Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper, often more than once. Whether you are aiming for a grade C or an A*, mastering this topic is non-negotiable. This guide explains all four solution methods, the discriminant, graph sketching, and common exam traps, with fully worked examples throughout.

一元二次方程几乎出现在每一份 IGCSE 数学试卷中，而且往往不止一题。无论你的目标是 C 还是 A*，掌握这一考点都是必不可少的。本指南将详细讲解四种解法、判别式、图像画法以及常见考试陷阱，并配有完整的例题解答。

1. What Is a Quadratic Equation? | 什么是一元二次方程

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

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. The condition a ≠ 0 is essential: if a = 0, the equation becomes linear (bx + c = 0), not quadratic. For example, 3x² − 5x + 2 = 0 is quadratic, but 2x + 7 = 0 is not.

一元二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a ≠ 0。这里 a 是 x² 的系数，b 是 x 的系数，c 是常数项。条件 a ≠ 0 至关重要：如果 a = 0，方程就变成一次方程 bx + c = 0，而不是二次方程。例如，3x² − 5x + 2 = 0 是二次方程，而 2x + 7 = 0 不是。

2. Rearranging into Standard Form | 化为标准形式

Many equations are not given in standard form. Before applying any solution method, you must expand brackets, collect like terms, and move every term to one side so the right-hand side is zero.

许多方程并非直接以标准形式给出。在应用任何解法之前，你必须先去括号、合并同类项，并将所有项移到等号的一侧，使右侧为零。

Example: Solve 2(x + 1)(x − 3) = 4x − 1.

例题：解方程 2(x + 1)(x − 3) = 4x − 1。

First expand: 2(x² − 2x − 3) = 4x − 1, giving 2x² − 4x − 6 = 4x − 1. Then move all terms to the left: 2x² − 8x − 5 = 0. Now the equation is in standard form with a = 2, b = −8, c = −5.

首先展开：2(x² − 2x − 3) = 4x − 1，即 2x² − 4x − 6 = 4x − 1。然后将所有项移到左边：2x² − 8x − 5 = 0。此时方程即为标准形式，其中 a = 2，b = −8，c = −5。

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic factorises neatly. The key principle is the zero product property: if AB = 0, then A = 0 or B = 0.

当方程可以整洁地因式分解时，因式分解法是最快的解法。其核心原理是零乘积性质：如果 AB = 0，那么 A = 0 或 B = 0。

Worked Example: Solve x² − 7x + 12 = 0.

示例：解 x² − 7x + 12 = 0。

Find two numbers that multiply to +12 and add to −7. These are −3 and −4. Hence:

找出两个数，它们相乘等于 +12，相加等于 −7。这两个数是 −3 和 −4。因此：

(x − 3)(x − 4) = 0

Using the zero product property, x − 3 = 0 or x − 4 = 0, so x = 3 or x = 4. Always check by substituting back.

根据零乘积性质，x − 3 = 0 或 x − 4 = 0，所以 x = 3 或 x = 4。记得代回原方程检验。

Special cases:

特殊情况：

Difference of two squares: x² − 9 = (x + 3)(x − 3) = 0, so x = ±3.
Perfect square: x² − 6x + 9 = (x − 3)² = 0, so x = 3 (repeated root).

平方差公式：x² − 9 = (x + 3)(x − 3) = 0，所以 x = ±3。
完全平方：x² − 6x + 9 = (x − 3)² = 0，所以 x = 3（重根）。

4. Solving by Completing the Square | 配方法

Completing the square rewrites x² + bx in the form (x + p)² + q. This method works for every quadratic and is especially useful for finding turning points of graphs.

配方法将 x² + bx 改写为 (x + p)² + q 的形式。这种方法适用于所有二次方程，在求图像顶点时尤其有用。

Worked Example: Solve x² + 6x + 5 = 0 by completing the square.

示例：用配方法解 x² + 6x + 5 = 0。

Take half of 6, which is 3, and square it to get 9. Write the first two terms as a perfect square, then adjust the constant:

取 6 的一半，即 3，平方得 9。将前两项写成完全平方形式，再调整常数项：

(x + 3)² − 9 + 5 = 0 → (x + 3)² − 4 = 0

Now (x + 3)² = 4. Take square roots: x + 3 = ±2. Therefore x = −3 + 2 = −1 or x = −3 − 2 = −5.

于是 (x + 3)² = 4。两边开平方：x + 3 = ±2。因此 x = −3 + 2 = −1 或 x = −3 − 2 = −5。

In general, for x² + bx + c = 0, the completed square form is (x + b/2)² − (b/2)² + c = 0.

一般地，对于 x² + bx + c = 0，配方形式为 (x + b/2)² − (b/2)² + c = 0。

5. The Quadratic Formula | 求根公式

The quadratic formula solves any quadratic equation, including those that do not factorise. It is derived by completing the square on the general form ax² + bx + c = 0.

求根公式可以解任意一元二次方程，包括无法因式分解的方程。它通过对一般形式 ax² + bx + c = 0 配方推导而来。

x = (−b ± √(b² − 4ac)) / 2a

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

示例：解 2x² + 3x − 2 = 0。

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

这里 a = 2，b = 3，c = −2。代入公式：

x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √25) / 4 = (−3 ± 5) / 4

This gives x = (2)/4 = 1/2 or x = (−8)/4 = −2. Always write the discriminant (b² − 4ac) clearly in your working; examiners award method marks for it.

由此得 x = 2/4 = 1/2 或 x = −8/4 = −2。在答题过程中务必清楚写出判别式 b² − 4ac，考官会为此给方法分。

6. The Discriminant | 判别式

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

判别式 Δ = b² − 4ac 可以告诉你二次方程有多少个实数根，而无需完整求解。

Discriminant value	Nature of roots	Graph interpretation
Δ > 0	Two distinct real roots	Curve crosses x-axis twice
Δ = 0	One repeated real root	Curve touches x-axis once
Δ < 0	No real roots	Curve does not meet x-axis

判别式的值	根的性质	图像含义
Δ > 0	两个不相等的实数根	抛物线与 x 轴相交两次
Δ = 0	一个重根（两个相等实根）	抛物线与 x 轴相切一次
Δ < 0	没有实数根	抛物线与 x 轴无交点

Example: For x² − 2x + 1 = 0, Δ = (−2)² − 4(1)(1) = 0, so there is exactly one root, x = 1. For x² + 2x + 5 = 0, Δ = 4 − 20 = −16 < 0, so there are no real roots.

例题：对于 x² − 2x + 1 = 0，Δ = (−2)² − 4(1)(1) = 0，所以只有一个根，x = 1。对于 x² + 2x + 5 = 0，Δ = 4 − 20 = −16 < 0，所以没有实数根。

7. Quadratic Word Problems | 二次方程应用题

In exam word problems, you first need to translate the information into a quadratic equation. Once set up, solve using any method, then decide which root makes sense in context.

在考试应用题中，你首先需要将题目信息转化为二次方程。列出方程后，用任意方法求解，再判断哪个根在情境中合理。

Worked Example: A rectangle has length (x + 3) cm and width x cm. Its area is 40 cm². Find the value of x.

示例：一个长方形的长为 (x + 3) 厘米，宽为 x 厘米，面积为 40 平方厘米。求 x 的值。

Area = length × width, so x(x + 3) = 40. Expand: x² + 3x = 40. Rearrange: x² + 3x − 40 = 0. Factorise: (x + 8)(x − 5) = 0. Hence x = −8 or x = 5.

面积 = 长 × 宽，所以 x(x + 3) = 40。展开得 x² + 3x = 40，整理得 x² + 3x − 40 = 0。因式分解：(x + 8)(x − 5) = 0。因此 x = −8 或 x = 5。

Since a width cannot be negative, x = 5 cm. Always reject the negative root in geometric contexts and state this in your answer.

由于宽度不能为负数，所以 x = 5 厘米。在几何情境中务必舍去负根，并在答案中注明这一点。

8. Quadratic Graphs and Roots | 二次函数图像与根

The solutions of ax² + bx + c = 0 correspond to the x-intercepts of the graph y = ax² + bx + c. The graph is a parabola, opening upward if a > 0 and downward if a < 0.

方程 ax² + bx + c = 0 的解对应于图像 y = ax² + bx + c 与 x 轴的交点。该图像是抛物线，当 a > 0 时开口向上，当 a < 0 时开口向下。

Key features to find when sketching:

画图时需要确定的关键特征：

Roots: solve the quadratic equation to find x-intercepts.
y-intercept: substitute x = 0 to get y = c.
Axis of symmetry: x = −b/(2a).
Turning point: substitute the axis of symmetry into y to find the vertex.

根：解二次方程得到 x 轴交点。
y 轴截距：令 x = 0，得 y = c。
对称轴：x = −b/(2a)。
顶点：将对称轴 x 值代入 y 得到顶点坐标。

Example: For y = x² − 4x + 3, factorising gives roots x = 1 and x = 3. The axis of symmetry is x = 2, and substituting gives y = −1, so the vertex is (2, −1).

例题：对于 y = x² − 4x + 3，因式分解得根 x = 1 和 x = 3。对称轴为 x = 2，代入得 y = −1，所以顶点为 (2, −1)。

9. Common Mistakes and Exam Tips | 常见错误与应试技巧

Read the question carefully: some questions specifically require factorisation, while others ask for answers to 2 decimal places or in surd form. Choosing the wrong method can lose method marks.

仔细审题：有些题目明确要求用因式分解法，有些则要求保留两位小数或写成根式形式。用错方法可能丢掉方法分。

Mistake	Correction
Forgetting factor the equation must equal zero	Always rearrange to ax² + bx + c = 0 before factorising
Sign errors when substituting into formula	Write b = −7 carefully as −(−7) = +7
Dropping the ± symbol	Always give both roots
Not checking answers in word problems	Reject roots that violate the context

常见错误	正确做法
忘记因式分解前必须先使方程等于零	因式分解前先化为 ax² + bx + c = 0
代入公式时符号出错	例如 b = −7 时写成 −(−7) = +7
漏掉 ± 符号	务必写出两个根
应用题中不检验答案	舍去不符合实际情况的根

Exam tip: If you have a scientific calculator, you can check roots quickly by using the polynomial solver, but always show full algebraic working on paper for method marks.

考试技巧：如果你有科学计算器，可以用多项式求解功能快速检验根，但试卷上务必展示完整的代数过程，以获得方法分。

10. Practice Questions | 巩固练习

Try these questions on your own before reading the solutions.

请先独立完成以下练习，再对照答案。

Q1. Solve x² − 5x − 14 = 0 by factorisation.

练习1：用因式分解法解 x² − 5x − 14 = 0。

Q2. Solve 3x² + 4x − 5 = 0 using the quadratic formula, giving answers to 2 decimal places.

练习2：用求根公式解 3x² + 4x − 5 = 0，答案保留两位小数。

Q3. Find the value of k for which x² + kx + 9 = 0 has exactly one real root.

练习3：求 k 的值，使 x² + kx + 9 = 0 只有一个实数根。

Solutions:

解答：

Q1: (x − 7)(x + 2) = 0, so x = 7 or x = −2.

练习1：(x − 7)(x + 2) = 0，所以 x = 7 或 x = −2。

Q2: a = 3, b = 4, c = −5. Then x = (−4 ± √(16 + 60))/6 = (−4 ± √76)/6 = (−4 ± 8.718)/6. This gives x ≈ 0.79 or x ≈ −2.12 (to 2 d.p.).

练习2：a = 3，b = 4，c = −5。则 x = (−4 ± √(16 + 60))/6 = (−4 ±

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Solving Quadratic Equations | 解一元二次方程