📚 Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学：攻克二次方程
A quadratic equation is one of the most important algebraic topics in the IGCSE Mathematics syllabus. It appears regularly in Paper 2 and Paper 4, often linked to graphs, problem solving and real-life applications. This revision guide covers the key methods, common pitfalls and exam techniques you need to master quadratics with confidence.
二次方程是 IGCSE 数学大纲中最重要的代数主题之一。它经常出现在 Paper 2 和 Paper 4 中，并与图像、问题解决和实际应用相结合。本复习指南涵盖关键方法、常见错误和考试技巧，帮助你自信地掌握二次方程。
1. What Is a Quadratic Expression? | 什么是二次式
and a ≠ 0. If it is set equal to zero, we call it a quadratic equation: ax² + bx + c = 0.
关于 x 的二次式的一般形式为 ax² + bx + c，其中 a、b、c 是常数且 a ≠ 0。如果让其等于 0，就得到二次方程：ax² + bx + c = 0。
ed a parabola. You must be able to recognise quadratic expressions, expand them, factorise them and solve the associated equations.
x 的最高次数是 2，因此 y = ax² + bx + c 的图像总是一条称为抛物线的曲线。你必须能够识别二次式、展开二次式、因式分解二次式并解相应的方程。
2. Expanding Double Brackets | 双括号展开
To expand (x + p)(x + q), multiply each term in the first bracket by each term in the second bracket. A common memory aid is FOIL: First, Outer, Inner, Last.
展开 (x + p)(x + q) 时，将第一个括号中的每一项分别乘以第二个括号中的每一项。常用的记忆方法是 FOIL：首项、外项、内项、末项。
(x + p)(x + q) = x² + (p + q)x + pq
Example: Expand (x + 3)(x − 2).
例如：展开 (x + 3)(x − 2)。
(x + 3)(x − 2) = x² − 2x + 3x − 6 = x² + x − 6
Always combine the middle terms carefully. In this example −2x + 3x simplifies to +x, which is a common place where sign errors occur.
一定要仔细合并中间项。在这个例子中，−2x + 3x 化简为 +x，这是经常出现符号错误的地方。
3. Factorising Quadratics of the Form x² + bx + c | 因式分解 x² + bx + c
umbers m and n such that m + n = b and mn = c. Then the factorised form is (x + m)(x + n).
对于 x² + bx + c，找出两个数 m 和 n，使 m + n = b 且 mn = c。那么因式分解后的形式为 (x + m)(x + n)。
d two numbers with sum 5 and product 6. They are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).
例如：因式分解 x² + 5x + 6。我们需要两个数，和为 5，积为 6。它们是 2 和 3，所以 x² + 5x + 6 = (x + 2)(x + 3)。
rs with sum −1 and product −12. The numbers are 3 and −4, so x² − x − 12 = (x + 3)(x − 4).
例如：因式分解 x² − x − 12。我们需要两个数，和为 −1，积为 −12。这两个数是 3 和 −4，所以 x² − x − 12 = (x + 3)(x − 4)。
Always expand your answer mentally to check that the middle term is correct. This will catch most sign errors before they cost marks.
始终在心中重新展开答案，检查中间项是否正确。这可以在丢分前发现大多数符号错误。
4. Factorising Quadratics of the Form ax² + bx + c | 因式分解 ax² + bx + c
When a ≠ 1, factorising is slightly harder. One reliable method is to split the middle term. Multiply a and c, then find two numbers whose product is ac and whose sum is b.
当 a ≠ 1 时，因式分解稍难一些。一种可靠的方法是拆分中间项。先求 a 与 c 的乘积，然后找出两个数，使它们的积为 ac，和为 b。
= 6. We need two numbers with product 6 and sum 7: they are 6 and 1. Split 7x into 6x + x.
例如：因式分解 2x² + 7x + 3。这里 ac = 2 × 3 = 6。我们需要两个数，积为 6，和为 7：它们是 6 和 1。把 7x 拆成 6x + x。
2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)
This method works for any factorisable quadratic. Practise it until the steps become automatic.
这种方法适用于任何可因式分解的二次式。反复练习，直到这些步骤变得熟练。
5. Solving Quadratic Equations by Factorisation | 因式分解法解二次方程
n A = 0 or B = 0. Set each bracket equal to zero and solve the resulting linear equations.
二次方程因式分解后，利用以下性质：若 AB = 0，则 A = 0 或 B = 0。令每个括号等于 0，再解出得到的一次方程。
Example: Solve x² − 5x + 6 = 0.
例如：解方程 x² − 5x + 6 = 0。
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3
After solving, substitute your answers back into the original equation to verify them. This is quick and very effective in an exam.
求解后，将答案代回原方程进行验证。这在考试中既快又非常有效。
6. Completing the Square | 配方法
Completing the square rewrites a quadratic as a squared term plus a constant. For x² + bx + c, the completed-square form is:
配方法将二次式改写为一个平方项加一个常数。对于 x² + bx + c，配方后的形式为：
x² + bx + c = (x + b/2)² − (b/2)² + c
Example: Write x² + 6x + 5 in completed-square form. Here b = 6, so b/2 = 3.
例如：将 x² + 6x + 5 写成配方的形式。这里 b = 6，所以 b/2 = 3。
x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4
)² − 4 = 0, so (x + 3)² = 4. Taking square roots gives x + 3 = ±2, hence x = −1 or x = −5.
要解 x² + 6x + 5 = 0，先写成 (x + 3)² − 4 = 0，所以 (x + 3)² = 4。开平方得 x + 3 = ±2，因此 x = −1 或 x = −5。
If a ≠ 1, factor a out of the first two terms before completing the square. This is essential for IGCSE questions that ask for the turning point or the minimum value.
如果 a ≠ 1，在配方前要先把 a 从前两项中提出来。对于要求顶点或最小值的 IGCSE 题目来说，这一点很重要。
7. The Quadratic Formula | 求根公式
ves any quadratic equation ax² + bx + c = 0, even when the quadratic cannot be factorised.
求根公式可以解任意二次方程 ax² + bx + c = 0，即使二次式不能因式分解也可以。
x = (−b ± √b² − 4ac) ÷ 2a
Example: Solve 2x² + 3x − 2 = 0 using the quadratic formula. Here a = 2, b = 3, c = −2.
例如：用求根公式解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。
x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2) = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4
This gives x = 2/4 = ½ or x = −8/4 = −2. Always simplify your final answers fully.
由此得到 x = 2/4 = ½ 或 x = −8/4 = −2。最后的答案一定要化简到最简形式。
8. The Discriminant and the Nature of Roots | 判别式与根的性质
e root: D = b² − 4ac. It tells us how many real roots the equation has without solving it.
判别式是求根公式中根号下的部分：D = b² − 4ac。它可以在不求解方程的情况下告诉我们方程有多少个实数根。
D = b² − 4ac
| Discriminant | Nature of roots | 根的性质 |
|---|---|---|
| D > 0 | Two distinct real roots | 两个不相等的实根 |
| D = 0 | One repeated real root | 一个重根（两个相等实根） |
| D < 0 | No real roots | 没有实数根 |
x + 5 = 0, D = 4² − 4 × 1 × 5 = 16 − 20 = −4. Since D < 0, the equation has no real roots.
例如：对于 x² + 4x + 5 = 0，D = 4² − 4 × 1 × 5 = 16 − 20 = −4。因为 D < 0，所以该方程没有实数根。
9. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有一个最低点。如果 a < 0，抛物线开口向下，有一个最高点。
erm c, because when x = 0, y = c. The x-intercepts are the real roots of ax² + bx + c = 0.
y 轴截距是常数项 c，因为当 x = 0 时，y = c。x 轴截距是 ax² + bx + c = 0 的实数根。
The axis of symmetry is the vertical line through the turning point. Its equation is:
对称轴是经过顶点的竖直线。它的方程为：
x = −b ÷ 2a
You may be asked to sketch a quadratic graph, so label the intercepts, axis of symmetry and turning point clearly.
题目可能会要求你画二次图像，所以要清楚地标出截距、对称轴和顶点。
10. Turning Points and Symmetry | 顶点与对称性
ordinate is x = −b ÷ 2a. Substitute this value into the equation to find the y-coordinate.
顶点是抛物线改变方向的点。它的 x 坐标为 x = −b ÷ 2a。将这个值代入方程即可求出 y 坐标。
6) ÷ (2 × 1) = 3. Then y = 3² − 6 × 3 + 8 = 9 − 18 + 8 = −1. The turning point is (3, −1).
例如：求 y = x² − 6x + 8 的顶点。x 坐标为 −(−6) ÷ (2 × 1) = 3。然后 y = 3² − 6 × 3 + 8 = 9 − 18 + 8 = −1。顶点为 (3, −1)。
² + k, the turning point is simply (h, k). This is why completing the square is so useful.
如果二次式以配方形式 y = (x − h)² + k 给出，则顶点就是 (h, k)。这就是配方法如此有用的原因。
11. Applications and Word Problems | 应用题
Quadratic equations often appear in real-life problems involving areas, products of numbers, projectile motion and business profit. Start by defining a variable and forming an equation.
二次方程经常出现在涉及面积、数字乘积、抛体运动和商业利润的实际问题中。首先设定一个变量并建立方程。
two consecutive positive integers is 42. Let the smaller integer be n. Then n(n + 1) = 42,
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