Quadratic Equations | 二次方程

📚 Quadratic Equations | 二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, geometry, and real-world problem solving. This article will guide you through the key methods of solving quadratic equations, help you understand their structure, and show you how to apply them with confidence.

二次方程是IGCSE数学中最重要的话题之一。它们出现在代数、图像、几何以及现实生活问题中。本文将引导你掌握解二次方程的主要方法，帮助你理解其结构，并自信地加以应用。

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

A quadratic equation is an equation that can be written in the standard form:

ax² + bx + c = 0

where a, b and c are constants, and a ≠ 0. The highest power of x is 2, which gives it the name “quadratic” (from the Latin word “quadratus” meaning square).

二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中a、b和c是常数，且a ≠ 0。x的最高次数是2，因此被称为“二次”（quadratic，源自拉丁语“quadratus”，意为“平方”）。

For example, 2x² + 5x – 3 = 0 is a quadratic equation, while x + 3 = 5 is linear because its highest power is 1.

例如，2x² + 5x – 3 = 0 是二次方程，而 x + 3 = 5 是一次方程（线性方程），因为其最高次数是1。

2. The Standard Form | 标准形式

Always rearrange a quadratic equation into the standard form ax² + bx + c = 0 before solving. This makes the coefficients clear and allows you to apply the correct method.

在解二次方程之前，一定要先将其整理成标准形式 ax² + bx + c = 0。这可以使系数清晰，并帮助你选择正确的方法。

Consider the equation 3x² + 2 = 7x. To write it in standard form, subtract 7x from both sides:

考虑方程 3x² + 2 = 7x。要写成标准形式，需要将两边都减去 7x：

3x² – 7x + 2 = 0

Now we can identify a = 3, b = -7 and c = 2.

现在我们可以确定 a = 3，b = -7，c = 2。

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer factors. You need to find two numbers that multiply to give ac and add to give b.

当二次方程具有简单的整数因式时，因式分解是最快的方法。你需要找到两个数，它们的乘积等于ac，和等于b。

Let’s solve x² + 5x + 6 = 0. Here a = 1, so we look for two numbers that multiply to 6 and add to 5: 2 and 3.

我们来解 x² + 5x + 6 = 0。这里 a = 1，所以我们找两个数，它们相乘得6，相加得5：即2和3。

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

For the product to be zero, one of the factors must be zero:

要使乘积为零，其中一个因式必须为零：

x + 2 = 0 or x + 3 = 0

Therefore x = -2 or x = -3.

因此 x = -2 或 x = -3。

When a ≠ 1, you may need to use the method of splitting the middle term. For example, solve 2x² + 7x + 3 = 0. Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: 1 and 6.

当 a ≠ 1 时，你可能需要运用拆项法。例如，解 2x² + 7x + 3 = 0。将a和c相乘：2 × 3 = 6。找到两个数，相乘得6，相加得7：即1和6。

2x² + 1x + 6x + 3 = 0

Factor in pairs: x(2x + 1) + 3(2x + 1) = 0, so (x + 3)(2x + 1) = 0. Thus x = -3 or x = -½.

两两分组提取公因式：x(2x + 1) + 3(2x + 1) = 0，所以 (x + 3)(2x + 1) = 0。因此 x = -3 或 x = -½。

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

When factorisation is difficult or impossible, use the quadratic formula. It works for any quadratic equation in the form ax² + bx + c = 0.

当因式分解困难或不可能时，应使用二次公式。它适用于任何形式为 ax² + bx + c = 0 的二次方程。

x = (−b ± √(b² − 4ac)) / (2a)

Here the symbol ± means you need to calculate both the plus and minus versions.

这里的符号 ± 表示你需要分别计算加号和减号两种情况。

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

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

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4

Since √40 = 2√10, we get x = (4 ± 2√10) / 4 = (2 ± √10) / 2. So the two solutions are:

由于 √40 = 2√10，我们得到 x = (4 ± 2√10) / 4 = (2 ± √10) / 2。因此两个解为：

x = (2 + √10) / 2 or x = (2 – √10) / 2

Always write the answer in the simplest exact form, unless the question asks for a decimal approximation.

除非题目要求小数近似值，否则答案应写成最简精确形式。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This is useful for solving equations and also for finding the turning point of a quadratic graph.

配方法将二次式重新写成 a(x + p)² + q 的形式。这有助于解方程，也有助于求二次图像的顶点。

For the quadratic x² + 6x + 1 = 0, first take half of 6, which is 3, and write:

对于二次式 x² + 6x + 1 = 0，首先取6的一半，即3，然后写出：

(x + 3)² – 9 + 1 = 0

Simplify: (x + 3)² – 8 = 0. Then (x + 3)² = 8.

化简得：(x + 3)² – 8 = 0。然后 (x + 3)² = 8。

Take the square root of both sides: x + 3 = ±√8, so x = -3 ± 2√2.

两边开平方：x + 3 = ±√8，所以 x = -3 ± 2√2。

If the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 5 = 0, start with 2(x² + 4x) + 5 = 0, then complete the square inside the bracket:

如果x²的系数不是1，应先提取该系数。对于 2x² + 8x + 5 = 0，先写作 2(x² + 4x) + 5 = 0，然后在括号内配方：

2[(x + 2)² – 4] + 5 = 0

So 2(x + 2)² – 8 + 5 = 0, which gives 2(x + 2)² = 3, leading to x = -2 ± √(3/2).

因此 2(x + 2)² – 8 + 5 = 0，即 2(x + 2)² = 3，从而得到 x = -2 ± √(3/2)。

6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: Δ = b² – 4ac. It tells us how many real solutions the equation has.

判别式是二次公式根号下的部分：Δ = b² – 4ac。它告诉我们方程有多少个实数解。

If b² – 4ac > 0, there are two distinct real roots.

如果 b² – 4ac > 0，方程有两个不相等实数根。
If b² – 4ac = 0, there is one repeated real root.

如果 b² – 4ac = 0，方程有一个二重实数根。
If b² – 4ac < 0, there are no real roots (the solutions are complex, but at IGCSE level we usually say there are no real solutions).

如果 b² – 4ac < 0，方程没有实数根（在IGCSE阶段通常说无实数解）。

For example, for the equation x² – 4x + 2 = 0, Δ = 16 – 8 = 8 > 0, so it has two real roots. For x² + 2x + 5 = 0, Δ = 4 – 20 = -16 < 0, so it has no real roots.

例如，对于方程 x² – 4x + 2 = 0，Δ = 16 – 8 = 8 > 0，因此有两个实数根。对于 x² + 2x + 5 = 0，Δ = 4 – 20 = -16 < 0，因此没有实数根。

7. Quadratic Graphs | 二次函数图像

The graph of a quadratic function y = ax² + bx + c is a curve called a parabola. If a > 0, it opens upwards (U-shaped); if a < 0, it opens downwards (n-shaped).

二次函数 y = ax² + bx + c 的图像是一条叫做“抛物线”的曲线。如果 a > 0，开口向上（U形）；如果 a < 0，开口向下（∩形）。

The roots of the equation ax² + bx + c = 0 are the x-intercepts of the graph. The y-intercept is c, and the turning point (minimum or maximum) can be found using the completed square form.

方程 ax² + bx + c = 0 的根就是图像与x轴的交点。与y轴的交点是c，顶点（最小值或最大值）可以用配方法得到。

For y = x² – 6x + 5, we can write y = (x – 3)² – 4. The turning point is (3, -4), and the y-intercept is 5. The roots are found by solving (x – 3)² – 4 = 0, which gives x = 1 or x = 5.

对于 y = x² – 6x + 5，我们可以写成 y = (x – 3)² – 4。顶点是 (3, -4)，与y轴的交点是5。解方程 (x – 3)² – 4 = 0 可得根为 x = 1 或 x = 5。

Sketching the graph requires the roots, the y-intercept, and the turning point. Remember that the axis of symmetry is the vertical line through the turning point, with equation x = -b/(2a).

草图需要标出根、y截距和顶点。注意对称轴是经过顶点的竖直线，其方程为 x = -b/(2a)。

8. Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities are often tested in IGCSE. The first step is to rearrange the inequality so that one side is zero, then solve the corresponding quadratic equation to find the critical values.

IGCSE经常考查二次不等式。第一步是将不等式整理成一边为零的形式，然后解对应的二次方程以找到临界值。

For example, solve x² – 3x – 4 < 0. First solve x² - 3x - 4 = 0, which factors as (x - 4)(x + 1) = 0, giving critical values x = -1 and x = 4.

例如，解不等式 x² – 3x – 4 < 0。首先解方程 x² - 3x - 4 = 0，因式分解为 (x - 4)(x + 1) = 0，得到临界值 x = -1 和 x = 4。

Since the coefficient of x² is positive, the parabola opens upward. The inequality is negative (below the x-axis) between the two roots. Therefore the solution is -1 < x < 4.

因为x²的系数为正，抛物线开口向上。不等式小于零（图像位于x轴下方）在两个根之间。因此解集为 -1 < x < 4。

If the inequality had been x² – 3x – 4 ≥ 0, the solution would be x ≤ -1 or x ≥ 4. Always check the direction of the inequality and whether the boundary values are included.

如果不等式是 x² – 3x – 4 ≥ 0，则解为 x ≤ -1 或 x ≥ 4。务必检查不等号的方向，以及临界值是否包含在内。

9. Word Problems | 应用题

Many quadratic equations arise from real-life situations. The key is to translate the words into an equation, then solve it and interpret the answer in context.

许多二次方程来自现实生活情境。关键是将文字转化为方程，然后求解，并结合情境解释答案。

Example: A rectangle has length 3 cm longer than its width. Its area is 28 cm². Find the width.

例如：一个矩形的长比宽长3厘米，面积为28平方厘米。求宽。

Let the width be x cm. Then the length is (x + 3) cm. The area is x(x + 3) = 28.

设宽为x厘米，则长为(x + 3)厘米。面积为 x(x + 3) = 28。

x² + 3x – 28 = 0

Factorise: (x + 7)(x – 4) = 0, so x = -7 or x = 4. Since width cannot be negative, the width is 4 cm.

因式分解得：(x + 7)(x – 4) = 0，所以 x = -7 或 x = 4。因为宽不能为负，所以宽为4厘米。

Always check whether your solutions make sense. Reject any negative or impossible values based on the problem context.

务必检查解是否符合实际。根据题目情境，舍去负数或不合理的值。

10. Common Mistakes and Revision Tips | 常见错误与复习建议

Students often make the following mistakes when solving quadratics:

学生在解二次方程时经常犯以下错误：

Forgetting to rearrange the equation into standard form ax² + bx + c = 0 before factoring.

在因式分解前忘记将方程整理成标准形式 ax² + bx + c = 0。
Using the quadratic formula incorrectly: mixing up negative signs, or forgetting that b² means b is squared including its sign.

错误使用二次公式：混淆负号，或忘记 b² 是指b的平方（包括符号）。
When completing the square, forgetting to expand and simplify correctly when a ≠ 1.

配方法时，在 a ≠ 1 的情况下忘记正确展开和化简。
In inequalities, writing the wrong direction of the inequality, especially when multiplying or dividing by a negative number.

在不等式中，尤其当乘以或除以负数时写错不等号方向。
Not checking the discriminant before attempting to find real roots.

在求实数根之前没有先检查判别式。

To revise effectively, practise factorising simple quadratics until it becomes automatic. Then move on to the quadratic formula and completing the square. Use the discriminant to classify the roots. Finally, attempt word problems and graph sketches to apply your skills in different contexts.

为了高效复习，请反复练习简单二次方程因式分解，直到熟练为止。然后过渡到二次公式和配方法。用判别式判断根的情况。最后通过应用题和图像草图在不同情境中运用你的技能。

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Quadratic Equations | 二次方程