📚 Solving Quadratic Equations: Factorising, Formula & Completing the Square | 解二次方程：因式分解、求根公式与配方法
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, geometry, trigonometry and even probability problems. Mastering the three main methods of solution - factorising, the quadratic formula and completing the square - will allow you to handle any exam question with confidence. This revision guide will walk you through each method step by step with worked examples and common pitfalls to avoid.
二次方程是IGCSE数学中最重要的主题之一。它们出现在代数、几何、三角甚至概率问题中。掌握三种主要解法——因式分解、求根公式和配方法——将使你能够自信地处理任何考试题目。本复习指南将逐步讲解每种方法，附有例题和需要避免的常见陷阱。
1. What Is a Quadratic Equation? | 什么是二次方程？
a, b and c are constants and a ≠ 0. The highest power of x is 2, which is why it is called "quadratic" (from the Latin quadratus, meaning square).
二次方程是任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数，且 a ≠ 0。x 的最高次数是 2，这就是为什么它被称为"二次"（quadratic 来自拉丁语 quadratus，意为平方）。
ng, always rearrange the equation so that the right-hand side is 0. The values of x that make the equation true are called roots or solutions.
如果 a = 0，方程就变成一次方程，因此 a ≠ 0 的条件至关重要。例如，3x² − 5x + 2 = 0 是二次方程，而 2x − 7 = 0 是一次方程。动手求解前，务必把方程整理成右边为 0 的形式。使方程成立的 x 值称为根或解。
2. Factorising Method | 因式分解法
The factorising method works when the quadratic expression can be written as the product of two linear factors. For an expression of the form x² + bx + c, look for two numbers whose product is c and whose sum is b.
因式分解法适用于二次表达式可以写成两个一次因式的乘积的情况。对于形如 x² + bx + c 的表达式，找两个数，使其乘积为 c、和为 b。
zero, at least one factor must be zero, so x + 2 = 0 or x + 3 = 0. Hence x = −2 or x = −3.
以 x² + 5x + 6 = 0 为例，我们需要两个乘积为 6、和为 5 的数：它们是 2 和 3，因为 2 × 3 = 6，2 + 3 = 5。所以 x² + 5x + 6 = (x + 2)(x + 3)，方程变为 (x + 2)(x + 3) = 0。若两个因式的乘积为零，则至少有一个因式为零，所以 x + 2 = 0 或 x + 3 = 0，即 x = −2 或 x = −3。
grouping: 2x(x + 1) + 3(x + 1) = 0, so (2x + 3)(x + 1) = 0. This gives x = −3/2 or x = −1.
对于 a ≠ 1 的表达式，例如 2x² + 5x + 3 = 0，先把 a 和 c 相乘：2 × 3 = 6。找两个乘积为 6、和为 5 的数：它们是 2 和 3。把中间项重写：2x² + 2x + 3x + 3 = 0，再分组因式分解：2x(x + 1) + 3(x + 1) = 0，所以 (2x + 3)(x + 1) = 0，得 x = −3/2 或 x = −1。
3. Difference of Two Squares and Perfect Squares | 平方差与完全平方
x + a). For example, x² − 16 = 0 becomes (x − 4)(x + 4) = 0, so x = ±4. Notice there is no "middle term" because the two factors cancel each other out.
考试中经常出现两种特殊形式。平方差公式为 x² − a² = (x − a)(x + a)。例如 x² − 16 = 0 变为 (x − 4)(x + 4) = 0，因此 x = ±4。注意此时没有"中间项"，因为两个因式互相抵消。
ated root, x = 3. This happens when the discriminant is zero, which we will examine later.
完全平方三项式为 x² ± 2ax + a² = (x ± a)²。例如 x² − 6x + 9 = 0 可以写成 (x − 3)² = 0。这个方程只有一个重根，即 x = 3。这种情况出现在判别式为零时，我们稍后会详细讨论。
4. The Quadratic Formula | 求根公式
he quadratic formula. For any quadratic equation ax² + bx + c = 0, the roots are given by:
当因式分解困难或无法进行时，应使用求根公式。对于任意二次方程 ax² + bx + c = 0，其根由下式给出：
x = (−b ± √b² − 4ac) / 2a
this formula. Here a = 2, b = 3, c = −2. Substitute carefully, paying attention to signs:
我们用该公式解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入时要格外小心符号：
x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2)
x = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4
Since √25 x = −2. Always check your solutions by substituting them back into the original equation.
因为 √25 = 5，所以 x = (−3 + 5)/4 = 1/2 或 x = (−3 − 5)/4 = −2。解为 x = 1/2 和 x = −2。务必把解代回原方程验证。
5. Completing the Square | 配方法
nd then add and subtract this number. For x² + 6x + 2 = 0, half of 6 is 3, and 3² = 9. So:
配方法将 x² + bx 改写成一个完全平方加上一个常数。关键步骤是取 b 的一半，将其平方，然后加上再减去这个数。对于 x² + 6x + 2 = 0，6 的一半是 3，3² = 9。于是：
x² + 6x + 9 − 9 + 2 = 0
(x + 3)² − 7 = 0
Now solve: (x + 3)² = 7, so x + 3 = ±√7, and therefore x = −3 ± √7. This gives the two exact roots x = −3 + √7 and x = −3 − √7, which can then be approximated with a calculator.
现在求解：(x + 3)² = 7，所以 x + 3 = ±√7，因此 x = −3 ± √7。这给出两个精确根 x = −3 + √7 和 x = −3 − √7，之后可以用计算器求近似值。
² − 4, so 2[(x + 2)² − 4] + 5 = 0, giving 2(x + 2)² = 3 and (x + 2)² = 3/2. Hence x = −2 ± √3/2.
如果 a ≠ 1，先把 a 提出来。对于 2x² + 8x + 5 = 0，写成 2(x² + 4x) + 5 = 0。在括号内配方：x² + 4x = (x + 2)² − 4，所以 2[(x + 2)² − 4] + 5 = 0，即 2(x + 2)² = 3，(x + 2)² = 3/2，因此 x = −2 ± √3/2。
6. The Discriminant | 判别式
The expression b² − 4ac inside the quadratic formula is called the discriminant. It tells us how many real roots a quadratic equation has without solving it completely.
求根公式中的 b² − 4ac 被称为判别式。它无需完全求解，就能告诉我们二次方程有多少个实数根。
| Discriminant Δ = b² − 4ac | Number of real roots | Graph interpretation |
| Δ > 0 | Two distinct real roots | The curve crosses the x-axis at two points |
| Δ = 0 | One repeated real root | The curve touches the x-axis at one point |
| Δ < 0 | No real roots | The curve never meets the x-axis |
so this equation has no real roots. Be careful: a negative discriminant still gives two complex roots, but these are not required at IGCSE level.
例如 x² − 2x + 5 = 0 的判别式为 Δ = (−2)² − 4 × 1 × 5 = 4 − 20 = −16，为负数，因此此方程没有实数根。注意：判别式为负时方程仍有复数根，但IGCSE阶段不要求掌握。
7. Solving Word Problems | 列方程解应用题
Word problems require you to translate a real-life situation into a quadratic equation. Start by defining a variable, then use the given information to form an equation. Finally, solve it and reject any solution that makes no sense in context.
应用题要求你把实际情境转化为二次方程。首先定义一个变量，然后用题目给出的信息列方程，最后求解并舍去在情境中无意义的解。
) = 0, so x = −8 or x = 5. Since a width cannot be negative, x = 5 cm. The length is 8 cm.
例题：一个长方形的长比宽长 3 cm，面积为 40 cm²，求宽。设宽为 x cm，则长为 (x + 3) cm。面积 = 长 × 宽，所以 x(x + 3) = 40。展开得 x² + 3x − 40 = 0。因式分解：(x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。因为宽不能为负，x = 5 cm，长为 8 cm。
8. Graphs and Roots | 图像与根
s the original equation. A quick sketch of the curve can help you check whether your solutions are reasonable.
ax² + bx + c = 0 的根恰好是函数 y = ax² + bx + c 的图像与 x 轴相交或相切处的 x 坐标，因为令 y = 0 就得到原方程。快速画出曲线草图有助于判断解是否合理。
, y = (x − 2)² − 3 has vertex (2, −3), and the equation (x − 2)² − 3 = 0 has roots x = 2 ± √3.
配方法还能揭示顶点。如果 y = a(x − h)² + k，则顶点为 (h, k)，对称轴为竖直线 x = h。例如 y = (x − 2)² − 3 的顶点为 (2, −3)，方程 (x − 2)² − 3 = 0 的根为 x = 2 ± √3。
9. Common Mistakes | 常见错误
Forgetting to rearrange the equation so that one side equals 0 before factorising or using the formula. The standard form is essential.
在因式分解或使用公式之前忘记把方程整理成一边为 0 的形式。标准形式至关重要。
sign when taking a square root. Remember that x² = 9 has two solutions: x = 3 and x = −3.
开平方时丢掉 ± 号。记住 x² = 9 有两个解：x = 3 和 x = −3。
the quadratic formula. Use brackets carefully, as in b² − 4ac with c = −2 becomes b² + 8.
把负的 c 值错误地代入求根公式。代入时要小心使用括号，例如 c = −2 时 b² − 4ac 应变为 b² + 8。
Dividing both sides by a variable such as x when x could be zero. This can silently remove a valid root.
在 x 可能为零时两边同时除以变量 x。这会悄悄丢掉一个有效根。
Confusing (x + a)² with x² + a². The middle term 2ax must not be forgotten.
混淆 (x + a)² 与 x² + a²。中间项 2ax 不可遗忘。
10. Practice Questions | 练习题
Try these questions on your own after reviewing each method. The answers are given below so you can check your working immediately.
复习完每种方法后，请独立完成以下练习。答案附在后面，方便你立即核对过程。
Solve by factorising: x² − 7x + 12 = 0.
用因式分解法求解：x² − 7x + 12 = 0。
ng the quadratic formula: 3x² + 2x − 8 = 0. Give your answers correct to 2 decimal places.
用求根公式求解：3x² + 2x − 8 = 0。答案精确到两位小数。
Solve by completing the square: x² + 8x + 3 = 0. Leave your answers in exact surd form.
用配方法求解：x² + 8x + 3 = 0。答案保留精确根式形式。
Find the value of k such that kx² + 6x + k = 0 has exactly one real root.
求 k 的值，使方程 kx² + 6x + k = 0 恰好有一个实数根。
Answers: 1. x = 3 or x = 4. 2. x ≈ 1.33 or x ≈ −2.00. 3. x = −4 + √13 or x = −4 − √13. 4. Set the discriminant to zero: 6² − 4k² = 0, so k² = 9, giving k = ±3.
答案：1. x = 3 或 x = 4。2. x ≈ 1.33 或 x ≈ −2.00。3. x = −4 + √13 或 x = −4 − √13。4. 令判别式为零：6² − 4k² = 0，所以 k² = 9，即 k = ±3。
With regular practice, factorising, the quadratic formula and completing the square become reliable tools. Remember to check the discriminant first, choose the most efficient method for each question, and always show clear working. Good luck with your IGCSE revision!
通过规律练习，因式分解、求根公式和配方法会成为可靠的解题工具。记得先看判别式，为每道题选择最高效的方法，并且始终写出清晰的过程。祝你IGCSE复习顺利！
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