Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

In the Edexcel IGCSE Mathematics syllabus, quadratic equations are a fundamental topic that appears in many forms. Mastering them is essential for success in both Paper 1 and Paper 2.

在 Edexcel IGCSE 数学大纲中，一元二次方程是一个基础且重要的主题，以多种形式出现在考试中。掌握它对于在 Paper 1 和 Paper 2 中取得好成绩至关重要。

1. What is a Quadratic Equation? | 什么是一元二次方程？

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable is 2.

一元二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。未知数的最高次数为 2。

For example, 2x² – 3x + 1 = 0 is a quadratic equation, while x³ – 2x = 0 is not.

例如，2x² – 3x + 1 = 0 是一元二次方程，而 x³ – 2x = 0 不是。

2. Expanding and Factorising Quadratic Expressions | 展开和因式分解二次表达式

To solve quadratics, you need to be fluent in expanding brackets and factorising. The general expansion is (x + p)(x + q) = x² + (p+q)x + pq.

要解一元二次方程，你需要熟练地进行括号展开和因式分解。一般展开式为 (x + p)(x + q) = x² + (p+q)x + pq。

For example, (x + 3)(x – 2) = x² + x – 6, because 3 + (-2) = 1 and 3 × (-2) = -6.

例如，(x + 3)(x – 2) = x² + x – 6，因为 3 + (-2) = 1，3 × (-2) = -6。

Factorising is the reverse process. To factorise x² + 5x + 6, find two numbers whose sum is 5 and product is 6. They are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).

因式分解是相反的过程。要因式分解 x² + 5x + 6，找到两个数，其和为 5，积为 6。它们是 2 和 3，所以 x² + 5x + 6 = (x + 2)(x + 3)。

3. Solving by Factorisation | 因式分解法求解

If a quadratic equation can be factorised, set each factor equal to zero. For the equation x² – 4x + 3 = 0, we factorise to get (x – 1)(x – 3) = 0.

如果一个一元二次方程可以被因式分解，就让每个因式等于零。对于方程 x² – 4x + 3 = 0，因式分解得到 (x – 1)(x – 3) = 0。

Then x – 1 = 0 or x – 3 = 0, so x = 1 or x = 3.

然后 x – 1 = 0 或 x – 3 = 0，所以 x = 1 或 x = 3。

If ab = 0, then a = 0 or b = 0.

如果 ab = 0，则 a = 0 或 b = 0。

When the coefficient of x² is not 1, such as 2x² + 5x – 3 = 0, you need to factorise using the method of splitting the middle term. Multiply a and c: 2 × (-3) = -6. Find two numbers with sum 5 and product -6: 6 and -1. Rewrite: 2x² + 6x – x – 3 = 0, then factor by grouping: 2x(x + 3) – 1(x + 3) = 0, giving (2x – 1)(x + 3) = 0. Thus x = ½ or x = -3.

当 x² 的系数不是 1 时，例如 2x² + 5x – 3 = 0，你需要使用拆中项法进行因式分解。将 a 和 c 相乘：2 × (-3) = -6。找到和为 5 且积为 -6 的两个数：6 和 -1。改写：2x² + 6x – x – 3 = 0，然后分组分解：2x(x + 3) – 1(x + 3) = 0，得到 (2x – 1)(x + 3) = 0。因此 x = ½ 或 x = -3。

4. Solving by the Quadratic Formula | 公式法求解

Not all quadratic equations can be factorised easily. The quadratic formula works for any quadratic equation:

并非所有一元二次方程都能轻松因式分解。公式法适用于所有一元二次方程：

x = (-b ± √(b² – 4ac)) / 2a

For example, solve 2x² + 3x – 1 = 0. Here a = 2, b = 3, c = -1. Substitute into the formula:

例如，解 2x² + 3x – 1 = 0。这里 a = 2，b = 3，c = -1。代入公式得：

x = (-3 ± √(3² – 4×2×(-1))) / (2×2) = (-3 ± √17) / 4

Therefore x = (-3 + √17)/4 or x = (-3 – √17)/4. These are exact values, often required in the exam.

因此 x = (-3 + √17)/4 或 x = (-3 – √17)/4。这些是精确值，考试中经常需要写出。

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, graphing, and finding the turning point.

配方法将二次式写成 a(x + p)² + q 的形式。这有助于求解、作图以及找到顶点。

For x² + 6x + 2 = 0, take half of 6, which is 3, and write (x + 3)² – 9 + 2 = 0, so (x + 3)² – 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7, hence x = -3 ± √7.

对于 x² + 6x + 2 = 0，取 6 的一半为 3，写成 (x + 3)² – 9 + 2 = 0，即 (x + 3)² – 7 = 0。然后 (x + 3)² = 7，所以 x + 3 = ±√7，因此 x = -3 ± √7。

If the coefficient of x² is not 1, first factor it out. For 2x² – 8x + 1 = 0, write 2(x² – 4x) + 1 = 0, then complete the square inside: 2[(x – 2)² – 4] + 1 = 0, giving 2(x – 2)² – 8 + 1 = 0, so 2(x – 2)² = 7, and (x – 2)² = 7/2. Thus x = 2 ± √(7/2).

如果 x² 的系数不是 1，先将其提取出来。对于 2x² – 8x + 1 = 0，写成 2(x² – 4x) + 1 = 0，然后在括号内配方：2[(x – 2)² – 4] + 1 = 0，得到 2(x – 2)² – 8 + 1 = 0，所以 2(x – 2)² = 7，即 (x – 2)² = 7/2。因此 x = 2 ± √(7/2)。

6. The Discriminant | 判别式

The expression b² – 4ac is called the discriminant. It tells us the number of real roots.

表达式 b² – 4ac 称为判别式。它告诉我们实根的个数。

If b² – 4ac > 0, the equation has two distinct real roots.

如果 b² – 4ac > 0，方程有两个不同的实根。
If b² – 4ac = 0, the equation has exactly one real root (a repeated root).

如果 b² – 4ac = 0，方程有一个实根（重根）。
If b² – 4ac < 0, the equation has no real roots.

如果 b² – 4ac < 0，方程没有实根。

For example, for x² – 4x + 4 = 0, b² – 4ac = 16 – 16 = 0, so it has one repeated root, x = 2.

例如，对于 x² – 4x + 4 = 0，b² – 4ac = 16 – 16 = 0，所以有一个重根 x = 2。

7. Solving Quadratics by Drawing Graphs | 图像法求解

The solutions of ax² + bx + c = 0 are the x-coordinates of the points where the graph y = ax² + bx + c crosses the x-axis (where y = 0).

ax² + bx + c = 0 的解，是抛物线 y = ax² + bx + c 与 x 轴交点的 x 坐标（此时 y = 0）。

To solve graphically, plot the parabola and read off the x-axis intercepts. If the graph does not intersect the x-axis, the equation has no real solutions.

要图像求解，画出抛物线并读出与 x 轴的交点。如果图像不与 x 轴相交，则方程没有实数解。

For example, the graph of y = x² – 2x – 3 crosses the x-axis at x = -1 and x = 3, so the equation x² – 2x – 3 = 0 has solutions x = -1 and x = 3.

例如，y = x² – 2x – 3 的图像与 x 轴交于 x = -1 和 x = 3，所以方程 x² – 2x – 3 = 0 的解为 x = -1 和 x = 3。

8. The Turning Point and Line of Symmetry | 顶点与对称轴

For a quadratic y = ax² + bx + c, the turning point (vertex) is found by completing the square or using the formula x = -b/(2a).

对于二次函数 y = ax² + bx + c，顶点可以通过配方法或公式 x = -b/(2a) 求出。

For y = x² – 4x + 1, completing the square gives y = (x – 2)² – 3, so the vertex is (2, -3) and the line of symmetry is x = 2.

对于 y = x² – 4x + 1，配方法得到 y = (x – 2)² – 3，所以顶点为 (2, -3)，对称轴为 x = 2。

If a > 0, the turning point is a minimum. If a < 0, it is a maximum.

如果 a > 0，顶点是最小值点；如果 a < 0，顶点是最大值点。

9. Word Problems Involving Quadratics | 涉及二次方程的应用题

Many exam problems require setting up a quadratic equation from a real-life situation. Read the problem carefully, define a variable, and translate the conditions into an equation.

许多考试题目要求从实际情况中建立二次方程。仔细阅读题目，设变量，并将条件转化为方程。

Example: The area of a rectangle is 40 cm². Its length is 3 cm longer than its width. Let the width be x cm. Then the length is (x + 3) cm, so x(x + 3) = 40, which gives x² + 3x – 40 = 0. Factorise: (x + 8)(x – 5) = 0, so x = -8 or x = 5. Since a width cannot be negative, x = 5, and the length is 8 cm.

例：一个矩形的面积为 40 cm²。它的长比宽长 3 cm。设宽为 x cm，则长为 (x + 3) cm，因此 x(x + 3) = 40，即 x² + 3x – 40 = 0。因式分解得 (x + 8)(x – 5) = 0，所以 x = -8 或 x = 5。因为宽度不能为负，所以 x = 5，长为 8 cm。

10. Common Mistakes and Exam Tips | 常见错误和考试技巧

Always rearrange the equation into the form ax² + bx + c = 0 before factorising or using the formula.

在使用因式分解或公式法之前，务必把方程整理成 ax² + bx + c = 0 的形式。
Do not forget to divide by the coefficient of x² when using the formula.

使用公式法时，别忘了除以 x² 的系数。
Check your solutions by substituting them back into the original equation.

将解代回原方程中进行检验。
When a quadratic has no real roots, do not try to factorise it over real numbers.

当二次方程没有实数根时，不要尝试在实数范围内因式分解。
When using a calculator, avoid rounding intermediate answers; give exact values where required.

使用计算器时，避免对中间结果进行四舍五入；在需要时给出精确值。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Quadratic Equations | 解一元二次方程