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Solving Quadratic Equations | 解二次方程

2026-08-30 03:36 · 本文部分内容由AI辅助创作

📚 Solving Quadratic Equations | 解二次方程

A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in almost every paper, from factorisation to graphing and problem-solving. Mastering this topic will significantly boost your score.

二次方程是IGCSE数学中最重要的话题之一。它几乎出现在每一份试卷中,从因式分解到作图再到应用题。掌握好这个知识点将显著提高你的分数。


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:

二次方程是指可以写成标准形式的方程:

ax² + bx + c = 0, where a ≠ 0

The letters a, b and c are constants (numbers), and x is the unknown variable. The term ax² makes it quadratic; if a = 0, the equation becomes linear.

字母 a、b 和 c 是常数(数字),x 是未知变量。ax² 项使其成为二次方程;如果 a = 0,方程就变成了一次方程(线性方程)。

Before solving, always rearrange your equation so that one side equals zero. This is the first step in every method.

在求解之前,始终将方程整理成一边等于零的形式。这是所有解法的第一步。


2. Solving by Factorisation | 用因式分解法求解

Factorisation is the fastest method when the quadratic has simple integer factors. We look for two numbers whose product equals ac and whose sum equals b (for ax² + bx + c).

当二次方程有简单的整数因式时,因式分解是最快的方法。我们寻找两个数,使它们的乘积等于 ac,和等于 b(对于 ax² + bx + c)。

Worked Example 1 | 示例 1: Solve x² + 5x + 6 = 0.

= 0. By the zero-product rule, either x + 2 = 0 or x + 3 = 0. Therefore x = −2 or x = −3.

找两个数相乘为6、相加为5:这两个数是2和3。因此 (x + 2)(x + 3) = 0。根据零乘积法则,x + 2 = 0 或 x + 3 = 0。所以 x = −2 或 x = −3。

Worked Example 2 | 示例 2: Solve 2x² + 7x + 3 = 0.

Here ac = 0. Group: x(2x + 1) + 3(2x + 1) = 0, giving (2x + 1)(x + 3) = 0. Hence x = −½ or x = −3.

这里 ac = 2 × 3 = 6。我们需要乘积为6、和为7的两个数:1和6。拆中间项:2x² + x + 6x + 3 = 0。分组:x(2x + 1) + 3(2x + 1) = 0,得到 (2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。


3. The Quadratic Formula | 二次公式法

When factorisation is difficult or impossible, use the quadratic formula. It works for every quadratic equation.

当因式分解困难或无法进行时,使用二次公式。它适用于所有二次方程。

x = (−b ± √b² − 4ac) / (2a)

lutions directly. The symbol ± means you calculate twice: once with plus, once with minus.

这个公式直接给出解。符号 ± 表示要计算两次:一次用加号,一次用减号。

Worked Example | 示例: Solve 2x² − 4x − 3 = 0.

Here a = 2, b = −4, c = −3. Substitute: x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4. Since √40 = 2√10, we get x = (4 ± 2√10) / 4 = 1 ± (√10)/2. These are the two exact solutions.

这里 a = 2,b = −4,c = −3。代入:x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4。因为 √40 = 2√10,所以 x = (4 ± 2√10) / 4 = 1 ± (√10)/2。这就是两个精确解。


4. Completing the Square | 配方法

Completing the square rewrites the quadratic in a perfect-square form. This method is useful for finding turning points and for solving equations without the formula.

配方法将二次方程改写为完全平方形式。这一方法对于求顶点坐标以及在不使用公式的情况下解方程非常有用。

Procedure | 步骤: For x² + bx + c = 0, write:

(x + b/2)² = (b/2)² − c

Worked Example | 示例: Solve x² + 6x + 2 = 0.

Write (x + 3)² − 9 + 2 = 0 → (x + 3)² = 7. Take square roots: x + 3 = ±√7, so x = −3 ± √7.

写成 (x + 3)² − 9 + 2 = 0 → (x + 3)² = 7。两边开平方:x + 3 = ±√7,所以 x = −3 ± √7。


5. The Discriminant | 判别式

under the square root: Δ = b² − 4ac. It tells us the nature of the roots without solving.

判别式是二次公式中根号下的部分:Δ = b² − 4ac。它无需解方程就能告诉我们根的性质。

Value of Δ | Δ的值 Nature of Roots | 根的性质
Δ > 0 Two distinct real roots | 两个不同的实数根
Δ = 0 One repeated root | 一个重根(两个相等的实数根)
Δ < 0 No real roots | 无实数根

Worked Example | 示例: + 4 = 0, Δ = 2² − 4 × 3 × 4 = 4 − 48 = −44 < 0. Therefore the equation has no real roots.

示例:对于 3x² + 2x + 4 = 0,Δ = 2² − 4 × 3 × 4 = 4 − 48 = −44 < 0。因此该方程没有实数根。


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6. Graphs of Quadratic Functions | 二次函数的图像

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upward (a smile shape). If a < 0, it opens downward (a frown shape).

y = ax² + bx + c 的图像是一条抛物线。如果 a > 0,开口向上;如果 a < 0,开口向下。

Sketching the graph uses these features: the intercepts, the vertex, and the direction of opening. You do not need to plot every point.

画草图时利用这些特征:截距、顶点和开口方向。不需要逐点描图。


7. Finding the Turning Point | 求顶点(极值点)

The turning point (vertex) of a quadratic is the maximum or minimum point on the graph. Use completing the square to find it exactly.

二次函数的顶点(极值点)是图像上的最高点或最低点。使用配方法可以精确求出。

Worked Example | 示例: Find the turning point of y = x² − 4x + 5.

− 4 + 5 = (x − 2)² + 1. Since the bracket squared is never negative, the minimum value of y is 1, occurring when x = 2. Thus the vertex is (2, 1).

配方:y = (x − 2)² − 4 + 5 = (x − 2)² + 1。因为平方项永不为负,所以 y 的最小值为1,此时 x = 2。因此顶点坐标为 (2, 1)。


8. Word Problems | 应用题

Many exam questions set up a quadratic equation from a real-life context. You will need to translate the words into symbols, solve, and then interpret the answer.

许多考题会从实际情境中建立二次方程。你需要将文字转化为符号,解方程,然后解释答案。

Worked Example | 示例: A rectangle has length (x + 3) cm and width (x − 1) cm. Its area is 15 cm². Find x.

15. Expand and rearrange: x² + 2x − 3 = 15 → x² + 2x − 18 = 0. Use the formula: x = (−2 ± √4 + 72) / 2 = (−2 ± √76) / 2 = (−2 ± 2√19) / 2 = −1 ± √19. Since length cannot be negative, x = −1 + √19 ≈ 3.36 cm.

面积方程:(x + 3)(x − 1) = 15。展开并整理:x² + 2x − 3 = 15 → x² + 2x − 18 = 0。使用公式:x = (−2 ± √4 + 72) / 2 = (−2 ± √76) / 2 = (−2 ± 2√19) / 2 = −1 ± √19。由于长度不能为负,x = −1 + √19 ≈ 3.36 cm。


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

Avoid these frequent errors to secure full marks on quadratic questions.

避免以下常见错误,确保在二次方程题目上拿到满分。


10. Practice Questions | 练习题目

Try these questions on your own, then check the answers below.

先独立完成以下题目,再对照答案。

Answers | 答案:

= −3 (difference of two squares). Q2: (x + 5)(x − 2) = 0 → x = −5 or x = 2. Q3: x = (−5 ± √33 9. Q4: y = (x + 4)² − 9, so vertex is (−4, −9). Q5: Δ = 16 − 16 = 0, so one repeated root.

Q1:x = 3 或 x = −3(平方差公式)。Q2:(x + 5)(x − 2) = 0 → x = −5 或 x = 2。Q3:x = (−5 ± √33) / 4 ≈ 0.19 或 −2.69。Q4:y = (x + 4)² − 9,顶点为 (−4, −9)。Q5:Δ = 16 − 16 = 0,所以有一个重根。


11. Summary and Revision Checklist | 总结与复习清单

You should now be able to handle any quadratic equation problem in the IGCSE exam. Use this checklist to confirm your readiness.

现在你应该能够应对IGCSE考试中任何二次方程问题。使用这份清单确认你是否已准备就绪。

Review this guide before your exam, practise past paper questions on quadratics, and you will gain confidence. Remember: every quadratic can be solved, but choosing the fastest method saves valuable exam time.

考试前复习本指南,练习历年真题中的二次函数题目,你就会充满信心。记住:每个二次方程都能解出,但选择最快的方法可以节省宝贵的考试时间。


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