📚 Mastering Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 掌握二次方程：因式分解、配方法与求根公式
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, and real-life modelling, and they act as a bridge between algebraic manipulation and coordinate geometry. In this revision guide, you will learn how to solve quadratic equations by factorising, by completing the square, and by using the quadratic formula. You will also learn how to decide which method is fastest for a given question, how the discriminant predicts the number of roots, and how roots connect to the graph of a parabola.
二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、图像和现实建模中，也是代数运算与坐标几何之间的桥梁。在本复习指南中，你将学习如何通过因式分解、配方法和求根公式来解二次方程。你还将学会判断某道题用哪种方法最快、判别式如何预测根的个数，以及根与抛物线图像之间的联系。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be written in the standard form a x² + b x + 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 a degree-two polynomial equation.
二次方程是任何可以写成标准形式 a x² + b x + c = 0 的方程，其中 a、b 和 c 是常数，且 a ≠ 0。变量 x 的最高次数为 2，因此它被称为二次多项式方程。
The name 'quadratic' comes from the Latin word quadratus t a quadratic does not have to contain all three terms: b or c (or even both) can be zero.
二次方程的名称来自拉丁语 quadratus，意为正方形。这提醒我们其关键特征是 x² 项。例子包括 x² - 5x + 6 = 0、2x² + 3x - 2 = 0 和 4x² - 9 = 0。注意，二次方程不必同时含有三项：b 或 c（甚至两者）都可以为零。
Quadratic equations can have two real roots, one repeated real root, or no real roots. The method you choose depends on the structure of the equation and the instruction in the question.
二次方程可以有两个实根、一个重根或没有实根。你选择的方法取决于方程的结构以及题目中的要求。
- A quadratic with two real roots crosses the x-axis twice. | 有两个实根的二次方程与 x 轴有两个交点。
- One repeated root means the curve just touches the x-axis. | 一个重根意味着曲线恰好与 x 轴相切。
- No real roots means the curve never meets the x-axis. | 没有实根意味着曲线永远不与 x 轴相交。
2. Standard Form and Rearranging | 标准形式与方程整理
s essential to place all terms on one side and leave zero on the other side before factorising or applying the formula. Many exam mistakes come from trying to factorise when the equation is not equal to zero.
标准形式为 a x² + b x + c = 0。在因式分解或使用公式之前，必须将所有项移到一边，另一边留零。许多考试错误都源于在方程不等于零时尝试因式分解。
the coefficient of x² is negative, you can multiply the whole equation by -1 to make it positive; this does not change the roots.
例如，将 x² + 5x = 6 整理为 x² + 5x - 6 = 0。这里 a = 1，b = 5，c = -6。如果 x² 的系数为负，你可以将整个方程乘以 -1 使其为正；这不会改变方程的根。
easier to solve once factorised. Rearranging correctly is the foundation for all later steps.
务必检查方程是否已完全简化。例如，3x² - 6x = 0 变为 3x(x - 2) = 0，因式分解后就容易解得多了。正确整理方程是所有后续步骤的基础。
a x² + b x + c = 0, a ≠ 0
3. Solving by Factorisation: Common Factors and Simple Trinomials | 因式分解法：公因式与简单三项式
his rule is the reason we can turn one quadratic equation into two simple linear equations.
因式分解依赖于零积法则：若 A × B = 0，则 A = 0 或 B = 0。正是这条法则让我们能把一个二次方程变成两个简单的一次方程。
If (x - p)(x - q) = 0, then x = p or x = q
that p + q = b (the sum) and p × q = c (the product). Then x² + b x + c = (x + p)(x + q).
对于简单三项式 x² + b x + c，你要找两个数 p 和 q，使 p + q = b（和）且 p × q = c（积）。于是 x² + b x + c = (x + p)(x + q)。
with sum -5 and product 6, namely -2 and -3. So (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
例：解 x² - 5x + 6 = 0。我们需要两个数，其和为 -5、积为 6，即 -2 和 -3。所以 (x - 2)(x - 3) = 0，得 x = 2 或 x = 3。
, so x = 3/2 or x = -3/2. Recognising this pattern saves a great deal of time in the exam.
平方差是特殊情况：4x² - 9 = 0 变为 (2x - 3)(2x + 3) = 0，所以 x = 3/2 或 x = -3/2。在考试中识别这个模式能节省大量时间。
4. Factorising Harder Quadratics When a ≠ 1 | 当 a ≠ 1 时更复杂的因式分解
When the coefficient of x² is not 1, factorising needs a little more care. The most reliable method is the 'split the middle term' method, sometimes called the ac method.
当 x² 的系数不为 1 时，因式分解需要更谨慎。最可靠的方法是「拆中间项」法，有时也称为 ac 法。
- Step 1: write the equation in the form a x² + b x + c = 0. | 第一步：把方程写成 a x² + b x + c = 0 的形式。
- Step 2: calculate a × c. | 第二步：计算 a × c。
- Step 3: find two numbers with product a × c and sum b. | 第三步：找两个数，其积为 a × c，和为 b。
- Step 4: split the middle term using those two numbers. | 第四步：用这两个数拆分中间项。
- Step 5: factorise in pairs, then take out the common bracket. | 第五步：分组分解，再提取公因式括号。
- Step 6: apply the zero product rule. | 第六步：应用零积法则。
oups as 2x(x + 2) - 1(x + 2) = 0, giving (x + 2)(2x - 1) = 0. Therefore x = -2 or x = 1/2.
例：解 2x² + 3x - 2 = 0。这里 a × c = 2 × (-2) = -4。积为 -4、和为 3 的两个数是 4 和 -1。于是 2x² + 4x - x - 2 = 0，分组为 2x(x + 2) - 1(x + 2) = 0，得 (x + 2)(2x - 1) = 0。因此 x = -2 或 x = 1/2。
ts mentally to check: (x + 2)(2x - 1) = 2x² + 3x - 2, which matches the original equation.
务必在脑中展开括号检查：(x + 2)(2x - 1) = 2x² + 3x - 2，与原方程一致。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q, which reveals the vertex of the parabola and prepares the equation for solving with surds.
配方法将二次式改写为 (x + p)² + q 的形式，这能直接显示抛物线的顶点，并为用根式求解做好准备。
x² + b x + c = (x + b/2)² - (b/2)² + c
x + 3)² - 8. The number inside the bracket is always half of the coefficient of x, and you subtract the square of that number to keep the expression balanced.
例如，x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8。括号里的数总是 x 系数的一半，再减去这个数的平方以保持表达式不变。
the first two terms: 2x² - 8x + 5 = 2(x² - 4x) + 5 = 2[(x - 2)² - 4] + 5 = 2(x - 2)² - 3.
当 a ≠ 1 时，先把系数 a 从前两项提出：2x² - 8x + 5 = 2(x² - 4x) + 5 = 2[(x - 2)² - 4] + 5 = 2(x - 2)² - 3。
2. Completing the square is therefore the standard tool for maximum and minimum problems.
这一形式立即告诉你最小值为 -3，在 x = 2 时取得。因此配方法是求解最大值与最小值问题的标准工具。
6. Solving Equations by Completing the Square | 用配方法解方程
To solve by completing the square, rearrange the equation so that the completed square equals a number, then take square roots. Remember that taking a square root gives both a positive and a negative value.
用配方法解方程时，先把方程整理成「完全平方式 = 一个数」，再开平方。记住开平方会同时得到正负两个值。
ete the square: (x + 3)² - 9 + 2 = 0, so (x + 3)² = 7. Taking square roots gives x + 3 = ± √7, hence x = -3 + √7 or x = -3 - √7.
例：解 x² + 6x + 2 = 0。配方得 (x + 3)² - 9 + 2 = 0，即 (x + 3)² = 7。开平方得 x + 3 = ± √7，因此 x = -3 + √7 或 x = -3 - √7。
(x + 3)² = 7 ⟹ x = -3 ± √7
- 8x + 5 = 0. From the previous section, 2(x - 2)² - 3 = 0, so (x - 2)² = 3/2 and x = 2 ± √3/2. Since √3/2 = √6 / 2, we can also write x = 2 ± √6 / 2.
a ≠ 1 的例子：解 2x² - 8x + 5 = 0。由上节可知 2(x - 2)² - 3 = 0，所以 (x - 2)² = 3/2，x = 2 ± √3/2。由于 √3/2 = √6 / 2，也可写成 x = 2 ± √6 / 2。
If the question says 'leave your answer in surd form', do not convert to decimals. If it asks for answers correct to 2 decimal places, evaluate the surd only at the very end.
如果题目要求「以根式形式作答」，就不要化成小数。如果要求「精确到两位小数」，则只在最后一步计算根式的值。
7. The Quadratic Formula | 求根公式
factorises. It is derived by completing the square on the general form a x² + b x + c = 0.
求根公式可以解任何二次方程，无论它能否因式分解。它是对一般形式 a x² + b x + c = 0 配方后推导出来的。
x = (-b ± √b² - 4ac) / (2a)
a = 2, b = 3, c = -2. Substituting gives x = (-3 ± √(3² - 4 × 2 × (-2))) / (2 × 2) = (-3 ± √9 + 16 -3 ± 5) / 4. So x = 2/4 = 1/2 or x = -8/4 = -2, which agrees with the factorised solution.
例：解 2x² + 3x - 2 = 0，其中 a = 2，b = 3，c = -2。代入得 x = (-3 ± √(3² - 4 × 2 × (-2))) / (2 × 2) = (-3 ± √9 + 16) / 4 = (-3 ± 5) / 4。所以 x = 2/4 = 1/2 或 x = -8/4 = -2，与因式分解的结果一致。
Example that does not factorise: solve x² + 4x + 1 = 0. Here x = (-4 ± √16 - 4) / 2 = (-4 ± √12) / 2 = -2 ± √3. Since √12 = 2√3, always simplify the surd.
无法因式分解的例子：解 x² + 4x + 1 = 0。这里 x = (-4 ± √16 - 4) / 2 = (-4 ± √12) / 2 = -2 ± √3。由于 √12 = 2√3，一定要把根式化简。
- Write the formula first, then substitute inside brackets. | 先写出公式，再用括号代入。
- Use brackets for negative values of b and c. | 当 b、c 为负数时要用括号。
- Divide the whole numerator by 2a, not just part of it. | 整个分子都要除以 2a，而不是只除一部分。
8. The Discriminant and the Number of Roots | 判别式与根的个数
The expression under the square root, b² - 4ac, is called the discriminant, often written as Δ. It tells you how many real roots the equation has before you solve it.
根号下的表达式 b² - 4ac 称为判别式，通常记作 Δ。它能在你求解之前告诉你方程有几个实根。
Δ = b² - 4ac
| Value of Δ | Δ 的值 | Number of real roots | 实根个数 | Graph behaviour | 图像表现 |
| Δ > 0 | Two distinct
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