", round at the very end only.
√8 可化简为 2√2，所以精确根为 x = -3 + 2√2 与 x = -3 - 2√2，约等于 0.17 和 -6.17（保留两位小数）。如果题目要求"以根式形式给出答案"，就写精确的根式；如果要求"保留三位有效数字"，只在最后一步取近似值。
Completing the square is the method of choice when the question asks for an exact answer involving surds, or when the coefficient b is even and the arithmetic stays tidy. When b is odd or the numbers are messy, the quadratic formula is usually quicker.
当题目要求用根式给出精确答案，或 b 为偶数、运算比较整洁时，配方法是最佳选择。当 b 为奇数、数字比较杂乱时，求根公式通常更快。
8. Completing the Square and the Vertex | 配方法与抛物线顶点
q) and the line of symmetry is x = -p. For y = x² + 6x + 1 = (x + 3)² - 8, the vertex is (-3, -8).
配方形式不只是解题工具，还能直接给出抛物线的顶点。若 y = a(x + p)² + q，则顶点（转折点）为 (-p, q)，对称轴为 x = -p。对于 y = x² + 6x + 1 = (x + 3)² - 8，顶点就是 (-3, -8)。
ola opens downwards and the vertex is a maximum. Because a perfect square can never be negative, the smallest value of (x + p)² is 0, which happens at x = -p.
a 的正负决定顶点是最小值还是最大值。若 a > 0，抛物线开口向上，顶点为最小值；若 a < 0，开口向下，顶点为最大值。由于完全平方不可能为负，(x + p)² 的最小值就是 0，此时 x = -p。
y = a(x + p)² + q ⇒ vertex (-p, q), axis of symmetry x = -p
9. The Quadratic Formula | 求根公式
on, whether or not it factorises. For ax² + bx + c = 0 with a ≠ 0, the roots are x = (-b ± √b² - 4ac) / (2a). You should memorise this formula, and you will find it on the formula sheet in some IGCSE papers, but relying on memory is safer and faster.
求根公式可以解任何二次方程，无论它能否因式分解。对于 a ≠ 0 的 ax² + bx + c = 0，其根为 x = (-b ± √b² - 4ac) / (2a)。这个公式必须背下来；部分 IGCSE 试卷的公式表上会给出它，但靠记忆更快也更稳妥。
x = (-b ± √b² - 4ac) / (2a)
Example: solve 2x² + 7x + 3 = 0. Here a = 2, b = 7, c = 3, so b² - 4ac = 49 - 24 = 25 and √25 giving x = -2/4 = -½ or x = -12/4 = -3, which matches the factorising result in Section 5.
例：解 2x² + 7x + 3 = 0。此时 a = 2，b = 7，c = 3，所以 b² - 4ac = 49 - 24 = 25，√25 = 5。于是 x = (-7 ± 5) / 4，得 x = -2/4 = -½ 或 x = -12/4 = -3，与第 5 节因式分解的结果一致。
efore typing them into a calculator. For a = 1, b = -5, c = 6, the calculation is x = (5 ± √25 - 24) / 2 = (5 ± 1) / 2, so x = 3 or x = 2. Writing -b as 5, not -5, is the crucial step.
代入时要格外小心：负值在计算器里要加括号。若 a = 1，b = -5，c = 6，则 x = (5 ± √25 - 24) / 2 = (5 ± 1) / 2，所以 x = 3 或 x = 2。把 -b 写成 5 而不是 -5，是关键的一步。
10. The Discriminant | 判别式
The expression under the square root, b² - 4ac, is called the discriminant and is usually written as Δ. Its value tells you how many real roots the equation has, without solving it. This is a very common exam question, often worth two or three marks on its own.
根号下的表达式 b² - 4ac 称为判别式，通常记作 Δ。它的值可以在不解方程的情况下告诉我们方程有多少个实根。这是非常常见的考题，往往单独占两到三分。
| Discriminant 判别式 | Number of real roots 实根个数 | Graph 图像 |
| Δ > 0 | Two distinct real roots 两个不同实根 | Crosses the x-axis twice 与 x 轴交于两点 |
| Δ = 0 | One repeated real root 一个重根 | Touches the x-axis 与 x 轴相切 |
| Δ < 0 | No real roots 无实根 | Does not meet the x-axis 与 x 轴不相交 |
Δ = b² - 4ac
-3 < 0, so the equation has no real solutions, which is a perfectly acceptable answer.
对于 x² - 6x + 9，Δ = 36 - 36 = 0，所以有一个重根 x = 3，曲线刚好与 x 轴相切。对于 x² + x + 1，Δ = 1 - 4 = -3 < 0，方程无实数解，这也是一个完全正确的答案。
11. Choosing the Best Method | 选择最合适的方法
Three methods, three situations. Factorising is fastest when the roots are integers or simple fractions. Completing the square is required when the question says "by completing the square" or asks for the vertex and the minimum or maximum value. The quadratic formula works every time and is the safe fallback.
三种方法对应三种情形。当根是整数或简单分数时，因式分解最快。当题目明确要求"用配方法"或要求求顶点与最值时，必须用配方法。求根公式则永远可行，是保底的安全选择。
If the numbers are small and tidy, try factorising first; if nothing works in about 20 seconds, switch to the formula. | 如果数字小而整齐，先试因式分解；若 20 秒左右试不出，就改用求根公式。
If the coefficient of x is even and the question mentions surds, completing the square is usually expected. | 如果 x 的系数为偶数且题目涉及根式，通常期望用配方法。
If a quadratic has non-integer roots such as 1.53 and -0.87, only the formula (or completing the square) can give them exactly. | 如果二次方程的根不是整数，例如 1.53 和 -0.87，只有求根公式（或配方法）能给出精确解。
Whatever method you use, check both roots by substituting them back into the original equation. | 无论用哪种方法，都要把两个根代回原方程检验。
A useful check comes from Vieta' the roots is -b/a and the product is c/a. If your two answers do not satisfy both, one of them is wrong.
一个有用的检验工具是韦达定理：对于 ax² + bx + c = 0，两根之和为 -b/a，两根之积为 c/a。如果算出的两个根不满足这两条，说明其中至少有一个是错的。
12. Common Exam Errors and Final Tips | 常见考试错误与应试建议
, but if you divide both sides by x you lose the root x = 0 completely. Never divide by an unknown.
最常见的错误是求解前没有把方程整理成标准形式。把 x² = 5x 因式分解为 x(x - 5
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