Solving Quadratic Equations: A Comprehensive Guide | 二次方程的求解方法详解

📚 Solving Quadratic Equations: A Comprehensive Guide | 二次方程的求解方法详解

Quadratic equations are fundamental to A-Level Mathematics and appear across algebra, calculus, and coordinate geometry. Mastering the various methods of solving them is essential for exam success and for building a solid mathematical foundation.

二次方程是 A-Level 数学的核心内容，贯穿代数、微积分和坐标几何。熟练掌握各类求解方法，不仅对考试至关重要，更是构建坚实数学基础的关键一步。

1. What Is a Quadratic Equation | 什么是二次方程

A quadratic equation is a polynomial equation of degree two, which can be written in the standard form:

二次方程是最高次数为二的整式方程，其标准形式可写成：

a x² + b x + c = 0, a ≠ 0

Here, a, b and c are real constants, and x is the unknown variable. The condition a ≠ 0 is crucial because if a = 0, the equation becomes linear, not quadratic.

其中 a、b、c 为实数常数，x 为未知数。条件 a ≠ 0 至关重要，因为若 a = 0，方程将退化为一次方程，而非二次方程。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic has simple integer or rational roots. The idea is to express the quadratic as a product of two linear factors.

当二次方程具有简单的整数根或有理数根时，因式分解法往往是最快捷的方法。其核心思想是将二次式写成两个一次因式的乘积。

For a quadratic written as x² + bx + c, find two numbers p and q such that p + q = b and pq = c. Then:

对于形式为 x² + bx + c 的二次式，寻找两个数 p 和 q，满足 p + q = b 且 pq = c，则：

x² + bx + c = (x + p)(x + q)

Example: Solve x² − 5x + 6 = 0. Since (−2) + (−3) = −5 and (−2) × (−3) = 6, the factorisation is (x − 2)(x − 3) = 0. Hence x = 2 or x = 3.

例：解方程 x² − 5x + 6 = 0。因为 (−2) + (−3) = −5，且 (−2) × (−3) = 6，所以因式分解为 (x − 2)(x − 3) = 0，因此 x = 2 或 x = 3。

For the general form ax² + bx + c, factorisation may require decomposition: find two numbers whose product is ac and whose sum is b, then split the middle term.

对于一般形式 ax² + bx + c，因式分解可能需使用十字相乘法或拆项法：找到两个数，其乘积为 ac，和为 b，再据此拆分中间项。

3. Solving by Completing the Square | 配方法

Completing the square transforms the quadratic into a perfect square plus a constant. This method is elegant, universal, and also gives the coordinates of the vertex of the parabola directly.

配方法将二次式化为一个完全平方项加上一个常数的形式。这种方法既简洁又通用，还能直接得到抛物线顶点的坐标。

For x² + bx + c, the process is:

对于 x² + bx + c，步骤如下：

x² + bx + c = (x + b/2)² − (b/2)² + c

Example: Solve x² + 6x + 5 = 0.

例：解方程 x² + 6x + 5 = 0。

(x + 3)² − 9 + 5 = 0 → (x + 3)² = 4 → x + 3 = ±2 → x = −1 or x = −5

When the coefficient of x² is not 1, first factor out a from the first two terms, then complete the square inside the brackets.

当 x² 的系数不为 1 时，先将 a 从含 x 的前两项中提出，再在括号内配平方。

4. The Quadratic Formula | 求根公式

The quadratic formula is derived by completing the square on the general form and provides a direct universal solution for any quadratic equation.

求根公式通过对一般形式配平方推导而来，为任何二次方程提供了直接而通用的解法。

x = [−b ± √(b² − 4ac)] / (2a)

Example: Solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Substitute: x = [4 ± √(16 + 24)] / 4 = [4 ± √40] / 4 = 1 ± (√10)/2.

例：解方程 2x² − 4x − 3 = 0。此时 a = 2，b = −4，c = −3。代入得：x = [4 ± √(16 + 24)] / 4 = [4 ± √40] / 4 = 1 ± (√10)/2。

This formula always works, even when the roots are complex numbers, and it is essential to memorise it accurately.

该公式恒有效，即便根为复数时也能使用，务必准确记忆。

5. The Discriminant | 判别式

The discriminant, denoted Δ, is the expression under the square root in the quadratic formula: Δ = b² − 4ac. It determines the number and type of roots without solving the full equation.

判别式用 Δ 表示，即求根公式中被开方的部分：Δ = b² − 4ac。它无需解完整方程即可确定根的个数与类型。

If Δ > 0: the equation has two distinct real roots.
If Δ = 0: the equation has exactly one real root (a repeated root).
If Δ < 0: the equation has no real roots; it has two complex conjugate roots.

若 Δ > 0：方程有两个不同的实数根。
若 Δ = 0：方程有一个实数根（重根）。
若 Δ < 0：方程无实数根，有两个共轭复数根。

Example: For x² − 3x + 5 = 0, Δ = 9 − 20 = −11 < 0, so there are no real roots.

例：对于 x² − 3x + 5 = 0，Δ = 9 − 20 = −11 < 0，因此没有实数根。

6. Roots and Coefficients (Vieta’s Formulas) | 根与系数的关系（韦达定理）

For a quadratic equation ax² + bx + c = 0 with roots α and β, the following relationships hold:

对于二次方程 ax² + bx + c = 0，设其两根为 α 和 β，则以下关系成立：

α + β = −b/a, αβ = c/a

These formulas allow us to check roots quickly and to construct a quadratic equation when the roots are known. For example, if the roots are 2 and −3, then the sum is −1 and the product is −6, giving the equation x² + x − 6 = 0.

这些公式可用于快速检验根的正误，也可以在已知根的情况下构造二次方程。例如，若两根为 2 和 −3，则和为 −1，积为 −6，对应的方程为 x² + x − 6 = 0。

Vieta’s formulas extend naturally to higher-degree polynomials and appear frequently in exam questions about roots.

韦达定理可自然推广至高次多项式，在涉及根的考题中频繁出现。

7. Deriving the Quadratic Formula | 求根公式的推导

Understanding the derivation of the quadratic formula helps you remember it and appreciate why it works. Starting from ax² + bx + c = 0:

理解求根公式的推导过程有助于记忆公式并领悟其原理。从 ax² + bx + c = 0 出发：

x² + (b/a)x = −c/a

(x + b/2a)² = −c/a + (b/2a)² = (b² − 4ac) / 4a²

x + b/2a = ± √(b² − 4ac) / 2a → x = [−b ± √(b² − 4ac)] / 2a

Notice that completing the square is not just a solving method — it is the very foundation of the quadratic formula.

注意，配方不仅是求解方法之一，更是求根公式的基础来源。

8. Special Cases: Missing Linear or Constant Terms | 特殊形式：缺项二次方程

When b = 0, the equation has the form ax² + c = 0 and can be solved directly by isolation:

当 b = 0 时，方程形如 ax² + c = 0，可直接通过移项开方求解：

x² = −c/a → x = ±√(−c/a)

When c = 0, the equation becomes ax² + bx = 0, which is solved by factoring out x:

当 c = 0 时，方程变为 ax² + bx = 0，通过提取公因式 x 求解：

x(ax + b) = 0 → x = 0 or x = −b/a

These special cases appear frequently in exam papers, and recognising them saves valuable time.

这些特殊形式在考试中常见，迅速识别可以节省宝贵时间。

9. Choosing the Right Method | 如何选择合适的方法

Different situations call for different strategies. The table below summarises the recommended approach.

不同情境需要不同策略。下表总结了建议的选择思路。

Situation	Recommended Method
Simple factorisable roots	Factorisation
Coefficient of x is even	Completing the square
Coefficients are messy or roots involve surds	Quadratic formula
Need vertex or turning point	Completing the square
Need to analyse the number of roots	Discriminant only

情境	推荐方法
可直接因式分解	因式分解法
一次项系数为偶数	配方法
系数复杂或根含根号	求根公式
需要顶点或极值	配方法
只需判断根的个数	仅用判别式

10. Applications in Problem Solving | 二次方程的实际应用

Quadratic equations appear in projectile motion, area problems, and optimisation. For example, the height h of an object thrown upward is modelled by h = −4.9t² + vt + h₀. Setting h = 0 gives a quadratic equation in time t.

二次方程广泛应用于抛体运动、面积问题与最优化问题。例如，物体上抛的高度可建模为 h = −4.9t² + vt + h₀。令 h = 0 就得到关于时间 t 的二次方程。

Worked Example: A rectangular field has area 100 m². Its length is 5 m more than its width. Find the width.

例题：一块矩形农田面积为 100 m²，长比宽多 5 m，求宽。

Let width = x. Then x(x + 5) = 100, so x² + 5x − 100 = 0. Using the formula: x = [−5 ± √(25 + 400)] / 2 = [−5 ± √425] / 2 ≈ 7.81 m (width cannot be negative).

设宽为 x，则 x(x + 5) = 100，即 x² + 5x − 100 = 0。用公式求解：x = [−5 ± √(25 + 400)] / 2 = [−5 ± √425] / 2 ≈ 7.81 m（宽度取正值）。

11. Common Mistakes and Tips | 常见错误与解题技巧

Examiners often report repetitive mistakes that cost candidates marks.

考官反馈中常见一些反复出现的错误，这些都会导致失分。

Forgetting the symbol ± when taking square roots.
Dividing both sides by x when x = 0 is a possible root (this only applies when c = 0, and x = 0 is actually a root).
Miscalculating the discriminant when b is negative; use parentheses.
Writing the quadratic formula incorrectly — always write it down first.
Omitting the final step of checking answers by substitution.

开方时忘记写 ± 号。
在 c = 0 时两边同时除以 x，导致丢失 x = 0 这个根。
当 b 为负数时判别式计算错误；建议使用括号。
求根公式记忆或书写错误——先写出公式再代入。
省略代入原方程检验答案的步骤。

Always substitute your solutions back into the original equation to verify them; it takes just a few seconds and prevents careless errors.

务必将解代回原方程验证，这只需几秒钟，却能避免粗心错误。

12. Summary and Practice Advice | 总结与练习建议

There are three main solving methods: factorisation, completing the square, and the quadratic formula. The discriminant tells you about the roots, and Vieta’s formulas link roots to coefficients. Each method has its place, and fluency comes from deliberate practice.

求解二次方程主要有三种方法：因式分解法、配方法和求根公式。判别式揭示根的性质，韦达定理建立了根与系数的联系。每种方法各有优势，熟练来自刻意练习。

To excel at A-Level, work through at least 20–30 quadratic equations of varying difficulty, including those with surd and complex roots. Use past paper questions to familiarise yourself with how quadratics are tested in combination with other topics.

要在 A-Level 中取得优异成绩，建议至少完成 20–30 道难度不同的二次方程练习，包括含根式和复根的情况。通过历年真题熟悉二次方程与其他知识点的综合考查方式。

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Solving Quadratic Equations: A Comprehensive Guide | 二次方程的求解方法详解