📚 Solving Quadratic Equations for IGCSE Edexcel Maths | IGCSE Edexcel 数学：解二次方程
efficiently and accurately is essential. This guide walks you through every tested method – factorisation, completing the square, and the quadratic formula – will be able to choose the fastest strategy for any quadratic you encounter in your exam.
二次方程是 IGCSE Edexcel 数学大纲中代数问题解决的基石。无论你是正在模拟抛体运动、优化面积，还是破解数字谜题，能够高效准确地求解形如 ax² + bx + c = 0 的方程都是必不可少的。本指南将带你逐一掌握所有考试中要求的方法——因式分解法、配方法和二次公式法，同时强调判别式的作用、常见陷阱以及解的图形意义。学完之后，你将能够在考试中针对任何二次方程选择最快捷的求解策略。
1. What Is a Quadratic Equation? | 什么是二次方程？
the standard form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The name ‘quadratic’ comes from the Latin word ‘ algebraic contexts, as well as in word problems involving area, number relationships and kinematics.
一元二次方程是任何可以整理成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是实数且 a ≠ 0。“二次”这一名称来源于拉丁语“quadratus”（意为正方形），因为未知数 x 的最高次幂是 2。在 IGCSE Edexcel 考试规范中，你会在纯代数的情境以及涉及面积、数字关系和运动学的文字题中遇到二次方程。
2. Standard Form and Coefficients | 标准形式与系数
are in descending powers of x. For example, 3x + 2x² = 7 should be rearranged to 2x² + 3x – 7 = 0. Here a = 2, b = 3 and c = – these values. Never forget that the sign belongs to the coefficient; in the expression x² – 5x + 6, b is –5, not 5.
在尝试解二次方程之前，你必须对方程进行整理，使等式一边为零，并且各项按 x 的降幂排列。例如，3x + 2x² = 7 应整理为 2x² + 3x – 7 = 0。这时 a = 2，b = 3，c = –7。正确识别 a、b、c 至关重要，因为二次公式和判别式都依赖于这些数值。永远不要忘记正负号属于系数本身；在表达式 x² – 5x + 6 中，b 是 –5，而不是 5。
Standard form: ax² + bx + c = 0 (a ≠ 0)
3. Solving by Factorisation | 因式分解法
actorisation only works smoothly when the roots are rational, but it is a favourite in non‑calculator IGCSE papers because it requires no complex arithmetic.
当二次方程可以表示为两个一次因式的乘积时，因式分解通常是最快的方法。对于方程 x² + 7x + 12 = 0，我们需要找到两个相乘得 12（c）且相加得 7（b）的数。这两个数是 3 和 4，因此我们可以写成 (x + 3)(x + 4) = 0。因式分解法只有在根为有理数时才能顺利使用，但它在 IGCSE 非计算器试卷中特别受欢迎，因为它不需要复杂的运算。
4. The Zero Product Property | 零乘积性质
ero, then at least one of the factors must be zero. Therefore, setting x + p = 0 gives x = –p, and setting x + q = 0 gives x = – are setting each bracket equal to zero to gain full marks in Edexcel structured questions.
一旦你得到了如 (x + p)(x + q) = 0 的因式分解形式，零乘积性质指出：如果两个因式的乘积为零，那么至少其中一个因式必须为零。因此，令 x + p = 0 得 x = –p，令 x + q = 0 得 x = –q。这一步骤在匆忙写出答案时常常被忽略；一定要清晰地展示你将每个括号设为零的过程，以便在 Edexcel 结构化问题中拿到全部分数。
If (A)(B) = 0, then A = 0 or B = 0.
5. Difference of Two Squares | 平方差公式
l quadratic pattern that IGCSE examiners love to test is the difference of two squares: a² – b² = (a – b)(a + b). For example, to solve x² – 25 = 0, recognise that 25 is 5², so the equation becomes (x – 5)(x + 5) = 0, giving x = 5 or x = –5. This technique also appears when the equation is not initially equal to zero; x² – 16 = 9 must be rearranged to x² – ch out for hidden differences of squares where both terms are perfect squares, such as 4x² – 49 = (2x – 7)(2x + 7).
IGCSE 考官喜欢考察的一种特殊二次方程模式是平方差公式：a² – b² = (a – b)(a + b)。例如，要解 x² – 25 = 0，识别出 25 是 5²，于是方程变为 (x – 5)(x + 5) = 0，得到 x = 5 或 x = –5。当方程最初不等于零时，这一技巧也会出现；x² – 16 = 9 必须先整理成 x² – 25 = 0，然后再进行因式分解。要留意那些隐藏的平方差，其中两项都是完全平方，例如 4x² – 49 = (2x – 7)(2x + 7)。
6. Completing the Square | 配方法
uation, even when factorisation fails. The goal is to rewrite ax² + bx + c in the form a(x – 0, you halve the coefficient of x to get 3, square it to obtain 9, and then write (x + 3)² – 9 + 1 = 0, which simplifies to (x + 3)² = 8. Solving gives x = –3 ± √8 = –3 ± 2√2. This method is also the key to deriving the quadratic formula and finding the vertex of a parabola, a skill linked directly to graphs in the syllabus.
配方法是一种功能强大的方法，因为它适用于任何二次方程，即使因式分解行不通。其目标是将 ax² + bx + c 重写成 a(x – h)² + k 的形式。对于像 x² + 6x + 1 = 0 这样简单的首一二次方程，你将 x 的系数折半得 3，平方得 9，然后写成 (x + 3)² – 9 + 1 = 0，化简为 (x + 3)² = 8。解出得 x = –3 ± √8 = –3 ± 2√2。这种方法也是推导二次公式和求抛物线顶点的关键，这一技能与大纲中的图像部分直接相关。
x² + bx = (x + b/2)² – (b/2)²
7. Quadratic Formula | 二次公式
o use it confidently. For any quadratic ax² + bx + c = 0, the solutions are given by x = [ –b ± √b² – 4ac es of a, b and c with careful attention to signs and arithmetic. For instance, solving 2x² – 4x – 3 = 0 yields x = [ –(–4) ± √((–4)² – 4×2×(–3)) ] / (2×2) = [ 4 ± √16 + 24 ] / 4 = [ 4 ± √40 ] / 4. Simplify √40 to 2√10, then divide each term by 2: x = 1 ± (√10)/2. Always show your substitution step in the exam to secure method marks.
二次公式是通用的求解工具，在 Edexcel IGCSE 公式表上提供，但你必须懂得如何自信地使用它。对于任何二次方程 ax² + bx + c = 0，解由 x = [ –b ± √b² – 4ac ] / 2a 给出。代入 a、b、c 的值时要格外注意正负号和运算。例如，解 2x² – 4x – 3 = 0 得 x = [ –(–4) ± √((–4)² – 4×2×(–3)) ] / (2×2) = [ 4 ± √16 + 24 ] / 4 = [ 4 ± √40 ] / 4。将 √40 简化为 2√10，然后各项除以 2：x = 1 ± (√10)/2。在考试中一定要展示代入步骤，以获得方法分。
8. Discriminant and Nature of Roots | 判别式与根的性质
The expression under the square root sign, D = b² – al root (a repeated root). If D < 0, there are no real roots, although the equation has two complex roots – E questions often ask you to find the value of k that gives equal roots; you simply set b² – 4ac = 0 and solve for k.
根号下的表达式 D = b² – 4ac 被称为判别式。它能在不实际求解的情况下揭示二次方程有多少个实数解。如果 D > 0，则有两个不同的实数根。如果 D = 0，则恰好有一个实数根（重根）。如果 D < 0，则没有实数根，尽管方程有两个复数根——这一概念在更高等级中有所涉及。IGCSE 题目常要求你求出使方程有相等实根的 k 值；你只需设 b² – 4ac = 0，然后解出 k 即可。
| Discriminant D = b² – 4ac | Nature of roots |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | One repeated real root |
| D < 0 | No real roots |
9. Graphical Interpretation | 图形解释
The solutions of a quadratic equation correspond to the x‑intercepts of the graph y = ax² + bx + c. If the parabola cuts the x‑axis at two points, the equation has two real distinct roots. If it just touches the x‑axis, it has a repeated root, and if it does not intersect the x‑ s. Completing the square gives the vertex (h, k), which helps sketch the graph. For y = x² – 4x + 3, rewriting as (x – 2)² – 1 shows the vertex is (2, –1) and the minimum value is –1. This deepens your understanding of why the discriminant acts as it does.
二次方程的解对应于图像 y = ax² + bx + c 的 x 轴截距。如果抛物线与 x 轴相交于两点，方程就有两个不同的实数根。如果只是刚好碰到 x 轴，则有一个重根；如果根本不与 x 轴相交，则没有实数解。配方法可以给出顶点 (h, k)，这有助于图像的草图绘制。对于 y = x² – 4x + 3，重写为 (x – 2)² – 1 后可知顶点为 (2, –1)，最小值为 –1。这能加深你对判别式作用原理的理解。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
One frequent error is forgetting to set the equation to zero before factorising. Students also mishandle signs when substituting into the quadratic formula, especially when b is negative or c is negative. Another pitfall is prematurely cancelling terms; remember that the quadratic formula deals with an entire numerator divided by 2a, so factorise before cancelling. In completing the square, always check your constant adjustment: (x + p)² expands to x² + 2px + p², so you must subtract p² to keep the expression equivalent. Finally, in Edexcel papers, unless asked for a specific form, you can choose the method you prefer – but always double‑check your solution by substituting back into the original equation.
一个常见的错误是在因式分解前忘记先将方程设为零。学生们在处理代入二次公式时的正负号也常出错，特别是当 b 为负或 c 为负时。另一个陷阱是过早地约分；记住二次公式是整个分子除以 2a，所以要先提取公因子再约分。在配方时，务必检查常数项的调整：(x + p)² 展开为 x² + 2px + p²，因此你必须减去 p² 以保持表达式等价。最后，在 Edexcel 试卷中，除非要求用特定形式，你可以选择自己喜欢的方法——但一定要通过代回原方程来检查你的解。
Always check: does your solution make ax² + bx + c equal zero?
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图