Solving Quadratic Equations | 一元二次方程求解

📚 Solving Quadratic Equations | 一元二次方程求解

Quadratic equations appear throughout the Edexcel IGCSE Mathematics syllabus. A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. Solving such equations is an essential skill for topics as varied as coordinate geometry, functions and problem solving.

在 Edexcel IGCSE 数学考纲中，一元二次方程无处不在。二次方程是指可以写成 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。解这类方程是学习坐标几何、函数和实际应用题等许多内容的关键技能。

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

A quadratic equation has a squared term as its highest power. The general form is ax² + bx + c = 0. Here are some examples: x² – 5x + 6 = 0, 2x² = 8, and x² = 3x.

二次方程是最高次数为 2 的方程，一般形式为 ax² + bx + c = 0。例如：x² – 5x + 6 = 0、2x² = 8 和 x² = 3x。

It is important to notice that a cannot be zero. If a = 0, the equation becomes linear, not quadratic. In the form above, b and c can be zero, but the x² term must exist.

需要注意 a 不能为 0。如果 a = 0，方程就变成一次方程，而不是二次方程。在上述形式中，b 和 c 都可以为 0，但 x² 项必须存在。

2. Expanding and Simplifying | 展开与化简

Before solving, quadratic expressions often need to be expanded. For example, (x + 3)(x – 2) is expanded as x² + x – 6. This skill is the reverse of factorisation and is used to check answers.

在求解之前，通常需要先将二次式展开。例如，(x + 3)(x – 2) 展开后得到 x² + x – 6。展开是因式分解的逆运算，也用于检验答案。

The distributive law is central to expansion. Multiply each term in the first bracket by each term in the second bracket. Then collect like terms.

展开的核心是乘法分配律。用第一个括号中的每一项乘以第二个括号中的每一项，然后合并同类项。

(x + p)(x + q) = x² + (p + q)x + pq

3. Factorising Quadratic Expressions | 二次式的因式分解

Factorising means writing a quadratic expression as a product of two binomial brackets. For x² + bx + c, we look for two numbers whose product is c and whose sum is b.

因式分解就是把一个二次式写成两个二项式的乘积。对于 x² + bx + c，我们需要找到两个数，它们的乘积为 c，和为 b。

For example, x² + 5x + 6 = (x + 2)(x + 3), because 2 × 3 = 6 and 2 + 3 = 5. For x² – 4, use the difference of two squares: (x – 2)(x + 2).

例如，x² + 5x + 6 = (x + 2)(x + 3)，因为 2 × 3 = 6，且 2 + 3 = 5。对于 x² – 4，可以使用平方差公式：(x – 2)(x + 2)。

When a ≠ 1, for example 2x² + 7x + 3, factorisation requires more care. Use methods such as grouping: split the middle term and factor by pairs.

当 a ≠ 1 时，例如 2x² + 7x + 3，因式分解需要更加细心。可以使用分组法：拆开中间项，再分组提取公因式。

4. Solving by Factorisation | 因式分解法解方程

Once the quadratic equation factorises, use the zero product property: if AB = 0, then either A = 0 or B = 0.

一旦二次方程能够因式分解，就可以使用零积性质：如果 AB = 0，那么必有 A = 0 或 B = 0。

Solve x² – 5x + 6 = 0. First factorise: (x – 2)(x – 3) = 0. Therefore x – 2 = 0 or x – 3 = 0, so x = 2 or x = 3.

解方程 x² – 5x + 6 = 0。首先因式分解得到 (x – 2)(x – 3) = 0。因此 x – 2 = 0 或 x – 3 = 0，所以 x = 2 或 x = 3。

Always rearrange the equation so one side is zero before factorising. For example, x² = 3x becomes x² – 3x = 0, then x(x – 3) = 0, giving x = 0 or x = 3.

在因式分解前，一定要把方程整理为一边为 0 的形式。例如 x² = 3x 应先改写为 x² – 3x = 0，然后 x(x – 3) = 0，得到 x = 0 或 x = 3。

5. Solving by Completing the Square | 配方法解方程

Completing the square rewrites x² + bx + c as (x + m)² + n. For x² + 6x, since (x + 3)² = x² + 6x + 9, we write x² + 6x = (x + 3)² – 9.

配方法将 x² + bx + c 改写为 (x + m)² + n。对于 x² + 6x，因为 (x + 3)² = x² + 6x + 9，所以 x² + 6x = (x + 3)² – 9。

This technique is useful when factorisation is difficult. For example, solve x² + 6x – 7 = 0 by writing (x + 3)² – 16 = 0. Then (x + 3)² = 16, so x + 3 = ±4, giving x = 1 or x = -7.

当因式分解困难时，配方法非常有用。例如，解 x² + 6x – 7 = 0，可写成 (x + 3)² – 16 = 0。于是 (x + 3)² = 16，所以 x + 3 = ±4，得到 x = 1 或 x = -7。

The general rule for x² + bx is to add and subtract (b/2)². This is also the foundation of the quadratic formula.

x² + bx 配方的一般规则是加减 (b/2)²。这也是二次公式的推导基础。

6. The Quadratic Formula | 二次公式

The quadratic formula solves any quadratic equation ax² + bx + c = 0. It is particularly useful when factorisation is impossible or time-consuming.

二次公式可以求解任意一元二次方程 ax² + bx + c = 0。当因式分解不可行或耗时较长时，它特别有用。

x = (-b ± √(b² – 4ac)) / 2a

For example, solve 2x² + 3x – 2 = 0. Here a = 2, b = 3 and c = -2. Substitute into the formula: x = (-3 ± √(9 + 16)) / 4 = (-3 ± 5) / 4, so x = 1/2 or x = -2.

例如解 2x² + 3x – 2 = 0。此时 a = 2，b = 3，c = -2。代入公式得 x = (-3 ± √(9 + 16)) / 4 = (-3 ± 5) / 4，因此 x = 1/2 或 x = -2。

Be careful with signs: when c is negative, b² – 4ac becomes b² + 4a|c|. Always write the exact expression first, then simplify.

请特别注意符号：当 c 为负数时，b² – 4ac 会变成 b² + 4a|c|。先把公式表达式完整写出来，再逐步化简。

7. The Discriminant | 判别式

The expression Δ = b² – 4ac is called the discriminant. It tells us how many real roots a quadratic equation has, without solving it fully.

表达式 Δ = b² – 4ac 称为判别式。它可以在不完整求解的情况下，告诉我们一元二次方程有几个实数根。

If Δ > 0, there are two distinct real roots.

若 Δ > 0，方程有两个不同的实数根。
If Δ = 0, there is exactly one repeated real root.

若 Δ = 0，方程有一个二重实数根。
If Δ < 0, there are no real roots.

若 Δ < 0，方程没有实数根。

For x² + 4x + 5 = 0, Δ = 16 – 20 = -4, so the equation has no real roots. The graph of y = x² + 4x + 5 does not cross the x-axis.

对于 x² + 4x + 5 = 0，Δ = 16 – 20 = -4，所以方程没有实数根。图像 y = x² + 4x + 5 与 x 轴不相交。

8. Roots and the Graph of a Quadratic | 方程的根与二次函数图像

For a quadratic function y = ax² + bx + c, the roots of the equation ax² + bx + c = 0 are the x-coordinates of the points where the parabola crosses or touches the x-axis.

对于二次函数 y = ax² + bx + c，方程 ax² + bx + c = 0 的根就是抛物线与 x 轴相交或相切点的横坐标。

If a > 0, the parabola opens upwards. If a < 0, it opens downwards. The vertex is at x = -b / (2a), and the y-coordinate reflects the minimum or maximum value.

当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。顶点横坐标为 x = -b / (2a)，其纵坐标对应函数的最小值或最大值。

You may be asked to sketch a quadratic graph. Mark the roots, the y-intercept c, and the axis of symmetry x = -b / (2a).

考试中可能要求画出二次函数草图。请标出根、y 轴截距 c 以及对称轴 x = -b / (2a)。

9. Word Problems | 应用题

Quadratic equations often arise in geometry and number problems. For example, the area of a rectangle is given by length × width. If one side is x and the other is x + 3, and the area is 28, then x(x + 3) = 28.

二次方程常出现在几何和数字问题中。例如，矩形面积等于长乘以宽。若一边为 x，另一边为 x + 3，面积为 28，则 x(x + 3) = 28。

Solve this by expanding: x² + 3x – 28 = 0. Factorise to get (x + 7)(x – 4) = 0, so x = 4 or x = -7. Since length must be positive, x = 4.

展开后得到 x² + 3x – 28 = 0，因式分解为 (x + 7)(x – 4) = 0，所以 x = 4 或 x = -7。由于长度必须为正，因此 x = 4。

Always reject any negative or impossible solutions in context. State your final answer with appropriate units.

在具体情境中，一定要舍去负数或不符合实际的解，并写出带单位的最终答案。

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

In the Edexcel IGCSE exam, always write the quadratic in the form ax² + bx + c = 0 before solving. Check whether a is 1 or not; choose the most efficient method.

在 Edexcel IGCSE 考试中，求解前务必先把方程整理成 ax² + bx + c = 0 的形式。观察 a 是否为 1，选择最高效的方法。

Common mistake	Correction
Dividing both sides by x	Factorise x out instead, to avoid losing the root x = 0
Forgetting the ± sign in the quadratic formula	Always include ± when taking a square root
Sign errors when c is negative	Substitute carefully: b² – 4ac with a negative c means plus

After solving, substitute your answers back into the original equation to check. Show all working: in Edexcel IGCSE, method marks are often as important as final answers.

解完后，把答案代回原方程检验。在 Edexcel IGCSE 中，方法分往往和最终答案分同等重要，所以请写出完整过程。

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Solving Quadratic Equations | 一元二次方程求解