📚 Mastering Quadratic Equations for IGCSE Mathematics | 精通 IGCSE 数学中的二次方程
Quadratic equations appear throughout the IGCSE Mathematics course, from algebraic manipulation to graph sketching and real-world modelling. This revision guide explains the key methods step by step, including factorising, completing the square, using the quadratic formula and interpreting the discriminant.
二次方程贯穿 IGCSE 数学课程的始终，涉及代数变形、函数图像和实际建模。本复习指南逐步讲解关键方法，包括因式分解法、配方法、求根公式以及判别式的解读。
1. Recognising a Quadratic Expression | 认识二次表达式
A quadratic expression in one variable has the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. The term ax² is called the quadratic term, bx is the linear term and c is the constant term.
一元二次表达式的一般形式为 ax² + bx + c，其中 a、b 和 c 是常数且 a ≠ 0。ax² 称为二次项，bx 称为一次项，c 为常数项。
If a were equal to zero, the expression would become bx + c, which is linear rather than quadratic. IGCSE questions often ask you to identify coefficients, especially when rearranging an equation into standard form.
如果 a 等于零，表达式就变为 bx + c，属于一次式而非二次式。IGCSE 题目常要求你识别系数，尤其是把方程整理为标准形式时。
2. Standard Form and Coefficients | 标准形式与系数
Before solving, always rearrange the equation so that one side is zero: ax² + bx + c = 0. This is called standard form.
求解前应始终把方程整理成一边为零的形式：ax² + bx + c = 0。这称为标准形式。
For example, 3x² – 7 = 2x should be rearranged to 3x² – 2x – 7 = 0. Here a = 3, b = -2 and c = -7.
例如，3x² – 7 = 2x 应整理为 3x² – 2x – 7 = 0。此时 a = 3，b = -2，c = -7。
3. Solving by Factorising | 因式分解法
Factorising is usually the fastest method when the quadratic has integer roots. We look for two numbers that multiply to ac and add to b.
当二次方程有整数根时，因式分解通常是最快的方法。我们寻找两个数，使它们相乘等于 ac、相加等于 b。
For x² + 5x + 6 = 0, we need two numbers with product 6 and sum 5: 2 and 3. So x² + 5x + 6 = (x + 2)(x + 3) = 0, giving x = -2 or x = -3.
对于 x² + 5x + 6 = 0，需要两个数乘积为 6、和为 5：2 和 3。因此 x² + 5x + 6 = (x + 2)(x + 3) = 0，得到 x = -2 或 x = -3。
If a ≠ 1, use the ac method or trial-and-error with brackets. Always check by expanding.
若 a ≠ 1，可使用 ac 方法或试括号法进行分解。务必通过展开来检验。
4. Completing the Square | 配方法
Completing the square is useful for deriving the vertex form and for equations that do not factorise neatly. The key identity is x² + bx = (x + b/2)² – (b/2)².
配方法适用于推导顶点式，以及难以整齐因式分解的方程。关键恒等式为 x² + bx = (x + b/2)² – (b/2)²。
For x² + 6x + 2 = 0, write x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7 = 0. Then x + 3 = ±√7, so x = -3 ± √7.
对于 x² + 6x + 2 = 0，写成 x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7 = 0。然后 x + 3 = ±√7，因此 x = -3 ± √7。
If the coefficient of x² is not 1, factor it out first before completing the square.
若 x² 的系数不是 1，需先将其提出，再进行配方。
5. The Quadratic Formula | 二次公式
The quadratic formula solves any quadratic equation in standard form. It is derived by completing the square on ax² + bx + c = 0.
求根公式可以求解任何标准形式的二次方程。它由对 ax² + bx + c = 0 进行配方法推导而来。
x = [-b ± √(b² – 4ac)] / 2a
Remember to substitute b, a and c with their signs. For 2x² – 3x – 5 = 0, we have a = 2, b = -3, c = -5.
代入时要保留 b、a、c 的符号。对于 2x² – 3x – 5 = 0，有 a = 2，b = -3，c = -5。
Then x = [3 ± √(9 + 40)] / 4 = [3 ± √49] / 4 = [3 ± 7] / 4, giving x = 5/2 or x = -1.
则 x = [3 ± √(9 + 40)] / 4 = [3 ± √49] / 4 = [3 ± 7] / 4，得到 x = 5/2 或 x = -1。
6. The Discriminant and Number of Roots | 判别式与根的数量
The discriminant is D = b² – 4ac. It determines the nature and number of real roots without solving the equation.
判别式为 D = b² – 4ac。它无需解方程即可判断实数根的性质和个数。
| Discriminant | Number of real roots |
| D > 0 | Two distinct real roots |
| D = 0 | One repeated real root |
| D < 0 | No real roots |
IGCSE exam questions often ask you to find the range of k for which a quadratic has two distinct real roots, so write b² – 4ac > 0 and solve the resulting inequality.
IGCSE 考试常要求找出使二次方程有两个不同实根的 k 的取值范围，因此应写出 b² – 4ac > 0，并解所得不等式。
7. Sketching Quadratic Graphs | 二次函数图像
A quadratic graph y = ax² + bx + c is a parabola. If a > 0, it opens upwards and has a minimum point; if a < 0, it opens downwards and has a maximum point.
二次函数图像 y = ax² + bx + c 是一条抛物线。若 a > 0，开口向上且有最小值点；若 a < 0，开口向下且有最大值点。
To sketch the graph, find the y-intercept at (0, c), the x-intercepts by solving ax² + bx + c = 0, and the turning point.
画图像时，需找到 y 轴截距 (0, c)、通过解 ax² + bx + c = 0 得到的 x 轴截距，以及顶点（转折点）。
Label all intercepts and the vertex clearly. Use the axis of symmetry to check the graph is symmetric.
清晰地标注所有截距和顶点。利用对称轴检查图像是否对称。
8. Vertex and Axis of Symmetry | 顶点与对称轴
The vertex of y = ax² + bx + c occurs at x = -b / (2a). The y-coordinate is found by substituting this x-value back into the equation.
y = ax² + bx + c 的顶点横坐标为 x = -b / (2a)。将 x 值代回原式即可求出纵坐标。
In completed-square form y = a(x – h)² + k, the vertex is (h, k) and the axis of symmetry is x = h. This form makes transformations easier.
在配方式 y = a(x – h)² + k 中，顶点为 (h, k)，对称轴为 x = h。这种形式更便于进行图像变换。
For y = (x – 3)² – 4, the vertex is (3, -4) and the graph is a upward parabola shifted 3 units right and 4 units down from y = x².
对于 y = (x – 3)² – 4，顶点为 (3, -4)，图像是 y = x² 向右平移 3 个单位、向下平移 4 个单位后得到的开口向上抛物线。
9. Quadratic Inequalities | 二次不等式
To solve ax² + bx + c > 0 or < 0, first sketch the parabola or use a sign table. The sign changes at the real roots.
解 ax² + bx + c > 0 或 < 0 时，首先画出抛物线草图或使用符号表。符号在实根处发生变化。
For x² – x – 6 < 0, factorise to (x - 3)(x + 2) < 0. Test intervals: x < -2, -2 < x < 3, x > 3. The solution is -2 < x < 3.
对于 x² – x – 6 < 0，因式分解为 (x - 3)(x + 2) < 0。检验区间：x < -2、-2 < x
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