📚 IGCSE Maths: Quadratic Equations Mastery | IGCSE 数学：二次方程完全攻略
Quadratic equations are a central part of the IGCSE Mathematics syllabus. They appear in algebra, graph sketching, coordinate geometry and real-life word problems. This revision guide covers every method you need: factorising, difference of two squares, completing the square, the quadratic formula, the discriminant, graph analysis and problem solving. Master these techniques and you will gain confidence in both Core and Extended papers.
二次方程是 IGCSE 数学考纲的核心内容。它们出现在代数、图像草图、坐标几何以及现实应用题中。本复习指南涵盖你需要掌握的所有方法：因式分解、平方差、配方法、二次公式、判别式、图像分析以及问题求解。掌握这些技巧，你将能够在核心卷和扩展卷中都充满信心。
1. What is a Quadratic Equation? | 什么是二次方程？
constants and a ≠ 0. If a were zero, the x² term would disappear and the equation would become linear.
二次方程是次数为 2 的多项式方程。它的变量最高次数是 x²。一般形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。如果 a 为零，x² 项会消失，方程就变成了线性方程。
Quadratic equations can have two real solutions, one repeated real solution, or no real solutions. In IGCSE, you will solve them by factorising, completing the square, using the quadratic formula, and reading solutions from graphs.
二次方程可以有两个实根、一个重实根或没有实根。在 IGCSE 中，你将通过因式分解、配方法、二次公式以及从图像上读取解来求解它们。
ax² + bx + c = 0, a ≠ 0
2. Standard Form and Coefficients | 标准形式与系数
er a is the coefficient of x², b is the coefficient of x, and c is the constant term. Before choosing a method, always rearrange the equation so that the right-hand side is exactly zero and the terms are in descending powers of x.
在方程 ax² + bx + c = 0 中，a 是 x² 的系数，b 是 x 的系数，c 是常数项。在选择方法之前，一定要先把方程整理成右边刚好为零、各项按 x 的降幂排列。
s 5x − 2 = −4x², rearrange it to 4x² + 5x − 2 = 0 before identifying a = 4, b = 5, c = −2.
例如，3x² − 7x + 2 = 0 中 a = 3，b = −7，c = 2。如果题目给出 5x − 2 = −4x²，应先将它整理为 4x² + 5x − 2 = 0，再确定 a = 4，b = 5，c = −2。
Being careful with signs at this stage prevents many substitution errors later. Write negative coefficients clearly and use brackets when substituting into formulas.
在这一步仔细处理符号可以避免后面许多代入错误。写清负系数，并在代入公式时使用括号。
3. Solving by Factorising | 因式分解法求解
+ s) = 0, then either px + q = 0 or rx + s = 0. This gives two linear equations to solve.
当二次方程可以写成两个括号乘积等于零时，我们使用零乘积性质。如果 (px + q)(rx + s) = 0，那么要么 px + q = 0，要么 rx + s = 0。这样就得到两个一次方程来求解。
ork, so x² + 5x + 6 = (x + 2)(x + 3). Setting each bracket to zero gives x = −2 or x = −3.
要对 x² + 5x + 6 进行因式分解，找两个数，它们相乘得 c = 6，相加得 b = 5。2 和 3 满足条件，因此 x² + 5x + 6 = (x + 2)(x + 3)。令每个括号为零，得到 x = −2 或 x = −3。
dle term or use systematic trial. 2x² + 7x + 3 = (2x + 1)(x + 3), giving x = −½ or x = −3.
对于 a ≠ 1 的二次式，例如 2x² + 7x + 3，可以使用拆分中项或系统试因式。2x² + 7x + 3 = (2x + 1)(x + 3)，解得 x = −½ 或 x = −3。
isation by expanding the brackets back: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3.
一定要通过展开括号来检查因式分解：(2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3。
4. Difference of Two Squares | 平方差公式
)(x − a). This appears frequently in IGCSE exams because it is quick to factorise once you recognise the pattern.
一种特殊的二次式模式是平方差：x² − a² = (x + a)(x − a)。这在 IGCSE 考试中经常出现，因为一旦你识别出这个模式，就能快速因式分解。
and x = 5. Another common case is 4x² − 9 = (2x + 3)(2x − 3), giving x = −3⁄2 or x = 3⁄2.
例如，x² − 25 = (x + 5)(x − 5)，所以 x² − 25 = 0 的解为 x = −5 和 x = 5。另一个常见例子是 4x² − 9 = (2x + 3)(2x − 3)，解得 x = −3⁄2 或 x = 3⁄2。
Remember that the sum of two squares, such as x² + 9, cannot be factorised using real numbers. Only the difference of squares has this neat factorised form.
记住，两个平方的和，例如 x² + 9，不能用实数进行因式分解。只有平方差才有这种简洁的因式分解形式。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression ax² + bx + c in the form a(x + p)² + q. This method is especially useful when the equation does not factorise easily and when you need to find the turning point of a graph.
配方法将二次式 ax² + bx + c 改写为 a(x + p)² + q 的形式。当方程不容易因式分解时，以及当你需要求图像顶点时，这种方法特别有用。
x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8. Setting this equal to zero gives x + 3 = ±√8, so x = −3 ± 2√2.
对于 x² + 6x + 1 = 0，取 b = 6 的一半，即 3，将它平方得 9。写作 x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。令其等于零，得 x + 3 = ±√8，所以 x = −3 ± 2√2。
: 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.
对于 a ≠ 1 的二次式，先将 a 从 x² 和 x 项中提出来。例如：2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。
x² + bx + c = (x
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