📚 Mastering Quadratics for Edexcel IGCSE Maths | 掌握Edexcel IGCSE数学的二次方程
Quadratics are one of the most important topics in Edexcel IGCSE Mathematics. You must be able to expand, factorise, solve, sketch and apply quadratic equations confidently. This revision guide covers every key skill you need for your exams, with clear worked examples and common pitfalls explained.
二次方程是Edexcel IGCSE数学中最重要的主题之一。你必须能够自信地展开、因式分解、求解、绘制并应用二次方程。本复习指南涵盖考试所需的每一项关键技能，并配有清晰的例题和常见错误的解释。
1. Expanding and Simplifying Quadratic Expressions | 展开与化简二次表达式
A quadratic expression is an expression where the highest power of the variable is 2. The general form is ax² + bx + c, where a, b and c are constants and a ≠ 0. Expanding means removing brackets by multiplying every term in one bracket by every term in the other bracket.
二次表达式是变量最高次数为2的表达式。一般形式为ax² + bx + c，其中a、b和c是常数，且a ≠ 0。展开意味着通过将一个括号中的每一项与另一个括号中的每一项相乘来去掉括号。
ves x² + 5x + 3x + 15, which simplifies to x² + 8x + 15. This method is often called FOIL: First, Outer, Inner, Last.
例如，展开(x + 3)(x + 5)时，需要计算x × x、x × 5、3 × x和3 × 5。得到x² + 5x + 3x + 15，化简为x² + 8x + 15。这种方法常称为FOIL法：首项、外项、内项、末项。
(ax + b)(cx + d) = acx² + (ad + bc)x + bd
ce, (a − b)² = a² − 2ab + b², and the difference of two squares, (a + b)(a − b) = a² − b².
你还应记住三种特殊的展开式：和的平方(a + b)² = a² + 2ab + b²，差的平方(a − b)² = a² − 2ab + b²，以及平方差(a + b)(a − b) = a² − b²。
Always collect like terms carefully after expanding. A common mistake is forgetting the middle term, especially when squaring a bracket such as (x + 4)². Remember that (x + 4)² is not x² + 16; it is x² + 8x + 16.
展开后一定要仔细合并同类项。一个常见错误是忘记中间项，特别是在计算(x + 4)²这类平方时。记住(x + 4)²不等于x² + 16，而等于x² + 8x + 16。
2. Factorising Quadratics | 因式分解
Factorising is the reverse of expanding. To factorise an expression means to write it as a product of brackets. For a quadratic of the form x² + bx + c, you need two numbers that multiply to give c and add to give b.
因式分解是展开的逆运算。因式分解一个表达式意味着把它写成括号相乘的形式。对于形如x² + bx + c的二次式，你需要找到两个数，它们相乘等于c，相加等于b。
+ 4). If the constant term is negative, one number is positive and the other is negative.
例如，分解x² + 7x + 12。数字3和4相乘等于12，相加等于7，所以x² + 7x + 12 = (x + 3)(x + 4)。如果常数项是负数，那么一个数为正，另一个数为负。
hod. The grouping method requires finding two numbers that multiply to a × c and add to b.
当x²的系数不为1时，因式分解会更难。对于像2x² + 7x + 3这样的二次式，你可以使用试错法或分组法。分组法需要找到两个数，它们相乘等于a × c，相加等于b。
+ x + 6x + 3, then group: x(2x + 1) + 3(2x + 1). Therefore 2x² + 7x + 3 = (2x + 1)(x + 3).
对于2x² + 7x + 3，a × c = 6，b = 7。两个数是1和6，因为1 + 6 = 7且1 × 6 = 6。改写中间项：2x² + x + 6x + 3，然后分组：x(2x + 1) + 3(2x + 1)。因此2x² + 7x + 3 = (2x + 1)(x + 3)。
Always check your factorisation by expanding the brackets. If expanding does not return the original expression, the factorisation is incorrect.
始终通过展开括号来检查你的因式分解。如果展开后不能回到原式，说明因式分解有误。
3. Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程
he equation equal zero, then factorise the left-hand side and set each factor equal to zero.
当二次方程写成ax² + bx + c = 0的形式时，可以用因式分解法求解。首先使方程等于零，然后分解左边部分，并令每个因式等于零。
ro, at least one number must be zero. Therefore x − 2 = 0 or x − 3 = 0, so x = 2 or x = 3.
例如，解方程x² − 5x + 6 = 0。它可以分解为(x − 2)(x − 3) = 0。如果两个数的乘积为零，那么至少有一个数必须为零。因此x − 2 = 0或x − 3 = 0，所以x = 2或x = 3。
ad write x² − 3x = 0 and factorise to get x(x − 3) = 0. The solutions are x = 0 and x = 3.
在因式分解之前，一定要把方程整理成标准形式。如果方程是x² = 3x，不要两边同时除以x。而应写成x² − 3x = 0，分解得到x(x − 3) = 0。解为x = 0和x = 3。
he factorisation is (x − 4)² = 0, then x = 4 is the only solution, called a repeated root.
方程可能有一个重根、两个不相等的根，或没有实数根。如果因式分解结果是(x − 4)² = 0，那么x = 4是唯一解，称为重根。
4. Solving by Completing the Square | 配方法
Completing the square is a powerful method that works for every quadratic equation, even when factorisation is difficult. The goal is to rewrite x² + bx + c in the form (x + p)² + q.
配方法是一种强大的工具，适用于所有二次方程，即使因式分解困难也可以使用。目标是将x² + bx + c改写成(x + p)² + q的形式。
x² + bx + c = (x + b/2)² − (b/2)² + c
ce (x + 3)² = x² + 6x + 9, subtract 9 to compensate. Therefore x² + 6x + 2 = (x + 3)² − 7.
例如，对x² + 6x + 2配方。取6的一半，即3，所以配方后以(x + 3)²开头。因为(x + 3)² = x² + 6x + 9，所以减去9加以补偿。因此x² + 6x + 2 = (x + 3)² − 7。
(x + 3)² − 7 = 0, add 7 to both sides, then take the square root of both sides: x + 3 = ±√7, so x = −3 ± √7.
要用配方法解二次方程，令配方后的表达式等于零并整理。对于(x + 3)² − 7 = 0，两边加7，然后两边开平方：x + 3 = ±√7，所以x = −3 ± √7。
). This is why completing the square is especially useful when sketching quadratic curves.
配方后的形式还可以揭示图像的顶点。对于y = (x + p)² + q，顶点在(−p, q)。这就是为什么在绘制二次曲线时配方法特别有用。
5. The Quadratic Formula | 求根公式
the equation does not factorise easily. You should memorise this formula for the Edexcel IGCSE exam.
求根公式可以给出任何二次方程ax² + bx + c = 0的解。当方程不容易因式分解时，它特别有用。你应该为Edexcel IGCSE考试记住这个公式。
x = (−b ± √b² − 4ac) / (2a)
2x² + 3x − 2 = 0. Here a = 2, b = 3 and c = −2. Substitute these values into the formula.
使用公式时，先从方程中确定a、b和c。例如，解2x² + 3x − 2 = 0。这里a = 2，b = 3，c = −2。将这些值代入公式。
First calculate b² − 4ac = 3² − 4 × 2 × (−2) = 9 + 16 = 25. Then x = (−3 ± √25) / 4. The two solutions are x = (−3 + 5) / 4 = 0.5 and x = (−3 − 5) / 4 = −2.
先计算b² − 4ac = 3² − 4 × 2 × (−2) = 9 + 16 = 25。然后x = (−3 ± √25) / 4。两个解分别为x = (−3 + 5) / 4 = 0.5和x = (−3 − 5) / 4 = −2。
lculator, remember to place the whole numerator in brackets. A common error is typing −3 ± √25 without brackets and dividing only part of the expression by 4.
使用计算器时，记得把整个分子放在括号里。一个常见错误是输入−3 ± √25时不加括号，导致只把表达式的一部分除以4。
6. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression b² − 4ac inside the quadratic formula is called the discriminant. It tells you how many real roots the quadratic equation has, without fully solving the equation.
求根公式中b² − 4ac这一部分称为判别式。它可以告诉你二次方程有多少个实数根，而无需完全解出方程。
| Discriminant | Nature of roots | Graph interpretation |
| b² − 4ac > 0 | Two distinct real roots | Curve crosses the x-axis twice |
| b² − 4ac = 0 | One repeated real root | Curve touches the x-axis once |
| b² − 4ac < 0 | No real roots | Curve does not intersect the x-axis |
the equation has one repeated root. Indeed, x² − 4x + 4 = (x − 2)², so the root is x = 2.
例如，方程x² − 4x + 4 = 0的判别式为(−4)² − 4 × 1 × 4 = 16 − 16 = 0。这确认方程有一个重根。事实上，x² − 4x + 4 = (x − 2)²，所以根为x = 2。
If the discriminant is a positive perfect square, the roots are rational and can usually be found by factorisation. If it is positive but not a perfect square, the roots are irrational and should be left in surd form.
如果判别式是一个正完全平方数，根是有理数，通常可以通过因式分解求出。如果是正数但不是完全平方数，根是无理数，应保留根式形式。
7. Quadratic Graphs: Key Features | 二次函数图像的关键特征
x² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
二次函数y = ax² + bx + c的图像是抛物线。如果a > 0，抛物线开口向上，有最低点。如果a < 0，开口向下，有最高点。
d the turning point. The y-intercept is found by setting x = 0, so it is the point (0, c).
每条二次曲线都有三个重要特征：y轴截距、根和顶点。y轴截距通过令x = 0得到，即点(0, c)。
bx + c = 0. If the quadratic factorises, the roots can be read directly from the factors.
根是曲线与x轴相交处的x坐标。通过解ax² + bx + c = 0求出。如果二次式可以因式分解，可以直接从因式中读出根。
e the square to write the function in the form y = (x + p)² + q, so the vertex is (−p, q).
顶点是曲线的最低点或最高点。其x坐标为x = −b/(2a)。或者，可通过配方法将函数写成y = (x + p)² + q的形式，则顶点为(−p, q)。
8. Drawing Quadratic Graphs | 绘制二次函数图像
To sketch a quadratic graph accurately, you should first find the y-intercept, the roots and the turning point. Plot these points, then draw a smooth symmetrical curve through them.
要准确绘制二次函数图像，你应先求出y轴截距、根和顶点。画出这些点，然后通过它们画一条平滑对称的曲线。
torise to find the roots: x² − 2x − 3 = (x − 3)(x + 1), so the roots are x = 3 and x = −1.
例如，绘制y = x² − 2x − 3的图像。y轴截距为(0, −3)。因式分解求根：x² − 2x − 3 = (x − 3)(x + 1)，所以根为x = 3和x = −1。
= 1² − 2 − 3 = −4. So the vertex is (1, −4). Since a = 1 > 0, the parabola opens upwards.
顶点位于两根之间的中点，即x = (−1 + 3)/2 = 1。代入x = 1得y = 1² − 2 − 3 = −4。因此顶点为(1, −4)。由于a = 1 > 0，抛物线开口向上。
When a table of values is required, choose x-values around the turning point and calculate y carefully. Plot at least five points, and check that the curve is symmetric about the vertical line through the vertex.
当题目要求列表取值时，选择顶点附近的x值并仔细计算y。至少描出五个点，并检查曲线关于通过顶点的竖直线对称。
9. Quadratic Inequalities | 二次不等式
e the quadratic and find its roots. Then use a number line or a sketch of the graph to decide which regions satisfy the inequality.
二次不等式涉及如x² − 3x − 10 < 0或2x² + 5x − 3 ≥ 0的表达式。要解它们，首先分解二次式并求根。然后使用数轴或函数图像草图判断哪些区间满足不等式。
s upwards, so the graph is below or on the x-axis between the roots. Therefore −2 ≤ x ≤ 5.
例如，解x² − 3x − 10 ≤ 0。因式分解：(x − 5)(x + 2) ≤ 0。根为x = 5和x = −2。抛物线开口向上，所以图像在两根之间位于x轴下方或与x轴相交。因此−2 ≤ x ≤ 5。
For the inequality x² − 2x − 3 > 0, factorise to get (x − 3)(x + 1) > 0. The graph is above the x-axis outside the roots, so the solution is x < −1 or x > 3.
对于不等式x² − 2x − 3 > 0，因式分解得(x − 3)(x + 1) > 0。图像在两根之外位于x轴上方，所以解为x < −1或x > 3。
A quick sign test can help: choose a value in each interval and check whether the product of the factors is positive or negative. This avoids mistakes when the leading coefficient is negative.
快速符号测试可以有所帮助：在每个区间内选择一个值，检查因式乘积为正还是为负。当首项系数为负时，这可以避免错误。
10. Applications and Exam Tips | 应用与考试提示
Quadratics appear in many real-world problems, including area, projectile motion and economics. A common question gives you a context and asks you to form a quadratic equation before solving it.
二次方程出现在许多实际问题中，包括面积、抛体运动和经济问题。常见题型是给出情境，要求你先建立二次方程再求解。
x² + 3x − 40 = 0, which factorises to (x + 8)(x − 5) = 0. Since x > 0, the width is 5 cm.
例如，矩形的面积等于长乘以宽。如果长比宽长3 cm，面积为40 cm²，设宽为x。长为x + 3，所以x(x + 3) = 40。整理得x² + 3x − 40 = 0，分解为(x + 8)(x − 5) = 0。由于x > 0，宽为5 cm。
Always reject negative or impossible solutions in context. If the question asks for a length or a time, the answer must be positive. State the final answer with the correct units.
在实际问题中，一定要舍去负数或不可能的根。如果题目问的是长度或时间，答案必须为正。最终
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