📚 IGCSE Maths: Quadratic Equations from Factorising to the Quadratic Formula | IGCSE 数学：二次方程从因式分解到求根公式
Quadratic equations are a central topic in IGCSE Mathematics. They appear in algebra, graphing and real-life modelling, so a confident command of solving methods is essential for top exam marks.
二次方程是 IGCSE 数学的核心主题。它们出现在代数、作图以及现实生活建模中，因此熟练掌握各种求解方法对于获得高分至关重要。
1. What is a Quadratic Equation? | 什么是二次方程
degree 2. Its standard form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是一个二次多项式方程。其标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。
ic expression. If a = 0, the equation becomes linear, so the condition a ≠ 0 is essential.
表达式 ax² + bx + c 称为二次表达式。如果 a = 0，方程就变成一次方程，因此 a ≠ 0 这一条件必不可少。
Quadratic equations can be written in different forms, such as expanded form, factorised form or completed-square form. Each form is useful for a different solving method.
二次方程可以写成不同的形式，例如展开式、因式分解式或配方形式。每种形式对应不同的求解方法。
2. Standard Form and Coefficients | 标准形式与系数
number a is the coefficient of x², b is the coefficient of x, and c is the constant term.
在标准形式 ax² + bx + c = 0 中，a 是 x² 的系数，b 是 x 的系数，c 是常数项。
a = 3, b = -5 and c = 2. Always take the sign of the coefficient together with the number.
例如，在 3x² - 5x + 2 = 0 中，a = 3、b = -5、c = 2。代入时一定要把系数的正负号一起带走。
nstance, 4x - x² = 3 becomes -x² + 4x - 3 = 0, or x² - 4x + 3 = 0 after multiplying by -1.
求解的第一步通常是把方程整理成标准形式。例如，4x - x² = 3 可化为 -x² + 4x - 3 = 0，或乘 -1 后得到 x² - 4x + 3 = 0。
3. Solving by Factorisation | 因式分解法求解
Factorisation is often the fastest method when the quadratic expression can be written as a product of two linear factors. We look for two numbers that multiply to give ac and add to give b.
当二次表达式可以写成两个一次因式的乘积时，因式分解法通常是最快的方法。我们需要找到两个数，它们的乘积为 ac，和为 b。
umbers with product 6 and sum 5. The numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).
对于 x² + 5x + 6 = 0，我们需要乘积为 6、和为 5 的两个数。这两个数是 2 和 3，因此 x² + 5x + 6 = (x + 2)(x + 3)。
2)(x + 3) = 0. Since one of the factors must be zero, the solutions are x = -2 or x = -3.
方程化为了 (x + 2)(x + 3) = 0。由于至少有一个因式必须为零，所以解为 x = -2 或 x = -3。
(x + 2)(x + 3) = 0 ⇒ x = -2 or x = -3
4. Zero Product Property | 零积性质
, then at least one of the terms must be zero. In symbols, if AB = 0, then A = 0 or B = 0.
零积性质指出，如果两个因式的乘积为零，那么至少有一个因式为零。用符号表示就是：若 AB = 0，则 A = 0 或 B = 0。
equations. After writing (x + 2)(x + 3) = 0, we set each bracket equal to zero separately.
这一性质就是因式分解能直接转化为一元一次方程的原因。写出 (x + 2)(x + 3) = 0 后，我们把每个括号分别设为零。
Be careful: the zero product property only works when the product equals zero. If the product equals 6, for example, we cannot say that one factor equals 6.
注意：零积性质只在乘积等于零时成立。如果乘积等于 6，就不能说其中一个因式等于 6。
5. Completing the Square | 配方法
Completing the square turns a quadratic expression into the form (x + p)² + q. This method is especially useful when factorisation is difficult or when we need to find the vertex of a parabola.
配方法将二次表达式转化为 (x + p)² + q 的形式。当因式分解比较困难，或需要求抛物线的顶点时，这一方法尤其有用。
x² + 6x as a perfect square. Take half of 6, square it to get 9, then add and subtract 9:
对于 x² + 6x + 5 = 0，先把 x² + 6x 配成完全平方。取 6 的一半，平方得 9，再加 9 减 9：
x² + 6x + 5 = (x² + 6x + 9) - 9 + 5 = (x + 3)² - 4
es (x + 3)² = 4. Taking square roots on both sides yields x + 3 = ±2, so x = -1 or x = -5.
令 (x + 3)² - 4 = 0，得到 (x + 3)² = 4。两边开平方得 x + 3 = ±2，因此 x = -1 或 x = -5。
6. The Quadratic Formula | 二次公式求根
adratic equation ax² + bx + c = 0, even when factorisation is not possible using integers.
二次公式可以求解任何形如 ax² + bx + c = 0 的二次方程，即使无法用整数进行因式分解。
The formula is:
公式如下：
x = (-b ± √b² - 4ac) ÷ (2a)
ave a = 2, b = 3 and c = -2. Substituting gives x = (-3 ± √(3² - 4 × 2 × (-2))) ÷ (2 × 2).
对于 2x² + 3x - 2 = 0，我们有 a = 2、b = 3、c = -2。代入后得到 x = (-3 ± √(3² - 4 × 2 × (-2))) ÷ (2 × 2)。
This simplifies to x = (-3 ± √25) ÷ 4, so x = (-3 + 5) ÷ 4 = 1/2 or x = (-3 - 5) ÷ 4 = -2.
化简得 x = (-3 ± √25) ÷ 4，所以 x = (-3 + 5) ÷ 4 = 1/2 或 x = (-3 - 5) ÷ 4 = -2。
7. Discriminant and Nature of Roots | 判别式与根的性质
c, is called the discriminant. It tells us how many real roots the quadratic equation has.
平方根内的表达式 D = b² - 4ac 称为判别式。它可以告诉我们二次方程有多少个实数根。
| Discriminant D | Nature of roots |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | One repeated real root |
| D < 0 | No real roots |
is twice. If it is zero, the graph touches the x-axis at one point. If it is negative, the graph does not meet the x-axis.
如果判别式为正，y = ax² + bx + c 的图像与 x 轴相交两次。如果为零，图像与 x 轴相切于一点。如果为负，图像与 x 轴没有交点。
8. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，存在一个最低点。如果 a < 0，抛物线开口向下，存在一个最高点。
0. These can be estimated from the graph or found exactly by factorisation, completing the square or the quadratic formula.
图像与 x 轴的交点就是方程 ax² + bx + c = 0 的实数根。这些交点可以从图像上估计，也可以通过因式分解、配方法或二次公式精确求出。
point can be found by completing the square. For y = (x + 3)² - 4, the vertex is (-3, -4).
通过配方法可以找到图像的顶点。例如对于 y = (x + 3)² - 4，顶点为 (-3, -4)。
9. Modelling and Word Problems | 建模与应用题
Many IGCSE questions ask you to form a quadratic equation from a real-life situation, such as area, projectile motion or product relationships.
许多 IGCSE 题目要求你根据现实情境建立二次方程，例如面积问题、抛体运动或乘积关系。
tegers is 132. Let the smaller integer be n. Then n(n + 1) = 132, giving n² + n - 132 = 0.
例题：两个连续正整数的乘积为 132。设较小的整数为 n，则 n(n + 1) = 132，整理得 n² + n - 132 = 0。
ng gives (n - 11)(n + 12) = 0. Since n is positive, n = 11, so the integers are 11 and 12.
因式分解得 (n - 11)(n + 12) = 0。因为 n 为正数，所以 n = 11，因此这两个整数是 11 和 12。
Always check that your solution makes sense in the original context, especially when negative or fractional roots appear.
一定要检查解是否符合原题的实际意义，尤其是出现负数或分数根时。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
One common mistake is forgetting to set the equation equal to zero before factorising. The zero product property only applies when one side is zero.
一个常见错误是在因式分解前没有先把方程化为等于零的形式。零积性质只有当等式一边为零时才成立。
Another mistake is losing a negative sign when identifying b and c in the quadratic formula. Always substitute values carefully with their signs.
另一个错误是在二次公式中识别 b 和 c 时丢失负号。代入时一定要小心带上符号。
example, 2x² + 4x = 0 can be solved as 2x(x + 2) = 0, giving x = 0 or x = -2 immediately.
解题时始终先寻找公因式。例如，2x² + 4x = 0 可以化为 2x(x + 2) = 0，立刻得到 x = 0 或 x = -2。
In exam-style questions, show every step clearly: standard form, factorised form, zero product property and final roots. This earns method marks even if a small arithmetic error occurs.
在考试类题目中，要清晰地写出每一步：标准形式、因式分解形式、零积性质以及最终的根。这样即使出现小的计算错误，也能获得方法分。
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