📚 Solving Quadratic Equations | 解一元二次方程
A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in almost every exam paper, either as a standalone question or as part of a larger problem involving graphs, areas, or kinematics. Mastering this topic is essential for achieving a high grade.
一元二次方程是 IGCSE 数学中最重要的话题之一。它几乎出现在每份试卷中，要么作为独立题目，要么作为涉及图像、面积或运动学等更大问题的一部分。掌握这一主题对于取得高分至关重要。
In this article, we will explore the standard form of a quadratic equation, the three main algebraic methods of solution — factorisation, completing the square, and the quadratic formula — as well as the discriminant and graphical interpretation. We will also look at common mistakes and real-world applications.
在本文中，我们将探讨一元二次方程的标准形式、三种主要的代数解法——因式分解法、配方法和求根公式法——以及判别式和图像解释。我们还会讨论常见错误和现实世界中的应用。
1. Standard Form | 标准形式
Every quadratic equation can be written in the standard form:
每个一元二次方程都可以写成标准形式：
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 crucial because if a = 0, the equation becomes linear, not quadratic.
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，因为如果 a = 0，方程就变成一次方程，而不是二次方程。
For example, 2x² + 5x − 3 = 0 has a = 2, b = 5, and c = −3. Always identify these three values before you choose a solution method.
例如，2x² + 5x − 3 = 0 中，a = 2，b = 5，c = −3。在选择解法之前，务必先确定这三个值。
Sometimes you must rearrange an equation into standard form first. For instance, 3x² = 7x − 2 becomes 3x² − 7x + 2 = 0 after moving all terms to one side.
有时你需要先将方程整理成标准形式。例如，3x² = 7x − 2 在将所有项移到一边后变成 3x² − 7x + 2 = 0。
2. Solving by Factorisation | 因式分解法
Factorisation is often the quickest method when the quadratic expression factorises neatly. The principle is simple: if the product of two factors is zero, then at least one of the factors must be zero. This is called the zero product property.
当二次表达式能够整齐地分解时，因式分解法往往是最快的方法。其原理很简单：如果两个因式的乘积为零，那么至少有一个因式必须为零。这称为零乘积性质。
Consider the equation x² + 5x + 6 = 0. We look for two numbers that multiply to give 6 and add to give 5. These numbers are 2 and 3, so we write:
考虑方程 x² + 5x + 6 = 0。我们寻找两个数，它们相乘得 6，相加得 5。这两个数是 2 和 3，因此我们写成：
(x + 2)(x + 3) = 0
Now we set each bracket equal to zero: x + 2 = 0 or x + 3 = 0. Therefore x = −2 or x = −3. The solution set is {−2, −3}.
现在我们令每个括号等于零：x + 2 = 0 或 x + 3 = 0。因此 x = −2 或 x = −3。解集为 {−2, −3}。
When the coefficient of x² is not 1, you may need to use the method of splitting the middle term or simply try pairs of factors. For example, 2x² + 7x + 3 = 0 factorises as (2x + 1)(x + 3) = 0, giving x = −½ or x = −3.
当 x² 的系数不为 1 时，你可能需要使用拆中项法或直接尝试因数对。例如，2x² + 7x + 3 = 0 分解为 (2x + 1)(x + 3) = 0，得到 x = −½ 或 x = −3。
3. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This method works for any quadratic equation, even when factorisation is difficult or impossible.
配方法将二次表达式改写为 a(x + p)² + q 的形式。这种方法适用于任何一元二次方程，即使因式分解很困难或不可能时也有效。
Let us solve x² + 6x + 2 = 0 by completing the square. First, take half of the coefficient of x, which is 6 ÷ 2 = 3, and write (x + 3)². Expanding (x + 3)² gives x² + 6x + 9, which is 7 more than our original expression. So we adjust:
让我们用配方法解 x² + 6x + 2 = 0。首先，取 x 系数的一半，即 6 ÷ 2 = 3，写成 (x + 3)²。展开 (x + 3)² 得到 x² + 6x + 9，这比原表达式多了 7。所以我们调整：
(x + 3)² − 7 = 0
Now solve: (x + 3)² = 7, so x + 3 = ±√7. Therefore x = −3 + √7 or x = −3 − √7.
现在求解：(x + 3)² = 7，所以 x + 3 = ±√7。因此 x = −3 + √7 或 x = −3 − √7。
When the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 5 = 0, factor out the 2: 2(x² + 4x) + 5 = 0, then complete the square inside: 2[(x + 2)² − 4] + 5 = 0, which simplifies to 2(x + 2)² − 3 = 0.
当 x² 的系数不为 1 时，先把它提取出来。对于 2x² + 8x + 5 = 0，提取 2：2(x² + 4x) + 5 = 0，然后在括号内配方：2[(x + 2)² − 4] + 5 = 0，化简为 2(x + 2)² − 3 = 0。
4. The Quadratic Formula | 求根公式法
The quadratic formula is a universal method that solves every quadratic equation, provided a ≠ 0. It is especially useful when factorisation fails or when the roots are irrational.
求根公式是一种通用方法，可以解出所有一元二次方程，前提是 a ≠ 0。当因式分解失败或根为无理数时，它尤其有用。
x = (−b ± √(b² − 4ac)) / 2a
Let us solve 3x² + 5x − 2 = 0 using the formula. Here a = 3, b = 5, c = −2. Substitute these values:
让我们用公式解 3x² + 5x − 2 = 0。这里 a = 3，b = 5，c = −2。代入这些值：
x = (−5 ± √(25 + 24)) / 6 = (−5 ± √49) / 6 = (−5 ± 7) / 6
This gives two roots: x = (−5 + 7) / 6 = 2/6 = 1/3, and x = (−5 − 7) / 6 = −12/6 = −2. So the solution set is {1/3, −2}.
这得到两个根：x = (−5 + 7) / 6 = 2/6 = 1/3，以及 x = (−5 − 7) / 6 = −12/6 = −2。所以解集为 {1/3, −2}。
You should memorise the formula and practise substituting carefully. A common error is forgetting the minus sign in front of b or misplacing the negative sign of c inside the square root.
你应该记住公式并仔细练习代入。一个常见错误是忘记 b 前面的负号，或弄错根号内 c 的负号。
5. The Discriminant | 判别式
The expression inside the square root, b² − 4ac, is called the discriminant and is denoted by Δ (Delta). It tells us the nature of the roots without solving the equation.
根号内的表达式 b² − 4ac 称为判别式，用 Δ（德尔塔）表示。它无需解方程就能告诉我们根的性质。
| Discriminant Δ | 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 | 一个重根（两个相等的实数根） |
| Δ < 0 | No real roots | 无实数根 |
For example, for x² − 4x + 4 = 0, the discriminant is (−4)² − 4 × 1 × 4 = 16 − 16 = 0, so the equation has exactly one repeated root. Indeed, it factorises as (x − 2)² = 0, giving x = 2.
例如，对于 x² − 4x + 4 = 0，判别式为 (−4)² − 4 × 1 × 4 = 16 − 16 = 0，所以方程恰好有一个重根。事实上，它分解为 (x − 2)² = 0，得到 x = 2。
For x² + 2x + 5 = 0, the discriminant is 4 − 20 = −16, which is negative. Therefore the equation has no real solutions. The graph of this function does not cross the x-axis.
对于 x² + 2x + 5 = 0，判别式为 4 − 20 = −16，是负数。因此方程没有实数解。该函数的图像不与 x 轴相交。
6. Graphical Interpretation | 图像解释
The graph of a quadratic function y = ax² + bx + c is a parabola. The solutions of ax² + bx + c = 0 correspond to the x-coordinates where the parabola crosses or touches the x-axis.
二次函数 y = ax² + bx + c 的图像是一条抛物线。ax² + bx + c = 0 的解对应于抛物线与 x 轴相交或相切处的 x 坐标。
If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point. This shape helps you sketch the graph quickly once you know the roots and the vertex.
如果 a > 0，抛物线开口向上，有最小值点。如果 a < 0，抛物线开口向下，有最大值点。一旦知道根和顶点，这种形状可以帮助你快速画出草图。
The vertex of the parabola has x-coordinate x = −b / (2a). You can find this by completing the square or using the symmetry of the roots. The y-coordinate is found by substituting this x-value back into the equation.
抛物线的顶点横坐标为 x = −b / (2a)。你可以通过配方法或利用根的对称性找到它。纵坐标通过将此 x 值代回方程求得。
7. Solving Equations that Require Rearrangement | 需要整理的方程
Many exam questions give you a quadratic equation that is not in standard form. You must first expand brackets, remove denominators, or collect all terms on one side before solving.
许多考试题目给出的方程并不是标准形式。你必须先展开括号、去掉分母，或将所有项移到一边，然后再求解。
For example, solve x(x + 3) = 10. Expanding gives x² + 3x = 10, then x² + 3x − 10 = 0. This factorises as (x + 5)(x − 2) = 0, so x = −5 or x = 2.
例如，解 x(x + 3) = 10。展开得 x² + 3x = 10，然后 x² + 3x − 10 = 0。分解为 (x + 5)(x − 2) = 0，所以 x = −5 或 x = 2。
Another common type involves fractions. For example, 6/x + x = 5. Multiply every term by x to get 6 + x² = 5x, then rearrange to x² − 5x + 6 = 0, which gives x = 2 or x = 3.
另一种常见类型涉及分数。例如，6/x + x = 5。每一项乘以 x 得 6 + x² = 5x，然后整理为 x² − 5x + 6 = 0，得到 x = 2 或 x = 3。
Always check your solutions by substituting them back into the original equation, especially when you multiplied by a variable, as this can introduce extraneous roots.
始终将解代回原方程进行检验，尤其是当你乘以一个变量时，因为这可能引入增根。
8. Word Problems | 应用题
Quadratic equations often arise from geometric or practical problems. A classic example is finding the dimensions of a rectangle given its area and perimeter.
一元二次方程通常来自几何或实际问题。一个经典的例子是已知面积和周长求矩形的尺寸。
Suppose a rectangle has an area of 24 cm² and its length is 5 cm longer than its width. Let the width be x cm. Then the length is (x + 5) cm, and the area equation is x(x + 5) = 24.
假设一个矩形的面积为 24 cm²，其长比宽长 5 cm。设宽为 x cm。则长为 (x + 5) cm，面积方程为 x(x + 5) = 24。
Expanding gives x² + 5x − 24 = 0, which factorises as (x + 8)(x − 3) = 0. Thus x = −8 or x = 3. Since a width cannot be negative, we reject x = −8. The width is 3 cm and the length is 8 cm.
展开得 x² + 5x − 24 = 0，分解为 (x + 8)(x − 3) = 0。因此 x = −8 或 x = 3。由于宽度不能为负，我们舍去 x = −8。宽为 3 cm，长为 8 cm。
In word problems, always interpret your answers in the context of the question. Reject any negative or unrealistic solutions and state your final answer with the correct units.
在应用题中，始终在问题情境中解释你的答案。舍去任何负数或不现实的解，并用正确的单位陈述最终答案。
9. Common Mistakes and Tips | 常见错误与建议
One of the most frequent mistakes in IGCSE exams is forgetting to set the equation to zero before factorising. If you factorise x² + 5x + 6 = 2 as (x + 2)(x + 3) = 2, you cannot conclude that x + 2 = 2 or x + 3 = 2. The zero product property only works when the product equals zero.
IGCSE 考试中最常见的错误之一是忘记在因式分解前将方程化为零。如果你把 x² + 5x + 6 = 2 分解为 (x + 2)(x + 3) = 2，你不能得出 x + 2 = 2 或 x + 3 = 2。零乘积性质只有在乘积等于零时才适用。
Another common error is incorrect sign handling in the quadratic formula. For the equation x² − 3x − 4 = 0, you must substitute b = −3 and c = −4 correctly. The discriminant becomes (−3)² − 4(1)(−4) = 9 + 16 = 25.
另一个常见错误是在求根公式中处理符号不当。对于方程 x² − 3x − 4 = 0，你必须正确代入 b = −3 和 c = −4。判别式变为 (−3)² − 4(1)(−4) = 9 + 16 = 25。
To avoid careless mistakes, follow a consistent routine: write down a, b, c; show all substitution steps; simplify carefully using a calculator if allowed; and check both roots in the original equation. Also, learn to recognise perfect squares like (x − 3)² = x² − 6x + 9, as these factorise instantly.
为避免粗心错误，请遵循一致的步骤：写下 a、b、c；写出所有代入步骤；如果允许使用计算器则仔细化简；并在原方程中检验两个根。同时，学会识别完全平方式，如 (x − 3)² = x² − 6x + 9，因为这些可以立即分解。
10. Summary and Exam Strategy | 总结与应试策略
To summarise, always identify whether the equation is in standard form, then choose the most efficient method. Factorisation is fastest for simple integer roots; completing the square works everywhere and also gives the vertex; the quadratic formula is the most reliable universal tool; and the discriminant tells you the nature of the roots upfront.
总结一下，始终先判断方程是否处于标准形式，然后选择最高效的方法。因式分解法对简单整数根最快；配方法适用于所有情况，还能给出顶点；求根公式是最可靠的通用工具；判别式则提前告诉你根的性质。
In the exam, use the discriminant as a quick sanity check. If the question asks for answers to 2 decimal places, use the quadratic formula and your calculator. If it asks for exact answers, prefer factorisation or completing the square.
在考试中，使用判别式作为快速合理性检查。如果题目要求答案保留到两位小数，请使用求根公式和计算器。如果要求精确答案，则优先使用因式分解法或配方法。
Finally, practise a wide variety of questions: equations with fractions, equations from graphs, and word problems. The more familiar you are with the patterns, the more confident you will be in the real exam. Remember that examiners reward clear working, so always show every step.
最后，练习各种类型的题目：含分数的方程、来自图像的方程以及应用题。你对这些题型越熟悉，在真正的考试中就会越自信。记住，阅卷老师会奖励清晰的解题过程，所以务必展示每一步。
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