📚 Mastering Quadratic Equations | 掌握一元二次方程
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from Paper 2 non-calculator questions to Paper 4 problem-solving tasks. Understanding how to expand, factorise, solve and graph quadratics is essential for securing a top grade at both Extended and Core levels.
一元二次方程贯穿IGCSE数学大纲，从Paper 2的无计算器题目到Paper 4的解题任务。理解如何展开、因式分解、求解和绘制二次函数图像，是在Extended和Core两个级别中取得高分的关键。
1. Expanding Double Brackets | 展开双括号
second bracket, then simplify. The general result is: (x + a)(x + b) = x² + (a + b)x + ab.
要展开 (x + a)(x + b)，请将第一个括号中的每一项乘以第二个括号中的每一项，然后化简。一般结果为：(x + a)(x + b) = x² + (a + b)x + ab。
(x + 3)(x + 5) = x² + 3x + 5x + 15 = x² + 8x + 15
Notice that 8 is the sum of 3 and 5, while 15 is the product of 3 and 5. This pattern is the key to factorising later.
注意8是3和5的和，而15是3和5的积。这个规律是后续因式分解的关键。
- For (ax + b)(cx + d), use FOIL: First, Outer, Inner, Last. | 对于 (ax + b)(cx + d)，使用FOIL法则：首项、外项、内项、末项。
- Be careful with negative signs: (x − 2)(x + 6) = x² + 4x − 12. | 小心负号：(x − 2)(x + 6) = x² + 4x − 12。
- Squaring a single bracket: (x + 4)² = x² + 8x + 16. | 单个括号平方：(x + 4)² = x² + 8x + 16。
2. Factorising Quadratics of the Form x² + bx + c | 分解 x² + bx + c 形式的二次式
Factorising is the reverse of expanding. For x² + bx + c, find two numbers that multiply to give c and add to give b.
因式分解是展开的逆运算。对于 x² + bx + c，找到两个数，使其乘积为c，和为b。
x² + 7x + 12 = (x + 3)(x + 4) because 3 × 4 = 12 and 3 + 4 = 7
Some special forms appear frequently in exam papers. Recognising them saves valuable time.
一些特殊形式在试卷中频繁出现。识别它们可以节省宝贵时间。
- Difference of two squares: x² − 9 = (x − 3)(x + 3). | 平方差：x² − 9 = (x − 3)(x + 3)。
- Perfect square: x² + 10x + 25 = (x + 5)². | 完全平方：x² + 10x + 25 = (x + 5)²。
- Common factor first: 2x² + 6x = 2x(x + 3). | 先提取公因式：2x² + 6x = 2x(x + 3)。
3. Solving Quadratic Equations by Factorisation | 用因式分解法求解一元二次方程
To solve a quadratic equation by factorising, first rearrange the equation so that one side is zero, then factorise and set each bracket equal to zero.
使用因式分解法求解一元二次方程时，先把方程整理为一边为零的形式，然后因式分解并将每个括号设为零。
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3
This method works because if the product of two expressions is zero, then at least one of them must be zero. This is called the zero-product property.
这种方法之所以有效，是因为如果两个表达式的乘积为零，那么其中至少有一个必须为零。这称为零乘积性质。
- Always write the final answer as "x = ... or x = ...". | 务必把最终答案写成"x = ... 或 x = ..."的形式。
- Check your solutions by substituting back into the original equation. | 将解代回原方程进行检验。
- If the coefficient of x² is not 1, factorise by finding two numbers that multiply to ac and add to b, then use grouping. | 如果x²的系数不为1，找到乘积为ac、和为b的两个数，然后用分组分解法。
4. The Quadratic Formula | 求根公式
solutions are given by the quadratic formula. You must memorise this formula for the exam.
对于任意一元二次方程 ax² + bx + c = 0，其解由求根公式给出。考试中你必须记住这个公式。
x = (−b ± √b² − 4ac) / 2a
Example: Solve 2x² + 5x − 3 = 0. Here a = 2, b = 5, c = −3.
示例：求解 2x² + 5x − 3 = 0。这里 a = 2，b = 5，c = −3。
x = (−5 ± √(25 − 4 × 2 × (−3))) / 4 = (−5 ± √49) / 4 = (−5 ± 7) / 4
= 1/2, or x = (−5 − 7) / 4 = −3. The ± sign means you must calculate two separate answers.
因此 x = (−5 + 7) / 4 = 1/2，或 x = (−5 − 7) / 4 = −3。± 符号意味着你必须计算出两个不同的答案。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. Take half the coefficient of x, square it, then adjust the constant term.
配方法将二次式改写为 (x + p)² + q 的形式。取x系数的一半并平方，然后调整常数项。
x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7
To solve x² + 6x + 2 = 0 by completing the square:
用配方法求解 x² + 6x + 2 = 0：
(x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7
The main advantages of this method are that it avoids factorisation errors, it works for any quadratic, and it directly gives the turning point of the corresponding graph.
这种方法的主要优点是避免因式分解错误，适用于任何二次方程，并且能直接给出对应图像的顶点坐标。
6. Sketching Quadratic Graphs | 绘制二次函数图像
The graph of y = ax² + bx + c is always a parabola. When a > 0 it opens upward (a 'happy' curve); when a < 0 it opens downward.
y = ax² + bx + c 的图像总是一条抛物线。当 a > 0 时开口向上（笑脸曲线）；当 a < 0 时开口向下。
| Form | 形式 | Key features | 关键特征 |
| y = ax² + bx + c | y-intercept at (0, c); roots from solving ax² + bx + c = 0 | y轴截距为(0, c)；根由解 ax² + bx + c = 0 得到 |
| y = a(x − h)² + k | Vertex (turning point) at (h, k) | 顶点在 (h, k) |
ertex. The axis of symmetry is the vertical line through the vertex, given by x = −b / 2a.
绘制二次函数图像：求y轴截距、根（若存在）以及顶点。对称轴是经过顶点的竖直线，其方程为 x = −b / 2a。
7. The Discriminant | 判别式
e discriminant. It tells us how many real roots the equation has without solving it fully.
表达式 Δ = b² − 4ac 称为判别式。无需完整求解方程，即可判断方程有多少个实数根。
- Δ > 0: two distinct real roots. | Δ > 0：两个不同的实数根。
- Δ = 0: one repeated real root (the curve touches the x-axis at one point). | Δ = 0：一个重根（图像与x轴相切于一点）。
- Δ < 0: no real roots (the curve does not intersect the x-axis). | Δ < 0：没有实数根（图像与x轴无交点）。
x² − 4x + 3 = 0, Δ = (−4)² − 4 × 1 × 3 = 16 − 12 = 4 > 0, so there are two distinct roots.
If a problem asks for equal roots, a tangent line, or a line touching a curve,
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