📚 Mastering Quadratic Equations for IGCSE Mathematics | 掌握IGCSE数学中的二次方程
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebraic manipulation to graph sketching and real-world modelling. A strong command of quadratics helps you solve equations accurately, interpret graphs with confidence, and tackle extended problem-solving questions under exam pressure. This article covers recognition, standard form, factorisation, completing the square, the quadratic formula, the discriminant, graphs and turning points, inequalities, applications, simultaneous equations, and the mistakes that cost the most marks.
二次方程贯穿IGCSE数学大纲，从代数变形到图像绘制再到实际问题建模，处处都会出现。牢固掌握二次方程，能帮助你准确解方程、自信解读图像，并在考试压力下从容应对扩展应用题。本文涵盖识别二次方程、标准形式、因式分解法、配方法、求根公式、判别式、图像与顶点、二次不等式、实际应用、联立方程，以及最容易丢分的常见错误。
1. Recognising Quadratic Equations | 识别二次方程
0. The condition a ≠ 0 is essential: if a = 0, the equation becomes linear rather than quadratic, and every method in this article would break down.
一元二次方程是未知数最高次数为 2 的方程，一般形式为 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。条件 a ≠ 0 极其关键：如果 a = 0，方程就退化为一次方程，本文所有方法都将失效。
s cubic. IGCSE questions often ask you to solve a quadratic equation, or to identify the coefficients a, b and c before substituting into a formula, so recognising the type of equation is the first skill to master.
例如，x² - 5x + 6 = 0 是二次方程，而 2x + 3 = 0 是一次方程，x³ - 4x = 0 是三次方程。IGCSE 题目常要求你解二次方程，或在使用公式前先识别系数 a、b、c，因此判断方程类型是必须掌握的第一步。
pressions such as (x + 3)(x - 5) = 0, 3 ÷ x = x + 2, and x⁴ - 5x² + 4 = 0 (in x²) all hide quadratics that must be revealed before solving.
你还必须能识别经过移项后隐藏的二次方程，例如 x² + 3 = 5x + 1，把所有项移到一边后变为 x² - 5x + 2 = 0。像 (x + 3)(x - 5) = 0、3 ÷ x = x + 2，以及 x⁴ - 5x² + 4 = 0（以 x² 为元）这类式子，都隐藏着需要先"揭露"再求解的二次结构。
- Quadratic: highest power of x is 2, for example 3x² - 7x + 2 = 0. | 二次：x 的最高次数为 2，例如 3x² - 7x + 2 = 0。
- Linear: highest power of x is 1, for example 5x - 4 = 0. | 一次：x 的最高次数为 1，例如 5x - 4 = 0。
- Hidden quadratic: rearrange first, then identify a, b and c. | 隐藏的二次：先移项整理，再确定 a、b、c。
2. Writing Quadratics in Standard Form | 将二次式写成标准形式
s on one side and zero on the other. This single habit makes factorisation, completing the square and formula substitution far more reliable, and it prevents the most common source of sign errors in exams.
求解之前，务必先把方程整理成 ax² + bx + c = 0 的标准形式，让所有项位于一边、另一边为零。这个习惯能让因式分解、配方法和代入公式都可靠得多，也能避免考试中最常见的符号错误。
ing 2x² and 7x from both sides. Always collect like terms carefully, keep the x² term positive where possible, and check every sign as you move terms across the equals sign.
例如，4x² - 8 = 2x² + 7x，两边同时减去 2x² 与 7x，可整理为 2x² - 7x - 8 = 0。务必仔细合并同类项，尽量让 x² 项系数为正，并在每一项跨过等号时逐一检查符号。
In standard form, a is the coefficient of x², b is the coefficient of x, and c is the constant term. Identifying these values correctly is especially important for the quadratic formula and the discriminant, where a single wrong sign changes the answer completely.
在标准形式中，a 是 x² 的系数，b 是 x 的系数，c 是常数项。正确识别这三个值对求根公式和判别式尤为重要，因为一个符号写错，答案就会完全不同。
| Equation | 方程 | a | b | c |
| x² - 5x + 6 = 0 | 1 | -5 | 6 |
| 2x² + 7x + 3 = 0 | 2 | 7 | 3 |
| 2x² - 7x - 8 = 0 | 2 | -7 | -8 |
3. Solving by Factorisation | 因式分解法求解
roduct of two brackets, then use the zero product rule: if A × B = 0, then A = 0 or B = 0. Each bracket gives one root.
当二次方程具有整数根或简单有理根时，因式分解通常是最快的方法。要解 ax² + bx + c = 0，先把表达式写成两个括号的乘积，再使用零积法则：若 A × B = 0，则 A = 0 或 B = 0。每个括号给出一个根。
give -5, so the equation becomes (x - 2)(x - 3) = 0 and the solutions are x = 2 and x = 3.
当 a = 1 时，寻找两个数，使它们的积为 c、和为 b。对于 x² - 5x + 6 = 0，-2 与 -3 的积为 6、和为 -5，于是方程写成 (x - 2)(x - 3) = 0，解为 x = 2 和 x = 3。
= -½ and x = -3. Always expand your brackets mentally to check that the middle term is correct.
当 a ≠ 1 时，可尝试 a 与 c 的因数对，或使用"拆中项"法。对于 2x² + 7x + 3 = 0，因式分解为 (2x + 1)(x + 3) = 0，解得 x = -½ 和 x = -3。写完括号后，务必心算展开一次，确认中间项正确。
ion is a difference of two squares, use the pattern a² - b² = (a + b)(a - b): x² - 9 = 0 becomes (x + 3)(x - 3) = 0.
两种特殊情形能节省大量时间。若各项有公因式，先提取：2x² - 8x = 0 化为 2x(x - 4) = 0，故 x = 0 或 x = 4。若方程是平方差，使用公式 a² - b² = (a + b)(a - b)：x² - 9 = 0 化为 (x + 3)(x - 3) = 0。
4. Completing the Square | 配方法
d when you need the vertex of a graph. The key identity is x² + bx = (x + b ÷ 2)² - (b ÷ 2)².
配方法把二次式改写为 (x + p)² + q = 0 的形式。当因式分解无法进行、答案需要精确形式、或需要求图像顶点时，它都极为有用。核心恒等式是 x² + bx = (x + b ÷ 2)² - (b ÷ 2)²。
x + 3)² - 9 + 1 = 0, which simplifies to (x + 3)² = 8. Taking square roots gives x + 3 = ±√8, so x = -3 + 2√2 or x = -3 - 2√2. Note that both the positive and negative square roots must be taken.
以 x² + 6x + 1 = 0 为例：6 的一半是 3，因此写成 (x + 3)² - 9 + 1 = 0，化简得 (x + 3)² = 8。开平方得 x + 3 = ±√8，所以 x = -3 + 2√2 或 x = -3 - 2√2。注意正负两个平方根都必须取到。
immediately reveals that the minimum value of the expression is -5, occurring when x = -2.
当 a ≠ 1 时，先从前两项提出 a。对于 2x² + 8x + 3，写成 2(x² + 4x) + 3 = 2[(x + 2)² - 4] + 3 = 2(x + 2)² - 5。这个形式立刻显示该表达式的最小值为 -5，在 x = -2 时取得。
x² + bx + c = (x + b ÷ 2)² + c - (b ÷ 2)²
5. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation, whether or not it factorises. It must be applied only after the equation is in standard form, and it is the safest method when the coefficients are awkward or the roots are irrational.
求根公式可以解任何二次方程，无论它能否因式分解。使用前必须先把方程化为标准形式；当系数复杂或根为无理数时，它是最稳妥的方法。
x = (-b ± √b² - 4ac) ÷ (2a)
Take 2x² - 7x - 8 = 0, where a = 2, b = -7 and c = -8. Substituting gives x = (7 ± √49 + 64 .91 to 3 significant figures. Show the substitution line clearly, because method marks are awarded even if the final value slips.
以 2x² - 7x - 8 = 0 为例，a = 2、b = -7、c = -8。代入得 x = (7 ± √49 + 64) ÷ 4 = (7 ± √113) ÷ 4。由于 √113 ≈ 10.63，两根约为 x ≈ 4.41 与 x ≈ -0.91（保留三位有效数字）。请清楚写出代入过程，因为即使最后数值出错，方法分仍然可以拿到。
nly at the final step. If the question asks for exact form, leave the answer as (7 ± √113) ÷ 4.
注意三个细节：本例中 b 为负，因此 -b 变成 +7；整个分子都要除以 2a，而不只是根号部分；只在最后一步才进行四舍五入。若题目要求精确形式，答案应保留为 (7 ± √113) ÷ 4。
6. The Discriminant and the Nature of Roots | 判别式与根的性质
It tells you how many real roots a quadratic has without solving it, which is exactly what many IGCSE questions require.
判别式是根号下的表达式，Δ = b² - 4ac。它能在不解方程的情况下告诉你二次方程有多少个实根，而这正是许多 IGCSE 题目所要求的。
| Discriminant | 判别式 | Nature of roots | 根的性质 | Graph | 图像 |
| b² - 4ac > 0 | Two distinct real roots | 两个不同实根 | Cuts x-axis twice | 与 x 轴交于两点 |
| b² - 4ac = 0 | One repeated real root | 一个重根 | Touches x-axis | 与 x 轴相切 |
| b² - 4ac < 0 | No real roots | 无实根 | Does not meet x-axis | 与 x 轴不相交 |
0, giving one repeated root x = -1. For x² + x + 1 = 0, Δ = 1 - 4 = -3 < 0, so there are no real roots.
对于 x² - 5x + 6 = 0，Δ = 25 - 24 = 1 > 0，因此有两个不同实根；由于 1 是完全平方数，两根都是有理数。对于 x² + 2x + 1 = 0，Δ = 4 - 4 = 0，得到重根 x = -1。对于 x² + x + 1 = 0，Δ = 1 - 4 = -3 < 0，因此没有实根。
ber to state clearly which condition you are using, and to check whether the question says "equal roots", "two distinct roots" or "no real roots".
非常常见的考题是：求使 kx² + 4x + 2 = 0 有等根的 k 值。令 b² - 4ac = 0，即 16 - 8k = 0，得 k = 2。请务必写清你使用的是哪个条件，并核对题目说的是"等根""两个不同实根"还是"无实根"。
7. Graphs of Quadratic Functions | 二次函数图像
. If a > 0 it opens upwards and has a minimum point; if a < 0 it opens downwards and has a maximum point. The larger the value of a, the narrower the curve.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，开口向上并有最低点；若 a < 0，开口向下并有最高点。a 的绝对值越大，曲线越窄。
= 0. The axis of symmetry lies exactly halfway between the roots and is given by x = -b ÷ (2a).
考试中最重要的是三个特征。令 x = 0 可得 y 轴截距 y = c。解 ax² + bx + c = 0 可得 x 轴截距（即根）。对称轴恰好位于两根的正中间，由 x = -b ÷ (2a) 给出。
To sketch, plot the y-intercept and the two roots, draw a smooth curve through them, and mark the turning point. Never join the points with straight lines, and never assume the curve crosses the x-axis if the discriminant is negative.
作图时，先标出 y 轴截距与两根，用平滑曲线连接，并标出顶点。切忌用折线连接各点；若判别式为负，也不要误以为曲线会与 x 轴相交。
8. Vertex, Axis of Symmetry and Turning Points | 顶点、对称轴与转折点
etry is the vertical line x = -p. This is the most efficient route in most exam questions.
配方法把 y = ax² + bx + c 化为顶点式 y = a(x + p)² + q。由此可知转折点为 (-p, q)，对称轴为竖直线 x = -p。在多数考题中，这是最快捷的路径。
If the question asks for the maximum or minimum value of the function, this single rewritten line answers it immediately.
对于 y = x² - 4x + 1，配方得 y = (x - 2)² - 3。因此顶点为 (2, -3)，对称轴为 x = 2，y 的最小值为 -3。若题目问函数的最大值或最小值，这一行改写即可直接作答。
ous; substituting this x-value back into the equation gives the y-coordinate of the turning point.
不求配方法也能找到顶点。顶点的横坐标是两根的平均值；当根不易求出时，可用 x = -b ÷ (2a)，再把这个 x 值代回方程即得转折点的纵坐标。
9. Solving Quadratic Inequalities | 二次不等式
A quadratic inequality is solved in three steps: find the roots of the corresponding equation, sketch the parabola, then read off the range of values that satisfy the inequality. The sketch prevents the classic error of writing a single interval when two are needed.
解二次不等式分三步：先求对应方程的根，再画出抛物线草图，最后从图上读出满足不等式的取值范围。画草图可以避免"需要两个区间却只写一个"这类经典错误。
pwards, so the expression is positive outside the roots: x < 2 or x > 3. For x² - 5x + 6 < 0, the expression is negative between the roots: 2 < x < 3.
以 x² - 5x + 6 > 0 为例，两根为 x = 2 和 x = 3，抛物线开口向上，故表达式在两根之外为正：x < 2 或 x > 3。若为 x² - 5x + 6 < 0，则表达式在两根之间为负：2 < x < 3。
ause the inequality is not strict. Always rearrange so that the right-hand side is zero before factorising, and use a number line to check your intervals.
注意不等号方向与边界
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