Quadratic Equations Complete Guide | 一元二次方程完全指南

📚 Quadratic Equations Complete Guide | 一元二次方程完全指南

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. They appear in almost every paper, from short multiple-choice questions to multi-step problems in Paper 4. Mastering this topic will strengthen your algebra skills and boost your overall grade significantly.

一元二次方程是 IGCSE 数学中考查频率最高的内容之一。几乎每份试卷都会出现，从简短的选择题到 Paper 4 中的多步骤综合题。掌握好这一内容，将强化你的代数能力，并显著提升总分。

1. What Is a Quadratic Equation? | 什么是一元二次方程

A quadratic equation is a polynomial equation of degree 2, which means the highest power of the unknown variable is exactly 2. The variable appears squared, but never cubed or raised to a higher power.

一元二次方程是次数为 2 的多项式方程，即未知数的最高次数正好为 2。变量以平方形式出现，绝不会出现立方或更高次幂。

The general form of a quadratic equation in one variable x is:

ax² + bx + c = 0

where a, b and c are constants, and a ≠ 0. If a = 0, the equation becomes linear (bx + c = 0), not quadratic. The value of a is called the leading coefficient.

其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，方程将退化为一次方程 bx + c = 0，而不是二次方程。a 被称为首项系数。

Here are some typical examples of quadratic equations:

以下是一些典型的一元二次方程示例：

x² − 5x + 6 = 0
2x² + 3x − 2 = 0
x² − 9 = 0
4x² = 16

2. Rearranging Equations into Standard Form | 将方程整理为标准形式

In an exam, a quadratic equation may not always be written in the standard form ax² + bx + c = 0. Sometimes you must expand brackets, collect like terms, or move all terms to one side before solving.

在考试中，一元二次方程不一定总以标准形式 ax² + bx + c = 0 出现。有时你需要先展开括号、合并同类项，或将所有项移到等式的一侧，然后再求解。

Example: Solve 3x² + 2 = 5x.

例：解方程 3x² + 2 = 5x。

First, move 5x to the left side:

首先，将 5x 移到等号左边：

3x² − 5x + 2 = 0

Now the equation is in standard form and ready to be solved by factorisation or the quadratic formula.

现在方程已经是标准形式，可以用因式分解法或求根公式来求解。

Always check whether the equation can be simplified before solving. For example, 4x² = 16 can be divided by 4 to give x² = 4.

求解前务必检查方程是否可以简化。例如，4x² = 16 可以两边同除以 4，得到 x² = 4。

3. Solving by Factorisation | 因式分解法

Factorisation is usually the fastest method when the quadratic expression can be written as the product of two linear factors. This method relies on the zero product property: if A × B = 0, then A = 0 or B = 0.

当二次表达式可以写成两个一次因式的乘积时，因式分解法通常是最快的方法。该方法依据零乘积性质：若 A × B = 0，则必有 A = 0 或 B = 0。

Steps for solving a quadratic equation by factorisation:

用因式分解法解一元二次方程的步骤：

Rearrange so that one side is 0.
Factorise the quadratic expression.
Set each factor equal to 0.
Solve the two resulting linear equations.

中文步骤：

移项，使等式一边为 0。
将二次表达式因式分解。
令每个因式分别等于 0。
解出两个一元一次方程。

Example: Solve x² − 5x + 6 = 0.

例：解方程 x² − 5x + 6 = 0。

We look for two numbers whose product is 6 and whose sum is −5. These numbers are −2 and −3.

我们寻找两个数，其乘积为 6，且其和为 −5。这两个数是 −2 和 −3。

x² − 5x + 6 = (x − 2)(x − 3) = 0

x − 2 = 0 → x = 2 or x − 3 = 0 → x = 3

So the solutions are x = 2 or x = 3. You should always substitute your answers back into the original equation to check them.

因此解为 x = 2 或 x = 3。你应始终将答案代回原方程进行检验。

4. Special Factorisation Patterns | 特殊因式分解模式

Two patterns appear very often in IGCSE papers and are worth recognising instantly.

两种特殊模式在 IGCSE 试卷中非常常见，值得你一眼识别。

Pattern 1: Difference of two squares

模式一：平方差公式

a² − b² = (a + b)(a − b)

Example: x² − 9 = (x + 3)(x − 3), so x = −3 or x = 3.

例：x² − 9 = (x + 3)(x − 3)，所以 x = −3 或 x = 3。

Pattern 2: Perfect square trinomial

模式二：完全平方式

a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)²

Example: x² − 6x + 9 = (x − 3)², so the only solution is x = 3, which is called a repeated root.

例：x² − 6x + 9 = (x − 3)²，因此唯一解为 x = 3，这被称为重根。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form (x + p)² + q. This is extremely useful for finding the turning point of a parabola and for solving equations that cannot be factorised easily.

配方法将二次表达式改写为 (x + p)² + q 的形式。这在求抛物线顶点时极为有用，也常用于解不易因式分解的方程。

Steps for completing the square on x² + bx + c:

对 x² + bx + c 进行配方的步骤：

Halve the coefficient of x: this gives p = b ÷ 2.
Write (x + p)² − p².
Add the constant c at the end.

中文步骤：

将 x 的系数取半：即 p = b ÷ 2。
写出 (x + p)² − p²。
最后加上常数 c。

Example: Solve x² + 6x + 5 = 0 by completing the square.

例：用配方法解方程 x² + 6x + 5 = 0。

The coefficient of x is 6, so half of it is 3:

x 的系数为 6，取其半为 3：

x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4

Now set the expression equal to 0:

令该表达式等于 0：

(x + 3)² − 4 = 0

(x + 3)² = 4 → x + 3 = ±2

x = −1 or x = −5

The completed square form also reveals the vertex of the parabola. For y = (x + 3)² − 4, the minimum point is (−3, −4).

配方形式还能直接给出抛物线的顶点。对于 y = (x + 3)² − 4，最小值点为 (−3, −4)。

6. The Quadratic Formula | 求根公式

When a quadratic cannot be factorised easily, or when the roots are irrational, use the quadratic formula. You will be given this formula in the IGCSE formula sheet, but memorising it saves time in the exam.

当二次方程不易因式分解，或根为无理数时，应使用求根公式。IGCSE 公式表会提供该公式，但记住它可以节省考试时间。

For the equation ax² + bx + c = 0:

对于方程 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) ÷ (2a)

Example: Solve 2x² + 3x − 2 = 0 using the formula.

例：用求根公式解方程 2x² + 3x − 2 = 0。

Here a = 2, b = 3, c = −2. Substitute into the formula:

这里 a = 2，b = 3，c = −2。代入公式：

x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2)

x = (−3 ± √(9 + 16)) ÷ 4 = (−3 ± √25) ÷ 4

x = (−3 + 5) ÷ 4 = 0.5 or x = (−3 − 5) ÷ 4 = −2

So the solutions are x = 0.5 or x = −2.

所以解为 x = 0.5 或 x = −2。

7. The Discriminant | 判别式

The expression b² − 4ac inside the square root is called the discriminant. It tells us how many roots a quadratic equation has without actually solving it.

根号内的表达式 b² − 4ac 被称为判别式。它可以在不求解的情况下告诉我们一元二次方程有多少个根。

Value of b² − 4ac	Number of real roots	Graph meaning
Positive ( > 0 )	Two distinct real roots	Parabola crosses the x-axis twice
Zero ( = 0 )	One repeated root	Parabola touches the x-axis once
Negative ( < 0 )	No real roots	Parabola never meets the x-axis

对应中文说明：判别式为正时有两个不同的实数根；判别式为零时有一个重根；判别式为负时没有实数根。

Example: How many roots does x² − 4x + 5 = 0 have?

例：方程 x² − 4x + 5 = 0 有几个根？

Here a = 1, b = −4, c = 5, so b² − 4ac = 16 − 20 = −4 < 0. Therefore the equation has no real roots.

这里 a = 1，b = −4，c = 5，所以 b² − 4ac = 16 − 20 = −4 < 0。因此该方程没有实数根。

8. Solving Quadratic Equations by Graphing | 图像法解一元二次方程

The graph of a quadratic function y = ax² + bx + c is a parabola. The solutions of the equation ax² + bx + c = 0 are the x-coordinates where the parabola crosses the x-axis. These points are called the roots or x-intercepts.

二次函数 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 轴交点处的 x 坐标。这些点被称为根或 x 截距。

Key features of a parabola:

抛物线的主要特征：

If a > 0, the parabola opens upwards (U shape) and has a minimum point.
If a < 0, the parabola opens downwards (∩ shape) and has a maximum point.
The axis of symmetry is the vertical line x = −b ÷ (2a).
The vertex lies on this axis of symmetry.

中文要点：当 a > 0 时，抛物线开口向上（U 形），有最小值点；当 a < 0 时，抛物线开口向下（∩ 形），有最大值点。对称轴是竖直线 x = −b ÷ (2a)，顶点位于对称轴上。

If the graph has no x-intercepts, then the corresponding equation has no real roots, which matches a negative discriminant.

若图像没有 x 截距，则对应方程没有实数根，这与判别式为负的情形一致。

9. Sum and Product of Roots | 根与系数的关系

For a quadratic equation ax² + bx + c = 0 with roots α and β, there are two useful relationships known as Vieta’s formulas:

对于根为 α 和 β 的一元二次方程 ax² + bx + c = 0，存在两个非常有用的关系，称为韦达定理：

Sum of roots: α + β = −b ÷ a

Product of roots: α × β = c ÷ a

Example: Given that the roots of 2x² + 6x − 8 = 0 are α and β, find α + β and αβ.

例：已知方程 2x² + 6x − 8 = 0 的两根为 α 和 β，求 α + β 与 αβ。

Solution: Here a = 2, b = 6, c = −8, so:

解：这里 a = 2，b = 6，c = −8，因此：

α + β = −6 ÷ 2 = −3 and αβ = −8 ÷ 2 = −4

These relationships can also be used to construct a quadratic equation from given roots. If the required roots are r and s, the equation is:

这些关系也可用于根据已知根来构造二次方程。若所需的根为 r 和 s，则方程为：

x² − (r + s)x + rs = 0

10. Word Problems Involving Quadratics | 二次方程应用题

In IGCSE Paper 4, quadratics often appear in word problems. The key is to translate the English sentence into an algebraic equation, solve it, and then check whether each answer makes sense in the context of the problem.

在 IGCSE Paper 4 中，二次方程常以应用题的形式出现。关键在于将题目中的文字表述转化为代数方程，解得结果后再检查每个答案是否符合实际问题背景。

Example: A rectangle has length 4 cm greater than its width. Its area is 45 cm². Find the width.

例：一个长方形的长比宽多 4 cm，面积为 45 cm²。求宽。

Let the width be x cm. Then the length is (x + 4) cm. Since area = length × width:

设宽为 x cm，则长为 (x + 4) cm。因为面积 = 长 × 宽：

x(x + 4) = 45 → x² + 4x − 45 = 0

Factorise:

因式分解：

(x + 9)(x − 5) = 0 → x = −9 or x = 5

Since the width cannot be negative, we reject x = −9. Therefore the width is 5 cm.

由于宽度不可能为负数，我们舍去 x = −9。因此宽为 5 cm。

Always check answers that involve lengths, distances, times or counts — reject any negative or zero values that do not fit the physical situation.

凡是涉及长度、距离、时间或数量的题目，务必检查答案——不符合实际情况的负数或零值应舍去。

11. Exam Tips and Common Mistakes | 考试技巧与常见错误

Students lose marks on quadratic questions for predictable reasons. Avoiding these mistakes can quickly raise your score.

学生在二次方程题目上丢分的原因通常是有规律可循的。避免这些错误能迅速提高分数。

Do not divide by x: Dividing both sides by x removes a valid root. For example, x² = 3x has solutions x = 0 and x = 3. If you divide by x, you lose the root x = 0.
切勿两边同除以 x：这样做会丢掉一个有效根。例如 x² = 3x 的解为 x = 0 和 x = 3。若两边除以 x，就会丢掉 x = 0。
Always rearrange to zero: Never try to factorise before moving all terms to one side.
务必移项归零：在把所有项移到同一侧之前，不要急于因式分解。
Check signs carefully: When b or c is negative, substituting into the quadratic formula is error-prone. Write every step.
注意符号：当 b 或 c 为负数时，代入求根公式极易出错。务必写出每一步。
Write ‘or’ correctly: Use ‘or’ between the two roots, not ‘and’, because x cannot equal both values at the same time.
正确使用 ‘or’：两个根之间用 ‘or’，而不是 ‘and’，因为 x 不可能同时等于两个值。

A common exam question type asks you to form a quadratic equation from a given graph or to find the coordinates of the turning point. For y = ax² + bx + c, the x-coordinate of the turning point is x = −b ÷ (2a).

一种常见考题要求你从给定图像写出二次方程，或求顶点坐标。对于 y = ax² + bx + c，顶点 x 坐标为 x = −b ÷ (2a)。

12. Practice Questions | 练习巩固

Try these questions on your own before checking the answers. They reflect the style of real IGCSE questions.

请先独立完成以下练习，再对照答案。这些题目贴近真实 IGCSE 考题风格。

Question 1: Solve x² − 7x + 10 = 0 by factorisation.

问题 1：用因式分解法解方程 x² − 7x + 10 = 0。

Answer: (x − 2)(x − 5) = 0 → x = 2 or x = 5.

答案：(x − 2)(x − 5) = 0 → x = 2 或 x = 5。

Question 2: Solve x² − 10x + 20 = 0 by completing the square. Give your answers in surd form.

问题 2：用配方法解方程 x² − 10x + 20 = 0，结果以根式形式给出。

Answer: (x − 5)² − 25 + 20 = 0 → (x − 5)² = 5 → x = 5 ± √5.

答案：(x − 5)² − 25 + 20 = 0 → (x − 5)² = 5 → x = 5 ± √5。

Question 3: Use the discriminant to determine whether 3x² + 2x + 5 = 0 has real roots.

问题 3：用判别式判断方程 3x² + 2x + 5 = 0 是否有实数根。

Answer: b² − 4ac = 4 − 60 = −56 < 0, so there are no real roots.

答案：b² − 4ac = 4 − 60 = −56 < 0，因此没有实数根。

Question 4: The sum of the squares of two consecutive positive integers is 85. Find the integers.

问题 4：两个连续正整数的平方和为 85。求这两个整数。

Answer: Let the integers be n and n + 1. Then n² + (n + 1)² = 85 → 2n² + 2n − 84 = 0 → n² + n − 42 = 0 → (n + 7)(n − 6) = 0 → n = 6 since n is positive. The integers are 6 and 7.

答案：设两整数为 n 和 n + 1。则 n² + (n + 1)² = 85 → 2n² + 2n − 84 = 0 → n² + n − 42 = 0 → (n + 7)(n − 6) = 0 → 因 n 为正整数，取 n = 6。两整数为 6 和 7。

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Quadratic Equations Complete Guide | 一元二次方程完全指南