📚 Solving Quadratic Equations | 解二次方程：IGCSE 核心考点精讲
A quadratic equation is one of the most important topics in the Edexcel IGCSE Mathematics syllabus. It appears in almost every exam paper, either directly or as part of word problems, graphs, and curve sketching. In this article, we will break down the key methods for solving quadratic equations, step by step, with worked examples and common pitfalls explained in both English and Chinese.
二次方程是 Edexcel IGCSE 数学大纲中最重要的考点之一。几乎所有试卷都会直接或间接地考查它，包括应用题、函数图像以及曲线草图等。本文将分步骤讲解解二次方程的几种核心方法，配合例题和常见错误分析，帮助你在考试中稳步得分。
1. What Is a Quadratic Equation? | 什么是二次方程
A quadratic equation is an equation of the form ax² + bx + c = 0 ariable x is 2. For example, 2x² − 5x + 3 = 0 and x² − 4 = 0 are both quadratic equations.
二次方程是指形如 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a ≠ 0。变量 x 的最高次数为 2。例如，2x² − 5x + 3 = 0 和 x² − 4 = 0 都是二次方程。
In the IGCSE syllabus, you are expected to solve quadratic equations by three main methods: factorisation, completing the square, and using the quadratic formula. You should also understand the discriminant and the relationship between roots and the graph.
在 IGCSE 大纲中，你需要掌握三种主要解法：因式分解法、配方法和二次公式法。同时，你还需要理解判别式以及根与函数图像之间的关系。
2. Expanding Double Brackets | 展开双括号
(p + q)x + pq. Multiply each term in the first bracket by each term in the second bracket.
在因式分解之前，你必须熟练展开两个括号相乘。规律是：(x + p)(x + q) = x² + (p + q)x + pq。即用第一个括号中的每一项去乘第二个括号中的每一项。
Example: (x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15.
例如：(x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15。
When the coefficient of x² is not 1, use the "FOIL" method carefully. For example, (2x + 1)(3x − 4) = 6x² − 8x + 3x − 4 = 6x² − 5x − 4.
当 x² 的系数不为 1 时，要小心使用“逐项相乘”的方法。例如，(2x + 1)(3x − 4) = 6x² − 8x + 3x − 4 = 6x² − 5x − 4。
3. Factorising Quadratics | 因式分解二次式
ve c and add to give b. If the two numbers are p and q, then x² + bx + c = (x + p)(x + q).
因式分解是展开的逆运算。对于形式为 x² + bx + c 的二次式，找到两个数，使其乘积为 c，和为 b。若这两个数为 p 和 q，则 x² + bx + c = (x + p)(x + q)。
wo numbers with product 12 and sum 7. These are 3 and 4, so x² + 7x + 12 = (x + 3)(x + 4).
例如：因式分解 x² + 7x + 12。我们需要两个乘积为 12、和为 7 的数，即 3 和 4，所以 x² + 7x + 12 = (x + 3)(x + 4)。
ative. For example, x² − 2x − 15 = (x − 5)(x + 3), because (−5) × 3 = −15 and −5 + 3 = −2.
如果 c 为负数，则一个数为正、一个数为负。例如，x² − 2x − 15 = (x − 5)(x + 3)，因为 (−5) × 3 = −15，且 −5 + 3 = −2。
(2x + 1)(x + 3). Check: 2x × x = 2x², 2x × 3 = 6x, 1 × x = x, 1 × 3 = 3, and 6x + x = 7x.
当 x² 的系数不为 1 时，例如 ax² + bx + c，可以使用“拆中项”法，或者直接尝试因子配对。例如，2x² + 7x + 3 = (2x + 1)(x + 3)。验证：2x × x = 2x²，2x × 3 = 6x，1 × x = x，1 × 3 = 3，且 6x + x = 7x。
4. Solving by Factorisation | 用因式分解法求解
If a quadratic equation can be factorised, the quickest way to solve it is to set each factor equal to zero. This works because if the product of two expressions is zero, then at least one of them must be zero.
如果一个二次方程可以因式分解，最快的解法是令每个因式等于零。这是因为如果两个表达式的乘积为零，那么其中至少一个必须为零。
0. Factorise: (x − 2)(x − 3) = 0. Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.
例如：解方程 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0。因此 x − 2 = 0 或 x − 3 = 0，所以 x = 2 或 x = 3。
² = 3x + 4. Rearrange: x² − 3x − 4 = 0. Factorise: (x − 4)(x + 1) = 0, so x = 4 or x = −1.
有时方程并非标准形式。你首先需要将其整理为 ax² + bx + c = 0。例如，解 x² = 3x + 4。整理得 x² − 3x − 4 = 0。因式分解得 (x − 4)(x + 1) = 0，所以 x = 4 或 x = −1。
5. Completing the Square | 配方法
nding the turning point of a parabola. The general form is: x² + bx = (x + b/2)² − (b/2)².
配方法适用于任何二次方程，并且特别适合用来求抛物线的顶点。一般形式为：x² + bx = (x + b/2)² − (b/2)²。
en complete the square: (x + 3)² − 9 = −1, so (x + 3)² = 8. Taking square roots: x + 3 = ±√8 = ±2√2, so x = −3 ± 2√2.
例如：用配方法解 x² + 6x + 1 = 0。首先改写为 x² + 6x = −1。然后配方：(x + 3)² − 9 = −1，所以 (x + 3)² = 8。开平方：x + 3 = ±√8 = ±2√2，因此 x = −3 ± 2√2。
+ 5 = 0, giving 2(x − 2)² − 8 + 5 = 0, so 2(x − 2)² = 3, and (x − 2)² = 3/2. Hence x = 2 ± √3/2.
当 x² 的系数不为 1 时，先把系数从 x² 和 x 项中提出。例如，解 2x² − 8x + 5 = 0。写成 2(x² − 4x) + 5 = 0，然后对括号内配方：2[(x − 2)² − 4] + 5 = 0，即 2(x − 2)² − 8 + 5 = 0，所以 2(x − 2)² = 3，故 (x − 2)² = 3/2，因此 x = 2 ± √3/2。
6. The Quadratic Formula | 二次公式
dratic formula solves any quadratic equation ax² + bx + c = 0. The solutions are given by:
二次公式可以求解任何二次方程 ax² + bx + c = 0。解由下式给出：
x = (−b ± √b² − 4ac) / (2a)
Always write the equation in the form ax² + bx + c = 0 first, then substitute a, b and c.
这个公式在 Edexcel IGCSE 的公式表中可以找到，但你必须知道何时以及如何使用它。务必先将方程写成 ax² + bx + c = 0 的形式，然后代入 a、b 和 c。
Solve 2x² + 3x − 5 = 0 using the quadratic formula. Here a = 2, b = 3, c = −5. Substitute:
例如：用二次公式解 2x² + 3x − 5 = 0。这里 a = 2，b = 3，c = −5。代入得：
x = (−3 ± √(3² − 4×2×(−5))) / (2×2) = (−3 ± √9 + 40) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4
So x = (4)/4 = 1 or x = (−10)/4 = −2.5. The solutions are x = 1 and x = −2.5.
所以 x = 4/4 = 1 或 x = −10/4 = −2.5。解为 x = 1 和 x = −2.5。
7. The Discriminant | 判别式
The expression b² − 4ac inside the quadratic formula is called the discriminant. It tells us the number and type of roots without solving the full equation.
二次公式中的表达式 b² − 4ac 称为判别式。它无需完全求解方程即可告诉我们根的数量和类型。
ere is one repeated real root (two equal roots). If b² − 4ac < 0, there are no real roots.
若 b² − 4ac > 0，则有两个不同的实数根；若 b² − 4ac = 0，则有一个重根（两个相等根）；若 b² − 4ac < 0，则没有实数根。
efore the equation has one repeated root. Indeed, (x − 2)² = 0, so x = 2 is a double root.
例如：对于方程 x² − 4x + 4 = 0，判别式为 (−4)² − 4×1×4 = 16 − 16 = 0。因此方程有一个重根。事实上，(x − 2)² = 0，所以 x = 2 是二重根。
You may also be asked to find the value of k for which a quadratic has equal roots. Set the discriminant equal to zero and solve for k.
有时你可能会被要求找出使二次方程有等根的 k 值。此时令判别式等于零，然后解出 k 即可。
8. Word Problems | 应用题
Quadratic equations often appear in problem-solving questions. You need to translate a real-world situation into a quadratic equation, solve it, and then check which solution makes sense.
二次方程常出现在实际应用题中。你需要将现实情境转化为二次方程，求出解，然后检查哪个解在题目中合理。
(x + 6)(x − 4) = 0. So x = −6 or x = 4. Since width cannot be negative, the width is 4 cm.
例如：一个矩形的面积为 24 cm²，长比宽多 2 cm。求宽。设宽为 x cm，则长为 x + 2，且 x(x + 2) = 24。得到 x² + 2x − 24 = 0。因式分解得 (x + 6)(x − 4) = 0，所以 x = −6 或 x = 4。因为宽不能为负数，所以宽为 4 cm。
Always check your answers by substituting back into the original context. Discard any unrealistic, for example negative length or time, unless the problem allows it.
务必把答案代回原题情境中验证。去除不切实际的解，例如负的长度或时间，除非题目允许。
9. Graphs of Quadratic Functions | 二次函数图像
The graph of a quadratic function is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
二次函数的图像是抛物线。若 a > 0，抛物线开口向上，有最小值点；若 a < 0，抛物线开口向下，有最大值点。
ning point is at (p, q). Alternatively, for y = ax² + bx + c, the x-coordinate is −b/(2a).
顶点的 x 坐标可以通过配方得到：y = (x − p)² + q，则顶点坐标为 (p, q)。或者，对于 y = ax² + bx + c，x 坐标为 −b/(2a)。
The roots of the equation are the x-intercepts of the graph. If the discriminant is negative, the graph does not cross the x-axis.
方程的根就是图像与 x 轴的交点。若判别式为负，则图像与 x 轴没有交点。
4x + 3 = (x − 2)² − 1 has a minimum at (2, −1) and crosses the x-axis at x = 1 and x = 3.
例如，y = x² − 4x + 3 = (x − 2)² − 1 的图像在 (2, −1) 处有最小值，并与 x 轴交于 x = 1 和 x = 3。
10. Common Mistakes | 常见错误
be factorised as x(x) = 9; instead, write x² − 9 = 0 and factorise as (x − 3)(x + 3) = 0.
常见错误之一是在因式分解或使用公式前忘记将方程整理为 ax² + bx + c = 0。例如，x² = 9 不应写成 x(x) = 9，而应写为 x² − 9 = 0，再因式分解为 (x − 3)(x + 3) = 0。
or is losing the ± sign when taking square roots. If x² = 16, then x = ±4, not just x = 4.
另一个错误是在开平方时丢失 ± 号。若 x² = 16，则 x = ±4，而不仅仅是 x = 4。
In factorisation, sign errors are very common. Always check your factors by expanding them back. Also, when the coefficient of x² is not 1, do not forget to consider all factor pairs.
在因式分解中，符号错误非常常见。务必通过展开来检查你的因式。另外，当 x² 的系数不为 1 时，不要忘记考虑所有因子配对。
11. Practice Questions | 练习题
Try these questions on your own, then check the answers below.
请独立尝试以下题目，然后核对下方的答案。
Solve x² − 9x + 18 = 0. (Factorise)
Solve 2x² − 7x + 3 = 0. (Factorise or formula)
Solve x² + 4x − 7 = 0, giving your answer in surd form. (Completing the square)
Find the value of k such that kx² + 6x + 1 = 0 has equal roots.
Answers: 1) x = 3 or x = 6. 2) x = ½ or x = 3. 3) x = −2 ± √11. 4) k = 9.
If you got all four correct, you are ready for the exam. If not, review the sections above and try again.
如果你四题全部正确，那么你已经准备好迎接考试了。如果还有错误，请复习对应章节并重新尝试。
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