📚 Mastering Quadratic Equations | IGCSE数学精讲：一元二次方程
Quadratic equations appear in virtually every IGCSE Mathematics paper, whether you sit the Core or Extended syllabus. From factorising to the quadratic formula, these skills underpin algebra, graphs, and many examination problems. This article gives you a complete, step-by-step guide, with worked examples, common pitfalls, and examiner-style advice.
一元二次方程几乎出现在每一份IGCSE数学试卷中，无论你参加Core（核心）还是Extended（拓展）大纲。从因式分解到求根公式，这些技能是代数、函数图像及众多考题的基础。本文将为你提供一份完整的分步指南，包含典型例题、常见错误和考官风格的备考建议。
1. Expanding Double Brackets | 展开双括号
Before solving a quadratic equation, you must be confident expanding products of two linear expressions. Use the distributive law, often remembered with the acronym FOIL: First, Outer, Inner, Last. Each term in the first bracket multiplies every term in the second bracket.
在解一元二次方程之前，你必须熟练展开两个一次表达式的乘积。运用分配律，通常用首字母缩写FOIL来记忆：First（第一个）、Outer（外侧）、Inner（内侧）、Last（最后一个）。第一个括号中的每一项都要与第二个括号中的每一项相乘。
Worked example: expand (x + 3)(x + 5).
例题：展开 (x + 3)(x + 5)。
x·x + x·5 + 3·x + 3·5 = x² + 5x + 3x + 15 = x² + 8x + 15
Notice the pattern: the coefficient of x is the sum of the two numbers (3 + 5 = 8), and the constant term is their product (3 × 5 = 15). This pattern is the key to factorising quadratics, which we explore next.
请注意这个规律：x的系数等于两个数字之和（3 + 5 = 8），常数项等于它们的乘积（3 × 5 = 15）。这个规律正是因式分解一元二次式的关键，我们接下来就讨论它。
2. Factorising Quadratics with a = 1 | 因式分解首项系数为1的二次式
A quadratic expression of the form x² + bx + c can often be written as (x + p)(x + q), where p + q = b and p × q = c. You simply look for two numbers with the correct sum and product.
形如 x² + bx + c 的二次式通常可以写成 (x + p)(x + q)，其中 p + q = b，p × q = c。你只需找到两个和与积都正确的数字即可。
Worked example: factorise x² + 7x + 12.
例题：因式分解 x² + 7x + 12。
We need two numbers that multiply to 12 and add to 7. Testing factor pairs of 12: 1 and 12 (sum 13), 2 and 6 (sum 8), 3 and 4 (sum 7). The pair 3 and 4 works.
我们需要找到两个数，乘积为12且和为7。逐一检验12的因数对：1和12（和为13），2和6（和为8），3和4（和为7）。其中3和4满足条件。
x² + 7x + 12 = (x + 3)(x + 4)
Careful with negative numbers. Factorise x² – 5x + 6. The constant term is positive, so both numbers have the same sign. They must multiply to +6 and add to -5, so both are negative: -2 and -3.
处理负数时要格外小心。因式分解 x² – 5x + 6。常数项为正，说明两数同号。它们相乘得+6且相加得-5，因此两数均为负数：-2和-3。
x² – 5x + 6 = (x – 2)(x – 3)
Always expand your answer mentally to check whether it matches the original expression.
养成心算展开结果的习惯，以检验是否与原式一致。
3. Factorising Quadratics with a ≠ 1 | 因式分解首项系数不为1的二次式
When the coefficient of x² is not 1, factorising requires more thought. The standard form is ax² + bx + c with a ≠ 1. One reliable method is to find two numbers that multiply to a × c and add to b, then split the middle term.
当x²的系数不为1时，因式分解需要更多的思考。标准形式为 ax² + bx + c，其中 a ≠ 1。一种可靠的方法是：找到两个数，其乘积等于 a × c，和等于 b，然后用它们拆分中间项。
Worked example: factorise 2x² + 7x + 3.
例题：因式分解 2x² + 7x + 3。
Here a = 2, b = 7, c = 3, so a × c = 6. Find two numbers multiplying to 6 and adding to 7: those are 1 and 6. Split the middle term and factor by grouping:
这里 a = 2，b = 7，c = 3，所以 a × c = 6。找到两个相乘得6、相加得7的数：即1和6。拆分中间项并分组因式分解：
2x² + 7x + 3 = 2x² + 1x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)
If a is negative, first factor out -1 to make the leading coefficient positive; this usually simplifies the working considerably.
如果a为负数，先提出-1使首项系数变为正数，这样通常能大大简化计算。
4. Difference of Two Squares | 平方差公式
The difference of two squares is one of the most useful identities in algebra. For any numbers a and b:
平方差公式是代数中最有用的恒等式之一。对任意数a和b：
a² – b² = (a + b)(a – b)
This allows you to factorise any expression that is a perfect square minus another perfect square. For example, x² – 9 = x² – 3² = (x + 3)(x – 3).
这个公式可以让你轻松分解任何“一个完全平方减另一个完全平方”的式子。例如，x² – 9 = x² – 3² = (x + 3)(x – 3)。
It also works when the first term has a coefficient: 4x² – 25 = (2x)² – 5² = (2x + 5)(2x – 5). Be careful: the sum of two squares, such as x² + 9, cannot be factorised over real numbers using this rule.
当首项含有系数时公式同样适用：4x² – 25 = (2x)² – 5² = (2x + 5)(2x – 5)。请注意：两个平方的和，例如 x² + 9，在实数范围内不能使用这个公式进行因式分解。
5. Solving Quadratic Equations by Factorising | 用因式分解法解一元二次方程
The zero-product property states that if the product of two expressions is zero, then at least one of them must be zero. This allows us to turn a factorised quadratic into two simple linear equations.
零乘积性质指出：如果两个表达式的乘积为零，那么至少有一个表达式为零。这使我们能够将一个分解后的二次方程转化为两个简单的一次方程。
Worked example: solve x² – 5x + 6 = 0.
例题：解方程 x² – 5x + 6 = 0。
First factorise: x² – 5x + 6 = (x – 2)(x – 3). Therefore:
先因式分解：x² – 5x + 6 = (x – 2)(x – 3)。因此：
(x – 2)(x – 3) = 0 ⇒ x – 2 = 0 or x – 3 = 0 ⇒ x = 2 or x = 3
Always rearrange the equation into the standard form ax² + bx + c = 0 before factorising. Solving 2x² = 3x – 1 by dividing both sides by x would lose the root x = 0 (if it were present); instead, bring everything to one side.
在因式分解之前，务必将方程整理成标准形式 ax² + bx + c = 0。如果直接在 2x² = 3x – 1 两边同时除以x，就会丢失可能的根x = 0；正确做法是把所有项移到同一边。
6. Completing the Square | 配方法
Completing the square rewrites x² + bx + c in the form (x + p)² + q. To do this, take half the coefficient of x, square it, and adjust the constant term.
配方法将 x² + bx + c 改写成 (x + p)² + q 的形式。做法是：取x系数的一半，将其平方，然后调整常数项。
Worked example: solve x² + 6x + 2 = 0 by completing the square.
例题：用配方法解方程 x² + 6x + 2 = 0。
Half of 6 is 3, and 3² = 9. Rewrite the expression:
6的一半是3，3² = 9。重写表达式：
x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7
Now solve the equation (x + 3)² – 7 = 0:
现在解方程 (x + 3)² – 7 = 0：
(x + 3)² = 7 ⇒ x + 3 = ±√7 ⇒ x = -3 ± √7
This method works for all quadratics, even when the roots are surds or when the discriminant is negative (yielding no real roots). It is also essential for finding the turning point of a quadratic graph.
这种方法对所有二次方程都适用，即使根是无理数，甚至判别式为负（无实根）时也能处理。它也是求二次函数图像顶点坐标的关键工具。
7. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation ax² + bx + c = 0 directly. It is the most general method and is especially useful when factorisation is difficult or impossible.
求根公式可以直接求解任何一元二次方程 ax² + bx + c = 0。它是最通用的方法，尤其适用于难以因式分解甚至无法因式分解的情况。
x = (−b ± √(b² − 4ac)) / 2a
Worked example: solve 2x² – 4x – 3 = 0 using the formula.
例题：用求根公式解方程 2x² – 4x – 3 = 0。
Here a = 2, b = -4, c = -3. Substitute into the formula:
这里 a = 2，b = -4，c = -3。代入公式：
x = (−(−4) ± √((−4)² − 4 × 2 × (−3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4
Simplify √40 = 2√10, so x = (4 ± 2√10) / 4 = 1 ± √10/2. On the calculator paper, give decimal answers to three significant figures if asked.
化简 √40 = 2√10，因此 x = (4 ± 2√10) / 4 = 1 ± √10/2。在使用计算器的试卷中，如果题目要求，请给出保留三位有效数字的小数答案。
8. The Discriminant | 判别式
The discriminant, denoted Δ, is the expression b² – 4ac found under the square root in the quadratic formula. It tells us how many real roots a quadratic equation has.
判别式记为Δ，即求根公式中根号下的表达式 b² – 4ac。它能告诉我们一元二次方程有多少个实数根。
| 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 | 一个相等的实数根（重根） |
| Δ < 0 | No real roots | 无实数根 |
For example, the equation x² + 2x + 5 = 0 has Δ = 2² – 4 × 1 × 5 = 4 – 20 = -16, which is negative. Hence it has no real roots; the graph never crosses the x-axis.
例如，方程 x² + 2x + 5 = 0 的判别式 Δ = 2² – 4 × 1 × 5 = 4 – 20 = -16，为负数。因此它没有实数根；其图像与x轴没有交点。
9. Equations Reducible to Quadratics | 可化为二次方程的方程
Some equations do not look quadratic at first but can be transformed into one. A common case is an algebraic fraction equation; multiply through by the common denominator to clear denominators.
有些方程初看不是二次方程，但可以转化为二次方程。常见的情况是含代数分式的方程；两边乘以公分母即可消去分母。
Worked example: solve x + 6/x = 5, where x ≠ 0.
例题：解方程 x + 6/x = 5，其中 x ≠ 0。
Multiply both sides by x:
两边同乘以x：
x² + 6 = 5x ⇒ x² – 5x + 6 = 0 ⇒ (x – 2)(x – 3) = 0 ⇒ x = 2 or x = 3
Another case uses substitution. Solve x⁴ – 5x² + 4 = 0 by letting u = x². Then u² – 5u + 4 = 0, so (u – 1)(u – 4) = 0, giving u = 1 or u = 4. Hence x = ±1 or x = ±2.
另一种情况使用换元法。解 x⁴ – 5x² + 4 = 0 时，令 u = x²。于是 u² – 5u + 4 = 0，即 (u – 1)(u – 4) = 0，得到 u = 1 或 u = 4。因此 x = ±1 或 x = ±2。
10. Word Problems Leading to Quadratics | 由应用题建立二次方程
Examiners love to test quadratics through word problems. The key is to define a variable, translate the conditions into an equation, solve it, and then critically check which solutions make sense in context.
考官非常喜欢通过应用题来考查二次方程。关键在于设变量、把条件转化为方程、求解，然后结合实际问题情境判断哪些解是合理的。
Worked example: The length of a rectangle is 3 cm longer than its width, and its area is 40 cm². Find the dimensions.
例题：一个长方形的长比宽多3 cm，面积为40 cm²。求它的长和宽。
Let the width be w cm. Then the length is (w + 3) cm. Area gives:
设宽为 w cm，则长为 (w + 3) cm。由面积可得：
w(w + 3) = 40 ⇒ w² + 3w – 40 = 0 ⇒ (w + 8)(w – 5) = 0 ⇒ w = -8 or w = 5
Since width cannot be negative, reject w = -8. Therefore the width is 5 cm
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