Solving Quadratic Equations: Factorisation, Formula & Completing the Square | 解一元二次方程：因式分解、求根公式与配方法

📚 Solving Quadratic Equations: Factorisation, Formula & Completing the Square | 解一元二次方程：因式分解、求根公式与配方法

Quadratic equations appear in nearly every IGCSE Mathematics paper, whether as a standalone question, a word problem, or a hidden step inside a coordinate geometry or calculus question. Mastering the three core methods – factorisation, the quadratic formula, and completing the square – is essential for full marks.

一元二次方程几乎出现在每一份 IGCSE 数学试卷中，无论是独立题目、应用题，还是坐标几何与微积分题目中的隐含步骤。掌握因式分解、求根公式与配方法这三大核心方法是拿满分的必备技能。

This article breaks down each method step by step, compares their advantages, and highlights the exam traps that cause students to lose marks. Worked examples follow the exact notation and layout that examiners reward.

本文逐步拆解每种方法，比较它们的适用场景，并指出导致学生失分的常见陷阱。所有示例均采用考官认可的标准写法与格式。

1. The General Form | 一般形式

A quadratic equation is any equation that can be written in the general form:

一元二次方程是指可以写成如下一般形式的方程：

ax² + bx + c = 0, where a ≠ 0

Here a is the coefficient of x², b is the coefficient of x, and c is the constant term. If a = 0, the equation becomes linear, not quadratic.

其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。如果 a = 0，方程就变成了一次方程，而非二次方程。

For example, 2x² – 3x + 1 = 0 has a = 2, b = -3, c = 1.
例如，在 2x² – 3x + 1 = 0 中，a = 2，b = -3，c = 1。
The equation must be rearranged into the general form before applying any method.
在使用任何方法之前，必须先整理方程，使其符合一般形式。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the expression has simple integer factors. The principle relies on the zero-product property: if the product of two expressions is zero, then at least one of them must be zero.

当表达式含有简单的整数因子时，因式分解是最快捷的方法。其原理基于零乘积性质：若两个表达式的乘积为零，则其中至少有一个为零。

Step-by-step process:

分步流程：

Rearrange the equation into the form ax² + bx + c = 0.
将方程整理为 ax² + bx + c = 0 的形式。
Factorise the left-hand side into two brackets.
将等号左边因式分解为两个括号的形式。
Set each bracket equal to zero and solve the resulting linear equations.
令每个括号分别等于零，然后求解所得的一次方程。

Worked Example 1: Solve x² – 5x + 6 = 0.

示例 1：解方程 x² – 5x + 6 = 0。

Find two numbers that multiply to 6 and add to -5: these are -2 and -3. Therefore:

找两个数相乘得 6 且相加得 -5：这两个数是 -2 和 -3。因此：

(x – 2)(x – 3) = 0

Set each bracket to zero:

令每个括号等于零：

x – 2 = 0 → x = 2 or x – 3 = 0 → x = 3

The solution set is x = 2 or x = 3. Always substitute back to check your answer.

解集为 x = 2 或 x = 3。务必代回原方程检验答案。

3. Difference of Two Squares | 平方差公式

A special case of factorisation is the difference of two squares. Any expression of the form x² – p² factorises as:

因式分解的一个特殊情况是平方差公式。任何形如 x² – p² 的表达式都可以分解为：

x² – p² = (x – p)(x + p)

This pattern appears frequently in IGCSE papers and can simplify otherwise difficult equations.

这个规律在 IGCSE 考试中频繁出现，可以简化原本较为复杂的方程。

Worked Example 2: Solve 4x² – 9 = 0.

示例 2：解方程 4x² – 9 = 0。

Recognise that 4x² = (2x)² and 9 = 3², so:

注意到 4x² = (2x)² 且 9 = 3²，因此：

(2x – 3)(2x + 3) = 0

Setting each bracket to zero gives:

令每个括号等于零可得：

2x – 3 = 0 → x = 3⁄2 or 2x + 3 = 0 → x = -3⁄2

Common mistake	Students often write x² = 9⁄4, then forget the ± sign. Always include both positive and negative roots.
常见错误	学生常直接写出 x² = 9⁄4，然后忘记 ± 号。务必同时写出正负两个根。

4. The Quadratic Formula | 求根公式

When factorisation is difficult or impossible, the quadratic formula provides a universal method. The formula is derived from completing the square and works for every quadratic equation.

当因式分解困难或无法进行时，求根公式提供了通用的解法。该公式由配方法推导而来，适用于所有一元二次方程。

x = (-b ± √(b² – 4ac)) / 2a

Worked Example 3: Solve 2x² + 5x – 3 = 0.

示例 3：解方程 2x² + 5x – 3 = 0。

Here a = 2, b = 5, c = -3. Substitute into the formula:

此处 a = 2，b = 5，c = -3。代入公式：

x = (-5 ± √(5² – 4 × 2 × (-3))) / (2 × 2)

Simplify inside the square root first:

先化简根号内的部分：

x = (-5 ± √(25 + 24)) / 4 = (-5 ± √49) / 4

Since √49 = 7, the two solutions are:

因为 √49 = 7，所以两个解为：

x = (-5 + 7) / 4 = 1⁄2 or x = (-5 – 7) / 4 = -3

Write answers exactly unless the question asks for a decimal approximation.

除非题目要求小数近似值，否则应写出精确答案。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form p(x + q)² + r. This method is essential for finding turning points of parabolas and for deriving the quadratic formula itself.

配方法将二次表达式改写为 p(x + q)² + r 的形式。这种方法对于求抛物线的顶点以及推导求根公式本身都至关重要。

The procedure for x² + bx + c:

x² + bx + c 的配方法步骤如下：

Halve the coefficient of x: this gives q.
将 x 的系数除以 2：得到 q。
Write (x + q)², then subtract q² and add c.
写出 (x + q)²，然后减去 q² 并加上 c。

Worked Example 4: Solve x² + 6x + 2 = 0 by completing the square.

示例 4：用配方法解方程 x² + 6x + 2 = 0。

Half of 6 is 3, so (x + 3)² = x² + 6x + 9. Therefore:

6 的一半是 3，所以 (x + 3)² = x² + 6x + 9。因此：

x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7

Set this equal to zero and solve:

令其等于零并求解：

(x + 3)² – 7 = 0 → (x + 3)² = 7

x + 3 = ±√7 → x = -3 ± √7

This form also reveals the vertex of the parabola y = x² + 6x + 2 is at (-3, -7).

这一形式也直接揭示出抛物线 y = x² + 6x + 2 的顶点坐标为 (-3, -7)。

6. The Discriminant | 判别式

The expression b² – 4ac inside the quadratic formula is called the discriminant, denoted by Δ. It determines the nature of the roots without solving the full equation.

求根公式中的 b² – 4ac 称为判别式，记作 Δ。它无需解完整方程即可判断根的性质。

Discriminant Δ = b² – 4ac	Nature of roots	根的性质
Δ > 0	Two distinct real roots	两个不相等的实数根
Δ = 0	One repeated real root	一个重根（两个相等实数根）
Δ < 0	No real roots	无实数根

Worked Example 5: Find the value of k such that x² + kx + 9 = 0 has one repeated root.

示例 5：求使方程 x² + kx + 9 = 0 有重根的 k 值。

Set Δ = 0:

令 Δ = 0：

k² – 4 × 1 × 9 = 0 → k² – 36 = 0 → k = ±6

Both k = 6 and k = -6 produce a repeated root. Many students forget the negative value, so always consider ±.

k = 6 和 k = -6 都会产生重根。许多学生会遗漏负值，所以务必考虑 ±。

7. Choosing the Best Method | 选择最佳方法

Different questions call for different approaches. Use this decision guide to save time in the exam.

不同题目需要不同解法。使用以下决策指南可以在考试中节省时间。

If the question says “factorise”, you must use factorisation – no other method is accepted for that part.
如果题目要求 “factorise”（因式分解），则必须使用因式分解法——这部分不接受其他方法。
If the coefficient of x² is 1 and the constant term has simple integer factors, try factorisation first.
如果 x² 的系数为 1 且常数项有简单整数因子，优先尝试因式分解。
If the coefficients are awkward decimals or large primes, use the quadratic formula.
如果系数是难处理的小数或较大的质数，使用求根公式。
If the question asks for the turning point or minimum/maximum value, use completing the square.
如果题目要求顶点坐标或最小值/最大值，使用配方法。

Exam tip: Show substitution into the formula clearly. Examiners award method marks for correct substitution even if your final arithmetic is wrong.

考试技巧：清晰展示代入公式的过程。即使最终计算有误，考官也会为正确的代入步骤给方法分。

8. Common Exam Mistakes | 常见考试错误

After marking thousands of IGCSE scripts, examiners consistently report the same errors. Avoid these to protect your marks.

在批改数以千计的 IGCSE 试卷后，考官们一致反映存在相同的错误。避免这些错误以保住你的分数。

Mistake \| 错误	Correct approach \| 正确做法
Forgetting to rearrange to = 0 before factorising	Always move all terms to one side first
解因式分解前忘记将方程整理为 = 0	务必先将所有项移到等号一边
Dropping the ± in x = ±√p	Write ± explicitly in every step
在 x = ±√p 中漏掉 ±	每一步都明确写出 ±
Sign errors when b is negative in the formula	Rewrite x = (-(-5) ± …) clearly
公式中 b 为负时符号出错	清楚写出 x = (-(-5) ± …)
Dividing both sides by x, losing the root x = 0	Factorise x out instead of dividing
两边同时除以 x，丢失 x = 0 这个根	应提取公因式 x，而非直接除以 x

9. Practice Problems | 练习巩固

Try these problems without a calculator where possible. Solutions are shown below each question.

请尽可能不用计算器完成以下练习。每题下方附有解答。

Question 1: Solve x² – 7x + 12 = 0 by factorisation.

练习 1：用因式分解法解 x² – 7x + 12 = 0。

Numbers -3 and -4 multiply to 12 and add to -7:

-3 和 -4 相乘得 12 且相加得 -7：

(x – 3)(x – 4) = 0 → x = 3 or x = 4

Question 2: Solve 3x² + x – 2 = 0 using the quadratic formula.

练习 2：用求根公式解 3x² + x – 2 = 0。

x = (-1 ± √(1² – 4 × 3 × (-2))) / 6 = (-1 ± √25) / 6

x = (-1 + 5) / 6 = 2⁄3 or x = (-1 – 5) / 6 = -1

Question 3: Write x² – 8x + 5 in completed square form.

练习 3：将 x² – 8x + 5 写成配方法的形式。

(x – 4)² – 16 + 5 = (x – 4)² – 11

Question 4: Determine the nature of roots of 2x² – 3x + 5 = 0.

练习 4：判断 2x² – 3x + 5 = 0 的根的性质。

Δ = (-3)² – 4 × 2 × 5 = 9 – 40 = -31 < 0 → no real roots

If you got all four correct, you are ready for exam-style quadratics questions. If not, revisit the relevant section before moving on.

如果你四题全对，说明你已经准备好应对考试风格的二次方程题目。若未全对，请在继续之前重新阅读相关章节。

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Solving Quadratic Equations: Factorisation, Formula & Completing the Square | 解一元二次方程：因式分解、求根公式与配方法