📚 Mastering Quadratic Equations for Cambridge IGCSE Mathematics | 剑桥 IGCSE 数学：攻克二次方程
Quadratic equations form a central topic in the Cambridge IGCSE Mathematics syllabus, appearing in both Core and Extended papers. They link algebra, graphing, and real-world problem solving. This article reviews the key methods, common pitfalls, and exam strategies you need to master.
二次方程是剑桥 IGCSE 数学课程的核心内容，在核心卷和拓展卷中都会出现。它将代数、图像和实际问题解决紧密结合。本文梳理了关键方法、常见错误和考试策略，帮助你彻底掌握这一专题。
1. What is a Quadratic Equation? | 什么是二次方程？
n be written in the standard form ax² + bx + c = 0, where a, b, c are constants and a ≠ 0.
二次方程是含有一个未知数的二次多项式方程，未知数的最高次数为 2，通常可写成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。
st one must be rearranged into standard form before solving, which gives 2x² - 5x + 1 = 0.
常见例子包括 3x² - 7x + 2 = 0、-x² + 4 = 0 以及 2x² = 5x - 1。最后一个方程需要先化为标准形式，得到 2x² - 5x + 1 = 0。
In IGCSE Mathematics, the unknown is usually x, but any letter can be used. The key test is whether the highest power of the variable is 2 and the equation contains an equals sign.
在 IGCSE 数学中，未知数通常用 x 表示，但也可以用其他字母。判断是否为二次方程的关键是变量的最高次数是否为 2，以及方程中是否含有等号。
2. Standard Form and Key Vocabulary | 标准形式与关键术语
5x - 1 becomes 2x² - 5x + 1 = 0. This makes the coefficients a, b and c easy to identify.
在求解之前，你必须能将方程化为标准形式。例如，2x² = 5x - 1 可化为 2x² - 5x + 1 = 0。这样更容易识别系数 a、b 和 c。
The table below summarises the key vocabulary used in Cambridge IGCSE Mathematics questions.
下表总结了剑桥 IGCSE 数学题目中常用的关键术语。
| Term | Meaning | 中文解释 |
| Quadratic term | ax² | 二次项 |
| Linear term | bx | 一次项 |
| Constant term | c | 常数项 |
| Root / solution | A value of x that makes the equation true | 方程的根或解 |
| Coefficient | a, b or c | 系数 |
member that a cannot be zero, because if a = 0 the equation becomes linear, not quadratic.
请记住，a 不能为零，因为如果 a = 0，方程就变成了一元一次方程，而不是二次方程。
3. Solving by Factorisation | 因式分解法
et each factor equal to zero. This works because if a product equals zero, at least one factor must be zero.
因式分解法的关键是把二次方程写成 (px + q)(rx + s) = 0，然后令每个因式分别等于零。这是因为如果乘积为零，那么至少有一个因式为零。
6 = 0 factorises to (x - 2)(x - 3) = 0. Setting each factor to zero gives x = 2 or x = 3.
例如，x² - 5x + 6 = 0 可因式分解为 (x - 2)(x - 3) = 0。令每个因式等于零，得到 x = 2 或 x = 3。
x² - 5x + 6 = 0 → (x - 2)(x - 3) = 0 → x = 2 or x = 3
or instance, 2x² + 3x - 2 = 0 factorises to (2x - 1)(x + 2) = 0, giving x = 1/2 or x = -2.
当首项系数不为 1 时，你需要尝试不同的组合。例如，2x² + 3x - 2 = 0 可分解为 (2x - 1)(x + 2) = 0，得到 x = 1/2 或 x = -2。
Always check your factorisation by expanding the brackets. If expansion returns the original quadratic, your factors are correct.
务必通过展开括号来检查因式分解是否正确。如果展开后回到原二次方程，说明因式分解无误。
4. The Quadratic Formula | 二次公式
When factorisation is difficult, use the quadratic formula. It is derived from completing the square and works for every quadratic equation.
当因式分解困难时，可以使用二次公式。它由配方法推导而来，适用于所有二次方程。
x = (-b ± √b² - 4ac) / (2a)
rd form ax² + bx + c = 0. Substitute carefully, especially when coefficients are negative.
在该公式中，a、b 和 c 是标准形式 ax² + bx + c = 0 中的系数。代入时要特别小心，尤其是系数为负数时。
= 3, b = -4, c = -2. Substituting gives x = [4 ± √((-4)² - 4 × 3 × (-2))] / (2 × 3) = [4 ± √16 + 24] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = [2 ± √10] / 3.
例题：解方程 3x² - 4x - 2 = 0。这里 a = 3，b = -4，c = -2。代入公式得 x = [4 ± √((-4)² - 4 × 3 × (-2))] / (2 × 3) = [4 ± √16 + 24] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = [2 ± √10] / 3。
The quadratic formula is particularly useful for Extended-level questions where factorisation is not obvious or where answers must be given in surd form.
二次公式在拓展卷题目中特别有用，因为有些题目不易因式分解，或者要求将答案保留为根式形式。
5. Completing the Square | 配方法
Completing the square transforms a quadratic expression into the form a(x + p)² + q. This form reveals the vertex of the graph and allows you to solve equations.
配方法可将二次表达式转化为 a(x + p)² + q 的形式。这种形式能直接显示图像的顶点，也可用于解方程。
adratic with leading coefficient 1, the identity is x² + bx + c = (x + b/2)² - (b/2)² + c.
对于首项系数为 1 的二次式，恒等式为 x² + bx + c = (x + b/2)² - (b/2)² + c。
5 = 0, write (x + 3)² - 4 = 0, so (x + 3)² = 4, giving x + 3 = ±2, hence x = -1 or x = -5.
例题：x² + 6x + 5 = (x + 3)² - 9 + 5 = (x + 3)² - 4。解方程 x² + 6x + 5 = 0 时，写成 (x + 3)² - 4 = 0，于是 (x + 3)² = 4，得到 x + 3 = ±2，所以 x = -1 或 x = -5。
t first. For example, 2x² + 8x + 1 = 2(x² + 4x) + 1 = 2[(x + 2)² - 4] + 1 = 2(x + 2)² - 7.
如果首项系数不是 1，需要先把它提出来。例如，2x² + 8x + 1 = 2(x² + 4x) + 1 = 2[(x + 2)² - 4] + 1 = 2(x + 2)² - 7。
6. The Discriminant and Nature of Roots | 判别式与根的性质
4ac. It is the part under the square root in the quadratic formula and tells you how many real roots the quadratic has.
判别式定义为 Δ = b² - 4ac。它是二次公式中根号下的部分，可以告诉你二次方程有多少个实数根。
The table below summarises the three cases you must know for Cambridge IGCSE Mathematics.
下表总结了剑桥 IGCSE 数学必须掌握的三种情况。
| Condition | Nature of roots | 根的性质 |
| Δ > 0 | Two distinct real roots | 两个不相等的实数根 |
| Δ = 0 | One repeated real root | 两个相等的实数根（一个重根） |
| Δ < 0 | No real roots | 没有实数根 |
If Δ is a positive perfect square, the roots are rational and the quadratic factorises neatly. If Δ is positive but not a perfect square, the roots are irrational surds.
如果 Δ 是一个正的完全平方数，那么方程有两个有理数根，且二次式可以整齐地因式分解。如果 Δ 为正但不是完全平方数，那么根是无理数根式。
² - 6x + 9 = 0, Δ = (-6)² - 4 × 1 × 9 = 36 - 36 = 0, so there is one repeated root, x = 3.
例题：对于 x² - 6x + 9 = 0，Δ = (-6)² - 4 × 1 × 9 = 36 - 36 = 0，所以有一个重根，x = 3。
7. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
二次函数 y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上，有最低点；若 a < 0，抛物线开口向下，有最高点。
b/(2a). The vertex is the turning point and has coordinates (-b/(2a), f(-b/(2a))). The constant c gives the y-intercept.
对称轴的方程是 x = -b/(2a)。顶点就是转折点，坐标为 (-b/(2a), f(-b/(2a)))。常数项 c 是图像与 y 轴的交点纵坐标。
on ax² + bx + c = 0. They can be found by factorisation, formula or completing the square.
如果存在 x 轴截距，它们就是方程 ax² + bx + c = 0 的根。可以通过因式分解、二次公式或配方法求出。
, and the roots are x = 1 and x = 3, so the graph crosses the x-axis at (1, 0) and (3, 0).
例如，y = x² - 4x + 3 中 a = 1，b = -4，c = 3。对称轴为 x = 2，顶点为 (2, -1)，根为 x = 1 和 x = 3，因此图像在 (1, 0) 和 (3, 0) 处与 x 轴相交。
8. Transformations of y = x² | y = x² 的变换
The graph of y = x² can be shifted, stretched or reflected to produce any quadratic graph.
顶点式 y = a(x - h)² + k 使图像的变换更容易识别。y = x² 的图像可以通过平移、伸缩或对称变换得到任意二次函数图像。
rtex is (h, k). The value of h shifts the graph left or right, and k shifts it up or down.
在 y = a(x - h)² + k 中，顶点为 (h, k)。h 的值使图像左右平移，k 的值使图像上下平移。
If a is negative, the graph is reflected in the x-axis. If |a| > 1, the graph is vertically stretched; if |a| < 1, it is vertically compressed.
如果 a 为负数，图像关于 x 轴反射。如果 |a| > 1，图像在竖直方向上被拉伸；如果 |a| < 1，图像在竖直方向上被压缩。
(1, -3), opens upwards, and is vertically stretched by a factor of 2 compared with y = x².
例题：y = 2(x - 1)² - 3 的顶点为 (1, -3)，开口向上，与 y = x² 相比在竖直方向上拉伸为原来的 2 倍。
9. Applications and Word Problems | 实际应用与文字题
Quadratic equations often appear in real-world problems such as area, projectile motion, and number puzzles. The first step is to define the unknown and form a quadratic equation from the given conditions.
二次方程经常出现在面积、抛体运动和数字谜题等实际问题中。第一步是设出未知数，并根据已知条件建立二次方程。
)(n - 8) = 0, so n = -9 or n = 8. Since n is positive, n = 8, so the integers are 8 and 9.
例题：两个连续正整数的乘积是 72。设较小的整数为 n，则 n(n + 1) = 72，得到 n² + n - 72 = 0。因式分解为 (n + 9)(n - 8) = 0，故 n = -9 或 n = 8。由于 n 为正数，所以 n = 8，两个整数为 8 和 9。
(w + 8)(w - 5) = 0, so w = 5 because width cannot be negative. The length is therefore 8.
另一种常见题型
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