Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Welcome to this revision guide on quadratic equations, a key topic in IGCSE Mathematics. This article will help you understand the standard form, solution methods, and common exam questions step by step.

欢迎阅读本复习指南，重点讲解 IGCSE 数学中的核心内容——二次方程。本文将通过逐步讲解，帮助你掌握标准形式、解法以及常见考试题型。

Quadratic equations appear in many areas of mathematics, from simple area problems to projectile motion. Mastering them is essential for your IGCSE exam.

二次方程出现在数学的许多领域，从简单的面积问题到抛体运动。掌握二次方程对 IGCSE 考试至关重要。

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

A quadratic equation is a polynomial equation of degree 2. The highest power of the unknown variable is 2, and it can be written in a standard form.

二次方程是最高次数为 2 的多项式方程。未知数的最高幂是 2，并且可以写成标准形式。

ax² + bx + c = 0, where a ≠ 0

ax² + bx + c = 0，其中 a ≠ 0

Here, ‘a’, ‘b’ and ‘c’ are constants, and ‘a’ cannot be zero. If a = 0, the equation becomes linear, not quadratic.

这里的 ‘a’、’b’ 和 ‘c’ 是常数，且 ‘a’ 不能为零。如果 a = 0，方程就变为一次方程，而不是二次方程。

Examples of quadratic equations include x² + 5x + 6 = 0 and 2x² – 3x – 2 = 0.

二次方程的例子包括 x² + 5x + 6 = 0 和 2x² – 3x – 2 = 0。

2. Standard Form and Identifying Coefficients | 标准形式与识别系数

Before solving, you must rewrite the equation in the form ax² + bx + c = 0. This means moving all terms to one side and simplifying.

在解方程之前，你必须将方程改写为 ax² + bx + c = 0 的形式。这意味着把所有项移到等号一侧并化简。

For example, the equation x² = 3x + 4 can be rearranged as x² – 3x – 4 = 0.

例如，方程 x² = 3x + 4 可以变形为 x² – 3x – 4 = 0。

Once in standard form, you can identify a, b and c. In the example above, a = 1, b = -3 and c = -4.

一旦写成标准形式，你就可以识别 a、b 和 c。在上面的例子中，a = 1，b = -3，c = -4。

Be careful with negative signs. The sign is part of the coefficient.

注意负号。负号属于系数的一部分。

3. Solving by Factorisation | 因式分解法

Factorisation is often the first method you should try, especially when the equation has integer roots.

因式分解通常是首先尝试的方法，尤其是当方程有整数根时。

To factorise x² + bx + c, find two numbers that multiply to give ‘c’ and add to give ‘b’.

要分解 x² + bx + c，需要找到两个数，它们相乘等于 ‘c’，相加等于 ‘b’。

For example, solve x² + 5x + 6 = 0. The numbers 2 and 3 multiply to 6 and add to 5, so:

例如，解 x² + 5x + 6 = 0。数字 2 和 3 相乘得 6，相加得 5，因此：

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

If the product of two factors is zero, then one factor must be zero. So x + 2 = 0 or x + 3 = 0.

如果两个因式的乘积为零，则至少有一个因式为零。所以 x + 2 = 0 或 x + 3 = 0。

Therefore x = -2 or x = -3.

因此 x = -2 或 x = -3。

When the coefficient of x² is not 1, use the ‘ac’ method. Multiply a and c, find two numbers that multiply to ac and add to b, then split the middle term.

当 x² 的系数不是 1 时，可以使用 ‘ac’ 法。先将 a 和 c 相乘，找到两个数相乘等于 ac 且相加等于 b，然后拆分中间项。

For example, 2x² + 7x + 3 = 0. Here ac = 2 × 3 = 6. The numbers 6 and 1 multiply to 6 and add to 7.

例如，2x² + 7x + 3 = 0。这里 ac = 2 × 3 = 6。数字 6 和 1 相乘得 6，相加得 7。

2x² + 6x + x + 3 = 0

Then group: 2x(x + 3) + 1(x + 3) = 0, so (2x + 1)(x + 3) = 0.

然后分组：2x(x + 3) + 1(x + 3) = 0，所以 (2x + 1)(x + 3) = 0。

Hence x = -½ or x = -3.

因此 x = -½ 或 x = -3。

4. Solving by Completing the Square | 配方法

Completing the square is a powerful method that works for any quadratic equation, including those with irrational roots.

配方法是一种强大的解法，适用于任何二次方程，包括具有无理数根的方程。

Start with ax² + bx + c = 0. Write the equation as x² + (b/a)x = -c/a, then add the square of half the coefficient of x to both sides.

从 ax² + bx + c = 0 开始。将方程改写为 x² + (b/a)x = -c/a，然后在两边加上 x 系数一半的平方。

For x² + 6x + 2 = 0, first write x² + 6x = -2. Half of 6 is 3, and 3² = 9.

对于 x² + 6x + 2 = 0，先写成 x² + 6x = -2。6 的一半是 3，而 3² = 9。

x² + 6x + 9 = -2 + 9

Now the left side is a perfect square: (x + 3)² = 7.

现在左边是一个完全平方：(x + 3)² = 7。

Take the square root of both sides: x + 3 = ±√7.

两边开平方：x + 3 = ±√7。

So x = -3 ± √7.

所以 x = -3 ± √7。

Remember to include both the positive and negative square roots.

记得同时包含正负平方根。

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

The quadratic formula is a universal method. It can solve any quadratic equation, even when factorisation is difficult or impossible.

二次公式法是一种通用方法。它可以求解任何二次方程，即使因式分解困难或无法进行。

x = (-b ± √(b² – 4ac)) / 2a

To use this formula, first identify a, b and c from ax² + bx + c = 0, then substitute them carefully.

使用这个公式时，首先从 ax² + bx + c = 0 中识别出 a、b 和 c，然后仔细代入。

For example, solve 2x² – 4x – 3 = 0. Here a = 2, b = -4 and c = -3.

例如，解 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。

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

Simplify the discriminant: 16 + 24 = 40, so x = (4 ± √40) / 4.

化简判别式：16 + 24 = 40，所以 x = (4 ± √40) / 4。

This gives x = 1 ± √10 / 2, which can be written as x ≈ 2.58 or x ≈ -0.58.

由此得到 x = 1 ± √10 / 2，可写成 x ≈ 2.58 或 x ≈ -0.58。

Always check whether your answer is required in exact form or as a decimal. In IGCSE, exact values are often preferred.

始终检查题目要求精确值还是小数。在 IGCSE 中，通常更倾向于精确值。

6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula inside the square root: Δ = b² – 4ac.

判别式是二次公式中根号内的部分：Δ = b² – 4ac。

It tells us how many real roots a quadratic equation has, without solving it completely.

它告诉我们二次方程有多少个实数根，而无需完整求解。

If Δ > 0, the equation has two distinct real roots.

如果 Δ > 0，方程有两个不同的实数根。
If Δ = 0, the equation has exactly one real root (a repeated root).

如果 Δ = 0，方程正好有一个实数根（重根）。
If Δ < 0, the equation has no real roots (only complex roots).

如果 Δ < 0，方程没有实数根（只有复数根）。

For example, for x² – 4x + 4 = 0, Δ = 16 – 16 = 0, so there is one repeated root x = 2.

例如，对于 x² – 4x + 4 = 0，Δ = 16 – 16 = 0，所以有一个重根 x = 2。

For x² + 2x + 5 = 0, Δ = 4 – 20 = -16 < 0, so there are no real roots.

对于 x² + 2x + 5 = 0，Δ = 4 – 20 = -16 < 0，所以没有实数根。

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

Many IGCSE questions present quadratic equations in a real-world context. You must form the equation from the given information before solving.

许多 IGCSE 题目会在实际情境中给出二次方程。你必须先根据已知信息建立方程，然后再求解。

For example, the area of a rectangle is 20 cm². Its length is (x + 2) cm and its width is (x – 1) cm. Find x.

例如，一个矩形的面积是 20 cm²。它的长是 (x + 2) cm，宽是 (x – 1) cm。求 x。

Area = length × width, so (x + 2)(x – 1) = 20.

面积 = 长 × 宽，所以 (x + 2)(x – 1) = 20。

Expanding gives x² + x – 2 = 20, so x² + x – 22 = 0.

展开得 x² + x – 2 = 20，所以 x² + x – 22 = 0。

Use the quadratic formula: x = (-1 ± √(1 + 88)) / 2 = (-1 ± √89) / 2.

使用二次公式：x = (-1 ± √(1 + 88)) / 2 = (-1 ± √89) / 2。

Since x is a length, it must be positive. So x ≈ 4.22.

由于 x 代表长度，必须为正。所以 x ≈ 4.22。

Always reject any negative solution that does not make sense in the context.

始终舍弃在情境中不合理的负数解。

8. Sketching Quadratic Graphs | 二次函数图像

Understanding the graph of y = ax² + bx + c helps you visualise the roots and the vertex.

理解 y = ax² + bx + c 的图像有助于你直观地观察根和顶点。

If a > 0, the graph is a U-shaped parabola. If a < 0, it is an upside-down U.

如果 a > 0，图像是 U 形抛物线。如果 a < 0，则是倒 U 形。

The roots of the equation ax² + bx + c = 0 are the x-intercepts of the graph.

方程 ax² + bx + c = 0 的根就是图像与 x 轴的交点。

The x-coordinate of the vertex is given by x = -b / 2a. Substitute this into the equation to find the y-coordinate.

顶点的 x 坐标由 x = -b / 2a 给出。将这个值代入方程可求得 y 坐标。

For sketching, identify three main features: roots, y-intercept and vertex.

画图时，需要识别三个主要特征：根、y 截距和顶点。

For example, sketch y = x² – 4x + 3. Factorise to get y = (x – 1)(x – 3), so roots are x = 1 and x = 3.

例如，画 y = x² – 4x + 3 的草图。分解为 y = (x – 1)(x – 3)，所以根是 x = 1 和 x = 3。

The y-intercept is c = 3. The vertex has x = 4 / 2 = 2, and y = 2² – 4×2 + 3 = -1.

y 截距是 c = 3。顶点的 x = 4 / 2 = 2，y = 2² – 4×2 + 3 = -1。

9. Quadratic Inequalities | 二次不等式

Quadratic inequalities are solved by finding the roots first, then testing intervals on a number line.

二次不等式先求解根，然后在数轴上测试区间。

For example, solve x² – 3x – 4 > 0.

例如，解 x² – 3x – 4 > 0。

Factorise: (x – 4)(x + 1) > 0. The roots are x = 4 and x = -1.

因式分解：(x – 4)(x + 1) > 0。根是 x = 4 和 x = -1。

Test the intervals: x < -1, -1 < x < 4, and x > 4.

测试区间：x < -1，-1 < x < 4，以及 x > 4。

For x < -1, both factors are negative, so the product is positive. For -1 < x < 4, the product is negative. For x > 4, both factors are positive, so the product is positive.

当 x < -1 时，两个因式均为负，乘积为正。当 -1 < x < 4 时，乘积为负。当 x > 4 时，两个因式均为正，乘积为正。

Therefore the solution is x < -1 or x > 4.

因此解为 x < -1 或 x > 4。

Remember to decide whether the endpoints are included. Strict inequalities use open circles, while ≤ or ≥ use closed circles.

记住判断端点是否包含。严格不等式用空心圆圈，而 ≤ 或 ≥ 用实心圆圈。

10. Summary and Exam Tips | 总结与考试技巧

Quadratic equations are a guaranteed topic in IGCSE mathematics. You should be confident in all three methods: factorisation, completing the square and the quadratic formula.

二次方程是 IGCSE 数学中的必考内容。你应该对三种方法都充满信心：因式分解法、配方法和二次公式法。

Always rearrange the equation into standard form before identifying a, b and c.

在识别 a、b 和 c 之前，始终将方程化为标准形式。
Check if factorisation is possible first; it is usually the quickest method.

首先检查是否可以进行因式分解；这通常是最快的方法。
If the coefficients are messy or roots involve surds, use the quadratic formula.

如果系数复杂或根包含无理数，使用二次公式。
Use completing the square when you need to find the vertex of a parabola or derive the quadratic formula.

当你需要求抛物线顶点或推导二次公式时，使用配方法。
Always check your solutions by substituting them back into the original equation.

始终通过将解代入原方程来检查答案。

Pay attention to the discriminant: it can save you time and help you avoid errors in multiple-choice questions.

关注判别式：它可以节省时间，并帮助你在选择题中避免错误。

In word problems, define your variable clearly and reject any answer that is not physically meaningful.

在应用题中，清楚定义变量，并舍弃任何不符合实际意义的答案。

11. Practice Questions | 练习题

Try these questions on your own before checking the answers.

请在查看答案之前先自己尝试以下题目。

Question 1: Solve x² – 7x + 12 = 0 by factorisation.

练习题 1：用因式分解法解 x² – 7x + 12 = 0。

Question 2: Solve 2x² + 5x – 3 = 0 using the quadratic formula.

练习题 2：使用二次公式解 2x² + 5x – 3 = 0。

Question 3: Determine the number of real roots of x² – 2x + 5 = 0.

练习题 3：判断 x² – 2x + 5 = 0 的实数根个数。

Question 4: The product of two consecutive integers is 56. Find the integers.

练习题 4：两个连续整数的乘积是 56。求这两个整数。

Answers: 1. x = 3 or x = 4. 2. x = 0.5 or x = -3. 3. No real roots. 4. 7 and 8.

答案：1. x = 3 或 x = 4。2. x = 0.5 或 x = -3。3. 没有实数根。4. 7 和 8。

If you made mistakes, go back to the relevant section and try again. Consistent practice is the key to success.

如果做错了，请回到相关章节再试一次。持续练习是成功的关键。

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Solving Quadratic Equations | 解二次方程