📚 Solving Quadratic Equations & Inequalities | 二次方程与不等式求解
Quadratic equations and inequalities are a fundamental part of the IGCSE Mathematics syllabus. Whether you are aiming for a solid B or pushing for an A*, understanding how to manipulate and solve these expressions is non-negotiable. This revision guide breaks down the essential methods, common pitfalls, and exam-style applications to help you master this topic with confidence.
二次方程与不等式是IGCSE数学教学大纲中最核心的部分之一。无论你的目标是稳定取得B，还是冲击A*，掌握这些表达式的运算与求解方法都是必不可少的。本复习指南将拆解关键解法、常见易错点以及考试风格的应用题，帮助你自信掌握这一主题。
1. Expanding & Factorizing Quadratics | 展开与因式分解
Expanding involves removing brackets by multiplying terms. The general form for a binomial product is (x + a)(x + b) = x² + (a + b)x + ab. Factorizing is the reverse process: rewriting a quadratic expression into the product of two linear factors.
展开式通过乘法去掉括号。二项式乘积的一般形式为 (x + a)(x + b) = x² + (a + b)x + ab。因式分解是展开的逆过程：将二次表达式重写为两个一次因式的乘积。
Example: Expand and simplify (2x – 3)(x + 5).
示例：展开并化简 (2x – 3)(x + 5)。
We multiply each term in the first bracket by each term in the second: 2x × x = 2x², 2x × 5 = 10x, -3 × x = -3x, and -3 × 5 = -15. Combining the like terms gives us 2x² + 7x – 15.
我们将第一个括号中的每一项与第二个括号中的每一项相乘：2x × x = 2x²，2x × 5 = 10x，-3 × x = -3x，以及 -3 × 5 = -15。合并同类项后得到 2x² + 7x – 15。
When factorizing x² + bx + c, we look for two numbers that multiply to ‘c’ and add to ‘b’.
在对 x² + bx + c 进行因式分解时，我们需要找到两个数，它们相乘等于常数项 ‘c’，相加等于一次项系数 ‘b’。
- Difference of squares: a² – b² = (a + b)(a – b)
- Perfect square: a² + 2ab + b² = (a + b)²
- 平方差公式：a² – b² = (a + b)(a – b)
- 完全平方公式：a² + 2ab + b² = (a + b)²
2. Solving by Factorizing | 因式分解法求解
The quickest way to solve a quadratic equation is often by factorizing, but it only works when the expression factorizes neatly. The core principle relies on the zero product property: if the product of two expressions is zero, then at least one of them must be zero.
求解二次方程最快的方法通常是因式分解法，但这种方法仅在表达式能够被整齐地分解时适用。其核心原理依赖于零乘积性质：如果两个表达式的乘积为零，那么其中至少有一个必须为零。
If AB = 0, then A = 0 or B = 0
如果 AB = 0，则 A = 0 或 B = 0
Example: Solve x² – 5x + 6 = 0.
示例：求解 x² – 5x + 6 = 0。
Factorizing the left-hand side, we get (x – 2)(x – 3) = 0. Applying the zero product property, either x – 2 = 0 or x – 3 = 0. Therefore, the solutions are x = 2 or x = 3.
对左边进行因式分解，我们得到 (x – 2)(x – 3) = 0。应用零乘积性质，要么 x – 2 = 0，要么 x – 3 = 0。因此，解为 x = 2 或 x = 3。
For quadratics where the coefficient of x² is greater than 1, such as 2x² + 5x – 3 = 0, we can factorize by grouping. Multiply the coefficient of x² (2) by the constant term (-3) to get -6. Find two numbers that multiply to -6 and add to 5, which are 6 and -1. Rewrite the equation as 2x² + 6x – x – 3 = 0, then group to get 2x(x + 3) – 1(x + 3) = 0. This gives us (2x – 1)(x + 3) = 0, leading to x = ½ or x = -3.
对于二次项系数大于1的二次方程，如 2x² + 5x – 3 = 0，我们可以使用分组分解法。将二次项系数 (2) 乘以常数项 (-3) 得到 -6。找到两个数相乘等于 -6 且相加等于 5，这两个数是 6 和 -1。将方程重写为 2x² + 6x – x – 3 = 0，然后分组得到 2x(x + 3) – 1(x + 3) = 0。这给出了 (2x – 1)(x + 3) = 0，从而得出 x = ½ 或 x = -3。
3. Completing the Square | 配方法
Completing the square is a powerful technique used to solve quadratic equations, find turning points of graphs, and derive the quadratic formula. It rewrites a quadratic in the form a(x + h)² + k.
配方法是一种强大的技巧，用于求解二次方程、找到图像的顶点坐标以及推导二次求根公式。它将二次式重写为 a(x + h)² + k 的形式。
For an expression like x² + bx, we add and subtract (b/2)² to make a perfect square.
对于像 x² + bx 这样的表达式，我们需要加上并减去 (b/2)² 以构成完全平方。
Example: Solve x² – 4x – 5 = 0 by completing the square.
示例：用配方法求解 x² – 4x – 5 = 0。
Take the coefficient of x, which is -4. Half of this is -2, and squaring it gives 4. Add and subtract 4 within the equation: x² – 4x + 4 – 4 – 5 = 0. This simplifies to (x – 2)² – 9 = 0. Rearranging gives (x – 2)² = 9. Taking the square root of both sides, we get x – 2 = ±3. Thus, x = 2 + 3 = 5 or x = 2 – 3 = -1.
取 x 的系数，即 -4。它的一半是 -2，平方后得到 4。在方程中加入并减去 4：x² – 4x + 4 – 4 – 5 = 0。化简为 (x – 2)² – 9 = 0。移项后得到 (x – 2)² = 9。对方程两边取平方根，得到 x – 2 = ±3。因此，x = 2 + 3 = 5 或 x = 2 – 3 = -1。
4. The Quadratic Formula | 二次求根公式
The quadratic formula is a universal method that solves any quadratic equation, even when it cannot be factorized easily. It is essential to memorize the formula and understand how to substitute values correctly.
二次求根公式是一种通用方法，可以解决任何二次方程，即使它无法被轻松因式分解。牢记公式并理解如何正确代值至关重要。
For ax² + bx + c = 0, x = (-b ± √(b² – 4ac)) / 2a
对于 ax² + bx + c = 0，x = (-b ± √(b² – 4ac)) / 2a
Example: Solve 2x² + 3x – 2 = 0 using the quadratic formula.
示例：使用二次求根公式求解 2x² + 3x – 2 = 0。
Here, a = 2, b = 3, and c = -2. Substituting these into the formula gives x = (-3 ± √(3² – 4 × 2 × -2)) / (2 × 2). The discriminant (the part under the square root) is 9 – (-16) = 25. Since √25 = 5, the equation becomes x = (-3 ± 5) / 4. The two solutions are x = (-3 + 5) / 4 = 2/4 = ½, and x = (-3 – 5) / 4 = -8/4 = -2.
此处，a = 2，b = 3，c = -2。将这些值代入公式得到 x = (-3 ± √(3² – 4 × 2 × -2)) / (2 × 2)。判别式（平方根内的部分）为 9 – (-16) = 25。由于 √25 = 5，方程变为 x = (-3 ± 5) / 4。两个解分别为 x = (-3 + 5) / 4 = 2/4 = ½，以及 x = (-3 – 5) / 4 = -8/4 = -2。
5. The Discriminant | 判别式
The discriminant, denoted as Δ (Delta) or simply the expression b² – 4ac, tells us about the nature of the roots of a quadratic equation without actually solving it. This is a highly tested concept in IGCSE.
判别式，表示为 Δ（德尔塔）或直接说表达式 b² – 4ac，能在不实际求解的情况下告诉我们二次方程根的性质。这是IGCSE考试中一个高频考点。
| Discriminant (Δ = b² – 4ac) | Nature of Roots |
| Δ > 0 | Two distinct real roots |
| Δ = 0 | Two equal real roots (one repeated root) |
| Δ < 0 | No real roots |
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