📚 Mastering Quadratic Equations: Solving and Sketching for IGCSE | IGCSE 二次方程：解方程与图像绘制
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebra and graphs to real-world modelling. This revision guide brings together the key methods you need to solve quadratic equations and to sketch their curves confidently under exam conditions.
二次方程贯穿 IGCSE 数学课程，从代数、图像到实际建模都会出现。本复习指南汇总了你需要掌握的核心方法，帮助你在考试中自信地解二次方程并绘制其图像。
1. Recognising a Quadratic Equation | 识别二次方程
ants and a ≠ 0. The highest power of the variable x is 2, which gives the equation its name.
二次方程是可以写成一般形式 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，这正是“二次”名称的由来。
ax² + bx + c = 0, where a ≠ 0
quation that can be rearranged into this pattern belongs to the same family of quadratics.
常见例子包括 x² - 5x + 6 = 0、2x² + 3x - 1 = 0 和 4x² - 9 = 0。任何能整理成这种形式的方程都属于二次方程家族。
2. Standard Form and Coefficients | 标准形式与系数
The coefficient a is the number multiplying x², the coefficient b is the number multiplying x, and c is the constant term. A missing term simply means that coefficient is 0.
解题前，应先把方程整理成 ax² + bx + c = 0 的形式。系数 a 是 x² 前面的数，系数 b 是 x 前面的数，c 是常数项。缺少某一项就表示该项系数为 0。
d form prevents sign errors when you substitute into factorised pairs or the quadratic formula.
例如，x² - 4 = 0 中 a = 1，b = 0，c = -4。把方程写成标准形式可以避免在代入因式对或求根公式时出现符号错误。
- 3x² + 7x - 2 = 0: a = 3, b = 7, c = -2
- -x² + 5x = 0: a = -1, b = 5, c = 0
- 2x² - 8 = 0: a = 2, b = 0, c = -8
3. Solving by Factorisation | 因式分解法
ind two numbers that multiply to 6 and add to 5: these are 2 and 3, so (x + 2)(x + 3) = 0.
当二次式可以分解为两个一次因式的乘积时，可使用因式分解法。对于 x² + 5x + 6 = 0，找出两个数，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。
-2 or x = -3. Always check by substituting the solutions back into the original equation.
利用零乘积性质，令每个因式等于零：x + 2 = 0 或 x + 3 = 0。解得 x = -2 或 x = -3。务必把解代回原方程进行检验。
If pq = 0, then p = 0 or q = 0
4. Completing the Square | 配方法
half of the coefficient of x, which is 3, and square it: (x + 3)² - 9 + 5 = (x + 3)² - 4.
配方法把 x² + bx + c 改写成顶点式 a(x + p)² + q。对于 x² + 6x + 5 这样的简单情形，取 x 系数的一半，即 3，再平方：(x + 3)² - 9 + 5 = (x + 3)² - 4。
Setting (x + 3)² - 4 = 0 gives (x + 3)² = 4, so x + 3 = ±√4 1 or x = -5. This method is especially useful for finding the vertex of a quadratic graph.
令 (x + 3)² - 4 = 0，得到 (x + 3)² = 4，所以 x + 3 = ±√4。因此 x = -3 ± 2，即 x = -1 或 x = -5。配方法在求二次函数图像顶点时特别有用。
x² + bx + c = (x + b/2)² - (b/2)² + c
5. The Quadratic Formula | 求根公式
ble, use the quadratic formula. For ax² + bx + c = 0, the solutions are given by x = [-b ± √b² - 4ac] / (2a). This formula works for every quadratic equation, including those with irrational or non-integer roots.
当因式分解困难或不可行时，使用求根公式。对于 ax² + bx + c = 0，解为 x = [-b ± √b² - 4ac] / (2a)。该公式适用于所有二次方程，包括具有无理根或非整数根的方程。
x = [-b ± √b² - 4ac] / (2a)
a = 2, b = -3, c = -4. Calculate b² - 4ac = (-3)² - 4(2)(-4) = 9 + 32 = 41. Thus x = [3 ± √41] / 4. Leave the answer in surd form unless a decimal is asked for.
例如，解 2x² - 3x - 4 = 0：a = 2，b = -3，c = -4。计算 b² - 4ac = (-3)² - 4(2)(-4) = 9 + 32 = 41。因此 x = [3 ± √41] / 4。除非题目要求小数，否则答案保留根式形式。
6. The Discriminant | 判别式
tinct real roots; if D = 0, it has one repeated real root; if D < 0, it has no real roots.
表达式 D = b² - 4ac 被称为判别式。它无需解方程就能判断根的性质。如果 D > 0，方程有两个不同实根；如果 D = 0，方程有一个重根；如果 D < 0，方程没有实根。
| Discriminant D | Nature of roots | 根的性质 |
|---|---|---|
| D > 0
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