📚 Solving Quadratic Equations for IGCSE Mathematics | IGCSE 数学：解一元二次方程
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from basic factorisation to real-life modelling. This article explains the main methods clearly so that you can choose the best approach in any exam question and show all the working needed for full marks.
一元二次方程贯穿 IGCSE 数学大纲，从基本因式分解到实际建模都会出现。这篇文章将清晰地讲解主要解法，帮助你在考试中选择最佳方法，并写出获得满分所需的完整步骤。
1. What is a Quadratic Equation? | 什么是一元二次方程？
where a, b and c are constants and a ≠ 0. The highest power of the unknown x is 2, which is why the word 'quadratic' is used.
在 IGCSE 数学中，一元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。未知数 x 的最高次数是 2，因此称为二次方程。
A quadratic equation usually has two solutions, called roots. These roots may be real and distinct, real and equal, or not real depending on the discriminant, which is discussed later.
一元二次方程通常有两个解，称为根。根据判别式的不同，这些根可能为两个不相等的实数根、两个相等的实数根，或者没有实数根。
2. Standard Form and Key Terms | 标准形式与关键术语
f x, and c is the constant term. If a = 1, the equation is often called a monic quadratic.
标准形式为 ax² + bx + c = 0。其中 a 是二次项系数，b 是一次项系数，c 是常数项。如果 a = 1，通常称为首一二次方程。
ax² + bx + c = 0, a ≠ 0
rrange the equation so one side is zero. For example, x² = 5x − 6 becomes x² − 5x + 6 = 0.
求解前，一定要把方程整理为一边等于零。例如，x² = 5x − 6 应改写为 x² − 5x + 6 = 0。
3. Solving by Factorisation | 因式分解法
be zero. Write the equation as (px + q)(rx + s) = 0, then set each bracket equal to zero.
当二次式容易因式分解时，我们可以使用零乘积性质：如果两个因式的乘积为零，那么至少有一个因式等于零。把方程写成 (px + q)(rx + s) = 0，然后令每个括号等于零。
ing gives (x − 2)(x − 3) = 0, so x − 2 = 0 or x − 3 = 0. The solutions are x = 2 or x = 3.
示例：解方程 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0，因此 x − 2 = 0 或 x − 3 = 0。解为 x = 2 或 x = 3。
s: (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6, which matches the original expression.
一定要通过展开括号来检验：(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6，与原式一致。
4. The Quadratic Formula | 二次公式法
equation ax² + bx + c = 0. It is derived from completing the square and can be stated as:
二次公式可以解任何一元二次方程 ax² + bx + c = 0。它由配方法推导而来，公式如下：
x = ( −b ± √b² − 4ac ) / (2a)
arefully, and calculate the square root part first. Remember to apply ± to get both roots.
从标准形式中确定 a、b、c，小心代入，并先计算平方根部分。记住要使用 ± 来得到两个根。
, c = −2. The discriminant is b² − 4ac = 3² − 4(2)(−2) = 9 + 16 = 25. Therefore x = ( −3 ± √25 ) / 4 = ( −3 ± 5 ) / 4, giving x = 1/2 or x = −2.
示例：解方程 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。判别式为 b² − 4ac = 3² − 4(2)(−2) = 9 + 16 = 25。因此 x = ( −3 ± √25 ) / 4 = ( −3 ± 5 ) / 4，得到 x = 1/2 或 x = −2。
5. Completing the Square | 配方法
ethod is especially useful for finding the vertex of a parabola and for solving quadratics that do not factorise neatly.
配方法将二次式写成 (x + p)² + q = 0 的形式。这种方法特别适用于求抛物线的顶点，以及求解不易因式分解的二次方程。
re it to get 9. Rewrite x² + 6x as (x + 3)² − 9, so the equation becomes (x + 3)² − 7 = 0.
对于 x² + 6x + 2 = 0，取 6 的一半 3，再平方得到 9。把 x² + 6x 写成 (x + 3)² − 9，因此方程变为 (x + 3)² − 7 = 0。
Then (x + 3)² = 7, so x + 3 = ±√7, giving x = −3 ± √7. These are the exact solutions; decimals can be found using a calculator if required.
于是 (x + 3)² = 7，所以 x + 3 = ±√7，得到 x = −3 ± √7。这些是精确解；如果需要，可以用计算器求出小数。
6. The Discriminant | 判别式
Δ = b² − 4ac. It tells us the nature of the roots without solving the equation completely.
判别式 Δ 定义为 Δ = b² − 4ac。它可以在不完全解方程的情况下告诉我们根的性质。
Δ = b² − 4ac
real roots. If Δ = 0, there is one repeated real root. If Δ < 0, there are no real roots.
如果 Δ > 0，有两个
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图