📚 Quadratic Equations: Methods and Exam Strategies | 二次方程：解法与考试策略
A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in almost every paper, either as a direct question or as part of a larger problem. Understanding how to solve and apply quadratic equations is therefore essential for achieving a high grade.
二次方程是 IGCSE 数学中最重要的考点之一。几乎每份试卷都会考到它，要么直接以二次方程的形式出现，要么作为更大问题的一部分。因此，理解和掌握二次方程的解法及应用，是取得高分的关键。
1. What Is a Quadratic Equation? | 什么是二次方程？
degree 2. Its general form is ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0.
二次方程是次数为 2 的多项式方程。它的一般形式是 ax² + bx + c = 0，其中 a、b 和 c 为常数，且 a ≠ 0。
he equation x + 1/x = 2 is not a quadratic equation because it contains a reciprocal term.
例如，2x² + 3x - 5 = 0 是一个二次方程，其中 a = 2，b = 3，c = -5。但 x² + 2x³ = 0 不是二次方程，因为 x 的最高次数是 3。同样地，x + 1/x = 2 也不是二次方程，因为它含有倒数项。
r the completed square form a(x + p)² + q = 0. Each form is useful for different purposes.
在 IGCSE 考试中，你们通常会遇到三种形式的二次方程：一般形式 ax² + bx + c = 0、因式分解形式 (px + q)(rx + s) = 0，以及配方法形式 a(x + p)² + q = 0。每种形式都有其不同的用途。
2. Solving by Factorization | 因式分解法
tors. The principle is based on the zero product property: if AB = 0, then A = 0 or B = 0.
当二次表达式能够分解为两个一次因式时，因式分解法是最直接的方法。其原理基于零积性质：如果 AB = 0，则 A = 0 或 B = 0。
y to give 6 and add to give 5. These numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).
要分解 x² + 5x + 6，我们需要找到两个数，它们乘积为 6，和为 5。这两个数是 2 和 3，因此 x² + 5x + 6 = (x + 2)(x + 3)。
Hence, the solution to x² + 5x + 6 = 0 is found by setting each bracket to zero:
因此，解方程 x² + 5x + 6 = 0 的方法是将每个括号设为零：
x + 2 = 0 → x = -2 或 x + 3 = 0 → x = -3
Always expand your factorised expression to verify it matches the original. This simple check prevents many careless errors. Remember that not every quadratic expression can be factorised using integers.
务必展开你的因式分解结果，检查是否与原式一致。这个简单的检查可以避免许多粗心错误。请记住，并非所有的二次表达式都能用整数进行因式分解。
3. Factorization with a Leading Coefficient | 首项系数不为 1 的因式分解
× 3 = 6. We then find two numbers that multiply to 6 and add to b = 7, which are 1 and 6.
当 a ≠ 1 时，例如 2x² + 7x + 3 = 0，我们将 a 和 c 相乘：2 × 3 = 6。然后找到两个数，乘积为 6，和为 b = 7，这两个数是 1 和 6。
Then we rewrite the middle term and factor by grouping:
然后改写中间项，并分组分解：
2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)
So the solutions are x = -½ and x = -3.
因此方程的解是 x = -½ 和 x = -3。
An alternative approach is trial and error: systematically test pairs of factors of a and c until the middle term matches. With practice, this becomes very quick for simple cases. Whichever method you choose, check that the sum of the inner and outer products equals bx.
另一种方法是试错法：系统地尝试 a 和 c 的因子对，直到中间项匹配。多加练习后，对于简单的情况这种方法会非常快捷。无论选择哪种方法，都要检查内项与外项乘积之和是否等于 bx。
4. Completing the Square | 配方法
x² + bx + c in the form a(x + p)² + q, where p = b ÷ (2a) and q adjusts the constant term.
配方法适用于任何二次方程。其核心思想是将 ax² + bx + c 改写为 a(x + p)² + q 的形式，其中 p = b ÷ (2a)，q 用于调整常数项。
r example, to solve x² + 6x + 5 = 0, we first note that x² + 6x = (x + 3)² - 9. Therefore:
例如，要解 x² + 6x + 5 = 0，我们首先注意到 x² + 6x = (x + 3)² - 9。因此：
(x + 3)² - 9 + 5 = 0 → (x + 3)² = 4
ing the square root of both sides gives x + 3 = ±2, so x = -3 + 2 = -1 or x = -3 - 2 = -5.
两边开平方得 x + 3 = ±2，因此 x = -3 + 2 = -1 或 x = -3 - 2 = -5。
Completing the square is particularly useful when the coefficient of x² is 1 and b is even, since the calculation stays neat. It is also the method that reveals the turning point of a parabola directly, which is why it is commonly tested in graph-related questions.
当 x² 的系数为 1 且 b 为偶数时，配方法特别有用，因为计算会保持简洁。配方法还能直接揭示抛物线的顶点坐标，这也是为什么它在图像类题目中经常被考查。
5. The Quadratic Formula | 求根公式
and works for every quadratic equation. For ax² + bx + c = 0, the solutions are given by:
求根公式由配方法推导而来，适用于所有二次方程。对于 ax² + bx + c = 0，解由以下公式给出：
x = (−b ± √b² − 4ac) ÷ (2a)
For example, solving 2x² + 4x - 3 = 0 with a = 2, b = 4, c = -3:
例如，解 2x² + 4x - 3 = 0，其中 a = 2，b = 4，c = -3：
x = (−4 ± √16 + 24) ÷ 4 = (−4 ± √40) ÷ 4 = (−4 ± 2√10) ÷ 4
ons, answers should be given to 3 significant figures unless the question specifies a different degree of accuracy. You should memor
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