📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are a cornerstone of the IGCSE Edexcel Mathematics syllabus. Understanding how to solve them opens the door to more advanced algebra and appears regularly in examinations. This article covers the main methods: factorisation, completing the square, and the quadratic formula, alongside the discriminant and graphical interpretation.
二次方程是IGCSE Edexcel数学大纲的基石。掌握如何解二次方程能为更高级的代数打开大门，并在考试中频繁出现。本文涵盖主要解法：因式分解法、配方法以及求根公式，连同判别式和图像解释。
1. What is a Quadratic Equation? | 什么是二次方程？
+ c = 0, where a, b, and c are constants and a ≠ 0. If a = 0, the equation becomes linear.
二次方程是一种多项式方程，其中变量的最高次幂为2。它可以写成 ax² + bx + c = 0 的形式，其中 a、b、c 为常数且 a ≠ 0。若 a = 0，方程退化为线性方程。
2. Standard Form of a Quadratic Equation | 二次方程的标准形式
s the coefficient of x² is positive, and the expression equals zero. For example, 3x² = 5x – 2 becomes 3x² – 5x + 2 = 0.
所有二次方程在求解前都应整理成标准形式 ax² + bx + c = 0。这确保 x² 的系数为正，且表达式等于零。例如，3x² = 5x – 2 应变为 3x² – 5x + 2 = 0。
3. Solving by Factorisation (Product-Sum Method) | 因式分解法（乘积-求和法）
for two numbers that multiply to c and add to b. For ax² + bx + c where a ≠ 1, we use the ‘ac method’ or grouping.
当二次式可分解为两个一次二项式的乘积时，因式分解法适用。对于 x² + bx + c = 0，我们寻找两个数，其积为 c 且和为 b。对于 a ≠ 1 的 ax² + bx + c，我们使用“ac方法”或分组分解。
- Find two numbers with product equal to ac and sum equal to b. / 找出两个数，其积等于 ac，其和等于 b。
- Split the middle term and factor by grouping. / 拆分一次项并通过分组进行因式分解。
- Rewrite as (px + q)(rx + s) = 0. / 重写为 (px + q)(rx + s) = 0 的形式。
4. Zero Product Property | 零乘积性质
The zero product property states that if the product of two factors is zero, then at least one of the factors must be zero. Hence, we set each bracket equal to zero and solve for x.
一旦二次式完成因式分解，方程变为 (mx + n)(px + q) = 0。零乘积性质指出，若两个因式的乘积为零，则至少其中一个因式为零。因此，我们令每个括号等于零并解出 x。
For example: (x – 2)(x + 5) = 0 → x – 2 = 0 or x + 5 = 0 → x = 2 or x = –5.
例如：(x – 2)(x + 5) = 0 → x – 2 = 0 或 x + 5 = 0 → x = 2 或 x = –5。
5. Examples of Factorising Quadratics | 因式分解二次式的例子
Example 1: Solve x² – 7x + 12 = 0. We need two numbers with product 12 and sum –7; they are –3 and –4. So, (x – 3)(x – 4) = 0, giving x = 3 or x = 4.
例1：解 x² – 7x + 12 = 0。我们需要两个数，积为 12，和为 –7；它们是 –3 和 –4。因此 (x – 3)(x – 4) = 0，得 x = 3 或 x = 4。
Example 2: Solve 2x² + 3x – 2 = 0. ac = –4. Numbers: 4 and –1 (sum = 3). Split: 2x² + 4x – x – 2 = 0 → 2x(x + 2) – 1(x + 2) = 0 → (2x – 1)(x + 2) = 0, so x = ½ or x = –2.
例2：解 2x² + 3x – 2 = 0。ac = –4。所求两数为 4 和 –1（和为 3）。拆分：2x² + 4x – x – 2 = 0 → 2x(x + 2) – 1(x + 2) = 0 → (2x – 1)(x + 2) = 0，故 x = ½ 或 x = –2。
6. Solving by Completing the Square | 配方法解二次方程
)². Steps: move c to right; add (b/2)² to both sides; factor left side; take square roots; solve for x.
该方法将二次式写成完全平方加常数的形式。对于 x² + bx + c = 0，我们加上并减去 (b/2)²。步骤：将 c 移至右边；两边加上 (b/2)²；左边进行因式分解；开平方根；解出 x。
For example: x² + 6x + 2 = 0 → x² + 6x = –2, add 9 → (x + 3)² = 7 → x + 3 = ±√7 → x = –3 ± √7.
例如：x² + 6x + 2 = 0 → x² + 6x = –2，加 9 → (x + 3)² = 7 → x + 3 = ±√7 → x = –3 ± √7。
Although less commonly tested, it is the foundation for deriving the quadratic formula.
尽管考得较少，但它是推导求根公式的基础。
7. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation ax² + bx + c = 0 directly:
x = –b ± √b² – 4ac / 2a
这一定式可直接求解任何二次方程 ax² + bx + c = 0：
x = –b ± √b² – 4ac / 2a
Memorise this formula, substitute the values of a, b and c from the standard form, and simplify. Pay careful attention to signs.
牢记此公式，将标准形式中的 a、b、c 值代入并简化。特别注意符号。
For example: 3x² – 5x – 2 = 0. Here a = 3, b = –5, c = –2. Substituting:
例如：3x² – 5x – 2 = 0。这里 a = 3，b = –5，c = –2。代入得：
x = 5 ± √(25 – 4×3×(–2)) / 6 = 5 ± √25 + 24 / 6 = 5 ± √49 / 6 = 5 ± 7 / 6
So x = (5 + 7)/6 = 2 or x = (5 – 7)/6 = –⅓.
因此 x = (5 + 7)/6 = 2 或 x = (5 – 7)/6 = –⅓。
8. Discriminant and Nature of Roots | 判别式与根的性质
The expression under the square root, Δ = b² – lled the discriminant. It determines the nature of the roots without solving the equation:
平方根号下的表达式 Δ = b² – 4ac 称为判别式。它无需解方程即可判断根的性质：
| Discriminant | Nature of roots | Graph |
|---|---|---|
| Δ > 0, perfect square | Two distinct real rational roots | Cuts x-axis twice at rational points |
| Δ > 0, not a square | Two distinct real irrational roots | Cuts x-axis twice at irrational points |
| Δ = 0 | One repeated real root | Touches x-axis at one point (vertex) |
| Δ < 0 | No real roots (complex roots) | Does not intersect x-axis |
判别式表格（中文对应）：
| 判别式 | 根的性质 | 图像 |
|---|---|---|
| Δ > 0，完全平方数 | 两个不相等的有理实根 | 与 x 轴交于两个有理点 |
| Δ > 0，非平方数 | 两个不等的无理实根 | 与 x 轴交于两个无理点 |
| Δ = 0 | 一个重实根 | 顶点与 x 轴相切于一点 |
| Δ < 0 | 无实根（复数根） | 不与 x 轴相交 |
9. Graphical Interpretation | 图像解释
abola. The solutions of ax² + bx + c = 0 are the x-intercepts of the parabola. The shape (∪ or ∩ ends on the sign of a. If a > 0, the parabola opens upwards; if a < 0, it opens downwards.
y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的解就是抛物线与 x 轴交点的横坐标。抛物线的形状（∪ 或 ∩）取决于 a 的符号。若 a > 0，抛物线开口向上；若 a < 0，则开口向下。
The discriminant tells us how many times the curve meets the x-axis. This links algebra to geometry, a common exam skill.
判别式告诉我们曲线与 x 轴的交点个数。这把代数与几何联系在一起，是常见的考试技能。
10. Common Mistakes & Tips | 常见错误与提示
- Forgetting to set the equation to zero before factorising.
在因式分解前忘记将方程设为零。 - Incorrect signs when finding the numbers for factorisation.
在寻找用于因式分解的数时弄错符号。 - Misapplying the quadratic formula – always use brackets when substituting negative values.
错误应用求根公式——代入负数时务必使用括号。 - Not simplifying surd answers fully.
没有将带根号的答案彻底化简。 - Confusing the discriminant with the formula itself.
将判别式与公式本身混淆。
Exam tip: always check your solutions by substituting them back into the original equation.
考试提示：务必将解代入原方程进行检验。
11. Exam-style Question | 模拟考题
Solve 2x² – 3x – 5 = 0 using the quadratic formula, and state the nature of the roots.
用求根公式解方程 2x² – 3x – 5 = 0，并说明根的性质。
a = 2, b = –3, c = –5. Discriminant Δ = (–3)² – 4(2)(–) = 9 + 40 = 49. Since 49 > 0 and a perfect square, there are two distinct rational roots.
a = 2，b = –3，c = –5。判别式 Δ = (–3)² – 4(2)(–5) = 9 + 40 = 49。因 49 > 0 且为完全平方数，故有两个不相等的有理根。
x = [3 ± √49] / (2×2) = (3 ± 7)/4 → x = 10/4 = 2.5 or x = –4/4 = –1.
x = [3 ± √49] / (2×2) = (3 ± 7)/4 → x = 10/4 = 2.5 或 x = –4/4 = –1。
12. Summary | 总结
Quadratic equations can be solved by factorisation (fastest when possible), completing the square, or the quadratic formula (universal). The discriminant reveals the number and type of roots without solving. Master these techniques, and you will tackle one of the most heavily weighted topics in IGCSE Mathematics confidently.
二次方程可通过因式分解法（最快，可行时首选）、配方法或求根公式（通用方法）来求解。判别式无需解方程即可揭示根的数目和类型。掌握这些技巧，你就能自信地应对IGCSE数学中权重最高的主题之一。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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