📚 Mastering Quadratic Equations | 掌握二次方程
Quadratic equations are a central topic in IGCSE Mathematics. Understanding how to solve and interpret them is essential for success in both Paper 2 and Paper 4. This guide provides a comprehensive review of the concepts, techniques, and common pitfalls you need to master.
二次方程是 IGCSE 数学的核心内容。理解如何求解和解释二次方程,对 Paper 2 和 Paper 4 的成功都至关重要。本指南全面复习相关概念、技巧和常见易错点,帮助你扎实掌握。
1. What is a Quadratic Equation? | 什么是二次方程?
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. Its standard form is:
二次方程是次数为 2 的多项式方程,即变量的最高次数为 2。其标准形式为:
ax² + bx + c = 0
Here, a, b and c are real numbers, and a ≠ 0. If a = 0, the equation becomes linear.
其中 a、b 和 c 是实数,且 a ≠ 0。如果 a = 0,方程就变成一次方程。
-
The equation is called “quadratic” because the highest exponent is 2, and “quad” historically relates to squares.
方程被称为“二次”是因为最高指数为 2,而 “quad” 在历史上与正方形相关。
-
A quadratic equation may have two distinct real roots, one repeated real root, or no real roots, depending on the discriminant.
二次方程可能有两个不同的实数根、一个重根或无实数根,这取决于判别式。
2. Expanding and Factorising | 展开与因式分解
Before solving a quadratic equation, you often need to expand or factorise quadratic expressions. Expanding means removing brackets, while factorising means writing the expression as a product of two linear factors.
在求解二次方程之前,通常需要展开或因式分解二次表达式。展开是指去掉括号,因式分解是指将表达式写成两个一次因式的乘积。
For example, the expression (x + 3)(x − 2) expands to x² + x − 6.
例如,表达式 (x + 3)(x − 2) 展开后得到 x² + x − 6。
Conversely, x² + x − 6 can be factorised back to (x + 3)(x − 2).
反过来,x² + x − 6 可以因式分解为 (x + 3)(x − 2)。
-
The product of two binomials (ax + b)(cx + d) expands according to the distributive law.
两个二项式 (ax + b)(cx + d) 的乘积按分配律展开。
-
When factorising x² + bx + c, look for two integers whose sum is b and whose product is c.
因式分解 x² + bx + c 时,寻找两个整数,使其和为 b、积为 c。
(x + 3)(x − 2) = x² + x − 6
3. Solving by Factorisation | 因式分解法求解
The factorisation method uses the zero-product property: if the product of two expressions is zero, then at least one of the factors must be zero.
因式分解法利用零乘积性质:若两个表达式的乘积为零,则至少一个因子必须为零。
Example: Solve x² − 4x − 5 = 0.
示例:解方程 x² − 4x − 5 = 0。
-
Factorise: (x − 5)(x + 1) = 0
因式分解:(x − 5)(x + 1) = 0
-
Set each factor to zero: x − 5 = 0 or x + 1 = 0
令每个因子为零:x − 5 = 0 或 x + 1 = 0
-
Solve: x = 5 or x = −1
求解:x = 5 或 x = −1
Always check your solutions by substituting them back into the original equation.
务必通过代回原方程来检验解。
Not all quadratic expressions can be factorised easily using integers; in those cases, use the quadratic formula.
并非所有二次表达式都能用整数轻松因式分解;此时应使用求根公式。
4. The Quadratic Formula | 求根公式
The quadratic formula gives the roots of any quadratic equation ax² + bx + c = 0. It is derived from the method of completing the square and is valid for all values of a ≠ 0.
求根公式给出任意二次方程 ax² + bx + c = 0 的根。它由配方法推导而来,对所有 a ≠ 0 均适用。
x = (−b ± √(b² − 4ac)) / (2a)
To use the formula, identify a, b and c from the equation, substitute them into the formula, and simplify.
使用公式时,先识别方程中的 a、b 和 c,代入公式并化简。
Example: Solve 2x² + 3x − 2 = 0.
示例:解方程 2x² + 3x − 2 = 0。
Here a = 2, b = 3, c = −2. Then:
这里 a = 2,b = 3,c = −2。于是:
x = (−3 ± √(3² − 4×2×(−2))) / (2×2) = (−3 ± √25) / 4
So x = (−3 + 5)/4 = 0.5 or x = (−3 − 5)/4 = −2.
因此 x = (−3 + 5)/4 = 0.5 或 x = (−3 − 5)/4 = −2。
5. The Discriminant and Nature of Roots | 判别式与根的性质
The expression b² − 4ac inside the quadratic formula is called the discriminant, often denoted by Δ.
求根公式中的 b² − 4ac 称为判别式,通常用 Δ 表示。
| Discriminant (Δ) | Nature of roots | 图形含义 |
| Δ > 0 | Two distinct real roots | 抛物线交 x 轴于两个不同点 |
| Δ = 0 | One repeated real root | 抛物线切 x 轴于一点(顶点) |
| Δ < 0 | No real roots | 抛物线不交 x 轴 |
It is important to distinguish between “real roots” and “no real roots” because the quadratic formula involves the square root of the discriminant; if Δ is negative, the square root is not a real number.
区分“实数根”和“无实数根”很重要,因为求根公式包含判别式的平方根;若 Δ 为负,则平方根不是实数。
In IGCSE, you are usually only asked to work with real roots and to state the number of roots.
在 IGCSE 中,通常只要求处理实数根,并说明根的个数。
6. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form a(x − h)² + k. This form reveals the vertex of the parabola and is useful for proving the quadratic formula.
配方法将二次表达式写成 a(x − h)² + k 的形式。这种形式能显示抛物线的顶点,并用于推导求根公式。
Example: Complete the square for x² + 6x + 8.
示例:将 x² + 6x + 8 配方。
Take half of 6, square it: (6/2)² = 9. Then:
取 6 的一半,再平方:(6/2)² = 9。于是:
x² + 6x + 8 = (x + 3)² − 9 + 8 = (x + 3)² − 1
So the vertex of y = x² + 6x + 8 is at (−3, −1).
因此 y = x² + 6x + 8 的顶点为 (−3, −1)。
-
The general form is: x² + bx = (x + b/2)² − (b/2)².
一般形式为:x² + bx = (x + b/2)² − (b/2)²。
-
For an expression with a coefficient of x² that is not 1, factor out a first.
对于 x² 系数不为 1 的表达式,先提出 a。
7. Graphs of Quadratic Functions | 二次函数图像
The graph of a quadratic function y = ax² + bx + c is a parabola. The sign of a determines the direction of the curve: if a > 0, it opens upwards; if a < 0, it opens downwards.
二次函数 y = ax² + bx + c 的图像是抛物线。a 的符号决定曲线的开口方向:若 a > 0,开口向上;若 a < 0,开口向下。
| 特征 | 公式/说明 |
| 对称轴 (axis of symmetry) | x = −b/(2a) |
| 顶点 (vertex) | (−b/(2a), f(−b/(2a))) |
| y 截距 | (0, c) |
| x 截距(根) | 由求解 ax² + bx + c = 0 得到 |
-
If the discriminant is positive, there are two x-intercepts; if it is zero, the vertex touches the x-axis; if negative, there are no x-intercepts.
若判别式为正,则有两个 x 截距;若为零,则顶点接触 x 轴;若为负,则没有 x 截距。
-
You may be asked to sketch the graph, clearly labelling the vertex, intercepts, and axis of symmetry.
题目可能要求画示意图,并清晰标出顶点、截距和对称轴。
8. Vertex Form and Transformations | 顶点式与图像变换
The vertex form y = a(x − h)² + k makes it easy to read the vertex (h, k). It also shows how the graph is transformed from the basic parabola y = x².
顶点式 y = a(x − h)² + k 可以轻松读出顶点 (h, k)。它还显示图像如何从基本抛物线 y = x² 变换而来。
-
If a is positive and greater than 1, the parabola is stretched vertically; if between 0 and 1, it is compressed.
若 a 为正且大于 1,抛物线纵向拉长;若在 0 和 1 之间,则被压缩。
-
The term (x − h) shifts the graph horizontally: positive h moves it right, negative h moves it left.
项 (x − h) 使图像水平平移:正 h 向右移,负 h 向左移。
-
The constant k shifts the graph vertically: positive k moves it up, negative k moves it down.
常数 k 使图像垂直平移:正 k 向上移,负 k 向下移。
To convert from standard form to vertex form, use completing the square.
要将标准式化为顶点式,可以使用配方法。
9. Applications and Problem Solving | 应用与实际问题
Quadratic equations often model real-world situations such as projectile motion, area problems, and revenue optimization. You may need to form a quadratic equation from a word problem and then solve it.
二次方程常用来模拟现实情境,如抛体运动、面积问题和收益最优化。你可能需要从文字题中建立二次方程并求解。
Example: The area of a rectangle is 36 cm². Its length is 5 cm more than its width. Find the width.
示例:一个矩形的面积为 36 平方厘米,长比宽多 5 厘米,求宽。
Let the width be x cm. Then the length is (x + 5) cm. So x(x + 5) = 36, giving x² + 5x − 36 = 0.
设宽为 x 厘米,则长为 (x + 5) 厘米。于是 x(x + 5) = 36,即 x² + 5x − 36 = 0。
Factorise: (x + 9)(x − 4) = 0, so x = −9 (rejected) or x = 4. The width is 4 cm.
因式分解:(x + 9)(x − 4) = 0,所以 x = −9(舍去)或 x = 4。宽为 4 厘米。
-
Always interpret the solutions in the context of the problem and discard values that do not make sense (like negative lengths).
始终在问题情境中解读解,并舍去没有意义的解(如负长度)。
-
Show clearly how you form the equation and label your final answer with units.
清晰展示方程的建立过程,并在最终答案中注明单位。
10. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Many students lose marks on quadratic equations due to avoidable mistakes. Here are some common pitfalls to avoid.
许多学生在二次方程上因可避免的错误而失分。以下是一些常见陷阱,需注意避免。
-
Forgetting to set the equation to zero before factorising or using the formula. Quadratic equations must be in the form ax² + bx + c = 0.
在因式分解或使用求根公式前,忘记将方程整理为零。二次方程必须化为 ax² + bx + c = 0 的形式。
-
Sign errors when substituting negative values into the quadratic formula. Always use brackets when substituting.
将负值代入求根公式时出现符号错误。代入时一定要加括号。
-
Confusing the direction of the parabola when a is negative. Remember that a < 0 opens downwards.
当 a 为负时,混淆抛物线开口方向。记住 a < 0 开口向下。
-
Not checking whether a solution is valid in the original equation or context (e.g., rejecting negative lengths).
未检查解在原方程或实际情境中是否有效(如舍去负长度)。
For Paper 2, you may need to solve quadratic equations using a calculator’s polynomial solver if allowed, but you must still know the algebraic methods for non-calculator questions.
在 Paper 2 中,如果允许使用计算器,你可能需要利用计算器的多项式求解功能,但在非计算器题目中仍必须掌握代数方法。
11. Practice Questions | 练习题
Test your understanding with these short questions. Solve each equation and sketch the corresponding graph where possible.
用以下短题测试你的理解。解每个方程,并在可能的情况下画出相应图像。
-
Solve x² − 7x + 10 = 0 by factorisation.
用因式分解法解 x² − 7x + 10 = 0。
-
Use the quadratic formula to solve 3x² + x − 2 = 0.
用求根公式解 3x² + x − 2 = 0。
-
Find the discriminant of 4x² − 4x + 1 = 0 and state the number of real roots.
求 4x² − 4x + 1 = 0 的判别式,并说明实数根的个数。
-
Complete the square for x² − 6x + 11 and write the vertex of y = x² − 6x + 11.
将 x² − 6x + 11 配方,并写出 y = x² − 6x + 11 的顶点。
-
A right-angled triangle has hypotenuse 13 cm and one leg 5 cm. Find the length of the other leg (use a quadratic equation).
一个直角三角形斜边为 13 厘米,一条直角边为 5 厘米。求另一条直角边的长度(用二次方程)。
Answers: 1) x = 2, 5; 2) x = 0.5 or x = −2; 3) Δ = 0, one repeated root; 4) (x − 3)² + 2, vertex (3, 2); 5) 12 cm (by Pythagoras: x² + 25 = 169).
答案:1) x = 2, 5;2) x = 0.5 或 x = −2;3) Δ = 0,一个重根;4) (x − 3)² + 2,顶点 (3, 2);5) 12 厘米(根据勾股定理:x² + 25 = 169)。
12. Summary | 总结
In this review, we covered the standard form of a quadratic equation, factorisation, the quadratic formula, the discriminant, completing the square, graph sketching, transformations, real-world applications, and common pitfalls.
本复习涵盖了二次方程的标准形式、因式分解、求根公式、判别式、配方法、图像绘制、图像变换、实际应用和常见错误。
Remember to practise solving quadratic equations fluently using all methods, and always check your answers. With regular practice, you will approach any quadratic problem with confidence.
请记住流畅地使用所有方法练习解二次方程,并随时检查答案。通过定期练习,你将自信地应对任何二次方程问题。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022(同微信)
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导,国外大学本科硕士研究生博士课程论文辅导