📚 IGCSE Maths: Solving Quadratic Equations | IGCSE 数学：二次方程求解
Quadratic equations appear throughout the IGCSE Mathematics course, from algebra and graphs to area and motion problems. Mastering the main solution methods gives you a reliable toolkit for both Core and Extended papers.
二次方程贯穿 IGCSE 数学课程的各个部分，从代数、图像到面积和运动问题。掌握主要解法能为你应对 Core 和 Extended 试卷提供可靠的工具。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation in one variable is any equation that can be rearranged into the standard form
一元二次方程是任何可以整理成标准形式的方程
ax² + bx + c = 0
where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which is why it is called quadratic.
其中 a、b、c 是常数，且 a ≠ 0。变量 x 的最高次数是 2，因此称为二次方程。
For example, x² + 5x + 6 = 0, 2x² − 3x + 1 = 0 and x² − 9 = 0 are all quadratic equations.
例如，x² + 5x + 6 = 0、2x² − 3x + 1 = 0 和 x² − 9 = 0 都是二次方程。
2. Recognising Quadratic Equations | 识别二次方程
Not every equation that contains x² is already in standard form. You may need to expand brackets, collect like terms and move all terms to one side first.
并不是所有含有 x² 的方程都已经处于标准形式。你可能需要先展开括号、合并同类项，并把所有项移到一边。
For example, 3x(x − 2) = 5 expands to 3x² − 6x = 5, then becomes 3x² − 6x − 5 = 0 in standard form.
例如，3x(x − 2) = 5 展开后得到 3x² − 6x = 5，再整理成标准形式 3x² − 6x − 5 = 0。
Some equations may look quadratic but are actually linear after simplifying. For instance, x(x + 1) = x² + 3 simplifies to x = 3, so the x² terms cancel.
有些方程看起来是二次方程，但化简后实际上是一次方程。例如 x(x + 1) = x² + 3 化简为 x = 3，x² 项抵消了。
Always check whether a ≠ 0 after simplification before using a quadratic method.
化简后一定要检查 a ≠ 0，然后再使用二次方程的解法。
3. Solving by Factorising: The Zero Product Property | 因式分解法：零乘积性质
Factorising is often the fastest method when a quadratic equation has simple factors. It relies on the zero product property:
当二次方程有简单的因式时，因式分解法通常是最快的方法。它依赖于零乘积性质：
If pq = 0, then p = 0 or q = 0.
This means if the product of two expressions is zero, at least one of the expressions must be zero.
这意味着如果两个表达式的乘积为零，那么其中至少有一个表达式必须为零。
To solve by factorising, first write the equation in standard form, factorise the left side, then set each factor equal to zero and solve.
要用因式分解法求解，先把方程写成标准形式，对左边进行因式分解，然后令每个因式等于零并求解。
4. Factorising When a = 1 | 当 a = 1 时的因式分解
For a quadratic of the form x² + bx + c, look for two numbers p and q such that
对于形如 x² + bx + c 的二次式，寻找两个数 p 和 q，使得
p + q = b and pq = c.
Then the factorised form is (x + p)(x + q).
那么因式分解形式就是 (x + p)(x + q)。
Example: x² + 7x + 12 = (x + 3)(x + 4), because 3 + 4 = 7 and 3 × 4 = 12.
例如：x² + 7x + 12 = (x + 3)(x + 4)，因为 3 + 4 = 7 且 3 × 4 = 12。
To solve x² + 7x + 12 = 0, set x + 3 = 0 and x + 4 = 0. This gives x = −3 or x = −4.
要解 x² + 7x + 12 = 0，令 x + 3 = 0 和 x + 4 = 0。得到 x = −3 或 x = −4。
For x² − 5x + 6 = 0, the numbers are −2 and −3, so (x − 2)(x − 3) = 0 and x = 2 or x = 3.
对于 x² − 5x + 6 = 0，这两个数是 −2 和 −3，所以 (x − 2)(x − 3) = 0，得到 x = 2 或 x = 3。
5. Factorising When a ≠ 1 | 当 a ≠ 1 时的因式分解
When the coefficient of x² is not 1, you can use the grouping method. For ax² + bx + c, find two numbers whose product is ac and whose sum is b.
当 x² 的系数不是 1 时，可以使用分组法。对于 ax² + bx + c，找到两个数，使它们的乘积为 ac，和为 b。
Then split the middle term into those two terms, factor in pairs, and take out the common binomial factor.
然后把中间项拆成这两项，分组提取公因式，并提取公因式二项式。
Example: Solve 2x² + 7x + 3 = 0. Here ac = 2 × 3 = 6, and we need two numbers with product 6 and sum 7: they are 6 and 1.
例如：解 2x² + 7x + 3 = 0。这里 ac = 2 × 3 = 6，需要两个数的乘积为 6，和为 7：它们是 6 和 1。
Rewrite as 2x² + 6x + x + 3 = 0, then group: 2x(x + 3) + 1(x + 3) = 0, so (2x + 1)(x + 3) = 0.
改写成 2x² + 6x + x + 3 = 0，然后分组：2x(x + 3) + 1(x + 3) = 0，所以 (2x + 1)(x + 3) = 0。
The solutions are x = −1/2 or x = −3.
解为 x = −1/2 或 x = −3。
6. Completing the Square | 配方法
Completing the square is useful when factorising is not obvious. The key identity is
当因式分解不明显时，配方法很有用。关键恒等式是
x² + bx = (x + b/2)² − (b/2)²
This rewrites part of the quadratic as a perfect square, which can then be solved by taking square roots.
这会把二次式的一部分改写为完全平方，然后通过开平方来求解。
Example: Solve x² + 6x + 5 = 0. First move the constant: x² + 6x = −5.
例如：解 x² + 6x + 5 = 0。先移常数项：x² + 6x = −5。
Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4, so (x + 3)² = 4.
两边加上 (6/2)² = 9：x² + 6x + 9 = 4，所以 (x + 3)² = 4。
Take the square root of both sides: x + 3 = ±2, giving x = −1 or x = −5.
两边开平方：x + 3 = ±2，得到 x = −1 或 x = −5。
If the coefficient of x² is not 1, divide every term by that coefficient before completing the square.
如果 x² 的系数不是 1，在配方前先每一项都除以该系数。
7. The Quadratic Formula | 求根公式
The quadratic formula solves every quadratic equation in standard form. It is derived by completing the square on ax² + bx + c = 0.
求根公式可以求解标准形式下的任何二次方程。它是通过对 ax² + bx + c = 0 配方法推导出来的。
x = (−b ± √(b² − 4ac)) ÷ (2a)
Here a is the coefficient of x², b is the coefficient of x, and c is the constant term.
这里 a 是 x² 的系数，b 是 x 的系数，c 是常数项。
Example: Solve 2x² − 4x − 6 = 0. Here a = 2, b = −4, c = −6.
例如：解 2x² − 4x − 6 = 0。这里 a = 2，b = −4，c = −6。
Substitute into the formula: x = (4 ± √((−4)² − 4 × 2 × (−6))) ÷ (2 × 2) = (4 ± √(16 + 48)) ÷ 4 = (4 ± 8) ÷ 4.
代入公式：x = (4 ± √((−4)² − 4 × 2 × (−6))) ÷ (2 × 2) = (4 ± √(16 + 48)) ÷ 4 = (4 ± 8) ÷ 4。
This gives x = 3 or x = −1.
得到 x = 3 或 x = −1。
8. The Discriminant and Nature of Roots | 判别式与根的性质
The expression under the square root in the quadratic formula is called the discriminant and is written as Δ.
求根公式中平方根号下的表达式称为判别式，记作 Δ。
Δ = b² − 4ac
The discriminant tells you how many real roots the quadratic equation has before you solve it.
判别式可以在求解之前告诉你二次方程有多少个实数根。
| Discriminant | 判别式 | Nature of roots | 根的性质 |
|---|---|
| Δ > 0 | Two distinct real roots | 两个不同的实数根 |
| Δ = 0 | One repeated real root | 一个重复的实数根 |
| Δ < 0 | No real roots | 没有实数根 |
Example: For 3x² − 2x + 5 = 0, Δ = (−2)² − 4 × 3 × 5 = 4 − 60 = −56. Since Δ < 0, the equation has no real roots.
例如：对于 3x² − 2x + 5 = 0，Δ = (−2)² − 4 × 3 × 5 = 4 − 60 = −56。因为 Δ < 0，方程没有实数根。
9. Choosing the Best Method | 选择最佳解法
Different methods suit different questions. Factorising is fastest when the factors are obvious, completing the square is useful for finding vertex form or exact simplified answers, and the quadratic formula always works.
不同方法适用于不同题目。当因式很明显时，因式分解法最快；配方法适用于求顶点形式或精确化简的答案；求根公式始终有效。
| Method | 方法 | Best use | 最佳使用场景 |
|---|---|
| Factorising | 因式分解 | Simple integer factors | 简单的整数因式 |
| Completing the square | 配方法 | 更多咨询请联系16621398022（同微信）
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