📚 Mastering Quadratic Equations | 掌握二次方程
Quadratic equations are among the most frequently tested topics in IGCSE Mathematics. They appear in Paper 2 and Paper 4, in both short-answer and problem-solving questions. Understanding the different methods of solution — factorisation, completing the square, and the quadratic formula — is essential for top marks.
二次方程是IGCSE数学中考查频率最高的内容之一，在Paper 2和Paper 4中都会出现，既有简答题也有应用解答题。掌握因式分解、配方和求根公式这三种解法，是获取高分的关键。
1. What Is a Quadratic Equation? | 什么是二次方程？
and a is not equal to 0. The highest power of the variable is 2, which is why it is called 'quadratic'.
二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数是2，因此称为“二次”方程。
ighest power is 1. If a = 0, the equation becomes linear, so the condition a ≠ 0 is vital.
例如，3x² − 5x + 2 = 0 是二次方程，而 2x + 7 = 0 是一次方程，因为它的最高次数是1。如果 a = 0，方程就变成一次方程，所以 a ≠ 0 这一条件非常重要。
2. Solving by Factorisation | 因式分解法
Factorisation is often the quickest method. The idea is to rewrite ax² + bx + c as a product of two brackets, then use the rule that if two numbers multiply to zero, at least one must be zero.
因式分解通常是最快捷的方法。其核心是把 ax² + bx + c 改写为两个括号的乘积，然后利用“两个数相乘为零，则至少有一个为零”的规则。
iply to 6 and add to −5. These are −2 and −3. Hence (x − 2)(x − 3) = 0, so x = 2 or x = 3.
考虑 x² − 5x + 6 = 0。我们需要找两个数，它们相乘得6，相加得 −5。这两个数是 −2 和 −3。因此 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。
or by grouping: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0, so x = −½ or x = −3.
当 x² 的系数不为1时，例如 2x² + 7x + 3 = 0，我们要找 a × c = 6 的因数，且它们相加等于 b = 7。这两个数是6和1。改写：2x² + 6x + x + 3 = 0，然后分组分解：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0，所以 x = −½ 或 x = −3。
3. Solving by Completing the Square | 配方法
Completing the square rewrites x² + bx as (x + b/2)² − (b/2)². This method works for any quadratic and is especially useful when the equation cannot be factorised easily.
配方法将 x² + bx 改写为 (x + b/2)² − (b/2)²。这种方法适用于任何二次方程，尤其是在方程不容易因式分解时特别有用。
9 + 5 = 0, so (x + 3)² = 4. Taking square roots gives x + 3 = ±2, hence x = −1 or x = −5.
示例：解 x² + 6x + 5 = 0。首先写出 x² + 6x = (x + 3)² − 9。于是方程变为 (x + 3)² − 9 + 5 = 0，即 (x + 3)² = 4。两边开平方得 x + 3 = ±2，因此 x = −1 或 x = −5。
ent. For example, 2x² − 8x + 3 = 0 becomes x² − 4x + 1.5 = 0 before completing the square.
当首项系数不是1时，先把整个方程除以该系数。例如，2x² − 8x + 3 = 0 先变为 x² − 4x + 1.5 = 0，然后再配方。
4. The Quadratic Formula | 求根公式
ormula can solve every quadratic equation. For ax² + bx + c = 0, the solution is given by:
求根公式可以解所有二次方程。对于 ax² + bx + c = 0，解为：
x = (−b ± √b² − 4ac) / 2a
Example: Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substitute into the formula:
示例：解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入公式：
x = (−3 ± √9 + 16) / 4 = (−3 ± 5) / 4
x = (−3 − 5)/4 = −2. Always write the answer as two separate solutions where appropriate.
因此 x = (−3 + 5)/4 = ½，或 x = (−3 − 5)/4 = −2。在需要时，务必把答案写成两个独立的解。
5. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant. It tells us how many real roots the equation has without solving it fully.
求根公式中根号内的表达式 b² − 4ac 称为判别式。它可以在不解方程的情况下告诉我们方程有多少个实数根。
| Discriminant Δ = b² − 4ac | Nature of roots |
| Δ > 0 | Two distinct real roots |
| Δ = 0 | One repeated real root |
| Δ < 0 | No real roots |
er hand, 4x² − 4x + 1 = 0 has Δ = 16 − 16 = 0, so it has exactly one repeated root, x = ½.
例如，x² + 2x + 5 = 0 的判别式 Δ = 4 − 20 = −16，因此没有实数根。另一方面，4x² − 4x + 1 = 0 的判别式 Δ = 16 − 16 = 0，因此只有一个重根 x = ½。
6. Sum and Product of Roots | 根的和与积
a. This is very useful for checking answers and for solving problems without finding the roots.
对于根为 α 和 β 的二次方程 ax² + bx + c = 0，两根之和为 α + β = −b/a，两根之积为 αβ = c/a。这在检查答案以及不求根直接解题时非常有用。
um 8/2 = 4 and product 6/2 = 3. The roots are 1 and 3, and indeed 1 + 3 = 4 and 1 × 3 = 3.
例如，方程 2x² − 8x + 6 = 0 的两根之和为 8/2 = 4，两根之积为 6/2 = 3。实际根为1和3，确实有 1 + 3 = 4 且 1 × 3 = 3。
uation from given roots. If the roots are p and q, the equation is x² − (p + q)x + pq = 0.
这一关系还允许我们从已知根构造二次方程。若根为 p 和 q，则方程为 x² − (p + q)x + pq = 0。
7. Graphs of Quadratic Functions | 二次函数图像
. When a > 0, the parabola opens upwards; when a < 0, it opens downwards. The x-intercepts are exactly the real roots of the equation.
y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。抛物线与 x 轴的交点正是方程的实数根。
equation to find the y-coordinate. The axis of symmetry is the vertical line x = −b/(2a).
抛物线顶点的 x 坐标为 x = −b/(2a)。把这个值代入方程即可求出 y 坐标。对称轴是直线 x = −b/(2a)。
intercepts are 1 and 3, and the y-intercept is 3. Plotting these points gives a clear parabola.
对于 y = x² − 4x + 3，对称轴为 x = 2，顶点为 (2, −1)。x 轴截距为1和3，y 轴截距为3。标出这些点就能画出清晰的抛物线。
8. Solving Word Problems | 应用题
Many exam questions present a real-world situation that leads to a quadratic equation. The key steps are: define a variable, build the equation from the information given, solve it, and then interpret the answer in context.
许多考试题目给出一个实际情境，最终会得到一个二次方程。关键步骤是：设定未知数、根据已知信息建立方程、求解，然后结合情境解释答案。
= 5 or x = −8. Since a width cannot be negative, the width is 5 cm and the length is 8 cm.
示例：一个长方形的长比宽多3厘米，面积为40平方厘米。设宽为 x 厘米，则长为 (x + 3) 厘米，于是 x(x + 3) = 40，即 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，所以 x = 5 或 x = −8。由于宽度不能为负数，所以宽为5厘米，长为8厘米。
Always check that your final answer satisfies the original problem. Negative lengths, times or counts are usually rejected in practical contexts.
务必检查最终的答案是否符合原题的实际意义。在实际情境中，长度、时间或数量为负数通常要舍去。
9. Common Exam Mistakes | 常见考试错误
or using the formula. For example, x² = 3x + 4 must be rewritten as x² − 3x − 4 = 0 first.
一个常见错误是忘记先将方程整理成 ax² + bx + c = 0 的形式，再进行因式分解或使用求根公式。例如，x² = 3x + 4 必须先改写为 x² − 3x − 4 = 0。
- Sign errors when substituting negative values into the quadratic formula. Use brackets carefully.
- 在把负值代入求根公式时出现符号错误。要仔细使用括号。
- Dropping the ± sign when taking square roots. Always write x = ±√k.
- 开平方时漏掉 ± 符号。务必写成 x = ±√k。
- Stopping after factorising without solving for x. The brackets must each be set to zero.
- 因式分解后没有继续求解 x。每个括号都必须分别等于零。
- Dividing both sides by a variable term such as x, which loses one of the solutions.
- 两边同时除以含变量的项（如 x），导致丢失一个解。
A good habit is to substitute your solutions back into the original equation to verify them. This takes a few seconds and can save many marks.
一个好习惯是把解代回原方程进行验证。这只需要几秒钟，却能避免大量丢分。
10. Practice Questions | 练习题
Try these questions yourself before checking the answers. They cover all the methods discussed above.
请先自己尝试以下题目，再对照答案。它们涵盖了上面讨论的所有方法。
- Solve x² − 7x + 12 = 0 by factorisation. | 用因式分解法解 x² − 7x + 12 = 0。
- Solve x² + 8x + 6 = 0 by completing the square, giving your answers to 2 decimal places. | 用配方法解 x² + 8x + 6 = 0，答案保留两位小数。
- Solve 3x² − 5x − 2 = 0 using the quadratic formula. | 用求根公式解 3x² − 5x − 2 = 0。
- Find the value of k for which kx² + 4x + 2 = 0 has exactly one real root. | 求 k 的值，使 kx² + 4x + 2 = 0 恰好有一个实数根。
r x = 4. 2. x = −1.35 or x = −6.65. 3. x = 2 or x = −1/3. 4. k = 2, since Δ = 16 − 8k = 0.
答案：1. x = 3 或 x = 4。2. x = −1.35 或 x = −6.65。3. x = 2 或 x = −1/3。4. k = 2，因为 Δ = 16 − 8k = 0。
11. Summary | 总结
Mastering quadratic equations requires knowing when to use each method. Factorisation is fastest when simple integer factors exist. Completing the square is reliable and reveals the vertex. The quadratic formula always works. The discriminant tells you the nature of the roots before you solve.
掌握二次方程的关键是知道何时使用哪种方法。当存在简单的整数因数时，因式分解最快。配方法可靠且能直接显示顶点。求根公式则永远有效。判别式能在求解前判断根的性质。
Practise each method until it becomes automatic. With regular revision, quadratic equations will become one of the easiest sections of the IGCSE Mathematics paper. Remember to always check your answers and write your final solutions clearly.
反复练习每种方法，直到熟练掌握。通过定期复习，二次方程将成为IGCSE数学试卷中最简单的部分之一。记住要始终检查答案，并清晰写出最终解。
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