📚 Mastering Quadratic Equations for IGCSE Maths: Factorisation to Formula | 掌握 IGCSE 数学二次方程：从因式分解到求根公式
Quadratic equations appear throughout the IGCSE Maths syllabus, from pure algebra to geometry, mensuration and even coordinate geometry. Students often lose marks because they rush into solving without first writing the equation in standard form or because they forget that a quadratic can have two solutions.
二次方程贯穿 IGCSE 数学的多个板块，包括纯代数、几何、测量以及坐标几何。很多学生丢分是因为还没有把方程写成标准形式就急于求解，或者忘记二次方程可能有两个解。
1. What is a Quadratic Equation? | 什么是二次方程？
nd c are constants and a ≠ 0. If a = 0, the equation becomes linear rather than quadratic.
二次方程是次数为 2 的多项式方程，变量最高次幂是 x²。一般形式为 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。如果 a = 0，方程就变成了一次方程而不是二次方程。
In IGCSE questions, you may see quadratic equations written in different orders or with brackets. Always rearrange them into standard form before trying to solve them.
在 IGCSE 题目中，二次方程可能以不同顺序或带括号的形式出现。解题前一定要先把它整理成标准形式。
2. Standard Form and Key Terms | 标准形式与关键术语
ficient of x², b is the coefficient of x, and c is the constant term. Identifying a, b and c correctly is essential for using the quadratic formula and the discriminant.
二次方程的标准形式是 ax² + bx + c = 0。其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。准确识别 a、b、c 对于使用求根公式和判别式非常重要。
a is the number multiplying x², such as 2 in 2x² + 3x − 5 = 0.
a 是 x² 的系数，例如在 2x² + 3x − 5 = 0 中，a = 2。
b is the number multiplying x, such as 3 in the same equation.
b 是 x 的系数，例如上述方程中的 b = 3。
c is the constant term, such as −5 in the same equation.
c 是常数项，例如上述方程中的 c = −5。
t signs belong to the coefficients. In 2x² − 3x − 5 = 0, b is −3 and c is −5, not 3 and 5.
注意符号属于系数本身。在 2x² − 3x − 5 = 0 中，b = −3，c = −5，而不是 3 和 5。
3. Solving by Factorisation | 因式分解法
linear factors. The method relies on the fact that if ab = 0, then either a = 0 or b = 0.
当二次方程可以写成两个一次因式的乘积时，因式分解通常是最快的方法。它的依据是：如果 ab = 0，那么 a = 0 或 b = 0。
To solve by factorisation, first write the equation in standard form. Then factorise the left-hand side into two brackets. Set each bracket equal to zero and solve for x.
用因式分解法求解时，先写出标准形式，然后把等号左边分解成两个括号。接着令每个括号等于零并解出 x。
3) = 0. Setting each factor to zero gives x + 2 = 0, so x = −2, and x + 3 = 0, so x = −3.
例如，x² + 5x + 6 = 0 可以因式分解为 (x + 2)(x + 3) = 0。令每个因式为零，得到 x + 2 = 0，因此 x = −2；x + 3 = 0，因此 x = −3。
nd add to 7, so 2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
IGCSE 题目有时要求分解 2x² + 7x + 3 这类表达式。找出两个数，它们相乘等于 ac，相加等于 b。这里 2 × 3 = 6，1 和 6 相乘得 6 且相加得 7，因此 2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。
4. Solving by the Quadratic Formula | 求根公式法
the quadratic formula will always work. For ax² + bx + c = 0, the solutions are given by:
当因式分解难以进行或无法用简单整数分解时，求根公式总能解决问题。对于 ax² + bx + c = 0，其解为：
x = (−b ± √b² − 4ac) ÷ 2a
Identify a, b and c carefully, substitute them into the formula, and simplify. Pay special attention to negative signs: if b is negative, then −b becomes positive, and if c is negative, then −4ac becomes positive.
仔细识别 a、b、c，代入公式并化简。特别注意负号：如果 b 是负数，那么 −b 就是正数；如果 c 是负数，那么 −4ac 就是正数。
e, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substituting into the formula gives:
例如，解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入公式得：
x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2) = (−3 ± √9 + 16) ÷ 4 = (−3 ± √25) ÷ 4
Therefore x = (−3 + 5) ÷ 4 = 1/2 or x = (−3 − 5) ÷ 4 = −2.
因此 x = (−3 + 5) ÷ 4 = 1/2，或者 x = (−3 − 5) ÷ 4 = −2。
5. Completing the Square | 配方法
Completing the square is a powerful technique that rewrites a quadratic in the form (x + p)² + q. It is especially useful for finding the vertex of a quadratic graph and for solving equations that cannot be factorised neatly.
配方法是一种重要的技巧，它把二次式写成 (x + p)² + q 的形式。它特别适用于求二次函数图像的顶点，以及解无法简单因式分解的方程。
For x² + bx, the completed square form is (x + b/2)² − (b/2)². This works because expanding (x + b/2)² gives x² + bx + (b/2)², so we must subtract (b/2)² to keep the expression unchanged.
对于 x² + bx，配方形式为 (x + b/2)² − (b/2)²。这是因为展开 (x + b/2)² 得到 x² + bx + (b/2)²，因此要减去 (b/2)² 以保持原式不变。
= −2. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 7. This gives (x + 3)² = 7, so x + 3 = ±√7, giving x = −3 ± √7.
例如，用配方法解 x² + 6x + 2 = 0。先把常数项移到右边：x² + 6x = −2。两边同时加上 (6/2)² = 9：x² + 6x + 9 = 7。于是得到 (x + 3)² = 7，所以 x + 3 = ±√7，最终 x = −3 ± √7。
6. Discriminant and Nature of Roots | 判别式与根的性质
= b² − 4ac. It tells you the number and type of roots without solving the equation fully.
二次方程 ax² + bx + c = 0 的判别式为 Δ = b² − 4ac。通过判别式，无需完整解方程就能判断根的个数和类型。
Δ = b² − 4ac
| 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 |
IGCSE questions often ask you to determine whether a quadratic has two real roots, one repeated root, or no real roots. Always compute b² − 4ac rather than trying to guess from the graph.
IGCSE 题目经常要求判断二次方程是有两个实数根、一个重复根还是没有实数根。一定要计算 b² − 4ac，而不是根据图像猜测。
7. Graphical Interpretation | 图像解释
x-coordinates of the points where the parabola crosses the x-axis. These are called the roots or x-intercepts.
y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的解就是抛物线与 x 轴交点的横坐标，这些交点称为根或 x 轴截距。
has a maximum point. The vertex or turning point has x-coordinate given by x = −b ÷ (2a).
如果 a > 0，抛物线开口向上并有最低点；如果 a < 0，抛物线开口向下并有最高点。顶点或转折点的横坐标为 x = −b ÷ (2a)。
ertex touches the x-axis exactly once. If Δ < 0, the parabola never touches the x-axis.
判别式决定抛物线与 x 轴的交点情况。如果 Δ > 0，抛物线与 x 轴相交两次；如果 Δ = 0，顶点恰好接触 x 轴一次；如果 Δ < 0，抛物线与 x 轴没有交点。
8. Common IGCSE Exam Pitfalls | IGCSE 常见失分点
Many marks are lost in quadratic problems because of small but avoidable mistakes. One of the most common is forgetting to write the equation in standard form before identifying a, b and c.
在二次方程题目中，很多分数因小错误而丢失，但这些错误是可以避免的。最常见的错误之一是在识别 a、b、c 之前忘记把方程写成标准形式。
Forgetting that a ≠ 0. If a = 0, the quadratic formula cannot be applied.
忘记 a ≠ 0。如果 a = 0，就不能使用求根公式。
g the negative root when solving x² = 9. The correct solutions are x = ±3, not only x = 3.
解 x² = 9 时漏掉负根。正确的解是 x = ±3，而不只是 x = 3。
Making sign errors in the quadratic formula, especially when b or c is negative.
在求根公式中出现符号错误，尤其是当 b 或 c 为负数时。
Forgetting to set each factor equal to zero after factorising.
因式分解后忘记令每个因式等于零。
Not simplifying surds or fractions in the final answer.
最终答案没有化简根式或分数。
Always check your answers by substituting them back into the original equation. This takes a few seconds and can catch many silent errors.
一定要把答案代回原方程进行检查。这只需要几秒钟，却能发现许多不易察觉的错误。
9. Worked IGCSE-Style Example | 典型例题精讲
Question: The area of a rectangular garden is 60 m². Its length is 4 m more than its width. Find the dimensions of the garden.
题目：一个矩形花园的面积为 60 m²，它的长比宽多 4 m。求该花园的长和宽。
+ 4 metres. The area is x(x + 4) = 60. Expanding gives x² + 4x = 60, so x² + 4x − 60 = 0.
设宽为 x 米，则长为 x + 4 米。面积为 x(x + 4) = 60。展开得 x² + 4x = 60，因此 x² + 4x − 60 = 0。
0 and add to 4. The numbers 10 and −6 work, so (x + 10)(x − 6) = 0. Thus x = −10 or x = 6.
对方程进行因式分解：需要两个数相乘为 −60 且相加为 4。10 和 −6 满足条件，因此 (x + 10)(x − 6) = 0。所以 x = −10 或 x = 6。
Since a width cannot be negative, x = 6. The width is 6 m and the length is x + 4 = 10 m.
因为宽不能为负数，所以 x = 6。宽为 6 m，长为 x + 4 = 10 m。
10. Quick Revision Checklist | 快速复习清单
Before an IGCSE exam, make sure you can confidently do all of the following: write a quadratic in standard form, factorise simple and harder quadratics, use the quadratic formula correctly, complete the square, calculate and interpret the discriminant, and match a quadratic equation to its graph.
在 IGCSE 考试前，请确保你能熟练完成以下内容：将二次方程写成标准形式、分解简单和较复杂的二次式、正确使用求根公式、完成配方、计算并解释判别式，以及将二次方程与其图像对应起来。
Know the standard form ax² + bx + c = 0 and why a ≠ 0.
掌握标准形式 ax² + bx + c = 0 以及为什么 a ≠ 0。
Use factorisation when possible, otherwise use the quadratic formula.
能用因式分解时优先使用，否则使用求根公式。
Remember that the discriminant decides the number of real roots.
记住判别式决定实数根的个数。
Always check solutions by substitution.
始终通过代入检验答案。
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