📚 Quadratic Equations: Factorisation, Formula and Graphs | 二次方程：因式分解、求根公式与图像
Quadratic equations are a fundamental part of the IGCSE Mathematics syllabus. They appear in algebra, graphs, and word problems, and mastering them is essential for exam success. In this revision guide, we will explore the key methods, properties, and common pitfalls of quadratic equations.
二次方程是IGCSE数学课程的基础内容。它们出现在代数、图像和文字题中，掌握它们对考试成功至关重要。在本复习指南中，我们将探讨二次方程的关键方法、性质和常见错误。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The highest power of x is 2. For example, 2x² – 3x + 5 = 0 is a quadratic equation, while x³ – 2x = 0 is not.
二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。x 的最高次数为 2。例如，2x² – 3x + 5 = 0 是二次方程，而 x³ – 2x = 0 不是。
If a = 0, the equation becomes linear. Therefore, the condition a ≠ 0 is essential. In IGCSE problems, you may need to rearrange a given expression into this standard form before solving.
如果 a = 0，方程就变成了一次方程。因此，条件 a ≠ 0 是必不可少的。在IGCSE题目中，你可能需要先将给定表达式重新整理成标准形式，再求解。
(x – 3)(x + 2) = 0
The expanded form of the above is x² – x – 6 = 0, which is the standard form. Always check that your final equation is set equal to zero.
上面表达式的展开形式是 x² – x – 6 = 0，即标准形式。始终检查最终方程是否等于零。
2. Solving by Factorisation | 因式分解法求解
Factorisation is the quickest method when the quadratic expression factorises into two linear factors. The principle is: if the product of two expressions is zero, then at least one of them must be zero. This is known as the zero product property.
当二次表达式可以分解为两个一次因式时，因式分解法是最快捷的方法。其原则是：如果两个表达式的乘积为零，则其中至少一个必须为零。这被称为零乘积性质。
To solve 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. Thus, the factorised form is (x – 2)(x – 3) = 0.
要解 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，分解形式为 (x – 2)(x – 3) = 0。
Setting each factor to zero gives x – 2 = 0 or x – 3 = 0, so x = 2 or x = 3. Always check your solutions by substituting them back into the original equation.
令每个因式等于零，得到 x – 2 = 0 或 x – 3 = 0，因此 x = 2 或 x = 3。始终通过将解代回原方程进行验证。
For equations where the coefficient of x² is not 1, such as 2x² + 7x + 3 = 0, we use the method of grouping. Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: these are 1 and 6. Rewrite the middle term: 2x² + x + 6x + 3 = 0. Group and factorise: x(2x + 1) + 3(2x + 1) = 0, giving (2x + 1)(x + 3) = 0. Then x = -½ or x = -3.
对于 x² 系数不为 1 的方程，如 2x² + 7x + 3 = 0，我们使用分组法。将 a 与 c 相乘：2 × 3 = 6。寻找两个数，相乘得 6，相加得 7：这两个数是 1 和 6。重写中间项：2x² + x + 6x + 3 = 0。分组并因式分解：x(2x + 1) + 3(2x + 1) = 0，得到 (2x + 1)(x + 3) = 0。因此 x = -½ 或 x = -3。
3. The Quadratic Formula | 求根公式
When factorisation is difficult or impossible, we use the quadratic formula. For any quadratic equation ax² + bx + c = 0, the solutions are given by:
当因式分解困难或无法进行时，我们使用求根公式。对于任意二次方程 ax² + bx + c = 0，其解为：
x = (-b ± √(b² – 4ac)) / (2a)
This formula gives all real solutions, provided the discriminant b² – 4ac is non-negative. For example, solve 2x² – 4x – 3 = 0. Here a = 2, b = -4, c = -3. Substitute into the formula:
该公式给出所有实数解，前提是判别式 b² – 4ac 非负。例如，解 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。代入公式：
x = (4 ± √((-4)² – 4×2×(-3))) / (2×2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4
Simplifying, √40 = 2√10, so x = (4 ± 2√10) / 4 = 1 ± (√10)/2. The two solutions are approximately x ≈ 2.58 and x ≈ -0.58. Never forget the ± sign, as it gives two distinct solutions.
化简，√40 = 2√10，因此 x = (4 ± 2√10) / 4 = 1 ± (√10)/2。两个解约为 x ≈ 2.58 和 x ≈ -0.58。切勿遗漏 ± 号，因为它给出两个不同的解。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This is especially useful for finding the vertex of a parabola and solving equations without factorisation. Starting from x² + 6x + 5 = 0, we take half of the coefficient of x, which is 3, and square it to get 9. Then rewrite the equation as (x + 3)² – 9 + 5 = 0, simplifying to (x + 3)² – 4 = 0.
配方法将二次式重写为 a(x + p)² + q 的形式。这在求抛物线顶点和解方程时特别有用。从 x² + 6x + 5 = 0 开始，取 x 系数的一半，即 3，平方得到 9。然后将方程改写为 (x + 3)² – 9 + 5 = 0，化简为 (x + 3)² – 4 = 0。
Solving (x + 3)² – 4 = 0 gives (x + 3)² = 4, so x + 3 = ±2, hence x = -1 or x = -5. This method also reveals the minimum point of the curve y = (x + 3)² – 4, which is (-3, -4).
解 (x + 3)² – 4 = 0 得到 (x + 3)² = 4，所以 x + 3 = ±2，因此 x = -1 或 x = -5。该方法还揭示曲线 y = (x + 3)² – 4 的最低点为 (-3, -4)。
For equations with a coefficient of x² greater than 1, factor out that coefficient first. For example, 2x² + 8x – 1 = 0 becomes 2(x² + 4x) – 1 = 0, then 2[(x + 2)² – 4] – 1 = 0, giving 2(x + 2)² – 9 = 0.
对于 x² 系数大于 1 的方程，先提取该系数。例如，2x² + 8x – 1 = 0 变为 2(x² + 4x) – 1 = 0，然后 2[(x + 2)² – 4] – 1 = 0，得到 2(x + 2)² – 9 = 0。
5. The Discriminant | 判别式
The expression b² – 4ac is called the discriminant. It determines the nature of the roots of a quadratic equation without solving it fully. If b² – 4ac > 0, the equation has two distinct real roots. If b² – 4ac = 0, it has exactly one real root (a repeated root). If b² – 4ac < 0, there are no real roots.
表达式 b² – 4ac 被称为判别式。它无需完全求解即可确定二次方程根的性质。如果 b² – 4ac > 0，方程有两个不相等的实数根。如果 b² – 4ac = 0，方程恰有一个实数根（重根）。如果 b² – 4ac < 0，则没有实数根。
For example, in x² – 4x + 4 = 0, the discriminant is (-4)² – 4×1×4 = 16 – 16 = 0, so the equation has a repeated root x = 2. In x² + 2x + 5 = 0, the discriminant is 4 – 20 = -16, so there are no real solutions.
例如，在 x² – 4x + 4 = 0 中，判别式为 (-4)² – 4×1×4 = 16 – 16 = 0，因此方程有重根 x = 2。在 x² + 2x + 5 = 0 中，判别式为 4 – 20 = -16，因此没有实数解。
When a question asks for the number of roots or whether the curve intersects the x-axis, always use the discriminant. This saves time and avoids solving unnecessarily.
当题目询问根的个数或曲线是否与 x 轴相交时，始终使用判别式。这可以节省时间并避免不必要的求解。
| Discriminant b² – 4ac | Nature of roots |
| Positive (> 0) | Two distinct real roots |
| Zero (= 0) | One repeated real root |
| Negative (< 0) | No real roots |
上面的表格总结了判别式的三种情况。请记住，对于IGCSE，通常只考实数根的情况。
6. Solving Quadratic Equations by Graphs | 用图像解二次方程
The solutions of a quadratic equation ax² + bx + c = 0 are the x-coordinates of the points where the graph of y = ax² + bx + c crosses the x-axis. For example, the curve y = x² – 2x – 3 crosses the x-axis at x = -1 and x = 3, so the equation x² – 2x – 3 = 0 has solutions x = -1 and x = 3.
二次方程 ax² + bx + c = 0 的解是二次函数 y = ax² + bx + c 的图像与 x 轴交点的横坐标。例如，曲线 y = x² – 2x – 3 与 x 轴交于 x = -1 和 x = 3，因此方程 x² – 2x – 3 = 0 的解为 x = -1 和 x = 3。
If the curve does not cross the x-axis, the equation has no real roots. If it just touches the x-axis at one point, the equation has one repeated root. Drawing a accurate graph is time-consuming, but it is useful for checking approximate solutions or when an exact method is not obvious.
如果曲线不与 x 轴相交，则方程没有实数根。如果曲线与 x 轴相切于一点，则方程有一个重根。绘制精确图像比较费时，但可用于检查近似解或当精确方法不明显时。
For approximate solutions, you may be given a graph and asked to read off the roots. Alternatively, you may need to draw a straight line to solve a modified equation. For example, to solve x² – 2x – 4 = 0 graphically, you could sketch y = x² – 2x – 3 and then draw the line y = 1 to find where they intersect, since x² – 2x – 4 = 0 is equivalent to x² – 2x – 3 = 1.
对于近似解，你可能会被给出一张图并要求读出根。或者，你可能需要画一条直线来解修改后的方程。例如，要以图像解 x² – 2x – 4 = 0，你可以画出 y = x² – 2x – 3，然后画直线 y = 1，求它们的交点，因为 x² – 2x – 4 = 0 等价于 x² – 2x – 3 = 1。
7. Word Problems Leading to Quadratics | 二次方程文字题
Many IGCSE problems translate real-life situations into quadratic equations. Common contexts include area, product of consecutive numbers, and projectile motion. The key steps are: read carefully, define a variable, form an equation, solve it, and then decide which solution is sensible.
许多IGCSE题目将现实情境转化为二次方程。常见情境包括面积、连续整数的乘积和抛体运动。关键步骤是：仔细阅读，定义变量，建立方程，解方程，然后判断哪个解合理。
Example: The length of a rectangle is 3 cm longer than its width, and its area is 40 cm². Let the width be x cm. Then the length is (x + 3) cm. The area equation is x(x + 3) = 40, which expands to x² + 3x – 40 = 0. Factorise: (x + 8)(x – 5) = 0, so x = -8 or x = 5. Since width cannot be negative, the width is 5 cm and the length is 8 cm.
例如：一个长方形的长比宽多 3 cm，面积为 40 cm²。设宽为 x cm，则长为 (x + 3) cm。面积方程为 x(x + 3) = 40，展开为 x² + 3x – 40 = 0。因式分解：(x + 8)(x – 5) = 0，因此 x = -8 或 x = 5。由于宽度不能为负，所以宽为 5 cm，长为 8 cm。
Always check whether a negative solution fits the context. In geometry problems, lengths are positive. In age problems, ages cannot be negative. Reject impossible answers and state your final conclusion clearly.
始终检查负解是否符合实际情境。在几何题中，长度为正。在年龄题中，年龄不能为负。舍弃不合理的答案，并清晰地陈述最终结论。
8. Common Mistakes and Exam Tips | 常见错误与考试技巧
One common mistake is forgetting to set the equation to zero before factorising. For example, x² = 3x, if you divide both sides by x, you lose the solution x = 0. Instead, rewrite as x² – 3x = 0 and factorise to get x(x – 3) = 0, so x = 0 or x = 3.
一个常见错误是忘记在因式分解前将方程设为零。例如，x² = 3x，如果两边同除以 x，你会丢失解 x = 0。相反，应改写为 x² – 3x = 0，因式分解得 x(x – 3) = 0，因此 x = 0 或 x = 3。
Another mistake is misapplying the quadratic formula. Always write down the values of a, b, and c first, and be careful with negative signs. Also, do not confuse the discriminant with the formula itself.
另一个错误是误用求根公式。始终先写出 a、b、c 的值，并小心负号。此外，不要将判别式与公式本身混淆。
When using the quadratic formula, simplify surds where possible. For example, √12 should be written as 2√3. In exam questions, you may be required to give answers in exact form or to a certain degree of accuracy.
使用求根公式时，应尽可能化简根式。例如，√12 应写成 2√3。在考试题目中，你可能需要给出精确形式或保留一定精确度的答案。
- Always check if factorisation is possible before using the formula.
- Set the equation equal to zero before starting.
- Substitute your answers back to verify they work.
- Read the question carefully to see whether answers should be exact or rounded.
中文的考试技巧列表：
- 在使用公式前，总是先尝试因式分解。
- 在开始前，将方程设为等于零。
- 将答案代回原方程验证。
- 仔细阅读题目，确定答案是精确形式还是四舍五入。
9. Worked Example: Mixed Method Exercise | 混合方法示例
Let us solve 3x² + 5x – 2 = 0 using all three methods. First, factorisation: we need two numbers that multiply to 3 × (-2) = -6 and add to 5. These are 6 and -1. Rewrite: 3x² + 6x – x – 2 = 0. Group: 3x(x + 2) – 1(x + 2) = 0, giving (3x – 1)(x + 2) = 0. Thus x = 1/3 or x = -2.
我们用三种方法解 3x² + 5x – 2 = 0。首先，因式分解：我们需要两个数，相乘得 3 × (-2) = -6，相加得 5。这两个数是 6 和 -1。重写：3x² + 6x – x – 2 = 0。分组：3x(x + 2) – 1(x + 2) = 0，得到 (3x – 1)(x + 2) = 0。因此 x = 1/3 或 x = -2。
Second, completing the square: First divide by 3: x² + (5/3)x – 2/3 = 0. Take half of 5/3, which is 5/6, and square it to get 25/36. Rewrite as (x + 5/6)² – 25/36 – 2/3 = 0. Combine constants: -25/36 – 24/36 = -49/36. So (x + 5/6)² = 49/36, hence x + 5/6 = ±7/6, giving x = 2/6 = 1/3 or x = -12/6 = -2.
其次，配方法：先将两边除以 3：x² + (5/3)x – 2/3 = 0。取 5/3 的一半即 5/6，平方得 25/36。改写为 (x + 5/6)² – 25/36 – 2/3 = 0。合并常数项：-25/36 – 24/36 = -49/36。因此 (x + 5/6)² = 49/36，所以 x + 5/6 = ±7/6，得到 x = 2/6 = 1/3 或 x = -12/6 = -2。
Third, the quadratic formula: For a = 3, b = 5, c = -2, substitute into x = (-5 ± √(25 + 24)) / 6 = (-5 ± √49) / 6 = (-5 ± 7) / 6. This gives x = 2/6 = 1/3 or x = -12/6 = -2. All three methods agree.
第三，求根公式：对于 a = 3，b = 5，c = -2，代入 x = (-5 ± √(25 + 24)) / 6 = (-5 ± √49) / 6 = (-5 ± 7) / 6。得到 x = 2/6 = 1/3 或 x = -12/6 = -2。三种方法结果一致。
10. Practice Questions and Summary | 练习与总结
Try these short questions on your own:
请尝试以下简短练习：
- Solve x² – 7x + 12 = 0 by factorisation.
- Solve 2x² + 3x – 5 = 0 using the quadratic formula.
- Determine the number of real roots of x² + 4x + 5 = 0.
- Complete the square for x² – 6x + 2.
中文版练习题：
- 用因式分解法解 x² – 7x + 12 = 0。
- 用求根公式解 2x² + 3x – 5 = 0。
- 判断 x² + 4x + 5 = 0 的实数根的个数。
- 对 x² – 6x + 2 进行配方。
Answers: 1) x = 3 or x = 4. 2) x = 1 or x = -2.5. 3) Discriminant = 16 – 20 = -4, so no real roots. 4) (x – 3)² – 7.
答案：1) x = 3 或 x = 4。2) x = 1 或 x = -2.5。3) 判别式 = 16 – 20 = -4，所以没有实数根。4) (x – 3)² – 7。
In summary, quadratic equations can be solved by factorisation, the quadratic formula, completing the square, or graphing. Each method has its advantages, and the best choice depends on the problem. Always set the equation to zero, check for factorisability first, and verify your solutions.
总之，二次方程可以通过因式分解、求根公式、配方法或图像法求解。每种方法各有优势，最佳选择取决于具体问题。始终将方程设为零，先检查是否可因式分解，并验证你的解。
With regular practice, you will become confident in recognising which method to use and avoiding the common traps. Keep a clear working layout and show all steps in the exam to earn method marks.
通过定期练习，你将能够自信地判断应使用哪种方法，并避免常见陷阱。在考试中保持清晰的解题步骤，展示完整过程以获得方法分。
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