📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, and problem-solving questions across every exam paper. Mastering the three main methods of solution — factorisation, completing the square, and the quadratic formula — will give you a solid foundation for higher-level study.
二次方程是 IGCSE 数学中最重要的考点之一，出现在代数、坐标几何以及各类应用题中。熟练掌握三种主要解法——因式分解法、配方法和求根公式——将为你后续的进阶学习打下坚实基础。
1. The Standard Form | 标准形式
A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The highest power of the variable is 2, which is why it is called 'quadratic' (from the Latin quadratus, meaning square).
二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数为 2，因此被称为 "二次" 方程。
For example, 3x² - 5x + 2 = 0 -5, and c = 2. If a = 0, the equation becomes linear, so the condition a ≠ 0 is essential.
例如，3x² - 5x + 2 = 0 是一个二次方程，其中 a = 3，b = -5，c = 2。如果 a = 0，方程就变成一次方程，因此 a ≠ 0 是必要条件。
2. Solving by Factorisation | 因式分解法
Factorisation is often the quickest method when the quadratic has simple integer coefficients. The idea is to write the left-hand side as a product of two linear factors, then use the zero product property: if p × q = 0, then either p = 0 or q = 0.
当二次方程的系数为简单整数时，因式分解法通常是最快的方法。其核心思想是将等号左边写成两个一次因式的乘积，然后运用零积性质：若 p × q = 0，则 p = 0 或 q = 0。
Consider the equation x² + 5x + 6 = 0. We need two numbers that multiply to 6 and add to 5. These numbers are 2 and 3. Therefore:
考虑方程 x² + 5x + 6 = 0。我们需要找到两个数，相乘为 6，相加为 5。这两个数是 2 和 3。因此：
(x + 2)(x + 3) = 0
Setting each factor to zero gives x = -2 or x = -3. Always check your solutions by substituting them back into the original equation.
令每个因式分别为零，得到 x = -2 或 x = -3。记得将解代回原方程进行检验。
When the coefficient of x² is not 1, the process is slightly more involved. For 2x² + 7x + 3 = 0 wo numbers that multiply to 6 and add to 7: these are 1 and 6. Then split the middle term:
当 x² 的系数不为 1 时，过程稍显复杂。对于 2x² + 7x + 3 = 0，先计算 a 与 c 的乘积：2 × 3 = 6。找到两个数相乘为 6、相加为 7：即 1 和 6。然后拆分中间项：
2x² + 6x + x + 3 = 0
2x(x + 3) + 1(x + 3) = 0
(2x + 1)(x + 3) = 0
Hence x = -½ or x = -3.
因此 x = -½ 或 x = -3。
3. Solving by Completing the Square | 配方法
Completing the square transforms a quadratic into the form (x + p)² = q, from which the solutions can be found by taking square roots. This method is especially useful when factorisation is not obvious.
配方法将二次方程转化为 (x + p)² = q 的形式，然后通过开平方求解。当因式分解不明显时，此方法尤为实用。
To complete the square for x² + 6x + 2 = 0, take half of the coefficient of x (which is 3) and write:
对 x² + 6x + 2 = 0 进行配方，取 x 系数的一半（即 3），写成：
(x + 3)² - 9 + 2 = 0
(x + 3)² - 7 = 0
(x + 3)² = 7
Taking the square root of both sides (remembering the ± sign) gives:
两边开平方（注意正负号）得到：
x + 3 = ±√7
x = -3 + √7 or x = -3 - √7
This also reveals the vertex of the parabola, which we will discuss later.
该方法还能直接揭示抛物线的顶点，我们稍后会讨论。
4. The Quadratic Formula | 求根公式
impossible. It is derived from completing the square on the general form ax² + bx + c = 0.
求根公式是适用于一切二次方程的通用方法，即使无法因式分解也能求解。它是对一般形式 ax² + bx + c = 0 配方推导而来的。
x = (-b ± √b² - 4ac) / 2a
Let us solve 3x² - 4x - 2 = 0 using this formula. Here a = 3, b = -4, c = -2.
我们用此公式求解 3x² - 4x - 2 = 0。其中 a = 3，b = -4，c = -2。
x = (4 ± √((-4)² - 4 × 3 × (-2))) / (2 × 3)
x = (4 ± √16 + 24) / 6
x = (4 ± √40) / 6
Since √40 = 2√10, we simplify to x = (2 ± √10) / 3. Leave answers in surd form unless the question asks for a decimal.
因为 √40 = 2√10，化简得 x = (2 ± √10) / 3。除非题目要求小数，否则保留根号形式即可。
5. The Discriminant | 判别式
The expression b² - 4ac inside the quadratic formula is called the discriminant. It tells us how many real roots a quadratic equation has, without solving it fully.
求根公式中的表达式 b² - 4ac 称为判别式。它无需完全求解，即可告诉我们二次方程有多少个实数根。
| Discriminant Value | Number of Real Roots | Graph Interpretation |
| b² - 4ac > 0 | Two distinct real roots | The parabola crosses the x-axis at two points |
| b² - 4ac = 0 | One repeated real root | The parabola touches the x-axis at one point |
| b² - 4ac < 0 | No real roots | The parabola does not intersect the x-axis |
For example, the equation x² + 2x + 5 = 0 riminant 2² - 4 × 1 × 5 = 4 - 20 = -16, which is negative. Therefore it has no real roots.
例如，方程 x² + 2x + 5 = 0 的判别式为 2² - 4 × 1 × 5 = 4 - 20 = -16，为负数，因此没有实数根。
In exam questions, you may be asked to find the range of k such that a quadratic has no real roots. Simply set the discriminant to be less than zero and solve the resulting inequality.
考试中常要求你求出 k 的取值范围，使得二次方程无实数根。只需令判别式小于零，并解所得不等式即可。
6. Graphs of Quadratic Functions | 二次函数的图像
rve called a parabola. The sign of a determines its orientation: if a > 0, the parabola opens upwards (a 'U' shape); if a < 0, it opens downwards (an 'n' shape).
y = ax² + bx + c 的图像是一条平滑曲线，称为抛物线。a 的正负决定开口方向：若 a > 0，抛物线开口向上（呈 "U" 形）；若 a < 0，开口向下（呈 "n" 形）。
+ bx + c = 0. If the discriminant is negative, the graph simply does not cross the x-axis.
图像与 x 轴的交点正是方程 ax² + bx + c = 0 的根。若判别式为负，图像则不与 x 轴相交。
xis of symmetry is the vertical line x = -b / 2a, and the vertex has x-coordinate -b / 2a.
图像与 y 轴的交点为 (0, c)，因为代入 x = 0 得 y = c。对称轴是竖直线 x = -b / 2a，顶点的横坐标也是 -b / 2a。
7. Solving by Formula from Graphs | 从图像求方程的解
= d on the same set of axes and reading off the x-coordinates of the intersection points.
有时题目给出 y = ax² + bx + c 的图像，要求你解相关方程如 ax² + bx + c = d。这时可在同一坐标系中画水平线 y = d，然后读取交点的横坐标即可。
0). Estimates from a graph are acceptable when answers are required to one decimal place.
例如，已知 y = x² - 4x + 1 的图像，要求解 x² - 4x + 1 = 0，只需读取图像与 x 轴（即 y = 0）的交点。当要求保留一位小数时，从图像估计的答案是允许的。
8. Quadratic Equations and the Turning Point | 二次方程与抛物线顶点
a y = ax² + bx + c can be found by completing the square. Writing the equation in the form y = a(x - h)² + k reveals the vertex directly at (h, k).
抛物线 y = ax² + bx + c 的顶点可通过配方法求得。将方程写成 y = a(x - h)² + k 的形式后，顶点坐标直接为 (h, k)。
Take y = x² - 6x + 11. Completing the square:
以 y = x² - 6x + 11 为例。配方：
y = (x - 3)² - 9 + 11 = (x - 3)² + 2
, this is a minimum point. This form is very useful in geometry and optimisation problems.
因此顶点为 (3, 2)，且 a = 1 > 0，该点为最小值点。这种形式在几何和优化问题中非常有用。
9. Word Problems Leading to Quadratics | 二次方程应用题
Many IGCSE questions require you to set up a quadratic equation from a real-world context. A classic example is a rectangle problem: the length of a rectangle is 3 cm more than its width, and its area is 70 cm². Find the dimensions.
许多 IGCSE 考题要求你从实际情境中建立二次方程。一个典型例子是矩形问题：一个矩形的长比宽多 3 cm，面积为 70 cm²，求其尺寸。
Let the width be x cm. Then the length is (x + 3) cm, so:
设宽为 x cm，则长为 (x + 3) cm，于是：
x(x + 3) = 70
x² + 3x - 70 = 0
(x + 10)(x - 7) = 0
th is 7 cm and the length is 10 cm. Always reject negative solutions when they represent physical quantities such as length, time, or distance.
所以 x = -10 或 x = 7。长度为负不合实际，舍去 x = -10。因此宽为 7 cm，长为 10 cm。当负数解代表长度、时间或距离等物理量时，务必舍去。
10. Common Mistakes and Exam Tips | 常见错误与应考技巧
the formula. Another is losing the ± sign when using completing the square or the formula.
一个常见错误是忘记将方程整理成标准形式 ax² + bx + c = 0，就直接因式分解或套用公式。另一个常见错误是在配方或使用求根公式时遗漏正负号。
Read the question carefully: if it asks for answers to two decimal places, you must use the quadratic formula or a calculator and round correctly. If it asks for exact answers, leave surds in simplified form.
仔细审题：若题目要求答案保留两位小数，必须用求根公式或计算器并正确四舍五入；若要求精确值，则保留化简后的根号形式。
- Always rearrange to ax² + bx + c = 0 first
- Check whether the question asks for exact or approximate answers
- When taking square roots, include both the positive and negative values
- Reject solutions that are meaningless in the context of the problem
- Verify your final answers by substituting into the original equation
- 务必先将方程化为 ax² + bx + c = 0 的形式
- 确认题目要求精确答案还是近似值
- 开平方时要同时取正负两个值
- 在问题情境中无意义的解要舍去
- 将最终答案代回原方程进行检验
Practising all three methods — factorisation, completing the square, and the quadratic formula — will help you recognise which approach is fastest for each type of question. In the exam, time is precious; choose the most efficient method and always show your working clearly, as method marks are awarded even if you make a small arithmetic slip.
熟练练习三种方法——因式分解法、配方法和求根公式——能帮助你在考试中快速判断哪种方法最合适。考试时间宝贵，请选择最高效的方法，并清晰地写出解题步骤，因为即使出现小计算错误，步骤分仍然可以获得。
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