Mastering Quadratic Equations | 掌握二次方程

📚 Mastering Quadratic Equations | 掌握二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper. From simple factorisation to the quadratic formula and the discriminant, this topic tests both your algebraic manipulation and your conceptual understanding. In this revision guide, we break down every core skill you need, with worked examples and common pitfalls explained clearly.

二次方程几乎出现在每一份 IGCSE 数学试卷中。从简单的因式分解到二次公式和判别式，这个考点既考查代数运算能力，也考查概念理解水平。在本复习指南中，我们将逐一拆解所有核心技能，并配以详细例题和常见易错点分析。

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

A quadratic equation is any equation that can be written in the general form ax² + bx + c = 0, where a, b and c are real numbers and a is not equal to 0. The highest power of the variable x is 2, which is why it is called a “quadratic” or “second-degree” equation.

二次方程是任何可以写成一般形式 ax² + bx + c = 0 的方程，其中 a、b、c 是实数，且 a 不等于 0。变量 x 的最高次幂是 2，因此它被称为“二次方程”或“二阶方程”。

Here are some examples of quadratic equations:

以下是二次方程的一些例子：

x² – 3x + 2 = 0
2x² + 5x – 1 = 0
-x² + 4x = 0

Notice that some coefficients may be zero, but the x² term must always be present. An equation such as x² = 9 is also quadratic because it can be rewritten as x² – 9 = 0.

注意，某些系数可以为零，但 x² 项必须始终存在。像 x² = 9 这样的方程也是二次方程，因为它可以改写为 x² – 9 = 0。

A common mistake is to think that any equation with an x² term is automatically quadratic. In fact, the equation must be reducible to the form ax² + bx + c = 0 with a ≠ 0; for example, x² + 5x = x² – 3 simplifies to 5x = -3, which is linear, not quadratic.

一个常见错误是认为任何含有 x² 项的方程都是二次方程。实际上，方程必须能化简为 ax² + bx + c = 0 且 a ≠ 0 的形式；例如 x² + 5x = x² – 3 可化简为 5x = -3，这是线性方程，而不是二次方程。

2. General Form and Key Terms | 一般形式与关键术语

The general form of a quadratic equation is written as ax² + bx + c = 0, and each part has a specific name:

二次方程的一般形式写作 ax² + bx + c = 0，每一部分都有特定名称：

Term \| 术语	Meaning \| 含义
a	Coefficient of x²; it must be non-zero. 二次项系数；必须非零。
b	Coefficient of x. 一次项系数。
c	Constant term. 常数项。
Root / Solution	The value(s) of x that satisfy the equation. 满足方程的 x 值。

A quadratic equation can have at most two distinct real roots. These roots are also known as the solutions or zeros of the equation because they make the expression equal to zero when substituted in.

二次方程最多有两个不同的实数根。这些根也被称为方程的解或零点，因为代入后可使整个表达式等于零。

When a question asks you to “solve” a quadratic equation, it wants you to find all such values of x. When it asks for “roots”, the same values are required. Be aware that some equations have only one repeated root, or no real roots at all.

当题目要求你“解”二次方程时，就是要你找出所有满足条件的 x 值。当题目要求“根”时，求的也是同样的值。请注意，有些方程只有一个重根，有些则完全没有实数根。

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic can be written as a product of two linear factors. For a quadratic with a = 1, we look for two numbers that multiply to give c and add to give b.

因式分解是当二次式可以写成两个一次因式乘积时最快的方法。对于 a = 1 的二次方程，我们寻找两个数，使它们相乘等于 c，相加等于 b。

Follow these steps:

按照以下步骤进行：

Write the equation in the form ax² + bx + c = 0.

将方程写成 ax² + bx + c = 0 的形式。
Find two numbers whose product is c and whose sum is b.

找出两个数，使乘积为 c，和为 b。
Write the factors as (x + p)(x + q) = 0.

将因式写成 (x + p)(x + q) = 0。
Use the zero product property: if the product is zero, then at least one factor must be zero.

使用零乘积性质：如果乘积为零，则至少有一个因式为零。

Worked example: solve x² + 5x + 6 = 0.

例题：解 x² + 5x + 6 = 0。

We need two numbers with product 6 and sum 5. These numbers are 2 and 3, because 2 × 3 = 6 and 2 + 3 = 5.

我们需要两个乘积为 6、和为 5 的数。这两个数是 2 和 3，因为 2 × 3 = 6，且 2 + 3 = 5。

(x + 2)(x + 3) = 0

Therefore x + 2 = 0 or x + 3 = 0, so x = -2 or x = -3. Always check your answers by substituting them back into the original equation.

因此 x + 2 = 0 或 x + 3 = 0，所以 x = -2 或 x = -3。务必把答案代回原方程进行检验。

When a ≠ 1, factorisation requires more care. For 2x² + 7x + 3 = 0, find two numbers with product 2 × 3 = 6 and sum 7; the numbers are 1 and 6. Split the middle term: 2x² + x + 6x + 3 = 0, then factor by grouping: x(2x + 1) + 3(2x + 1) = 0, giving (2x + 1)(x + 3) = 0, so x = -1/2 or x = -3.

当 a ≠ 1 时，因式分解需要更加小心。对于 2x² + 7x + 3 = 0，找出乘积为 2 × 3 = 6、和为 7 的两个数：1 和 6。拆开中间项：2x² + x + 6x + 3 = 0，再分组分解：x(2x + 1) + 3(2x + 1) = 0，得到 (2x + 1)(x + 3) = 0，所以 x = -1/2 或 x = -3。

Two special cases are worth memorising. The difference of squares a² – b² = (a – b)(a + b) means x² – 9 = (x – 3)(x + 3). A perfect square trinomial such as x² + 6x + 9 = (x + 3)² gives one repeated root, x = -3.

有两个特殊情形值得牢记。平方差公式 a² – b² = (a – b)(a + b) 意味着 x² – 9 = (x – 3)(x + 3)。完全平方三项式如 x² + 6x + 9 = (x + 3)² 会给出一个重根 x = -3。

4. Solving by the Quadratic Formula | 二次公式法

The quadratic formula works for every quadratic equation, even when factorisation is difficult or impossible. It is derived from completing the square and is given by:

二次公式适用于所有二次方程，即使因式分解困难或无法进行时也有效。它由配方法推导而来，公式如下：

x = (-b ± √(b² – 4ac)) / 2a

To use the formula, always write the equation in the general form first, then identify a, b and c carefully, including their signs.

使用公式时，务必先将方程写成一般形式，然后仔细确定 a、b、c 的值，包括它们的符号。

Worked example: solve 2x² + 3x – 2 = 0 using the quadratic formula.

例题：用二次公式解 2x² + 3x – 2 = 0。

Here a = 2, b = 3 and 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 = (-3 ± 5) / 4

So x = (-3 + 5) / 4 = 2/4 = 1/2, or x = (-3 – 5) / 4 = -8/4 = -2. The solution set is x = 1/2 or x = -2.

所以 x = (-3 + 5) / 4 = 2/4 = 1/2，或 x = (-3 – 5) / 4 = -8/4 = -2。解集为 x = 1/2 或 x = -2。

Examiners often award method marks for correctly substituting a, b and c, so show every line of working even if you are confident with mental arithmetic. A calculator may be allowed on some questions, but always write down the substitution first.

考官通常会给正确的代入 a、b、c 的过程分，所以即使你心算很有把握，也要写出每一步过程。有些题目允许使用计算器，但务必先写出代入过程。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic ax² + bx + c in the form a(x – h)² + k. This form directly reveals the vertex of the parabola and is also used to derive the quadratic formula.

配方法将二次式 ax² + bx + c 改写为 a(x – h)² + k 的形式。这种形式直接揭示抛物线的顶点，也用于推导二次公式。

For a quadratic with a = 1, take half the coefficient of x, square it, and add and subtract this value to maintain equality.

对于 a = 1 的二次式，取 x 系数的一半，将其平方，然后加上并减去这个值以保持等式不变。

Worked example: solve x² + 6x + 5 = 0 by completing the square.

例题：用配方法解 x² + 6x + 5 = 0。

Half of 6 is 3, and 3² = 9. Add and subtract 9:

6 的一半是 3，且 3² = 9。加上并减去 9：

x² + 6x + 9 – 9 + 5 = 0

(x + 3)² – 4 = 0

(x + 3)² = 4

Taking the square root of both sides gives x + 3 = ±2, so x = -1 or x = -5. Remember to include the ± sign when taking a square root.

两边开平方得到 x + 3 = ±2，所以 x = -1 或 x = -5。记住开平方时一定要写 ± 号。

When a ≠ 1, factor out a first. For example, 2x² – 8x + 5 = 0 becomes 2(x² – 4x) + 5 = 0. Complete the square inside the brackets: 2[(x – 2)² – 4] + 5 = 0, which simplifies to 2(x – 2)² – 3 = 0. This shows the vertex is at (2, -3).

当 a ≠ 1 时，先把 a 提出来。例如 2x² – 8x + 5 = 0 变为 2(x² – 4x) + 5 = 0。在括号内配方：2[(x – 2)² – 4] + 5 = 0，化简为 2(x – 2)² – 3 = 0。这表明抛物线的顶点在 (2, -3)。

6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root sign, b² – 4ac. It tells us how many real roots a quadratic equation has without solving it fully.

判别式是二次公式中根号下的部分，即 b² – 4ac。它告诉我们二次方程有多少个实数根，而无需完整求解。

The three cases are summarised below:

这三种情况总结如下：

Discriminant \| 判别式	Number of real roots \| 实数根个数	Graph meaning \| 图像含义
b² – 4ac > 0	Two distinct real roots 两个不同的实数根	Parabola crosses the x-axis twice 抛物线穿过 x 轴两次
b² – 4ac = 0	One repeated root 一个重根	Parabola touches the x-axis once 抛物线只接触 x 轴一次
b² – 4ac < 0	No real roots 没有实数根	Parabola does not meet the x-axis 抛物线与 x 轴不相交

Worked example: determine the nature of the roots of 3x² – 6x + 2 = 0.

例题：判断 3x² – 6x + 2 = 0 的根的性质。

Here a = 3, b = -6 and c = 2. The discriminant is:

这里 a = 3，b = -6，c = 2。判别式为：

Δ = (-6)² – 4 × 3 × 2 = 36 – 24 = 12

Since 12 > 0, the equation has two distinct real roots.

因为 12 > 0，所以方程有两个不同的实数根。

If the discriminant is not a perfect square, the roots are irrational and the quadratic cannot be factorised over integers. This is a quick way to decide which method to use in an exam.

如果判别式不是完全平方数，则根为无理数，该二次式无法在整数范围内因式分解。这是考试中快速判断使用哪种方法的好办法。

7. Graphical Interpretation | 图像解释

The graph of y = ax² + bx + c is a parabola. The value of a determines its shape: if a > 0, the parabola opens upward and has a minimum vertex; if a < 0, it opens downward and has a maximum vertex.

y = ax² + bx + c 的图像是一条抛物线。a 的值决定其形状：若 a > 0，抛物线开口向上，有最小值顶点；若 a < 0，抛物线开口向下，有最大值顶点。

The roots of the equation are the x-coordinates where the graph crosses the x-axis. When the discriminant is zero, the vertex lies exactly on the x-axis and the root is repeated.

方程的根是图像与 x 轴交点的 x 坐标。当判别式为零时，顶点恰好落在 x 轴上，根是重根。

The axis of symmetry of the parabola is the vertical line:

抛物线的对称轴是直线：

x = -b / 2a

The vertex lies on this axis. Its y-coordinate is found by substituting x = -b / 2a back into the equation.

顶点位于这条对称轴上。将其 x 坐标代入原方程即可求出顶点的 y 坐标。

Worked example: find the axis of symmetry and the vertex of y = x² – 4x + 3.

例题：求 y = x² – 4x + 3 的对称轴和顶点。

Here a = 1 and b = -4, so the axis of symmetry is x = -(-4) / (2 × 1) = 2. Substituting x = 2 gives y = 4 – 8 + 3 = -1. The vertex is (2, -1).

这里 a = 1，b = -4，所以对称轴为 x = -(-4) / (2 × 1) = 2。代入 x = 2 得 y = 4 – 8 + 3 = -1。顶点为 (2, -1)。

Completing the square gives the same result more directly: y = (x – 2)² – 1, so the vertex is (2, -1). Practice both methods, as exam questions may ask for the vertex or the minimum/maximum value in either form.

配方法也能更直接地得到相同结果：y = (x – 2)² – 1，所以顶点为 (2, -1)。两种方法都要多加练习，因为考试题目可能以任一形式要求你求出顶点或最小值/最大值。

8. Word Problems with Quadratics | 二次方程应用题

Many exam questions translate a real-world situation into a quadratic equation. You should read the problem carefully, define a variable, form the equation, solve it, and then check that the answer makes sense in the context.

许多考试题目会从实际情境中抽象出二次方程。你应仔细审题、设定变量、构造方程、求解，然后检验答案在情境中是否合理。

Worked example: the length of a rectangle is 3 cm more than its width, and its area is 40 cm². Find the width.

例题：一个长方形的长比宽多 3 cm，面积为 40 cm²。求宽。

Let the width be x cm. Then the length is (x + 3) cm, and the area gives:

设宽为 x cm，则长为 (x + 3) cm，由面积得：

x(x + 3) = 40

x² + 3x – 40 = 0

Factorise: (x + 8)(x – 5) = 0, so x = -8 or x = 5. Since a width cannot be negative, we reject x = -8. The width is 5 cm.

因式分解：(x

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Mastering Quadratic Equations | 掌握二次方程