Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in nearly every IGCSE Mathematics examination paper. Whether you sit the Core or the Extended paper, solving equations of the form ax² + bx + c = 0 is a core skill that supports many other topics, including graphs, coordinate geometry and problem solving. This article gives you a complete, exam-focused guide to the three standard solution methods, the discriminant, word problems and common traps, with fully worked examples throughout.

二次方程几乎出现在每份 IGCSE 数学试卷中。无论你参加 Core（核心）还是 Extended（扩展）试卷，掌握 ax² + bx + c = 0 形式方程的解法都是一项核心技能，它支撑着图像、坐标几何和实际问题解决等多个专题。本文将为你提供一份完整的、紧扣考点的指南，涵盖三种标准解法、判别式、应用题与常见陷阱，并配有完整的例题演示。

1. What Is a Quadratic Equation? | 什么是二次方程

A quadratic equation is any equation that can be rearranged into the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which is why it is called a quadratic equation. If a = 0, the x² term disappears and the equation becomes linear, so quadratic equations must always have a non-zero leading coefficient.

二次方程是可以整理为标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量 x 的最高次数为 2，因此称之为二次方程。若 a = 0，x² 项消失，方程退化为一次方程，所以二次方程的首项系数必须不为零。

In the IGCSE syllabus you must be able to solve quadratic equations by three methods: factorisation, the quadratic formula and completing the square. Factorisation is fastest when the expression factorises neatly; the quadratic formula always works, even for awkward coefficients; completing the square is especially useful for finding turning points and solving equations without a calculator. You should be comfortable choosing the most efficient method for each question.

在 IGCSE 考纲中，你需要掌握三种解二次方程的方法：因式分解法、求根公式法和配方法。当表达式能整齐分解时，因式分解法最快；求根公式法永远有效，即使系数很复杂；配方法在求拐点以及不用计算器解方程时特别有用。你应该能够针对每道题选择最高效的方法。

2. Solving by Factorisation | 因式分解法求解

The factorisation method is built on the zero product property: if the product of two real numbers is zero, then at least one of the numbers must be zero. In symbols, if p × q = 0, then p = 0 or q = 0. This only works when the product equals zero, so you must always rearrange the equation so that one side is zero before factorising.

因式分解法建立在零乘积性质之上：若两个实数的乘积为零，则至少有一个数必须为零。用符号表示，若 p × q = 0，则 p = 0 或 q = 0。这个性质仅在乘积等于零时才成立，因此在分解因式之前，你必须先将方程整理为一边为零的形式。

Example 1: Solve x² − 5x + 6 = 0.

例题 1：解方程 x² − 5x + 6 = 0。

• Step 1: Factorise the left-hand side. Find two numbers whose product is 6 and whose sum is −5. These numbers are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3).

  第一步：分解左侧因式。找出乘积为 6、和为 −5 的两个数。这两个数是 −2 和 −3，因此 x² − 5x + 6 = (x − 2)(x − 3)。

• Step 2: Apply the zero product property. Since (x − 2)(x − 3) = 0, either x − 2 = 0 or x − 3 = 0.

  第二步：应用零乘积性质。因为 (x − 2)(x − 3) = 0，所以 x − 2 = 0 或 x − 3 = 0。

• Step 3: Solve the two linear equations to get x = 2 or x = 3. Both values satisfy the original equation.

  第三步：解这两个一次方程，得到 x = 2 或 x = 3。两个值都满足原方程。

x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3

Example 2: Solve x² + 7x + 12 = 0. We need two numbers with product 12 and sum 7; these are 3 and 4. Hence (x + 3)(x + 4) = 0, so x = −3 or x = −4.

例题 2：解方程 x² + 7x + 12 = 0。我们需要乘积为 12、和为 7 的两个数，即 3 和 4。因此 (x + 3)(x + 4) = 0，所以 x = −3 或 x = −4。

3. Factorising Quadratics with a ≠ 1 | 二次项系数不为 1 的因式分解

When the coefficient of x² is not 1, for example 2x² + 7x + 3 = 0, the factorisation requires an extra step. The method below is known as the ‘multiply ac and split the middle term’ method.

当 x² 的系数不为 1 时，例如 2x² + 7x + 3 = 0，因式分解需要额外步骤。下面的方法被称为”相乘 ac 并拆分中间项”法。

• Step 1: Identify a = 2, b = 7 and c = 3. Calculate a × c = 2 × 3 = 6.

  第一步：确定 a = 2，b = 7，c = 3。计算 a × c = 2 × 3 = 6。

• Step 2: Find two numbers whose product is 6 and whose sum is b = 7. These numbers are 6 and 1.

  第二步：找出乘积为 6、和为 b = 7 的两个数。这两个数为 6 和 1。

• Step 3: Rewrite the middle term 7x as 6x + x, giving 2x² + 6x + x + 3 = 0.

  第三步：将中间项 7x 改写为 6x + x，得到 2x² + 6x + x + 3 = 0。

• Step 4: Factor by grouping. 2x(x + 3) + 1(x + 3) = 0, so (x + 3)(2x + 1) = 0.

  第四步：分组提取公因式。2x(x + 3) + 1(x + 3) = 0，因此 (x + 3)(2x + 1) = 0。

• Step 5: Solve: x + 3 = 0 or 2x + 1 = 0, so x = −3 or x = −1/2.

  第五步：求解：x + 3 = 0 或 2x + 1 = 0，所以 x = −3 或 x = −1/2。

2x² + 7x + 3 = 0 → (x + 3)(2x + 1) = 0 → x = −3 or x = −1/2

Example 3: Solve 6x² + x − 2 = 0. Here a × c = 6 × (−2) = −12. We need two numbers with product −12 and sum 1; these are 4 and −3. Splitting the middle term gives 6x² + 4x − 3x − 2 = 2x(3x + 2) − 1(3x + 2) = (3x + 2)(2x − 1) = 0. Therefore x = −2/3 or x = 1/2.

例题 3：解方程 6x² + x − 2 = 0。这里 a × c = 6 × (−2) = −12。我们需要乘积为 −12、和为 1 的两个数，即 4 和 −3。拆分中间项得 6x² + 4x − 3x − 2 = 2x(3x + 2) − 1(3x + 2) = (3x + 2)(2x − 1) = 0。因此 x = −2/3 或 x = 1/2。

4. Solving by the Quadratic Formula | 用求根公式求解

Some quadratic equations do not factorise easily, especially when the solutions involve surds. In such cases, the quadratic formula is the most reliable method. For any quadratic equation ax² + bx + c = 0, the solutions are given by the formula below.

有些二次方程不容易因式分解，尤其是当解含有根号时。在这种情况下，求根公式是最可靠的方法。对于任何二次方程 ax² + bx + c = 0，解由下面的公式给出。

x = (−b ± √(b² − 4ac)) / 2a

Example 4: Solve x² − 2x − 8 = 0 using the

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