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Mastering Quadratics for Edexcel IGCSE | 掌握二次函数:Edexcel IGCSE数学复习指南

2026-08-30 00:26 · 本文部分内容由AI辅助创作

📚 Mastering Quadratics for Edexcel IGCSE | 掌握二次函数:Edexcel IGCSE数学复习指南

Quadratics are one of the most heavily tested topics in the Edexcel IGCSE Mathematics exam. From factorising to the quadratic formula, from completing the square to sketching graphs, a solid grasp of quadratics can earn you a significant number of marks across both Paper 1 and Paper 2.

二次函数是 Edexcel IGCSE 数学考试中考查频率最高的主题之一。从因式分解到求根公式,从配方法到画图,扎实掌握二次函数能帮助你在 Paper 1 和 Paper 2 中拿到大量分数。


1. The General Form of a Quadratic | 二次函数的一般形式

A quadratic expression is written in the form ax² + bx + c, where a, b, and c are constants and a ≠ 0. The highest power of x is 2, which gives the graph its characteristic U-shape.

二次表达式写作 ax² + bx + c,其中 a、b、c 为常数,且 a ≠ 0。x 的最高次数是 2,这使图像呈现特有的 U 形。

When a > 0, the parabola opens upwards. When a < 0, it opens downwards. The value of c is the y-intercept.

当 a > 0 时,抛物线开口向上;当 a < 0 时,开口向下。c 的值就是 y 轴截距。


2. Expanding and Simplifying | 展开与化简

15. Remember: multiply each term in the first bracket by each term in the second bracket.

展开意味着去掉括号。例如,(x + 3)(x + 5) = x² + 8x + 15。记住:用第一个括号中的每一项去乘以第二个括号中的每一项。

For three terms such as (x + 1)(x + 2)(x + 3), first multiply two factors, then multiply the result by the third.

对于三项如 (x + 1)(x + 2)(x + 3),先乘两个因式,再将结果乘第三个因式。

(x + 3)² = x² + 6x + 9

You should also be comfortable expanding expressions like (2x − 1)(x + 4) = 2x² + 7x − 4.

你还应熟练展开如 (2x − 1)(x + 4) = 2x² + 7x − 4 这样的表达式。


3. Factorising Quadratics | 二次因式分解

Factorising is the reverse of expanding. To factorise x² + bx + c, find two numbers that multiply to c and add to b.

因式分解是展开的逆运算。要分解 x² + bx + c,需要找到两个数,它们相乘等于 c,相加等于 b。

e −2 and −3 because (−2) × (−3) = 6 and (−2) + (−3) = −5. So x² − 5x + 6 = (x − 2)(x − 3).

例如,分解 x² − 5x + 6。这两个数是 −2 和 −3,因为 (−2) × (−3) = 6 且 (−2) + (−3) = −5。所以 x² − 5x + 6 = (x − 2)(x − 3)。

When a ≠ 1, use the method of factors of a and c, or the 'cross method'. For 2x² + 7x + 3, write (2x + 1)(x + 3).

当 a ≠ 1 时,可使用系数 a 和 c 的因数法,或交叉相乘方法。对于 2x² + 7x + 3,可写成 (2x + 1)(x + 3)。

Always check by expanding your answer.

务必通过展开来检查你的答案。


4. Solving Quadratic Equations by Factorising | 利用因式分解解二次方程

equals zero, you can solve it by factorising. If (x − p)(x − q) = 0, then x = p or x = q.

如果二次方程等于零,你可以通过因式分解来求解。若 (x − p)(x − q) = 0,则 x = p 或 x = q。

Example: solve x² − 3x − 10 = 0. Factorise: (x − 5)(x + 2) = 0. Therefore x = 5 or x = −2.

示例:解 x² − 3x − 10 = 0。因式分解得 (x − 5)(x + 2) = 0。因此 x = 5 或 x = −2。

not in the form = 0, rearrange it first. For example, x² = 4x + 5 becomes x² − 4x − 5 = 0.

如果方程不是等于 0 的形式,先移项。例如,x² = 4x + 5 变为 x² − 4x − 5 = 0。


5. The Quadratic Formula | 求根公式

n factorising is difficult or impossible, use the quadratic formula. For ax² + bx + c = 0:

当因式分解困难或无法进行时,使用求根公式。对于 ax² + bx + c = 0:

x = (−b ± √b² − 4ac) / 2a

You must memorise this formula. It is not given on the Edexcel IGCSE formula sheet.

你必须记住这个公式。Edexcel IGCSE 的公式表中不会给出它。

Example: solve 2x² + 5x − 3 = 0. Here a = 2, b = 5, c = −3. Substitute:

示例:解 2x² + 5x − 3 = 0。这里 a = 2,b = 5,c = −3。代入:

x = (−5 ± √(25 − 4×2×(−3))) / 4 = (−5 ± √49) / 4 = (−5 ± 7) / 4

So x = 2/4 = 0.5 or x = −12/4 = −3.

因此 x = 2/4 = 0.5 或 x = −12/4 = −3。


6. The Discriminant | 判别式

the square root: Δ = b² − 4ac. It tells you how many real roots a quadratic equation has.

判别式是根号内的部分:Δ = b² − 4ac。它告诉我们二次方程有多少个实数根。

Δ = b² − 4acNumber of real roots实数根个数
Δ > 0Two distinct real roots两个不同的实数根
Δ = 0One repeated real root一个重根
Δ < 0No real roots没有实数根

If the question asks for 'roots', it usually means real roots. You can also use the discriminant to determine whether a quadratic intersects, touches, or misses the x-axis.

如果题目问“根”,通常指实数根。你也可以用判别式判断二次函数的图像与 x 轴相交、相切还是相离。


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7. Completing the Square | 配方法

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

配方法将二次式改写为 a(x + p)² + q 的形式。它有助于找到顶点(转折点)和解方程。

For x² + bx + c, take half of b, square it, and adjust:

对于 x² + bx + c,取 b 的一半,平方,然后调整:

x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7

If a ≠ 1, first factor a out of the x² and x terms before completing the square.

如果 a ≠ 1,先将 a 从 x² 和 x 项中提出来,再进行配方。


8. Solving by Completing the Square | 用配方法解方程

ed square form a(x + p)² + q = 0, you can solve it by rearranging and taking square roots.

一旦二次方程成为 a(x + p)² + q = 0 的形式,你可以通过移项并开平方来求解。

Example: solve (x + 3)² − 7 = 0.

示例:解 (x + 3)² − 7 = 0。

(x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7

This gives two irrational roots. In IGCSE exams, you may leave your answer in surd form.

这给出两个无理根。在 IGCSE 考试中,答案可以保留根号形式。


9. Sketching Quadratic Graphs | 二次函数图像的草图

To sketch a quadratic graph, you need three key features: the y-intercept, the roots (if any), and the turning point.

要画二次函数图像的草图,你需要三个关键特征:y 轴截距、根(如果有)以及转折点。

3. y-intercept at (0, 3). Complete the square: (x − 2)² − 1, so the minimum point is (2, −1).

示例:画 y = x² − 4x + 3 的草图。因式分解得 (x − 1)(x − 3),所以根在 x = 1 和 x = 3。y 轴截距为 (0, 3)。配方得 (x − 2)² − 1,所以最低点为 (2, −1)。


10. Applications: Word Problems | 应用:文字题

Quadratic equations often appear in geometric or real-life problems. For example, the area of a rectangle is given by a quadratic expression.

二次方程经常出现在几何或实际生活问题中。例如,矩形的面积由二次表达式给出。

Problem: A rectangle's length is 3 cm longer than its width, and its area is 28 cm². Find its width.

问题:一个矩形的长比宽长 3 cm,面积为 28 cm²。求它的宽。

xpand: x² + 3x − 28 = 0. Factorise: (x + 7)(x − 4) = 0. Since width is positive, x = 4 cm.

设宽为 x,则长为 x + 3,且 x(x + 3) = 28。展开得 x² + 3x − 28 = 0。因式分解得 (x + 7)(x − 4) = 0。由于宽为正数,x = 4 cm。

Always reject negative solutions for lengths, times, or counts.

对于长度、时间或数量,务必舍去负数解。


11. Common Exam Traps | 常见考试陷阱

Many marks are lost through careless errors. Watch out for these common mistakes:

很多分数因粗心错误而丢失。请注意以下常见错误:


12. Practice Quick Questions | 快速练习

Try these questions to test yourself. Answers are shown below.

试试这些题目来测试自己。答案在下方。

  1. Factorise x² + 7x + 12
  2. Solve 3x² − 5x − 2 = 0
  3. Write y = x² + 8x + 5 in completed square form
  4. Find the discriminant of 2x² + 3x + 1 and state the number of real roots
  5. Sketch the graph of y = x² − 2x − 3, labelling the roots, y-intercept, and turning point

real roots 5. Roots at x = 3 and x = −1; y-intercept at (0, −3); turning point at (1, −4)

答案:1. (x + 3)(x + 4) 2. x = 2 或 x = −1/3 3. (x + 4)² − 11 4. Δ = 1,两个实数根 5. 根为 x = 3 和 x = −1;y 截距为 (0, −3);转折点为 (1, −4)


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