📚 Mastering Quadratic Equations for Cambridge IGCSE Mathematics | 掌握剑桥 IGCSE 数学二次方程求解
Quadratic equations are one of the most important topics in the Cambridge IGCSE Mathematics syllabus. They appear regularly in Paper 1 and Paper 2, as well as in problem-solving questions involving area, motion, and optimisation. Understanding how to solve a quadratic equation accurately is essential for building confidence in algebra and for handling more advanced topics such as functions and graphs.
二次方程是剑桥 IGCSE 数学大纲中最重要的主题之一。它们经常出现在 Paper 1 和 Paper 2 中，也出现在涉及面积、运动和最优化的应用问题中。准确掌握二次方程的求解方法对于建立代数信心以及处理函数和图像等更高级主题至关重要。
1. What Is a Quadratic Equation? | 什么是二次方程？
and a ≠ 0. For example, x² - 3x + 2 = 0 and 2x² + 5x - 3 = 0 are both quadratic equations.
二次方程是一个次数为 2 的多项式方程，也就是说变量最高次幂是 2。它的一般形式包含 x² 项，并可写成标准形式 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。例如 x² - 3x + 2 = 0 和 2x² + 5x - 3 = 0 都是二次方程。
solutions, one solution, or no real solutions. This is because its graph is a parabola, and the solutions correspond to the points where the parabola crosses the x-axis.
与一次方程（如 2x + 3 = 7）不同，二次方程可能有两个解、一个解或没有实数解。这是因为它的图像是一条抛物线，而解对应于抛物线与 x 轴相交的点。
2. The Standard Form ax² + bx + c = 0 | 标准形式 ax² + bx + c = 0
en often multiplied by -1 to give x² - 3x + 4 = 0. In this case, a = 1, b = -3, and c = 4.
在求解二次方程之前，必须先将它整理成标准形式。这样才能正确识别系数 a、b 和 c。例如，方程 3x - x² = 4 应整理为 -x² + 3x - 4 = 0，然后通常乘以 -1 得到 x² - 3x + 4 = 0。此时 a = 1，b = -3，c = 4。
h is linear, not quadratic. Cambridge exam questions often test whether you can rearrange an equation and identify the coefficients before applying a method.
条件 a ≠ 0 至关重要。如果 a = 0，方程就变成 bx + c = 0，这是一次方程而不是二次方程。剑桥考试题目经常考查你是否能先整理方程并识别系数，然后再应用某种方法。
3. Solving by Factorisation | 因式分解法
ne factor must equal zero, so px + q = 0 or rx + s = 0. This gives the solutions directly.
当二次方程有简单的整数根时，因式分解通常是最快的方法。其思路是将 ax² + bx + c 写成两个一次括号的乘积：(px + q)(rx + s) = 0。如果乘积等于零，那么至少有一个因式为零，因此 px + q = 0 或 rx + s = 0。这样可以直接得到解。
use (-2) × (-3) = 6 and (-2) + (-3) = -5. Therefore the factorised form is (x - 2)(x - 3).
当 a = 1 时，你需要找到两个数，它们的乘积等于 c，和等于 b。例如，在 x² - 5x + 6 中，这两个数是 -2 和 -3，因为 (-2) × (-3) = 6 且 (-2) + (-3) = -5。因此因式分解形式为 (x - 2)(x - 3)。
When a ≠ 1, the method is similar but requires extra care with the coefficient of x². You can use the ac method or trial and error. For 2x² + 5x - 3, we need factors of 2 and -3 that combine to give 5. The correct factorisation is (2x - 1)(x + 3).
当 a ≠ 1 时，方法类似，但需要额外注意 x² 的系数。可以使用 ac 方法或试凑法。对于 2x² + 5x - 3，我们需要 2 和 -3 的因数组合来得到 5。正确的因式分解为 (2x - 1)(x + 3)。
4. Worked Example: Factorisation | 例题：因式分解法
Worked Example 1: Solve the equation x² - 7x + 10 = 0.
例题 1：解方程 x² - 7x + 10 = 0。
ct 10 and sum -7. The numbers are -2 and -5 because (-2) × (-5) = 10 and (-2) + (-5) = -7.
步骤 1：识别系数。这里 a = 1，b = -7，c = 10。找到乘积为 10、和为 -7 的两个数。这两个数是 -2 和 -5，因为 (-2) × (-5) = 10 且 (-2) + (-5) = -7。
: Write the factorised form: x² - 7x + 10 = (x - 2)(x - 5). Set each factor equal to zero.
步骤 2：写出因式分解形式：x² - 7x + 10 = (x - 2)(x - 5)。令每个因式等于零。
x - 2 = 0 or x - 5 = 0
on to obtain x = 2 or x = 5. Always check by substituting back into the original equation.
步骤 3：解每个一次方程得到 x = 2 或 x = 5。务必代回原方程检验。
Worked Example 2: Solve 2x² + 5x - 3 = 0 by factorisation.
例题 2：用因式分解法解 2x² + 5x - 3 = 0。
1)(x + 3). Expand to check: 2x² + 6x - x - 3 = 2x² + 5x - 3, which is correct. Therefore:
我们需要两个括号 (2x + p)(x + q)，使得 2q + p = 5 且 pq = -3。尝试 p = -1 和 q = 3，得到 (2x - 1)(x + 3)。展开验证：2x² + 6x - x - 3 = 2x² + 5x - 3，正确。因此：
2x - 1 = 0 or x + 3 = 0
So x = 1/2 or x = -3. The solutions are x = 1/2 and x = -3.
所以 x = 1/2 或 x = -3。方程的解为 x = 1/2 和 x = -3。
5. Completing the Square | 配方法
ted accordingly. This form reveals the vertex of the parabola and makes solving straightforward.
配方法是一种强大的技巧，适用于任何二次方程，即使因式分解困难或不可能时也能使用。其目标是将 ax² + bx + c 改写成 a(x + p)² + q 的形式，其中 p = b/(2a)，q 相应调整。这个形式能够揭示抛物线的顶点，并使求解变得直接。
ssential because (x + 3)² expands to x² + 6x + 9, so we must subtract 9 to keep the expression equal.
对于 a = 1 的简单情形，取 x 的系数，将其减半，再平方。对于 x² + 6x，6 的一半是 3，3² = 9，所以 x² + 6x = (x + 3)² - 9。这个恒等式非常关键，因为 (x + 3)² 展开为 x² + 6x + 9，因此必须减去 9 才能保持表达式相等。
16, which gives (x + 3)² = 16. Taking square roots yields x + 3 = ±4, so x = 1 or x = -7.
用配方法解方程时，首先将常数项移到等号另一边。对于 x² + 6x - 7 = 0，重写为 x² + 6x = 7。然后在两边同时加 9：x² + 6x + 9 = 16，得到 (x + 3)² = 16。取平方根得到 x + 3 = ±4，因此 x = 1 或 x = -7。
6. Worked Example: Completing the Square | 例题：配方法
Worked Example 3: Solve x² - 4x - 5 = 0 by completing the square.
例题 3：用配方法解 x² - 4x - 5 = 0。
Step 1: Move the constant term to the right-hand side: x² - 4x = 5.
步骤 1：将常数项移到等号右边：x² - 4x = 5。
of x, which is -4/2 = -2, and square it: (-2)² = 4. Add 4 to both sides: x² - 4x + 4 = 9.
步骤 2：取 x 的系数的一半，即 -4/2 = -2，再平方：(-2)² = 4。在两边同时加 4：x² - 4x + 4 = 9。
des: x - 2 = ±3. This gives two equations: x - 2 = 3 or x - 2 = -3. Hence x = 5 or x = -1.
步骤 3：将左边写成完全平方：(x - 2)² = 9。然后对两边取平方根：x - 2 = ±3。这给出两个方程：x - 2 = 3 或 x - 2 = -3。因此 x = 5 或 x = -1。
Completing the square is also the method used to derive the quadratic formula. In the Cambridge IGCSE, you may be asked to express a quadratic in the form a(x + p)² + q or to find the coordinates of the vertex from this form.
配方法也是推导求根公式所用的方法。在剑桥 IGCSE 中，你可能会被要求将二次函数表示为 a(x + p)² + q 的形式，或从这个形式求出顶点坐标。
7. The Quadratic Formula | 求根公式
g any quadratic equation. It states that for ax² + bx + c = 0, the solutions are given by:
求根公式是求解任何二次方程最通用的方法。它指出，对于 ax² + bx + c = 0，解为：
x = [-b ± √b² - 4ac] / (2a)
This formula is derived by completing the square on the general quadratic equation. It is especially useful when factorisation is not obvious, when the coefficients are large, or when the roots are irrational. You simply substitute the values of a, b, and c and simplify carefully.
该公式是通过对一般二次方程进行配方法推导出来的。当因式分解不明显、系数较大或根为无理数时，它尤其有用。你只需代入 a、b 和 c 的值并仔细化简即可。
b. For example, in x² - 3x - 2 = 0, a = 1, b = -3, and c = -2. The formula gives x = [3 ± √9 + 8] / 2 = [3 ± √17] / 2.
在使用公式之前，务必记住将方程写成标准形式。一个常见错误是在方程不等于零时使用公式，或者弄错 b 的符号。例如，在 x² - 3x - 2 = 0 中，a = 1，b = -3，c = -2。公式给出 x = [3 ± √9 + 8] / 2 = [3 ± √17] / 2。
8. Worked Example: Quadratic Formula | 例题：求根公式
Worked Example 4: Solve 2x² - 3x - 2 = 0 using the quadratic formula.
例题 4：使用求根公式解 2x² - 3x - 2 = 0。
Step 1: Identify a = 2, b = -3, c = -2. Substitute into the formula:
步骤 1：识别 a = 2，b = -3，c = -2。代入公式：
x = [-(-3) ± √((-3)² - 4 × 2 × (-2))] / (2 × 2)
(-3)² = 9, and -4 × 2 × (-2) = 16, so the expression under the root is 9 + 16 = 25. Thus:
步骤 2：化简根号内部：(-3)² = 9，且 -4 × 2 × (-2) = 16，因此根号下为 9 + 16 = 25。于是：
x = [3 ± √25] / 4 = [3 ± 5] / 4
solutions: x = (3 + 5)/4 = 8/4 = 2, and x = (3 - 5)/4 = -2/4 = -1/2. So x = 2 or x = -1/2.
步骤 3：这给出两个解：x = (3 + 5)/4 = 8/4 = 2，以及 x = (3 - 5)/4 = -2/4 = -1/2。所以 x = 2 或 x = -1/2。
Worked Example 5: Solve x² + 2x - 6 = 0, giving answers in surd form.
例题 5：解 x² + 2x - 6 = 0，答案用根式表示。
Here a = 1, b = 2, c = -6. Substitute: x = [-2 ± √4 + 24] / 2 = [-2 ± √28] / 2 = [-2 ± 2√7] / 2 = -1 ± √7.
这里 a = 1，b = 2，c = -6。代入：x = [-2 ± √4 + 24] / 2 = [-2 ± √28] / 2 = [-2 ± 2√7] / 2 = -1 ± √7。
9. The Discriminant and Nature of Roots | 判别式与根的性质
² - 4ac. It tells you about the nature of the roots without actually solving the equation.
判别式是求根公式中平方根号下的部分，用 Δ = b² - 4ac 表示。它无需实际解方程就能告诉你根的性质。
- If Δ > 0, the equation has two distinct real roots.
- 如果 Δ > 0，方程有两个不同的实数根。
- , the equation has exactly one real root, sometimes called a repeated root or double root.
- 如果 Δ = 0，方程恰好有一个实数根，有时称为重根或二重根。
- If Δ < 0, the equation has no real roots. The solutions would be complex numbers, which are not covered in the IGCSE syllabus.
- 如果 Δ < 0，方程没有实数根。解将是复数，这不在 IGCSE 大纲范围内。
ss than zero. Therefore the equation has no real solutions, and its graph does not cross the x-axis.
例如，考虑 x² - 4x + 5 = 0。这里 a = 1，b = -4，c = 5。判别式为 (-4)² - 4 × 1 × 5 = 16 - 20 = -4，小于零。因此该方程没有实数解，其图像不与 x 轴相交。
10. Choosing the Best Method | 选择最佳方法
In an exam, you should choose the most efficient method based on the equation given. The table below summarises the strengths of each method.
在考试中，你应该根据所给方程选择最有效的方法。下表总结了每种方法的优点。
| Method | Best used when | 方法 | 适用情况 |
|---|---|---|---|
| Factorisation | Roots are integers or simple fractions | 因式分解法 | 根为整数或简单分数 |
| Completing the square | Vertex form required or leading coefficient 1 | 配方法 | 需要顶点形式或首项系数为 1 |
| Quadratic formula | Any equation, especially non-factorable ones | 求根公式 | 任何方程，尤其是不可因式分解的 |
If the question asks you to give answers to a certain number of decimal places or significant figures, the quadratic formula is often the safest choice. If the equation has small coefficients and you can spot the factors quickly, factorisation saves time. Completing the square is particularly useful when the question explicitly asks for the vertex or the maximum/minimum value.
如果题目要求答案保留一定的小数位数或有效数字，求根公式通常是最安全的选择。如果方程系数较小且你能快速看出因式，因式分解法可以节省时间。当题目明确要求求顶点或最大值/最小值时，配方法特别有用。
11. Common Mistakes and Tips | 常见错误与建议
3x = 10, students may try to factorise x² - 3x instead of first writing x² - 3x - 10 = 0.
在剑桥 IGCSE 数学考试中，很多分数是因为解二次方程时可避免的错误而丢失的。一个常见的错误是在因式分解前忘记将方程设为零。例如，对于 x² - 3x = 10，学生可能试图对 x² - 3x 进行因式分解，而不是先写成 x² - 3x - 10 = 0。
Another common error is mishandling negative signs. In the quadratic formula, if b is negative, -b becomes positive. Also, when calculating b² - 4ac, remember that subtracting a negative product increases the value. Always double-check the signs of a, b, and c after rearranging.
另一个常见错误是处理负号不当。在求根公式中，如果 b 为负，-b 就变为正。此外，在计算 b² - 4ac 时，要记住减去一个负的乘积会使数值增大。重新整理后务必仔细检查 a、b、c 的符号。
te each value back into the original left-hand side to confirm it equals zero. This takes only a few seconds and can prevent elementary mistakes.
一个有用的技巧是通过代入来验算你的解。如果你解 x² + 5x + 6 = 0 得到 x = -2 和 x = -3，将每个值代回原方程左侧，确认它等于零。这只需几秒钟，却能防止低级错误。
12. Exam-Style Practice | 考试类型练习
Try these exam-style questions to test your understanding. Write full working, as Cambridge IGCSE awards method marks even if the final answer is incorrect.
尝试以下考试类型的题目来检验你的理解。请写出完整步骤，因为剑桥 IGCSE 即使最终答案错误也会给方法分。
Practice Question 1: Solve x² - 4x - 12 = 0 by factorisation.
练习题 1：用因式分解法解 x² - 4x - 12 = 0。
We need two numbers with product -12 and
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