Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear frequently across the Edexcel IGCSE Mathematics syllabus. Mastering their solution methods is essential not only for Paper 2 but also for problem-solving questions throughout the exam. This article breaks down every method you need, with step-by-step examples and common pitfalls.

二次方程在 Edexcel IGCSE 数学考纲中频繁出现。掌握其解法不仅对 Paper 2 至关重要，也贯穿整份试卷的应用题与难题。本文将详细拆解所有必考方法，配逐步例题与常见易错点。

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

A quadratic equation is an equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable is 2. Examples include x² – 5x + 6 = 0 and 2x² + 3x – 1 = 0.

二次方程是能写成形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数是 2。例如 x² – 5x + 6 = 0 和 2x² + 3x – 1 = 0。

If the equation is not in this standard form, you may need to expand brackets or rearrange terms first. For instance, (x – 1)(x + 2) = 4 must be expanded and rearranged to x² + x – 6 = 0.

如果方程不是标准形式，你需要先展开括号或移项。例如 (x – 1)(x + 2) = 4 必须先展开并整理为 x² + x – 6 = 0。

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2. Solving by Factorisation | 因式分解法

Factorisation is usually the fastest method when simple integer roots exist. The idea is to rewrite the quadratic as a product of two linear factors, then use the zero-product property: if AB = 0, then A = 0 or B = 0.

当方程存在简单整数根时，因式分解通常是最快的方法。其核心是将二次式写成两个一次因式的乘积，然后利用零乘积性质：若 AB = 0，则 A = 0 或 B = 0。

• Example: Solve x² – 5x + 6 = 0. Find two numbers that multiply to 6 and add to -5: they are -2 and -3. Hence (x – 2)(x – 3) = 0. So x = 2 or x = 3.

• 示例：解 x² – 5x + 6 = 0。找到两个数相乘为 6，相加为 -5：它们是 -2 和 -3。因此 (x – 2)(x – 3) = 0。所以 x = 2 或 x = 3。

For equations like ax² + bx + c with a ≠ 1, use the method of splitting the middle term. For 2x² + 7x + 3 = 0, multiply a by c: 2 × 3 = 6. Find factors of 6 that sum to 7: 1 and 6. Rewrite as 2x² + x + 6x + 3, then factor by grouping: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3) = 0, so x = -½ or x = -3.

对于 a ≠ 1 的方程如 ax² + bx + c，可使用“分中项”法。对于 2x² + 7x + 3 = 0，先计算 a × c：2 × 3 = 6。然后找和为 7 的因数：1 和 6。改写为 2x² + x + 6x + 3，再分组因式分解：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3) = 0，所以 x = -½ 或 x = -3。

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3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This method always works, reveals the turning point, and is required for deriving the quadratic formula. For x² + 6x + 5 = 0, take half of 6 (which is 3), write (x + 3)² – 9 + 5 = 0, simplify to (x + 3)² – 4 = 0, then solve: (x + 3)² = 4, so x + 3 = ±2, giving x = -1 or x = -5.

配方法将二次式改写为 (x + p)² + q 的形式。这种方法总可行，同时能揭示顶点坐标，也是推导二次公式的基础。对于 x² + 6x + 5 = 0，取 6 的一半（即 3），写成 (x + 3)² – 9 + 5 = 0，化简为 (x + 3)² – 4 = 0，然后求解：(x + 3)² = 4，所以 x + 3 = ±2，得 x = -1 或 x = -5。

When the coefficient of x² is not 1, factor it out first. For 2x² – 8x + 1 = 0, write 2(x² – 4x) + 1 = 0, complete the square inside: 2[(x – 2)² – 4] + 1 = 0, expand to 2(x – 2)² – 8 + 1 = 0, so 2(x – 2)² = 7, giving x = 2 ± √(7/2).

当 x² 的系数不为 1 时，先把该系数提出来。对于 2x² – 8x + 1 = 0，写成 2(x² – 4x) + 1 = 0，在括号内配方：2[(x – 2)² – 4] + 1 = 0，展开得 2(x – 2)² – 8 + 1 = 0，于是 2(x – 2)² = 7，所以 x = 2 ± √(7/2)。

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4. The Quadratic Formula | 二次公式

For any quadratic equation ax² + bx + c = 0, the solutions are given by the formula:

x = (-b ± √(b² – 4ac)) / (2a)

This formula is printed on the Edexcel formula sheet, but you must know how to substitute values correctly. Always write down the values of a, b and c first to avoid sign errors.

对于任意二次方程 ax² + bx + c = 0，其解由以下公式给出：

x = (-b ± √(b² – 4ac)) / (2a)

该公式印在 Edexcel 公式表中，但你必须学会正确代入。先写下 a、b、c 的值，避免符号错误。

• Example: Solve 2x² – 3x – 2 = 0. Here a = 2, b = -3, c = -2. Substitute: x = (3 ± √((-3)² – 4 × 2 × (-2))) / (2 × 2) = (3 ± √(9 + 16)) / 4 = (3 ± 5) / 4, giving x = 2 or x = -½.

• 示例：解 2x² – 3x – 2 = 0。其中 a = 2，b = -3，c = -2。代入：x = (3 ± √((-3)² – 4 × 2 × (-2))) / (2 × 2) = (3 ± √(9 + 16)) / 4 = (3 ± 5) / 4，得 x = 2 或 x = -½。

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5. The Discriminant b² – 4ac | 判别式 b² – 4ac

The discriminant tells us how many real roots a quadratic equation has, without solving it fully. Let Δ = b² – 4ac.

判别式能让我们不解方程就判断二次方程有多少实根。设 Δ = b² – 4ac。

• If Δ > 0: two distinct real roots (the curve crosses the x-axis twice).

• If Δ = 0: one repeated root (the curve touches the x-axis once).

• If Δ < 0: no real roots (the curve does not intersect the x-axis).

如果 Δ > 0：有两个不同实根（曲线与 x 轴相交两次）。

如果 Δ = 0：有一个重根（曲线与 x 轴相切一次）。

如果 Δ < 0：无实根（曲线不与 x 轴相交）。

Example: For 3x² + 2x + 1 = 0, Δ = 2² – 4 × 3 × 1 = 4 – 12 = -8, so there are no real roots. 示例：对于 3x² + 2x + 1 = 0，Δ = 2² – 4 × 3 × 1 = 4 – 12 = -8，因此无实根。

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6. Quadratic Graphs and Roots | 二次函数图象与根

The solutions to a quadratic equation are the x-coordinates where the graph y = ax² + bx + c crosses the x-axis. These are also called the x-intercepts or zeros of the function.

二次方程的解就是函数 y = ax² + bx + c 的图象与 x 轴交点的 x 坐标，也称作 x 截距或零点。

The axis of symmetry of the parabola is x = -b/(2a), and the vertex (turning point) lies on this axis. Using the completing-square form y = a(x + p)² + q, the vertex is (-p, q).

抛物线的对称轴为 x = -b/(2a)，顶点（转向点）在这条轴上。利用配方形式 y = a(x + p)² + q，顶点坐标为 (-p, q)。

If the discriminant is negative, the graph never touches the x-axis. If the coefficient a is positive, the parabola opens upwards; if a is negative, it opens downwards.

若判别式为负，图象不与 x 轴相交。若系数 a 为正，抛物线开口向上；若 a 为负，则开口向下。

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7. Solving Quadratic Equations by Factorising — Special Cases | 因式分解解二次方程——特殊情况

Some quadratics are missing the bx or c term. For example, x² – 9 = 0 is a difference of squares: (x – 3)(x + 3) = 0, so x = 3 or x = -3.

有些二次方程缺少 bx 项或 c 项。例如 x² – 9 = 0 是平方差：(x – 3)(x + 3) = 0，所以 x = 3 或 x = -3。

For x² – 4x = 0, take out the common factor x: x(x – 4) = 0, so x = 0 or x = 4. Similar square binomials like x² + 2x + 1 = (x + 1)² = 0 give a repeated root x = -1.

对于 x² – 4x = 0，提取公因式 x：x(x – 4) = 0，所以 x = 0 或 x = 4。类似完全平方二项式如 x² + 2x + 1 = (x + 1)² = 0 给出重根 x = -1。

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

Exam questions often put quadratics into real-world contexts. You need to translate the wording into an equation, solve it, and then check which answer makes sense.

考试题常将二次方程放入实际情境。你需要把文字转化为方程，解方程，然后检查哪个答案合理。

Example: A rectangle has length (x + 3) cm and width (x – 1) cm. Its area is 45 cm². Find x. The equation is (x + 3)(x – 1) = 45, so x² + 2x – 3 = 45 → x² + 2x – 48 = 0. Factorise: (x + 8)(x – 6) = 0, giving x = -8 or x = 6. Since a length cannot be negative, x = 6.

示例：一个长方形的长为 (x + 3) cm，宽为 (x – 1) cm，面积为 45 cm²，求 x。方程为 (x + 3)(x – 1) = 45，即 x² + 2x – 3 = 45 → x² + 2x – 48 = 0。因式分解得 (x + 8)(x – 6) = 0，得到 x = -8 或 x = 6。由于长度不能为负数，取 x = 6。

Other common contexts include projectile motion (height = -5t² + 20t + 2), number puzzles, and geometry. Always define the variable clearly and discard any meaningless negative answer when applied to a physical situation.

其它常见情境包括抛体运动（高度 = -5t² + 20t + 2）、数字谜题和几何问题。务必明确设变量，并在实际情境中舍弃无意义的负解。

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9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Students often lose marks in quadratic questions for avoidable reasons. Here are the most frequent pitfalls and the correct habits to develop.

学生在二次方程问题上常因可避免的原因失分。以下是最常见的陷阱以及应养成的正确习惯。

• Forgetting to rearrange to zero: Always move all terms to one side before factorising. For example, x² = 3x + 4 must become x² – 3x – 4 = 0.

• 忘记整理为零：因式分解前必须把所有项移项到一边。例如 x² = 3x + 4 必须先化为 x² – 3x – 4 = 0。

• Sign errors in the formula: If b is negative, substitute carefully. Use brackets: (-3)², not -3².

• 公式代入符号错误：当 b 为负数时要仔细代入。使用括号：(-3)²，不能写成 -3²。

• Dividing by a variable: Never divide both sides by x as you may lose the root x = 0. Factorise instead.

• 除以变量：切勿两边同除以 x，否则会丢失根 x = 0。应改为因式分解。

• Ignoring the ± symbol: When taking a square root, always include both positive and negative results unless the context says otherwise.

• 忽略 ± 符号：开平方时，除非情境另有要求，必须同时保留正根和负根。

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10. Choosing the Best Method | 如何选择最优解法

Not every method is equally efficient for every equation. Develop a strategy so you can answer quickly and accurately in the exam.

并非每种方法对所有方程都同样高效。建立一套策略，让你在考试中快速而准确地作答。

Situation | 情况	Recommended Method | 推荐方法
Simple integer roots, a = 1 | 简单整数根，a = 1	Factorisation | 因式分解
a ≠ 1 but factorisable | a ≠ 1 但可分解	Splitting middle term | 分中项法
Coefficient of x² is 1, b is even | x² 系数为 1，b 为偶数	Completing the square | 配方法
Any quadratic, especially with irrational roots | 任意二次式，尤其带无理根	Quadratic formula | 二次公式
Asking about number of roots | 询问根的个数	Discriminant | 判别式

In the exam, if factorisation is not obvious after a few seconds, switch to the formula or completing the square. Always check whether a question asks for exact answers or values to 3 significant figures.

考试中，如果几秒钟内看不出因式分解，就改用公式法或配方法。同时注意题目要求精确值还是保留 3 位有效数字。

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11. Exam-Style Worked Example | 考试风格例题精解

Let’s try a complete example that combines multiple skills. Solve the equation 2x² + 5x – 12 = 0 by factorisation and show the coordinates where the graph crosses the x-axis.

让我们尝试一道结合多种技能的完整例题。用因式分解解方程 2x² + 5x – 12 = 0，并写出图象与 x 轴交点的坐标。

First, multiply a and c: 2 × (-12) = -24. Find two numbers that multiply to -24 and add to 5: they are 8 and -3. Rewrite the middle term: 2x² + 8x – 3x – 12. Group: 2x(x + 4) – 3(x + 4) = (x + 4)(2x – 3) = 0. Hence x = -4 or x = 3/2. The x-axis crossing points are (-4, 0) and (1.5, 0).

首先计算 a × c：2 × (-12) = -24。找到两个数相乘为 -24，相加为 5：它们是 8 和 -3。改写中项：2x² + 8x – 3x – 12。分组：2x(x + 4) – 3(x + 4) = (x + 4)(2x – 3) = 0。所以 x = -4 或 x = 3/2。与 x 轴交点坐标为 (-4, 0) 和 (1.5, 0)。

Notice that if the question had asked for the vertex, you could complete the square or use x = -b/(2a) to find the axis of symmetry, then substitute back to get the y-coordinate.

注意，如果题目要求顶点，你可以用配方法或 x = -b/(2a) 求出对称轴，再代回原式得到 y 坐标。

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12. Practice Questions | 练习巩固

Here are three quick questions to test your understanding. Solve each by the method indicated, then check your answers below.

以下三道快速练习题用于检验理解。请按指定方法解答，然后对照下方答案。

• 1. Factorise and solve: x² – 7x + 10 = 0. | 因式分解并求解：x² – 7x + 10 = 0。

• 2. Use the quadratic formula: 3x² + 4x – 2 = 0 (give values to 2 decimal places). | 使用二次公式：3x² + 4x – 2 = 0（保留两位小数）。

• 3. Find the discriminant of 2x² + 4x + 1 = 0 and state the number of real roots. | 求 2x² + 4x + 1 = 0 的判别式，并说明实根个数。

Answers: 1. x = 2 or x = 5; 2. x ≈ 0.39 or x ≈ -1.72; 3. Δ = 8 > 0, so two distinct real roots.

答案：1. x = 2 或 x = 5；2. x ≈ 0.39 或 x ≈ -1.72；3. Δ = 8 > 0，所以有两个不同实根。

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