Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in almost every exam paper, either as a direct question or as a tool for solving more complex problems such as coordinate geometry and word problems. This article explains the key methods for solving quadratic equations — factorisation, completing the square, and the quadratic formula — along with the discriminant and common exam tips.

一元二次方程是 IGCSE 数学中最重要的内容之一。几乎每份试卷都会出现这类题目，要么直接考查，要么作为解坐标几何、应用题等更复杂问题的工具。本文将系统地讲解解一元二次方程的主要方法——因式分解法、配方法和求根公式法，并讨论判别式与常见考试技巧。

1. Standard Form and Definition | 标准形式与定义

A quadratic equation in one variable is an equation that can be written in the standard form:

一元二次方程是指可以写成如下标准形式的方程：

ax² + bx + c = 0

Here, a, b and c are constants, and a ≠ 0. The reason we require a ≠ 0 is that if a = 0, the equation becomes linear, not quadratic. The term ax² is called the quadratic term, bx is the linear term, and c is the constant term.

其中 a、b、c 均为常数，且 a ≠ 0。要求 a ≠ 0 的原因是：如果 a = 0，方程就变成了一次方程，而不是二次方程。ax² 称为二次项，bx 称为一次项，c 称为常数项。

For example, 3x² − 5x + 2 = 0 is in standard form with a = 3, b = −5 and c = 2. A solution of the equation is a value of x that makes the left-hand side equal to zero. A quadratic equation can have two distinct real roots, one repeated real root, or no real roots.

例如，3x² − 5x + 2 = 0 就是标准形式，其中 a = 3，b = −5，c = 2。方程的解是使左边等于零的 x 值。一元二次方程可能有两个不同的实数根、一个重根，也可能没有实数根。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method for solving a quadratic equation, especially when the roots are integers or simple fractions. The idea is to rewrite the quadratic expression as a product of two linear factors, then use the zero product property: if p × q = 0, then p = 0 or q = 0.

因式分解法通常是解一元二次方程最快的方法，尤其当根是整数或简单分数时。核心思想是将二次表达式改写为两个一次因式的乘积，然后利用零乘积性质：如果 p × q = 0，则 p = 0 或 q = 0。

Example 1: Solve x² − 5x + 6 = 0.

例 1：解方程 x² − 5x + 6 = 0。

We look for two numbers that multiply to 6 and add to −5. These numbers are −2 and −3. Therefore:

我们需要找到两个数，它们相乘等于 6，相加等于 −5。这两个数是 −2 和 −3。因此：

(x − 2)(x − 3) = 0

So x − 2 = 0 or x − 3 = 0, which gives x = 2 or x = 3.

所以 x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。

Example 2: Solve 2x² + 7x + 3 = 0.

例 2：解方程 2x² + 7x + 3 = 0。

Here the coefficient of x² is not 1. We need factors of 2 × 3 = 6 that add to 7. The numbers 1 and 6 work. We rewrite the middle term:

这里 x² 的系数不是 1。我们需要找到 2 × 3 = 6 的因数，且它们的和为 7。数字 1 和 6 满足条件。我们重写中间项：

2x² + x + 6x + 3 = 0

Now factor by grouping: x(2x + 1) + 3(2x + 1) = 0, which gives (2x + 1)(x + 3) = 0. Hence x = −½ or x = −3.

然后分组因式分解：x(2x + 1) + 3(2x + 1) = 0，即 (2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。

Always check whether the equation is in standard form before factorising. If it is not, first rearrange all terms to one side.

因式分解前务必确认方程是标准形式。如果不是，首先要将所有项移到一边。

3. Completing the Square | 配方法

Completing the square is a powerful algebraic technique that transforms a quadratic expression into the form a(x + p)² + q. This method not only solves equations but also reveals the vertex of a parabola and the minimum or maximum value of a quadratic function.

配方法是一种重要的代数技巧，它将二次表达式转化为 a(x + p)² + q 的形式。这种方法不仅能解方程，还能揭示抛物线的顶点以及二次函数的最小值或最大值。

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

例：用配方法解方程 x² + 6x + 2 = 0。

Take half of the coefficient of x, which is 6 ÷ 2 = 3, and square it to get 9. Then write:

取 x 系数的一半，即 6 ÷ 2 = 3，平方得 9。于是写为：

(x + 3)² − 9 + 2 = 0

Simplify: (x + 3)² − 7 = 0. Then (x + 3)² = 7. Taking square roots gives x + 3 = ±√7, so x = −3 ± √7.

化简得 (x + 3)² − 7 = 0。于是 (x + 3)² = 7。两边开平方得 x + 3 = ±√7，所以 x = −3 ± √7。

For the general expression x² + bx, we always add and subtract (b/2)². If the coefficient of x² is not 1, factor it out first before completing the square.

对于一般形式 x² + bx，我们总是加上并减去 (b/2)²。如果 x² 的系数不是 1，需要先将其提取出来，再进行配方。

4. The Quadratic Formula | 求根公式法

The quadratic formula is the most general method. It works for any quadratic equation, even when factorisation is difficult or impossible. For the equation ax² + bx + c = 0, the solutions are given by:

求根公式是最通用的方法。它对任何一元二次方程都适用，即使方程难以因式分解甚至无法因式分解。对于方程 ax² + bx + c = 0，解为：

x = [−b ± √(b² − 4ac)] / 2a

Here the symbol ± means that we take two values: one with a plus sign and one with a minus sign. Be careful to substitute the correct values of a, b and c, including their signs.

这里的 ± 表示取两个值：一个用加号，一个用减号。代入 a、b、c 的值时务必小心，包括它们的正负号。

Example: Solve 2x² − 4x − 3 = 0 using the quadratic formula.

例：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 ± √(16 + 24)] / 4 = [4 ± √40] / 4

Simplify √40 = 2√10, so x = (4 ± 2√10)/4. Therefore x = (2 ± √10)/2.

化简 √40 = 2√10，所以 x = (4 ± 2√10)/4。因此 x = (2 ± √10)/2。

It is essential to show the substitution clearly in your exam working. A correct formula with an arithmetic error will lose marks, while showing each step helps you earn method marks even if the final answer is wrong.

考试中必须清楚地写出代入过程。即使公式正确，运算错误也会失分；而写出详细步骤，即使最终答案有误，也能获得方法分。

5. The Discriminant and the Nature of Roots | 判别式与根的性质

The expression b² − 4ac inside the quadratic formula is called the discriminant, often denoted by the symbol Δ. It tells us about the nature of the roots without actually solving the equation.

求根公式中的 b² − 4ac 称为判别式，常用符号 Δ 表示。它可以在不解方程的情况下判断根的性质。

If b² − 4ac > 0, the equation has two distinct real roots.

如果 b² − 4ac > 0，方程有两个不同的实数根。
If b² − 4ac = 0, the equation has exactly one repeated real root (a double root).

如果 b² − 4ac = 0，方程有一个重根（即两个相等的实数根）。
If b² − 4ac < 0, the equation has no real roots (the roots are complex).

如果 b² − 4ac < 0，方程没有实数根（根为复数）。

Example: Determine the nature of the roots of x² − 3x + 5 = 0.

例：判断方程 x² − 3x + 5 = 0 的根的性质。

Here a = 1, b = −3, c = 5. The discriminant is b² − 4ac = 9 − 20 = −11 < 0, so the equation has no real roots.

这里 a = 1，b = −3，c = 5。判别式为 b² − 4ac = 9 − 20 = −11 < 0，因此方程没有实数根。

6. Choosing the Right Method | 选择合适的方法

The table below summarises the advantages and typical uses of each method. In an exam, factorisation is usually the fastest, but you should be fluent in all three.

下表总结了每种方法的优势和典型用途。在考试中，因式分解通常最快，但你需要熟练掌握所有三种方法。

Method 方法	Best Used When 适用情况	Example 示例
Factorisation 因式分解法	Roots are integers or simple fractions 根为整数或简单分数	x² − 7x + 12 = 0
Completing the Square 配方法	When the coefficient of x is even, or the vertex is needed x 系数为偶数，或需要求顶点	x² + 8x − 1 = 0
Quadratic Formula 求根公式法	Always works, especially when factorisation is hard 万能方法，尤其适合难以因式分解的情况	2x² + 3x − 7 = 0

If you are asked to solve “giving your answer to 2 decimal places,” use the quadratic formula and a calculator. If the question says “show your working,” factorisation or completing the square is usually expected.

如果题目要求“答案精确到两位小数”，应使用求根公式并借助计算器。如果题目要求“写出过程”，通常期望使用因式分解法或配方法。

7. Equations That Are Not in Standard Form | 非标准形式的方程

Sometimes the quadratic equation is not given as ax² + bx + c = 0. You must first expand brackets, collect like terms, and rearrange everything to one side before applying any solving method.

有时一元二次方程并不是以 ax² + bx + c = 0 的形式给出。你必须先展开括号、合并同类项，并将所有项移到一边，才能使用任何求解方法。

Example: Solve 3x(x + 1) = 2(x + 5).

例：解方程 3x(x + 1) = 2(x + 5)。

First expand both sides: 3x² + 3x = 2x + 10. Then bring all terms to the left: 3x² + x − 10 = 0.

先展开两边：3x² + 3x = 2x + 10。然后将所有项移到左边：3x² + x − 10 = 0。

Now factorise: (3x − 5)(x + 2) = 0, so x = 5/3 or x = −2.

因式分解得 (3x − 5)(x + 2) = 0，所以 x = 5/3 或 x = −2。

Do not divide by an expression containing x, because you might lose a root. Instead, always collect all terms on one side.

切勿两边同时除以含有 x 的表达式，否则可能丢根。应始终将所有项移到一边。

8. Word Problems Involving Quadratics | 一元二次方程应用题

Quadratic equations are often used to model real-life situations such as the area of a rectangle, the height of a projectile, or the revenue of a business. In such problems, you need to define a variable, form an equation, solve it, and then check whether the solution makes sense in the context.

一元二次方程常用于建立实际问题的数学模型，如矩形面积、抛射体高度或企业收入。解这类题需要定义变量、建立方程、求解，并检查解在实际情况中是否有意义。

Example: A rectangle has a length that is 3 cm longer than its width. Its area is 40 cm². Find the width.

例：一个矩形的长比宽长 3 cm，面积为 40 cm²。求宽。

Let the width be x cm. Then the length is (x + 3) cm. The area equation is x(x + 3) = 40, which gives x² + 3x − 40 = 0.

设宽为 x cm，则长为 (x + 3) cm。面积方程为 x(x + 3) = 40，即 x² + 3x − 40 = 0。

Factorise: (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since a width cannot be negative, we reject x = −8. The width is 5 cm.

因式分解得 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。宽度不能为负数，因此舍去 x = −8。答案是宽为 5 cm。

Always check the units and whether the answer is reasonable. Negative lengths, times or distances usually mean the solution is invalid.

始终检查单位和答案是否合理。负数长度、时间或距离通常意味着该解无效。

9. Common Mistakes to Avoid | 常见错误以避免

Many students lose unnecessary marks due to small but repeated errors. Here are the most common pitfalls in solving quadratic equations.

许多学生因为反复出现的小错误而丢分。以下是解一元二次方程时最常见的陷阱。

Incorrect sign substitution: When using the quadratic formula, forgetting to include the minus sign in b or c leads to wrong answers.

符号代入错误：使用求根公式时，忘记 b 或 c 的负号会导致答案错误。
Forgetting to rearrange: Solving x² = 5x as x = 5 is wrong because you divided by x. The correct way is x² − 5x = 0 → x(x − 5) = 0 → x = 0 or x = 5.

忘记移项：将 x² = 5x 直接解得 x = 5 是错误的，因为两边同除了 x。正确做法是 x² − 5x = 0 → x(x − 5) = 0 → x = 0 或 x = 5。
Losing the ± sign: When taking square roots, always remember that x² = k gives x = ±√k, not just x = √k.

漏掉 ± 号：开平方时，记住 x² = k 解得 x = ±√k，而不是只有 x = √k。
Misidentifying coefficients: Make sure the equation is in standard form before reading off a, b and c.

系数识别错误：在读取 a、b、c 之前，务必确认方程已经是标准形式。

10. Practice Questions | 练习题目

Try the following questions on your own before checking the solutions. These reflect typical IGCSE exam questions on quadratic equations.

请先独立完成以下练习，再对照答案。这些题目反映了 IGCSE 考试中关于一元二次方程的典型题型。

Q1: Solve x² − 2x − 15 = 0.

题目 1：解方程 x² − 2x − 15 = 0。

Q2: Solve x² + 5x − 2 = 0, giving your answers correct to 2 decimal places.

题目 2：解方程 x² + 5x − 2 = 0，答案精确到两位小数。

Q3: The length of a rectangle is 4 cm more than twice its width. The area is 30 cm². Find the width.

题目 3：矩形的长比宽的 2 倍还多 4 cm，面积为 30 cm²。求宽。

Solutions:

答案：

S1: (x − 5)(x + 3) = 0, so x = 5 or x = −3.

解 1：(x − 5)(x + 3) = 0，所以 x = 5 或 x = −3。

S2: Using the formula with a = 1, b = 5, c = −2 gives x = [−5 ± √25 + 8] / 2 = [−5 ± √33] / 2. Hence x ≈ 0.37 or x ≈ −5.37.

解 2：用求根公式，a = 1，b = 5，c = −2，得 x = [−5 ± √25 + 8] / 2 = [−5 ± √33] / 2。因此 x ≈ 0.37 或 x ≈ −5.37。

S3: Let width = x, length = 2x + 4. Then x(2x + 4) = 30 → 2x² + 4x − 30 = 0 → x² + 2x − 15 = 0 → (x + 5)(x − 3) = 0 → x = 3. Width is 3 cm.

解 3：设宽为 x，长为 2x + 4。则 x(2x + 4) = 30 → 2x² + 4x − 30 = 0 → x² + 2x − 15 = 0 → (x + 5)(x − 3) = 0 → x = 3。宽为 3 cm。

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Solving Quadratic Equations | 解一元二次方程