📚 Mastering Quadratic Equations and Functions | 掌握二次方程与函数
Quadratic equations and functions are one of the most heavily examined topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, graph sketching, and real-world modelling, so a confident understanding of factorising, the quadratic formula, completing the square, and the discriminant is essential for top grades.
二次方程与函数是 IGCSE 数学中最常考查的主题之一。它们出现在代数、坐标几何、图像绘制和实际建模中，因此扎实掌握因式分解、二次公式、配方法和判别式是取得高分的关键。
1. What is a Quadratic? | 什么是二次式
A quadratic expression is any algebraic expression that can be written in the standard form ax² + bx + c, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which is why the graph of a quadratic function is always a parabola.
二次式是任何可以写成标准形式 ax² + bx + c 的代数表达式，其中 a、b、c 为常数且 a ≠ 0。变量 x 的最高次数为 2，因此二次函数的图像总是抛物线。
In IGCSE questions, you may be asked to identify a quadratic expression, write it in descending powers of x, or recognise that a given graph is quadratic. Remember that if a = 0, the expression becomes linear bx + c, so the condition a ≠ 0 is crucial.
在 IGCSE 题目中，你可能需要识别二次式、按 x 的降幂书写表达式，或判断给定图像是否为二次函数。请记住，如果 a = 0，表达式就变成一次式 bx + c，因此 a ≠ 0 这一条件至关重要。
- Standard form: ax² + bx + c
- Coefficient a cannot be zero
- Graph shape is a smooth curve called a parabola
- The y-intercept is always at (0, c)
标准形式为 ax² + bx + c；系数 a 不能为零；图像是一条光滑的曲线，称为抛物线；y 轴截距始终在 (0, c)。
2. Expanding Double Brackets | 展开双括号
Expanding two binomials is a core algebraic skill that underpins factorisation, solving equations, and graph work. To expand (x + p)(x + q), multiply every term in the first bracket by every term in the second bracket. The FOIL method helps you remember the order: First, Outer, Inner, Last.
展开两个二项式是一项核心代数技能，它为因式分解、解方程和图像学习打下基础。要展开 (x + p)(x + q)，需要将第一个括号中的每一项分别乘以第二个括号中的每一项。FOIL 方法可以帮助你记住顺序：首项、外项、内项、末项。
For example, (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15. Notice how the coefficient of x is the sum p + q, and the constant term is the product pq.
例如，(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15。请注意，x 的系数是 p + q，常数项是 pq 的乘积。
Two special expansions save time in exams: (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b². Also remember the difference of squares: (a + b)(a − b) = a² − b².
两个特殊的展开式可以在考试中节省时间：(a + b)² = a² + 2ab + b² 和 (a − b)² = a² − 2ab + b²。还要记住平方差公式：(a + b)(a − b) = a² − b²。
- (x + p)(x + q) = x² + (p + q)x + pq
- (x + p)² = x² + 2px + p²
- (x + p)(x − p) = x² − p²
展开公式包括 (x + p)(x + q) = x² + (p + q)x + pq；(x + p)² = x² + 2px + p²；(x + p)(x − p) = x² − p²。
3. Factorising Quadratic Expressions | 因式分解二次式
Factorising a quadratic means writing it as a product of two linear brackets. For a quadratic of the form x² + bx + c, find two numbers whose product is c and whose sum is b. This is often the first step in solving quadratic equations.
因式分解二次式意味着将其写成两个一次括号的乘积。对于形如 x² + bx + c 的二次式，需要找到两个数，使它们的乘积为 c、和为 b。这通常是解二次方程的第一步。
For example, to factorise x² − 7x + 12, identify two numbers that multiply to give 12 and add to give −7. The numbers −3 and −4 satisfy (−3) × (−4) = 12 and (−3) + (−4) = −7, so x² − 7x + 12 = (x − 3)(x − 4).
例如，对 x² − 7x + 12 进行因式分解，需要找到两个相乘得 12、相加得 −7 的数。−3 和 −4 满足 (−3) × (−4) = 12 且 (−3) + (−4) = −7，因此 x² − 7x + 12 = (x − 3)(x − 4)。
When the leading coefficient is not 1, such as in 2x² + 7x + 3, use the splitting-the-middle-term method. Multiply a by c to get 2 × 3 = 6, find two numbers that multiply to 6 and add to 7, namely 6 and 1. Rewrite as 2x² + 6x + x + 3, then factorise in pairs to get (2x + 1)(x + 3).
当首项系数不为 1 时，例如 2x² + 7x + 3，可以使用拆中项法。先将 a 与 c 相乘得 2 × 3 = 6，找到两个相乘为 6、相加为 7 的数，即 6 和 1。将原式改写为 2x² + 6x + x + 3，然后分组提取公因式，得到 (2x + 1)(x + 3)。
Always check your factorisation by expanding the brackets. If the expanded form matches the original quadratic, your answer is correct.
始终通过展开括号来检查因式分解结果。如果展开后与原二次式一致，答案就是正确的。
4. Solving Quadratic Equations by Factorisation | 因式分解法解二次方程
To solve a quadratic equation by factorising, first rearrange the equation into the standard form ax² + bx + c = 0. Then factorise the left-hand side and apply the zero-product rule: if A × B = 0, then either A = 0 or B = 0, or both.
要通过因式分解法解二次方程，首先将方程整理成标准形式 ax² + bx + c = 0。然后对左边进行因式分解，并应用零积法则：若 A × B = 0，则 A = 0 或 B = 0，或两者同时为零。
For example, solve x² + 5x + 6 = 0. Factorising gives (x + 2)(x + 3) = 0. Setting x + 2 = 0 gives x = −2, and setting x + 3 = 0 gives x = −3. The two solutions are x = −2 and x = −3.
例如，解方程 x² + 5x + 6 = 0。因式分解得 (x + 2)(x + 3) = 0。令 x + 2 = 0 得 x = −2，令 x + 3 = 0 得 x = −3。因此两个解为 x = −2 和 x = −3。
- Rearrange the equation so one side is exactly 0
- Factorise the quadratic expression fully
- Set each bracket equal to zero
- Solve each resulting linear equation
- Substitute your answers back to check
先将方程一边化为 0；将二次式彻底因式分解；令每个括号分别等于零；解每个一次方程；最后将答案代回检验。
A common mistake is to divide both sides of an equation by x, which can lose the solution x = 0. Always bring all terms to one side and factorise instead of dividing by a variable.
常见错误是方程两边同除以 x，这可能会丢失 x = 0 这个解。务必把所有项移到一边并因式分解，而不要除以变量。
5. Completing the Square | 配方法
Completing the square rewrites x² + bx + c in the form (x + p)² + q, where p is half the coefficient of x. This form is extremely useful for finding the turning point of a parabola and for solving quadratics that cannot be factorised easily.
配方法将 x² + bx + c 改写为 (x + p)² + q 的形式，其中 p 是 x 系数的一半。这种形式对于求抛物线的顶点以及解不易因式分解的二次方程非常有用。
For x² + 6x + 2, take half of 6, which is 3, and square it to get 9. Add and subtract this value inside the expression: x² + 6x + 9 − 9 + 2 = (x + 3)² − 7. Therefore the completed square form is (x + 3)² − 7.
对于 x² + 6x + 2，取 6 的一半得到 3，将其平方得到 9。在表达式中加上并减去这个值：x² + 6x + 9 − 9 + 2 = (x + 3)² − 7。因此完全平方形式为 (
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