📚 Mastering Quadratic Equations for Edexcel IGCSE Math | Edexcel IGCSE 数学二次方程全解析
Quadratic equations form a cornerstone of the Edexcel IGCSE Mathematics syllabus. Whether factorising, completing the square or using the quadratic formula, a deep grasp of this topic will boost your confidence and exam performance. This article breaks down every essential concept step by step, linking theory to the kinds of questions that appear regularly on both Paper 1 and Paper 2.
二次方程是 Edexcel IGCSE 数学课程的核心内容之一。无论采用因式分解、配方法还是求根公式，扎实掌握这一主题都能极大提升你的解题信心和考试成绩。本文循序渐进地梳理每一个重要概念，把理论与卷一卷二中频繁出现的考题紧密连接。
1. Recognizing a Quadratic Equation | 识别二次方程
If a = 0 the equation becomes linear, so the x² term must be present to remain quadratic.
二次方程是指能写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。若 a = 0，则方程退化为一次方程，因此必须保留 x² 项才叫二次方程。
= 0. Always rearrange so that one side equals zero before deciding on a solution strategy.
Edexcel IGCSE 试题常将二次方程隐藏于几何或现实情境中。例如，矩形面积问题可能导出 x² + 5x − 24 = 0。务必在确定解法前将所有项移到等号一边并使方程等于零。
- Standard form: ax² + bx + c = 0
- If a = 0: not quadratic
- Rearranging key step
- x² coefficient cannot be zero
- 标准形式：ax² + bx + c = 0
- 若 a = 0：不是二次方程
- 移项是关键步骤
- x² 系数不能为零
2. Solving by Factorising | 因式分解法求解
at multiply to c and add to b. Then write (x + p)(x + q) = 0 and set each bracket to zero.
当二次式能分解为两个简单的一次因式时，因式分解法十分高效。对于首项系数为1的二次式 x² + bx + c，找出两个数，使其乘积为 c 且和为 b，然后将方程写成 (x + p)(x + q) = 0，再令每个括号等于零。
le term and factorise in pairs: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 a ≠ 1 时，如 2x² + 7x + 3，先将 a 与 c 相乘（2×3=6），找出两个乘积为 6 且和为 7 的数（6 和 1），然后将中间项拆开，分组分解：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
| Quadratic | Factors | Solutions |
| x² + 5x + 6 | (x+2)(x+3) | x = −2, −3 |
| x² − x − 12 | (x−4)(x+3) | x = 4, −3 |
| 3x² − 11x − 4 | (3x+1)(x−4) | x = −1/3, 4 |
3. Quadratic Formula | 二次方程求根公式
o messy, rely on the quadratic formula. For ax² + bx + c = 0, the solutions are x = [ −b ± √b² − 4ac ] / (2a). The expression under the square root, b² − 4ac, is called the discriminant.
当因式分解过于复杂时，可依靠求根公式。对于 ax² + bx + c = 0，解为 x = [ −b ± √b² − 4ac ] / (2a)。根号下的式子 b² − 4ac 称为判别式。
Always write down the values of a, b and c before substituting. Be extremely careful with signs: if any coefficient is negative, put it in brackets. Simplify the numerator fully before dividing by 2a, and remember you often get two distinct answers unless the discriminant equals zero.
代入前一定要先写出 a、b、c 的值。特别注意符号：如果系数为负，用括号括起来。先完全化简分子，再除以 2a，记住除非判别式等于零，通常你会得到两个不相同的实数解。
x = [ −b ± √b² − 4ac ] / (2a)
- Write a, b, c clearly
- Calculate discriminant first
- Split into '+' and '−' cases
- 先明确写出 a、b、c
- 先计算判别式的值
- 拆分为 '+' 和 '−' 两种情况
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This method is not only a solving tool but also the gateway to understanding the turning point of a parabola and the derivation of the quadratic formula itself.
配方法将二次式改写为 a(x + p)² + q 的形式。这种方法不仅是求解利器，更是理解抛物线顶点以及推导求根公式的钥匙。
x] + 5. Half of 4 is 2, square is 4, so 2[(x+2)² − 4] + 5 = 2(x+2)² − 8 + 5 = 2(x+2)² − 3.
从 x² + bx 入手，取 b 的一半，平方，然后在表达式里加上再减去这个平方值。以 2x² + 8x + 5 为例，先把前两项的系数 2 提出来：2[x² + 4x] + 5。4 的一半是 2，平方是 4，于是得到 2[(x+2)² − 4] + 5 = 2(x+2)² − 8 + 5 = 2(x+2)² − 3。
Once in completed square form, solving (x + p)² = k becomes straightforward: x = −p ± √k + p)² + q sits at (−p, q). This is frequently examined in the context of sketching graphs.
一旦化为完全平方形式，解 (x + p)² = k 就变得简单：x = −p ± √k。抛物线 y = a(x + p)² + q 的顶点坐标为 (−p, q)。这在画图题中出现频率极高。
5. The Discriminant and Nature of Roots | 判别式与根的性质
ly one repeated real root; Δ < 0 means no real roots (the graph never touches the x-axis).
判别式 Δ = b² − 4ac 能立刻告诉你二次方程有多少个实数解。Δ > 0 有两个相异实根；Δ = 0 有一个重根；Δ < 0 则无实根（图像与 x 轴无交点）。
Edexcel IGCSE often asks: "Find the value of k for which the equation has equal roots." n terms of k, and solve. This connects algebra with logical reasoning about intersections.
Edexcel IGCSE 常出类似“求 k 的值使方程有等根”的题目。令 b² − 4ac = 0，用含 k 的式子替换 a、b、c，再解方程。这能将代数知识与交点逻辑推理相结合。
- Δ > 0: two distinct real roots
- Δ = 0: one repeated root
- Δ < 0: no real roots
- Δ > 0：两个相异实根
- Δ = 0：一个重根
- Δ < 0：无实根
6. Graphs of Quadratic Functions | 二次函数的图像
The graph of y = ax² + bx + c is a parabola. If a > 0 the parabola opens upward (∪-shaped); if a < 0 it opens downward (∩-shaped). The y-intercept is (0, c). The x-intercepts are the roots of ax² + bx + c = 0.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上（∪形）；若 a < 0，则开口向下（∩形）。y 轴截距为 (0, c)。x 轴交点即是方程 ax² + bx + c = 0 的根。
pers love asking you to sketch a parabola, label the vertex, axis of symmetry and intercepts.
利用完全平方形式可求出顶点坐标。对称轴为 x = −b/(2a) 或通过 (x + p)² 得出 x = −p。IGCSE 试卷热衷于让你画出抛物线草图，并标注顶点、对称轴及各截距。
Symmetry axis: x = −b / (2a)
- a > 0: ∪ shape
- a < 0: ∩ shape
- y-intercept = c
- Vertex from completing square
- a > 0：∪形
- a < 0：∩形
- y 截距为 c
- 配方法得顶点
7. Solving Quadratic Inequalities | 解二次不等式
lve x² − 4x − 5 = 0 to get critical values −1 and 5, then test intervals on a number line.
形如 x² − 4x − 5 > 0 的二次不等式，最好先从对应方程入手。解 x² − 4x − 5 = 0 得到临界值 −1 和 5，然后在数轴上测试区间。
Sketch the parabola to see where it sits above or below the x-axis. For > 0, choose the outer regions; for < 0, choose the interval between the roots. Always present the final answer using set notation or inequalities as required by the question.
画出抛物线草图，观察哪些部分在 x 轴上方或下方。对于 > 0，选取根两侧的外部区域；对于 < 0，选取两根之间的区间。最终答案务必按照题目要求，用集合符号或不等式表示。
| Inequality | Critical values | Solution |
| x² − x − 6 > 0 | x = −2, 3 | x < −2 or x > 3 |
| x² + 4x ≤ 0 | x = −4, 0 | −4 ≤ x ≤ 0 |
8. Forming Quadratics from Roots | 已知根构造二次方程
elies on sum and product of roots, a topic that links algebra with symmetric functions and is explicitly tested in Edexcel IGCSE.
已知两根 α 和 β，二次方程可写为 x² − (α+β)x + αβ = 0。这里用到根的和与积，是代数与对称函数的联系点，也是 Edexcel IGCSE 明确的考点。
0. This skill is also essential when one root is given in surd form; the other root is often its conjugate.
例如，若两根为 3 和 −7，和为 −4，积为 −21，得到方程 x² + 4x − 21 = 0。当一根以根式形式给出时，另一根往往为其共轭根式，这项技能也至关重要。
x² − (sum of roots)x + (product of roots) = 0
- Sum α+β = −b/a
- Product αβ = c/a
- 和 α+β = −b/a
- 积 αβ = c/a
9. Simultaneous Equations – One Quadratic, One Linear | 联立方程：一次与二次
ar expression into the quadratic, rearrange to standard form, and solve for x. Then back-substitute to find y.
IGCSE 中常见 y = mx + c 与 y = ax² + bx + c 的联立。将一次表达式代入二次式，整理成标准形式，解出 x，再代回求出 y。
Always give solutions as coordinate pairs. Be prepared for two intersection points, one (tangent) or none (no intersection). The discriminant of the resulting quadratic tells the number of intersection points between line and parabola at a glance.
答案务必要写成坐标对。可能出现两个交点、一个切点或无交点的情况。由代入后所得二次方程的判别式，可一眼看出直线与抛物线的交点个数。
- Substitute linear into quadratic
- Solve for x
- Back-substitute for y
- 将一次式代入二次式
- 解 x
- 回代求 y
10. Real-Life Applications and Word Problems | 实际应用与文字题
Edexcel IGCSE contextualises quadratics through area, projectile motion, profit and number puzzles. Identify the unknown, express other quantities in terms of it, build an equation, solve and finally check whether all solutions make sense in context.
Edexcel IGCSE 通过面积、抛体运动、利润及数字谜题等情境考查二次方程的应用。设出未知数，用其表示其他量，建立方程，求解，最后检验所有解在语境中是否合理。
Negative lengths or times must be discarded. Always reread the question to decide which root is physically possible. This step tests your ability to interpret mathematical solutions back into a real-world scenario.
负值的长、宽、时间等应舍去。务必重读题目，判断哪一个根在实际情况中有意义。这一步考验你将数学解还原到现实场景的解读能力。
+1) = 72 → x² + x − 72 = 0 → (x+9)(x−8) = 0 → x = 8 (reject −9). The integers are 8 and 9.
示例：两个连续正整数的乘积为 72。设较小的数为 x，则 x(x+1) = 72 → x² + x − 72 = 0 → (x+9)(x−8) = 0 → x = 8（舍去 −9）。这两个数为 8 和 9。
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