📚 Solving Quadratic Equations | 解一元二次方程
Quadratic equations are one of the most essential topics in IGCSE Mathematics. A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. In this article, you will learn every method you need to solve such equations accurately, from factorisation to the quadratic formula, together with the discriminant, graphical interpretation and common exam pitfalls.
一元二次方程是 IGCSE 数学中最核心的考点之一。二次方程是指形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。在这篇文章中，你将系统学会求解此类方程的各种方法：从因式分解到求根公式，再到判别式、函数图像以及考试中常见的易错点。
1. Standard Form and Key Definitions | 标准形式与基本定义
Before solving, you must recognise the standard form. A quadratic equation is written as ax² + bx + c = 0, where a is the coefficient of x², b is the coefficient of x, and c is the constant term. The degree of the equation is 2 because the highest power of x is x². A quadratic equation can have two real roots, one repeated root, or no real roots, depending on the value of the discriminant.
在求解之前，必须先辨认标准形式。一元二次方程写作 ax² + bx + c = 0，其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。方程的次数为 2，因为 x 的最高次幂是 x²。根据判别式的取值，二次方程可以有两个不同的实数根、一个重根，或者没有实数根。
For example, the equation 2x² − 4x + 1 = 0 has a = 2, b = −4 and c = 1. Always rearrange any given equation into this form before applying the methods below.
例如，方程 2x² − 4x + 1 = 0 中 a = 2，b = −4，c = 1。使用下面的方法之前，务必将任给方程化为这种标准形式。
2. Expanding and Factorising Quadratics | 展开与因式分解
Factorisation is the reverse process of expanding brackets. When you expand (x + p)(x + q), you obtain x² + (p + q)x + pq. This tells us that to factorise x² + bx + c, we must find two numbers whose product is c and whose sum is b.
因式分解是去括号展开的逆运算。将 (x + p)(x + q) 展开，可以得到 x² + (p + q)x + pq。这说明：要将 x² + bx + c 因式分解，就要找到两个数，使它们的乘积等于 c，同时它们的和等于 b。
Consider x² + 5x + 6. We need two numbers with product 6 and sum 5. The numbers 2 and 3 work because 2 × 3 = 6 and 2 + 3 = 5. Therefore:
以 x² + 5x + 6 为例。我们需要找两个数，乘积为 6 且和为 5。2 和 3 满足条件，因为 2 × 3 = 6，且 2 + 3 = 5。因此：
x² + 5x + 6 = (x + 2)(x + 3)
When a ≠ 1, such as 2x² + 7x + 3, you may need to use the method of grouping or trial and improvement. Multiply a and c to get 2 × 3 = 6. Find two numbers with product 6 and sum 7: they are 1 and 6. Rewrite the middle term:
当 a ≠ 1 时，例如 2x² + 7x + 3，需要使用分组法或试凑法。先计算 a × c = 2 × 3 = 6。找到乘积为 6 且和为 7 的两个数：1 和 6。然后改写中间项：
2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)
Notice that the two groups share the common factor (2x + 1). Master this skill because factorisation is the fastest method when the equation has nice integer roots.
注意两个分组都含有公因式 (2x + 1)。请熟练掌握这一技能，因为当方程有整数根时，因式分解是最快的求解方法。
3. Solving by Factorisation | 因式分解法解方程
Once the quadratic expression is factorised, use the zero product property: if A × B = 0, then A = 0 or B = 0. To solve x² − 5x + 6 = 0, first factorise:
一旦二次式完成因式分解，就可以利用「零乘积性质」：若 A × B = 0，则 A = 0 或 B = 0。解 x² − 5x + 6 = 0，先因式分解：
x² − 5x + 6 = (x − 2)(x − 3) = 0
Now set each bracket equal to zero:
接下来令每个括号等于零：
x − 2 = 0 ⇒ x = 2, or x − 3 = 0 ⇒ x = 3
So the roots are x = 2 and x = 3. Always check your answers by substituting them back into the original equation. When the coefficient of x² is negative, multiply the whole equation by −1 first, which makes factorisation easier.
因此根为 x = 2 和 x = 3。务必把答案代回原方程检验。当 x² 的系数为负数时，先将方程两边乘以 −1，这样更便于因式分解。
4. Solving by Taking Square Roots | 直接开平方法
If the equation can be written in the form (x + m)² = k, you can solve it directly by taking square roots. From (x + m)² = k, we get x + m = ±√k. Remember the ± sign: it produces both solutions.
若方程能写成 (x + m)² = k 的形式，就可以直接开平方求解。由 (x + m)² = k 可得 x + m = ±√k。切记加上 ± 号：这样才能得到两个解。
For example, solve (2x − 1)² = 9:
例如，解 (2x − 1)² = 9：
2x − 1 = ±3 ⇒ 2x = 1 ± 3 ⇒ x = 2 or x = −1
This method is especially useful when the quadratic is already a perfect square, or when a question asks for exact answers involving surds. If k is negative, the equation has no real solutions.
当二次式本身是完全平方式，或题目要求带根号的精确答案时，这种方法尤其好用。若 k 为负数，则方程无实数解。
5. Completing the Square | 配方法
Completing the square rewrites x² + bx as (x + b/2)² − (b/2)². This is based on the identity (x + p)² = x² + 2px + p². For x² + 6x + 2 = 0, take half of 6, which is 3, and write:
配方法将 x² + bx 改写为 (x + b/2)² − (b/2)²。其依据是恒等式 (x + p)² = x² + 2px + p²。对于 x² + 6x + 2 = 0，取 6 的一半是 3，于是：
x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7 = 0
(x + 3)² = 7 ⇒ x + 3 = ±√7 ⇒ x = −3 ± √7
If a ≠ 1, factor out a first. For example, 2x² + 8x + 5 = 0 becomes 2(x² + 4x) + 5 = 0, then complete the square inside the bracket. Completing the square also reveals the turning point of a quadratic graph: y = (x + 3)² − 7 has vertex (−3, −7).
若 a ≠ 1，先把 a 提出来。例如 2x² + 8x + 5 = 0 可写成 2(x² + 4x) + 5 = 0，再在括号内配方。配方法还能直接给出二次函数图像的顶点：y = (x + 3)² − 7 的顶点为 (−3, −7)。
6. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation, including those that cannot be factorised. For ax² + bx + c = 0, the solutions are given by:
求根公式可以解任何一元二次方程，包括无法因式分解的方程。对于 ax² + bx + c = 0，解为：
x = (−b ± √(b² − 4ac)) / (2a)
Let us solve 2x² − 4x − 3 = 0 using the formula. Here a = 2, b = −4, c = −3. Substitute carefully:
我们用公式解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。代入时务必小心：
x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± (√10)/2
Do not forget to change the sign of b. If b = −4, then −b = 4. Also remember that the entire expression is divided by 2a, not just the first term. Leave your answer in exact surd form unless the question asks for a decimal.
注意不要忘记改变 b 的符号。若 b = −4，则 −b 就是 4。还要注意是整个表达式除以 2a，而不是只有第一项。若题目未要求小数，请保留精确根号形式。
7. The Discriminant | 判别式
The expression Δ = b² − 4ac inside the square root is called the discriminant. It tells us the nature of the roots without solving the equation completely.
根号内的表达式 Δ = b² − 4ac 称为判别式。它无需完全解方程，就能判断根的性质。
| Discriminant / 判别式 | Nature of Roots / 根的性质 |
| Δ > 0 | Two distinct real roots / 两个不相等的实数根 |
| Δ = 0 | One repeated real root / 一个二重实数根 |
| Δ < 0 | No real roots / 无实数根 |
For the equation 3x² + 6x + 3 = 0, Δ = 36 − 36 = 0, so the equation has exactly one repeated root. You can verify this by noting that 3x² + 6x + 3 = 3(x + 1)² = 0. Examination questions often ask you to find the range of k such that a quadratic equation has two distinct real roots; simply set b² − 4ac > 0 and solve the resulting inequality.
对于方程 3x² + 6x + 3 = 0，Δ = 36 − 36 = 0，所以方程只有一个重根。你可以验证：3x² + 6x + 3 = 3(x + 1)² = 0。考试常要求你求 k 的范围，使某二次方程有两个不同的实数根；只需令 b² − 4ac > 0，再解这个不等式即可。
8. Solving Word Problems | 列方程解应用题
Quadratic equations frequently appear in word problems. Read the question carefully, define a variable, form an equation using the given information, and then solve it. Finally, check whether both answers make sense in the context of the problem.
二次方程经常出现在应用题中。先仔细读题，设未知数，根据题中信息列方程，再求解。最后必须检验两个答案在题目背景下是否合理。
Example: A rectangle has length (x + 3) cm and width x cm. Its area is 28 cm². Find x. The area gives:
例：一个长方形的长为 (x + 3) cm，宽为 x cm，面积为 28 cm²。求 x。由面积得：
x(x + 3) = 28 ⇒ x² + 3x − 28 = 0 ⇒ (x + 7)(x − 4) = 0
x = −7 or x = 4
Since x is a length, it cannot be negative, so the width is 4 cm and the length is 7 cm. Always reject solutions that violate the physical constraints of the problem, such as negative lengths or times.
由于 x 表示长度，不能为负数，所以宽为 4 cm，长为 7 cm。务必舍去不符合实际问题条件的解，例如负的长度或时间。
9. Graphs of Quadratic Functions | 二次函数图像
The graph of y = ax² + bx + c is a parabola. When a > 0, the parabola opens upwards and has a minimum point; when a < 0, it opens downwards and has a maximum point. The roots of the equation ax² + bx + c = 0 are exactly the x-intercepts of the parabola.
函数 y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，抛物线开口向上，有最低点；当 a < 0 时，抛物线开口向下，有最高点。方程 ax² + bx + c = 0 的根恰好就是抛物线与 x 轴交点的横坐标。
Consider y = x² − 2x − 3. Factorising gives y = (x − 3)(x + 1), so the x-intercepts are x = 3 and x = −1. The axis of symmetry lies halfway between the roots, at x = 1. Substituting x = 1 gives y = 1 − 2 − 3 = −4, so the vertex is (1, −4). The y-intercept is (0, −3). These five key points allow you to sketch the parabola quickly and accurately.
以 y = x² − 2x − 3 为例。因式分解得 y = (x − 3)(x + 1)，因此与 x 轴的交点横坐标为 x = 3 和 x = −1。对称轴位于两根中点处，即 x = 1。代入 x = 1，得 y = 1 − 2 − 3 = −4，所以顶点为 (1, −4)。与 y 轴的交点为 (0, −3)。利用这五个关键点，就能快速而准确地画出抛物线。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Students often lose marks by making avoidable errors. Here are the most frequent ones, together with advice for avoiding them. First, always rearrange the equation into the standard form ax² + bx + c = 0 before solving, so that you use the correct signs for a, b and c.
学生常常因为可以避免的错误而丢分。以下是最常见的几个问题及应对建议。首先，求解前务必把方程化为标准形式 ax² + bx + c = 0，这样 a、b、c 的符号才不会出错。
Second, do not divide both sides by x. For example, in x² = 5x, dividing by x gives x = 5, but you lose the root x = 0. Instead, rearrange: x² − 5x = 0, then factorise to get x(x − 5) = 0, giving x = 0 or x = 5.
第二，不要两边同除以 x。例如在 x² = 5x 中，两边除以 x 得到 x = 5，却漏掉了根 x = 0。正确做法是移项：x² − 5x = 0，再因式分解得 x(x − 5) = 0，所以 x = 0 或 x = 5。
Third, when using the quadratic formula, write down the values of a, b and c explicitly before substitution. This reduces sign errors with negative coefficients. Fourth, always show every step clearly, because method marks are awarded even if the final answer is wrong. Finally, check your roots by substituting them into the original equation.
第三，使用求根公式前，先把 a、b、c 的值明确写出来，这能减少负系数带来的符号错误。第四，每一步都要清楚写出，因为即使最终答案错误，方法分仍然可以获得。最后，将根代回原方程进行检验。
By mastering factorisation, completing the square, the quadratic formula and the discriminant, you can solve any quadratic equation that appears in the IGCSE examination. Practise a variety of problems, pay attention to the context of word problems, and always check your answers. With consistent effort, quadratics will become one of your strongest topics.
通过掌握因式分解、配方法、求根公式和判别式，你就能解出 IGCSE 考试中出现的任意一元二次方程。多做不同类型的练习，注意应用题的实际背景，并经常检验答案。只要坚持训练，二次方程一定会成为你最有把握的考点之一。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply