📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear throughout the IGCSE Edexcel Maths syllabus, from basic algebraic manipulation to advanced problem-solving in geometry and physics. Understanding how to solve them efficiently using multiple methods is essential for exam success. This article breaks down the key techniques: factorising, completing the square, the quadratic formula, and graphical interpretation, all aligned with the Edexcel IGCSE specification.
二次方程贯穿 IGCSE Edexcel 数学考纲，从基础的代数运算到几何与物理中的高级问题解决都有涉及。高效掌握多种解法是考试成功的关键。本文详细讲解因式分解、配方法、二次公式以及图像解释等核心方法，完全贴合 Edexcel IGCSE 考试要求。
1. Recognising a Quadratic Equation | 识别二次方程
ants and a ≠ 0. The highest power of the variable x is 2. In IGCSE, you will encounter quadratics with integer, fractional and even negative coefficients.
二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数且 a ≠ 0。变量 x 的最高次数为 2。在 IGCSE 中，你会遇到整数、分数甚至负系数的二次方程。
2. Solving by Factorising | 因式分解法求解
3. This method is quick but only works when the quadratic factorises neatly with integers.
因式分解法是将 ax² + bx + c 表示为两个一次括号的乘积。例如，x² + 5x + 6 = 0 分解为 (x + 2)(x + 3) = 0。令每个括号等于零得到解 x = -2 和 x = -3。这种方法快速，但仅适用于能用整数顺利分解的二次方程。
3. Factorising When a > 1 | 当 a > 1 时的因式分解
e term, and factor by grouping: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 x² 系数不为 1 时，必须考虑 a 和 c 的因数组合。对于 2x² + 7x + 3，计算 a × c = 6，找出两个数乘积为 6 且和为 7（6 和 1），拆分中间项并分组分解：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
4. Difference of Two Squares | 平方差公式
A special quadratic form is x² – a², which factorises as (x + a)(x – a). For example, 9x² – 25 = (3x + 5)(3x – if terms are perfect squares before attempting to factorise; this saves time in the exam.
一种特殊的二次形式是 x² – a²，可分解为 (x + a)(x – a)。例如，9x² – 25 = (3x + 5)(3x – 5)。在尝试因式分解前务必检查各项是否为完全平方，这可以在考试中节省时间。
5. Completing the Square | 配方法
. For x² + 6x + 5, halve the coefficient of x to get 3, square it to get 9, write (x + 3)² – 9 + 5 = (x + 3)² – x = -1 or x = -5. This method works for any quadratic and leads to the quadratic formula.
配方法将 x² + bx + c 转化为 (x + p)² + q 的形式。对于 x² + 6x + 5，取 x 系数的一半得 3，平方得 9，写成 (x + 3)² – 9 + 5 = (x + 3)² – 4。令其等于零：(x + 3)² = 4，所以 x + 3 = ±2，解得 x = -1 或 x = -5。该方法适用于任何二次方程，并可推导出二次公式。
6. The Quadratic Formula | 二次公式法
The quadratic formula is x = [-b ± √b² – 4ac] / 2a. This formula solves any quadratic equation ax² + bx + c = 0. For 3x² – 4x – 2 = 0, a=3, b=-4, c=-2. Substituting: x = [4 ± √16 + 24] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = (2 ± √10) / 3. Always remember to simplify surds fully.
二次公式是 x = [-b ± √b² – 4ac] / 2a。该公式可解任何二次方程 ax² + bx + c = 0。对于 3x² – 4x – 2 = 0，a=3，b=-4，c=-2。代入得：x = [4 ± √16 + 24] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = (2 ± √10) / 3。务必始终将根式化简到最简。
7. The Discriminant and Number of Roots | 判别式与根的数量
The discriminant is the expression under the square root: D = b² – ex roots, not assessed at IGCSE). This helps predict solution types without fully solving.
判别式是平方根下的表达式：D = b² – 4ac。若 D > 0，有两个不同的实数根。若 D = 0，有一个重根。若 D < 0，无实数根（复数根，IGCSE 不考）。这有助于在不解出完整解的情况下预判解的类型。
8. Solving by Graphing | 图像法求解
sketch; the roots are where the curve crosses the x-axis. For example, the graph of y = x² – = 1 and x = 3, confirming solutions. The turning point gives the minimum or maximum value.
二次方程可通过找出抛物线 y = ax² + bx + c 与 x 轴交点的图像法求解。绘制图像或利用给定的草图；根是曲线与 x 轴相交的位置。例如，y = x² – 4x + 3 的图像与 x 轴交于 x = 1 和 x = 3，确认了解。顶点给出了最小值或最大值。
9. Word Problems Leading to Quadratics | 二次方程应用题
problems involve forming a quadratic from a real-life scenario. For instance, a rectangle' ts area is 40 m². Let width = x, then length = x + 3, so x(x + 3) = 40, leading to x² + 3x – 40 = 0. Solve by factorising: (x + 8)(x – 5) = 0, so x = 5 (width positive). Always check that solutions make sense in context.
许多 IGCSE 问题涉及从实际情境中建立二次方程。例如，一个矩形的长比宽多 3 m，面积为 40 m²。设宽为 x，则长为 x + 3，所以 x(x + 3) = 40，得到 x² + 3x – 40 = 0。因式分解得 (x + 8)(x – 5) = 0，所以 x = 5（宽度为正）。务必检查解在题目情境中是否有意义。
10. Solving Simultaneous Equations with a Quadratic | 联立方程中的二次方程
ion into the quadratic, rearrange into standard form, and solve. For y = 2x + 1 and y = x² – 2, equate: 2x + 1 = x² – 2 → x² – 2x – 3 = 0 → (x – 3)(x + 1) = 0. Find x then y. Two intersection points usually result.
Edexcel IGCSE 经常要求解决一个一次方程与一个二次方程联立的问题。将一次表达式代入二次方程，整理成标准形式并求解。对于 y = 2x + 1 和 y = x² – 2，令两者相等：2x + 1 = x² – 2 → x² – 2x – 3 = 0 → (x – 3)(x + 1) = 0。求出 x 再求 y。通常得到两个交点。
11. Common Mistakes to Avoid | 常见错误与规避
ll terms to one side. When using the formula, be careful with negative signs; b² means (b)² so negative signs disappear. Always simplify your final answers, especially surds and fractions.
注意在因式分解前不要忘记将方程设为零。切勿除以 x，除非考虑 x = 0 的可能性；而应将所有项移到一边。使用公式时，注意负号；b² 表示 (b)²，负号会消失。务必化简最终答案，尤其是根式和分数。
12. Exam Strategy and Practice | 考试策略与练习
In the Edexcel IGCSE, quadratic solving appears on both Paper 1 (no calculator) and Paper 2 (calculator allowed). For non-calculator, rely on factorising and completing the square; for calculator, the quadratic formula is efficient but still requires setting out steps to gain method marks. Practise past papers regularly and always check your solutions by substitution.
在 Edexcel IGCSE 中，二次方程求解出现在 Paper 1（无计算器）和 Paper 2（允许计算器）中。无计算器部分依靠因式分解和配方法；允许计算器时，二次公式虽然高效，但仍需写清步骤以获得方法分。定期练习往年真题，并始终通过代入检验解的正确性。
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