Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in nearly every paper, from short factorisation questions to multi-step problems involving areas, motion and profit. Mastering the techniques in this guide will give you a clear advantage in your exams.

一元二次方程是 IGCSE 数学中最重要的考点之一，几乎每张试卷都会出现——从简短的因式分解题，到涉及面积、运动与利润的多步骤综合题。掌握本指南中的方法与技巧，将帮助你在考试中取得明显优势。

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

A quadratic equation in one variable can be written in the general form ax² + bx + c = 0, where a, b and c are constants and a is not equal to zero. The term ax² is called the quadratic term, bx the linear term, and c the constant term.

含一个未知数的一元二次方程可以写成一般形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。ax² 称为二次项，bx 称为一次项，c 称为常数项。

ax² + bx + c = 0 , a ≠ 0

If a = 0, the equation becomes linear, not quadratic. This is why the condition a ≠ 0 is essential. For example, 3x² − 5x + 2 = 0 is quadratic, while 2x + 7 = 0 is linear.

若 a = 0，方程变成一次方程而非二次方程，因此 a ≠ 0 这一条件至关重要。例如 3x² − 5x + 2 = 0 是二次方程，而 2x + 7 = 0 是一次方程。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the equation has integer or simple rational roots. To solve by factorisation, rewrite the quadratic in the form (px + q)(rx + s) = 0, then set each bracket equal to zero.

当方程具有整数或简单有理根时，因式分解法通常是最快的方法。用因式分解法求解时，先将二次式改写成 (px + q)(rx + s) = 0 的形式，然后分别令每个括号等于零。

For x² + bx + c = 0 with a = 1, find two numbers whose sum is b and whose product is c.

对于 a = 1 的 x² + bx + c = 0，找出两个数，使它们的和为 b、积为 c。

Example 1: Solve x² − 5x + 6 = 0. We need two numbers with sum 5 and product 6, so they are 2 and 3. Therefore (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

例 1：解 x² − 5x + 6 = 0。需要两数之和为 5、积为 6，因此取 2 和 3。于是 (x − 2)(x − 3) = 0，得 x = 2 或 x = 3。

When a ≠ 1, you may need to factorise brackets such as (mx + n)(px + q).

当 a ≠ 1 时，可能需要分解形如 (mx + n)(px + q) 的括号。

Example 2: Solve 2x² + 7x + 3 = 0. Testing factors of 2 and 3, we get (2x + 1)(x + 3) = 0. Hence x = −1/2 or x = −3.

例 2：解 2x² + 7x + 3 = 0。试取 2 与 3 的因子组合，得 (2x + 1)(x + 3) = 0。因此 x = −1/2 或 x = −3。

3. Solving by Completing the Square | 配方法

Completing the square converts the quadratic into the form (x + p)² + q = 0. This method works for any quadratic, even when factorisation is not obvious.

配方法将二次式转化为 (x + p)² + q = 0 的形式。这种方法适用于任何二次方程，即使因式分解不明显也能求解。

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

例：用配方法解 x² + 6x + 2 = 0。

(x + 3)² − 9 + 2 = 0 → (x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7

The number inside the bracket is always half of b. Here b = 6, so half of 6 is 3, and we write (x + 3)². Then subtract 3², which is 9, to keep the expression unchanged.

括号内的数始终是 b 的一半。此处 b = 6，故取 3，写成 (x + 3)²，再减去 3² 即 9，以保持原式不变。

If a ≠ 1, first divide the whole equation by a, then complete the square. For example, 2x² + 8x − 5 = 0 becomes x² + 4x − 5/2 = 0 before completing the square.

当 a ≠ 1 时，先将整个方程除以 a，再配方。例如 2x² + 8x − 5 = 0 先化为 x² + 4x − 5/2 = 0，再进行配方。

4. The Quadratic Formula | 求根公式

The quadratic formula provides a universal algebraic solution. For ax² + bx + c = 0, the roots are given by the following formula.

求根公式提供了一种通用的代数解法。对于 ax² + bx + c = 0，根由以下公式给出。

x = (−b ± √(b² − 4ac)) / 2a

This formula is stated on the IGCSE formula sheet, but you must know how to substitute values correctly and simplify the result.

该公式印在 IGCSE 公式表上，但你必须能正确代入数值并化简结果。

Example: Solve 2x² − 4x − 3 = 0 using the formula. Here a = 2, b = −4, c = −3. Substituting gives x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2.

例：用公式法解 2x² − 4x − 3 = 0。此处 a = 2，b = −4，c = −3。代入得 x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2。

Always write down the values of a, b and c first to avoid sign errors.

做题时先写出 a、b、c 的值，避免符号错误。
Do not cancel incorrectly: every term inside the numerator must be divided by 2a.

化简时切勿错误约分：分子中的每一项都必须除以 2a。
Simplify surds completely, for example √40 = 2√10.

根式要化简到底，例如 √40 = 2√10。

5. The Discriminant | 判别式

The expression b² − 4ac inside the square root is called the discriminant, often written as Δ or simply D. It tells us how many real roots a quadratic equation has without solving it fully.

求根公式根号内的表达式 b² − 4ac 被称为判别式，常用 Δ 或 D 表示。不用完全求解，就可以判断二次方程实根的个数。

Discriminant	Nature of roots	判别式	根的性质
b² − 4ac > 0	Two distinct real roots	b² − 4ac > 0	两个不相等的实根
b² − 4ac = 0	One repeated real root	b² − 4ac = 0	两个相等的实根（一个重根）
b² − 4ac < 0	No real roots	b² − 4ac < 0	无实数根

If the question asks for ‘real roots’ or ‘no real roots’, the discriminant is usually the fastest route. For example, find k such that x² + kx + 9 = 0 has one repeated root. The discriminant must equal zero: k² − 36 = 0, so k = 6 or k = −6.

当题目提到“有实数根”或“无实数根”时，通常用判别式求解最快。例如，求 k 使 x² + kx + 9 = 0 有重根。令判别式为零：k² − 36 = 0，得 k = 6 或 k = −6。

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

The roots of ax² + bx + c = 0 are the x-coordinates of the points where the curve y = ax² + bx + c intersects the x-axis. This connection between algebra and graphs is frequently tested in IGCSE.

方程 ax² + bx + c = 0 的根，就是曲线 y = ax² + bx + c 与 x 轴交点的横坐标。这种代数与图像之间的联系在 IGCSE 中经常考查。

If a > 0, the parabola opens upwards (a U shape).

若 a > 0，抛物线开口向上（呈 U 形）。
If a < 0, the parabola opens downwards (an inverted U shape).

若 a < 0，抛物线开口向下（呈倒 U 形）。
The axis of symmetry is the vertical line x = −b / 2a.

对称轴是竖直线 x = −b / 2a。
The vertex (turning point) lies on the axis of symmetry.

顶点（转向点）位于对称轴上。

You can also read approximate roots directly from a graph. A question may give a sketch of y = x² − 4x + 3 and ask you to state the roots. Since the curve crosses the x-axis at x = 1 and x = 3, these are the roots.

你也可以直接从图像中读出根的近似值。题目可能会给出 y = x² − 4x + 3 的草图并要求写出根。由于曲线与 x 轴交于 x = 1 和 x = 3，因此这两个就是方程的根。

7. Word Problems | 应用题

Many IGCSE questions require setting up a quadratic equation from a real-world situation, then solving it and interpreting the answers. The key steps are: define x, form the equation and solve.

许多 IGCSE 题目需要从实际情境中建立二次方程，然后求解并解释答案。关键步骤是：设 x、列方程、解方程。

Example: A rectangle has length (x + 4) cm and width (x − 1) cm. Its area is 30 cm². Find x and hence the dimensions of the rectangle.

例：一个长方形的长为 (x + 4) cm，宽为 (x − 1) cm，面积为 30 cm²。求 x 及长方形的长和宽。

(x + 4)(x − 1) = 30 → x² + 3x − 4 = 30 → x² + 3x − 34 = 0

Using the quadratic formula with a = 1, b = 3 and c = −34 gives x = (−3 ± √145) / 2. Since x must be positive, x = (−3 + √145) / 2 ≈ 4.52. The length is about 8.52 cm and the width about 3.52 cm.

用求根公式，取 a = 1，b = 3，c = −34，得 x = (−3 ± √145) / 2。由于 x 必须为正，所以 x = (−3 + √145) / 2 ≈ 4.52。长约为 8.52 cm，宽约为 3.52 cm。

Always reject a negative root if it does not make sense in the context of the problem, and always state your final answer with the correct units.

如果负根在题目情境中没有意义，应将其舍去，并记得在最终答案中加上正确的单位。

8. Common Mistakes to Avoid | 常见错误避坑

Students often lose marks in quadratic questions due to small but avoidable errors. Here are the most common ones.

学生在二次方程题目中经常因为小而可避免的错误丢分。以下是最常见的几类问题。

Dividing both sides by x. This loses the root x = 0. For example, x² = 3x must be solved as x² − 3x = 0, then x(x − 3) = 0.

两边同除以 x。这会丢失 x = 0 这个根。例如 x² = 3x 应化为 x² − 3x = 0，再分解为 x(x − 3) = 0。
Moving terms with the wrong sign. When rewriting an equation into the form ax² + bx + c = 0, every term must change sign when crossing the equals sign.

移项时符号错误。将方程写成 ax² + bx + c = 0 时，跨越等号移项的每一项都必须变号。
Forgetting the ± sign when taking square roots. From (x + 3)² = 7, you must write x + 3 = ±√7.

开平方时忘记 ±。由 (x + 3)² = 7，必须写成 x + 3 = ±√7。
Choosing the wrong signs in factor brackets. For x² − 5x + 6, both brackets must use minus signs, giving (x − 2)(x − 3).

因式分解时括号内符号选错。对于 x² − 5x + 6，两个括号都应为负号，即 (x − 2)(x − 3)。
Approximating too early. Do not round surds until the final step.

过早取近似值。应在最后一步再对根号进行四舍五入。

9. Exam-Style Practice | 真题演练

Try these questions in the style of IGCSE papers. Cover each answer with your hand, attempt the question, then check against the solutions below.

请尝试以下 IGCSE 风格的题目。先用手遮住答案，独立作答，再对照下方解析核对。

Question 1: Solve x² − 2x − 15 = 0.

题目 1：解 x² − 2x − 15 = 0。

Question 2: Solve 3x² + 5x − 2 = 0 using the quadratic formula.

题目 2：用求根公式解 3x² + 5x − 2 = 0。

Question 3: The equation kx² + 4x + 1 = 0 has two distinct real roots. Find the range of values of k, given that k ≠ 0.

题目 3：方程 kx² + 4x + 1 = 0 有两个不相等实根。已知 k ≠ 0，求 k 的取值范围。

Question	Solution	题目	答案
1	(x − 5)(x + 3) = 0, so x = 5 or x = −3	1	(x − 5)(x + 3) = 0，即 x = 5 或 x = −3
2	x = (−5 ± √(25 + 24)) / 6 = (−5 ± 7) / 6, so x = 1/3 or x = −2	2	x = (−5 ± √(25 + 24)) / 6 = (−5 ± 7) / 6，即 x = 1/3 或 x = −2
3	16 − 4k > 0, so k < 4, and k ≠ 0, therefore k < 4 but k ≠ 0	3	16 − 4k > 0，即 k < 4，且 k ≠ 0，故 k < 4 且 k ≠ 0

In Question 3, the discriminant is 16 − 4k. Since there are two distinct real roots, we require 16 − 4k > 0, giving k < 4. The condition k ≠ 0 keeps the equation quadratic.

题目 3 中，判别式为 16 − 4k。因为有两个不相等实根，所以需要 16 − 4k > 0，即 k < 4。条件 k ≠ 0 保证方程仍为二次方程。

10. Key Summary | 核心总结

Quadratic equations can be solved by four main methods: factorisation, completing the square, the quadratic formula, and graphical sketching. Choose the most efficient method based on the question.

解一元二次方程有四种主要方法：因式分解法、配方法、求根公式法以及图像法。请根据题目特点选择最高效的方法。

Method	Best used when	方法	适用情形
Factorisation	Roots are integers or simple fractions	因式分解法	根为整数或简单分数
Completing the square	Finding the turning point	配方法	求顶点坐标时
Quadratic formula	Factorisation is not obvious	求根公式法	因式分解不明显时
Graphical method	Approximate roots or checking answers	图像法	求近似根或检验答案

Remember the discriminant b² − 4ac: positive means two real roots, zero means one repeated root, negative means no real roots. With regular practice, you will soon recognise which method to use and avoid careless errors.

请牢记判别式 b² − 4ac：大于零有两个实根，等于零有一个重根，小于零无实根。只要勤加练习，你很快就能判断该用哪种方法，并避免粗心失分。

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Solving Quadratic Equations | 解一元二次方程