📚 Mastering Quadratic Equations | 掌握二次方程
Quadratic equations are one of the most fundamental topics in IGCSE Mathematics. Whether you are aiming for a grade C or an A*, understanding how to solve and apply these equations is essential. This guide covers every method you need, with clear steps and worked examples.
二次方程是 IGCSE 数学中最基础且最重要的内容之一。无论你的目标是 C 等级还是 A*，掌握二次方程的解法与应用都至关重要。本指南将涵盖你需要的一切方法，并通过清晰步骤和例题帮助你理解。
1. What Is a Quadratic Equation? | 什么是二次方程？
and c are constants, and a ≠ 0. The graph of a quadratic function is a curve called a parabola, which opens upwards when a > 0 and downwards when a < 0.
二次方程是次数为 2 的多项式方程。其一般形式为 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。二次函数的图像是一条称为抛物线的曲线，当 a > 0 时开口向上，当 a < 0 时开口向下。
ax² + bx + c = 0 (a ≠ 0)
Every quadratic equation has at most two solutions, also called roots. These roots may be real and distinct, real and equal, or complex — but in IGCSE you will focus on real roots.
每个二次方程至多有两个解，也称为根。这些根可能是两个不同的实数根、两个相等的实数根，或复数根——但在 IGCSE 中你只需关注实数根。
2. Solving by Factorisation | 因式分解法
rewrite the equation in the form (px + q)(rx + s) = 0, then set each factor equal to zero.
当二次表达式容易分解时，因式分解是最快捷的方法。关键思路是将方程改写为 (px + q)(rx + s) = 0 的形式，然后令每个因式分别等于零。
Step-by-step process:
步骤详解：
- Step 1: Rearrange the equation so one side is zero.
- 第一步：整理方程，使一侧为零。
- Step 2: Factorise the quadratic expression.
- 第二步：对二次表达式进行因式分解。
- Step 3: Set each bracket equal to zero and solve.
- 第三步：令每个括号等于零并求解。
Example: Solve x² − 5x + 6 = 0.
示例：解 x² − 5x + 6 = 0。
Factorise: (x − 2)(x − 3) = 0. Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.
因式分解：(x − 2)(x − 3) = 0。因此 x − 2 = 0 或 x − 3 = 0，得 x = 2 或 x = 3。
3. Solving by Completing the Square | 配方法
Completing the square is a powerful technique that can solve any quadratic equation. It also helps you identify the vertex of the parabola. The goal is to express ax² + bx + c in the form a(x − h)² + k.
配方是一种强大的技巧，可以解任何二次方程。它还能帮助你确定抛物线的顶点。目标是将 ax² + bx + c 表达为 a(x − h)² + k 的形式。
For a quadratic x² + bx, add and subtract (b/2)² to complete the square.
对于 x² + bx，加上和减去 (b/2)² 来完成配方。
x² + bx = (x + b/2)² − (b/2)²
Worked example: Solve x² + 6x − 7 = 0 by completing the square.
例题：用配方法解 x² + 6x − 7 = 0。
Step 1: Move constant to the right side:
第一步：将常数移到右边：
x² + 6x = 7
Step 2: Add (6/2)² = 9 to both sides:
第二步：两边同时加上 (6/2)² = 9：
x² + 6x + 9 = 16 → (x + 3)² = 16
Step 3: Take square roots: x + 3 = ±4 → x = 1 or x = −7.
第三步：开平方：x + 3 = ±4 → x = 1 或 x = −7。
4. The Quadratic Formula | 二次公式
ormula gives both roots directly. This method always works, even when factorisation fails.
对于任何二次方程 ax² + bx + c = 0，公式可以直接给出两个根。即使因式分解无法使用，这个方法也始终有效。
x = (−b ± √b² − 4ac) / 2a
You are expected to memorise this formula for the IGCSE exam. The expression under the square root, b² − 4ac, is called the discriminant.
IGCSE 考试要求你熟记此公式。根号内的表达式 b² − 4ac 称为判别式。
Example: Solve 2x² − 3x − 2 = 0 using the formula.
示例：用公式解 2x² − 3x − 2 = 0。
Here a = 2, b = −3, c = −2. Substitute:
这里 a = 2，b = −3，c = −2。代入：
x = (3 ± √9 + 16) / 4 = (3 ± 5) / 4
Therefore x = 2 or x = −0.5.
因此 x = 2 或 x = −0.5。
5. The Discriminant (b² − 4ac) | 判别式
The discriminant tells us about the nature of the roots without solving the full equation — this is a favourite exam question type.
判别式可以让我们在不完整解方程的情况下判断根的性质——这是考试中非常常见的一类问题。
- If b² − 4ac > 0: two distinct real roots.
- 若 b² − 4ac > 0：有两个不同的实数根。
- If b² − 4ac = 0: two equal real roots (one repeated root).
- 若 b² − 4ac = 0：有两个相等的实数根（一个重根）。
- If b² − 4ac < 0: no real roots (complex roots).
- 若 b² − 4ac < 0：没有实数根（复数根）。
Example: Determine the nature of the roots of x² + 4x + 5 = 0.
示例：判断 x² + 4x + 5 = 0 的根的性质。
a = 1, b = 4, c = 5. Discriminant = 16 − 20 = −4 < 0. Therefore no real roots.
a = 1，b = 4，c = 5。判别式 = 16 − 20 = −4 < 0。因此没有实数根。
6. Sum and Product of Roots | 根的和与积
If α and β are the roots of ax² + bx + c = 0, then:
若 α 和 β 是 ax² + bx + c = 0 的两个根，则：
Sum of roots: α + β = −b/a
Product of roots: αβ = c/a
These relationships let you write a quadratic equation given its roots, or find symmetric expressions without solving the equation.
这些关系让我们能够在已知根的情况下写出二次方程，或者在不解方程的情况下求关于根的对称表达式。
Example: Find the quadratic equation with roots 3 and −2.
示例：求以 3 和 −2 为根的二次方程。
Sum = 1, product = −6. Equation: x² − (sum)x + (product) = 0 → x² − x − 6 = 0.
和 = 1，积 = −6。方程：x² − (和)x + (积) = 0 → x² − x − 6 = 0。
7. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities are solved by first finding the critical values (the roots of the corresponding quadratic equation), then testing intervals or sketching the parabola.
解二次不等式首先需要找到临界值（即对应二次方程的根），然后通过测试区间或画抛物线图来确定解集。
Example: Solve x² − x − 6 < 0.
示例：解 x² − x − 6 < 0。
= −2. Since the parabola opens upward, the inequality holds between the roots: −2 < x < 3.
因式分解得 (x − 3)(x + 2) < 0。根为 x = 3 和 x = −2。由于抛物线开口向上，不等式在两根之间成立：−2 < x < 3。
8. Graphs of Quadratic Functions | 二次函数图像
. The vertex (turning point) depends on a and the completed-square form. The y-intercept is at (0, c), and the x-intercepts are the roots.
y = ax² + bx + c 的图像是抛物线。顶点（转向点）取决于 a 以及配方法得到的形式。y 截距为 (0, c)，x 截距是方程的根。
| Key Feature | 关键特征 | How to Find | 如何求得 |
| Vertex x-coordinate | x = −b/2a |
| y-intercept | (0, c) |
| x-intercepts | Solve ax² + bx + c = 0 |
| Vertex y-coordinate | Substitute x into equation |
When a > 0 the vertex is a minimum point; when a < 0 it is a maximum point.
当 a > 0 时顶点是最小值点；当 a < 0 时顶点是最大值点。
9. Applications: Word Problems | 应用题
Quadratic equations often appear in area, motion and number-pattern problems. The key is to translate the words into a quadratic equation first.
二次方程常出现在面积、运动和数列模式问题中。关键在于先将文字转化为二次方程。
Example: The area of a rectangle is 63 cm². Its length is 2 cm more than its width. Find the width.
示例：一个矩形的面积为 63 cm²。它的长比宽多 2 cm。求宽。
2) = 63 → x² + 2x − 63 = 0 → (x + 9)(x − 7) = 0. Since width cannot be negative, x = 7 cm.
设宽为 x，长为 x + 2。则 x(x + 2) = 63 → x² + 2x − 63 = 0 → (x + 9)(x − 7) = 0。由于宽不能为负数，所以 x = 7 cm。
10. Common Mistakes to Avoid | 常见错误提醒
Students often lose marks for simple errors. Watch out for the following:
学生常因一些简单错误丢分。请注意以下几点：
- Forgetting to rearrange the equation to zero before factorising.
- 因式分解前忘记将方程整理为等于零的形式。
- Sign errors when applying the quadratic formula.
- 代入二次公式时出现符号错误。
- Forgetting that a square root has two values (positive and negative).
- 忘记平方根有两个值（正和负）。
- Not checking solutions in word problems (negative lengths are invalid).
- 在应用题中未检验解（负长度不符合实际）。
- Dropping a negative sign when moving terms across the equals sign.
- 移项时漏掉负号。
11. Exam Tips and Last-Minute Checklist | 备考技巧与考前清单
To maximise your marks, always choose the most efficient method for each question. Use the discriminant to check your roots quickly. Show all working clearly — even if your final answer is wrong, you may earn method marks.
为了最大限度地得分，请针对每道题选择最有效的方法。使用判别式快速检查你的根是否正确。清晰展示所有步骤——即使最终答案错误，你也可能获得方法分。
- Memorise the quadratic formula.
- 熟记二次公式。
- Practise completing the square until it feels automatic.
- 反复练习配方法，直到熟练掌握。
- Sketch the parabola to visualise inequalities and ranges.
- 画抛物线草图，形象地理解不等式和解集。
- Double-check signs in a, b and c before substituting.
- 代入前仔细检查 a、b、c 的符号。
12. Practice Questions | 练习题目
Here are some questions to test your understanding. Solve them without a calculator where possible:
以下练习用于检验你的理解。尽可能不用计算器求解：
- Solve x² − 7x + 10 = 0 by factorisation. | 用因式分解法解 x² − 7x + 10 = 0。
- Solve x² + 8x + 6 = 0 by completing the square. | 用配方法解 x² + 8x + 6 = 0。
- Use the formula to solve 3x² + 5x − 2 = 0. | 用公式法解 3x² + 5x − 2 = 0。
- Find the discriminant of x² − 6x + 9 = 0 and state the nature of roots. | 求 x² − 6x + 9 = 0 的判别式并判断根的性质。
- A right-angled triangle has legs x and x + 3 with hypotenuse 15. Find x. | 直角三角形直角边为 x 和 x + 3，斜边为 15，求 x。
Answers: 1) x = 2, 5. 2) x = −4 ± √10 3) x = 1/3, −2. 4) b² − 4ac = 0, two equal real roots. 5) x = 9 (since x² + (x+3)² = 225).
答案：1) x = 2, 5。2) x = −4 ± √10。3) x = 1/3, −2。4) b² − 4ac = 0，两个相等的实数根。5) x = 9（因为 x² + (x+3)² = 225）。
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