Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear in almost every Edexcel IGCSE Mathematics examination paper, whether at Foundation or Higher tier. They are also the gateway to many other topics, including graphs, inequalities, and coordinate geometry. This article explains the standard form of a quadratic equation, the four main solving methods, the discriminant, and the common pitfalls that lose students marks.

二次方程几乎出现在每一份 Edexcel IGCSE 数学试卷中，无论是基础级（Foundation）还是高级（Higher）都不可或缺。它也是学习图像、不等式和坐标几何等许多后续专题的基石。本文将系统讲解二次方程的标准形式、四种主要解法、判别式，以及学生最容易失分的常见陷阱。

1. Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation is any equation that can be written in the standard form shown below, where a, b and c are constants and a ≠ 0.

二次方程是指可以写成如下标准形式的方程，其中 a、b、c 为常数，且 a ≠ 0。

ax² + bx + c = 0

The highest power of the variable x is 2, which is why the equation is called “quadratic” (from the Latin word quadratus, meaning “square”). The condition a ≠ 0 is essential: if a = 0, the equation becomes bx + c = 0, which is simply a linear equation and no longer quadratic.

变量的最高次数为 2，因此这类方程被称为”二次”方程（quadratic 源自拉丁语 quadratus，意为”平方”）。条件 a ≠ 0 至关重要：如果 a = 0，方程退化为 bx + c = 0，变成一次方程，不再是二次方程。

Before solving, you must often rearrange the given equation into standard form. For example:

解方程之前，通常需要先将题目给出的方程整理成标准形式。例如：

x² – 5x + 6 = 0 is already in standard form with a = 1, b = -5, c = 6.
x² = 16 must be rewritten as x² – 16 = 0.
2x² + 3x = 2 must be rearranged to 2x² + 3x – 2 = 0.

x² – 5x + 6 = 0 已经是标准形式，其中 a = 1，b = -5，c = 6。
x² = 16 必须改写为 x² – 16 = 0。
2x² + 3x = 2 必须整理为 2x² + 3x – 2 = 0。

Always look carefully at the signs: a common mistake is to read b as positive when it is actually negative. Write the equation in the order ax² + bx + c = 0 before identifying the coefficients.

务必仔细观察符号：一个常见错误是把负的 b 误读为正数。在识别系数之前，一定要先把方程按 ax² + bx + c = 0 的顺序写整齐。

2. Solving by Factorisation | 因式分解法

When the quadratic expression can be factorised, this is usually the quickest method. The key principle is the zero product property: if two expressions multiply to give zero, then at least one of them must be zero.

当二次表达式可以因式分解时，这通常是最快的解法。核心原理是零积性质：若两个表达式的乘积为零，则其中至少有一个必须为零。

If A × B = 0, then A = 0 or B = 0.

Example 1: Solve x² – 5x + 6 = 0.

例 1：解方程 x² – 5x + 6 = 0。

Step 1: Find two numbers that multiply to 6 and add to -5. These numbers are -2 and -3.

步骤 1：找到两个数，乘积为 6，且和为 -5。这两个数是 -2 和 -3。

(x – 2)(x – 3) = 0

Step 2: Set each factor equal to zero: x – 2 = 0 or x – 3 = 0.

步骤 2：令每个因式等于零：x – 2 = 0 或 x – 3 = 0。

Step 3: State the solutions: x = 2 or x = 3.

步骤 3：写出解：x = 2 或 x = 3。

To factorise a monic quadratic x² + bx + c, find two numbers whose product is c and whose sum is b. For x² – 5x + 6, the numbers are -2 and -3 because (-2) × (-3) = 6 and (-2) + (-3) = -5.

分解首项系数为 1 的二次式 x² + bx + c 时，需要找到两个数，使它们的乘积为 c、和为 b。对于 x² – 5x + 6，这两个数是 -2 和 -3，因为 (-2) × (-3) = 6，且 (-2) + (-3) = -5。

The following special patterns are frequently tested:

下列特殊模式在考试中经常出现：

Pattern	Factorised form	Example
Difference of two squares: x² – a²	(x + a)(x – a)	x² – 9 = (x + 3)(x – 3)
Perfect square: x² + 2ax + a²	(x + a)²	x² + 8x + 16 = (x + 4)²
Perfect square: x² – 2ax + a²	(x – a)²	x² – 10x + 25 = (x – 5)²

For non-monic quadratics such as 2x² + 7x + 3, multiply the leading coefficient and the constant term: 2 × 3 = 6. Then find two numbers that multiply to 6 and add to 7, namely 1 and 6. Split the middle term and factorise by grouping:

对于首项系数不为 1 的二次式，例如 2x² + 7x + 3，先将首项系数与常数项相乘：2 × 3 = 6。然后找到两个数，乘积为 6、和为 7，即 1 和 6。拆开中间项并使用分组分解：

2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)

Always check your factorisation by expanding the brackets before writing your final answer.

写出最终答案前，务必通过展开括号来检验因式分解是否正确。

3. Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This form is extremely useful for finding the turning point of a graph and for solving equations when factorisation is not possible.

配方法将二次表达式改写为 a(x + p)² + q 的形式。这种形式在求图像顶点坐标以及无法因式分解时解方程都非常有用。

Worked example: Solve x² + 6x + 2 = 0 by completing the square.

例题：用配方法解方程 x² + 6x + 2 = 0。

Step 1: Halve the coefficient of x: 6 ÷ 2 = 3.

步骤 1：将 x 的系数除以 2：6 ÷ 2 = 3。

Step 2: Write x² + 6x as (x + 3)² – 9. The -9 is needed because (x + 3)² = x² + 6x + 9, which has an extra 9.

步骤 2：将 x² + 6x 改写为 (x + 3)² – 9。因为 (x + 3)² = x² + 6x + 9，多出了 9，所以要减去 9。

Step 3: Substitute back into the equation: (x + 3)² – 9 + 2 = 0, so (x + 3)² = 7.

步骤 3：代回原方程：(x + 3)² – 9 + 2 = 0，因此 (x + 3)² = 7。

Step 4: Take square roots of both sides, remembering the ± sign:

步骤 4：两边开平方，注意保留 ± 符号：

x + 3 = ±√7, so x = -3 ± √7

The solutions are x = -3 + √7 and x = -3 – √7. These are exact values, which is a major advantage over rounding from a decimal approximation.

解为 x = -3 + √7 和 x = -3 – √7。这些是精确值，比四舍五入的近似小数更有优势。

If the coefficient of x² is not 1, factor it out first. For example, 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 3. This form directly shows that the turning point is (-2, -3).

如果 x² 的系数不为 1，先将其提取出来。例如 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 3。这种形式直接给出顶点坐标为 (-2, -3)。

4. The Quadratic Formula | 求根公式

For any quadratic equation ax² + bx + c = 0, the solutions can always be found using the quadratic formula. This formula is printed in the Edexcel IGCSE formula booklet, so you do not need to memorise it, but you must practise substituting the values accurately.

对于任意二次方程 ax² + bx + c = 0，总可以使用求根公式求出解。该公式印在 Edexcel IGCSE 公式册中，不需要背诵，但必须通过练习学会准确代入数值。

x = (-b ± √(b² – 4ac)) / (2a)

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

例题：用求根公式解方程 2x² + 3x – 2 = 0。

Step 1: Identify the coefficients: a = 2, b = 3, c = -2.

步骤 1：确定系数：a = 2，b = 3，c = -2。

Step 2: Calculate the discriminant: b² – 4ac = 3² – 4 × 2 × (-2) = 9 + 16 = 25.

步骤 2：计算判别式：b² – 4ac = 3² – 4 × 2 × (-2) = 9 + 16 = 25。

Step 3: Substitute into the formula:

步骤 3：代入公式：

x = (-3 ± √25) / (2 × 2) = (-3 ± 5) / 4

Step 4: Evaluate both solutions: x = (-3 + 5) / 4 = 2 / 4 = 0.5, and x = (-3 – 5) / 4 = -8 / 4 = -2.

步骤 4：分别计算两个解：x = (-3 + 5) / 4 = 2 / 4 = 0.5，x = (-3 – 5) / 4 = -8 / 4 = -2。

When using the formula, take particular care with negative values of b and c. Write each substitution out fully to avoid sign errors. On a calculator paper, you may be asked to give answers to 1 or 2 decimal places; on a non-calculator paper, leave the answer in surd form such as (-3 ± √25) / 4 simplified.

使用公式时，要特别注意 b 和 c 为负数的情况。把每一步代写完整，避免符号错误。在允许使用计算器的试卷中，题目可能要求答案保留 1 或 2 位小数；在不允许使用计算器的试卷中，答案应保留根式形式，如 (-3 ± √25) / 4 的化简结果。

5. The Discriminant | 判别式

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

判别式是求根公式中根号内的部分：Δ = b² – 4ac。无需解方程，它就能告诉我们二次方程有多少个实数根。

Value of Δ	Number of real roots	Graph interpretation
Δ > 0	Two distinct real roots	The curve crosses the x-axis at two points
Δ = 0	One repeated real root	The curve touches the x-axis at the vertex
Δ < 0	No real roots	The curve does not touch the x-axis

判别式的值	实数根个数	图像含义
Δ > 0	两个不同的实数根	曲线与 x 轴有两个交点
Δ = 0	一个二重实数根	曲线在顶点处与 x 轴相切
Δ < 0	没有实数根	曲线不与 x 轴相交

Example 2: For x² – 4x + 4 = 0, we have a = 1, b = -4, c = 4, so Δ = (-4)² – 4 × 1 × 4 = 16 – 16 = 0. The equation has one repeated root. Indeed, x² – 4x + 4 = (x – 2)², so the only solution is x = 2.

例 2：对于 x² – 4x + 4 = 0，a = 1，b = -4，c = 4，所以 Δ = (-4)² – 4 × 1 × 4 = 16 – 16 = 0。该方程有一个二重根。事实上，x² – 4x + 4 = (x – 2)²，唯一解为 x = 2。

Exam questions often ask you to “find the set of values of k for which the equation has two distinct real roots”. In such problems, set Δ > 0 and solve the resulting inequality. Similarly, use Δ = 0 for a tangent line and Δ < 0 for no intersection.

考试中常见的问题是”求 k 的取值范围，使方程有两个不同的实数根”。此时令 Δ > 0，并解所得的不等式。类似地，相切时用 Δ = 0，无交点时用 Δ < 0。

6. Solving Quadratics Graphically | 图像法解二次方程

The real roots of ax² + bx + c = 0 correspond exactly to the x-coordinates of the points where the graph of y = ax² + bx + c crosses or touches the x-axis. This is because setting y = 0 is equivalent to solving the quadratic equation.

方程 ax² + bx + c = 0 的实数根恰好对应抛物线 y = ax² + bx + c 与 x 轴相交或相切点的横坐标。这是因为令 y = 0 就等价于解二次方程。

To solve a quadratic graphically:

用图像法解二次方程的步骤：

Draw the graph of y = ax² + bx + c over a suitable range of x values.
Read off the x-coordinates of the points where the graph meets the x-axis.
If the graph does not meet the x-axis, the equation has no real solutions.

在合适的 x 范围内画出 y = ax² + bx + c 的图像。
读出图像与 x 轴交点的横坐标。
如果图像不与 x 轴相交，则方程没有实数解。

You may also be asked to solve an equation such as x² – 2x – 3 = x + 1. In this case, draw the curve y = x² – 2x – 3 and the straight line y = x + 1 on the same axes; the x-coordinates of the intersection points are the solutions to the original equation. This works because at

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Solving Quadratic Equations | 解一元二次方程