📚 Mastering Quadratic Equations for IGCSE Maths | IGCSE 数学二次方程完全攻略
Quadratic equations are one of the most important topics in IGCSE Maths. They appear in algebra, graph work, problem solving and even in subjects like physics when motion is modelled. Mastering them gives you a strong foundation for further study.
二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、图像分析、应用题，甚至在物理中用于建立运动模型。掌握二次方程能为你后续的学习打下扎实基础。
1. What Is a Quadratic Equation? | 什么是二次方程？
stants and a ≠ 0. The condition a ≠ 0 is essential because if a were zero, the x² term would disappear and the equation would become linear.
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、c 为常数且 a ≠ 0。a ≠ 0 这一条件非常重要，因为如果 a 为零，x² 项就会消失，方程就变成一次方程。
The highest power of the unknown x is 2. This means a quadratic equation can have up to two real solutions, also called roots, although it can also have one repeated root or no real roots at all.
未知数 x 的最高次数是 2。这意味着二次方程最多可以有两个实数解，也就是根，但也可能只有一个重根，或者没有实数根。
- Standard form: ax² + bx + c = 0 | 标准形式：ax² + bx + c = 0
- Example: 2x² + 5x – 3 = 0 | 示例：2x² + 5x – 3 = 0
- a ≠ 0 is required | 必须满足 a ≠ 0
2. Solving by Factorising | 因式分解法求解
Factorising is usually the fastest method when the quadratic expression can be written as the product of two linear factors. The key idea is the zero product rule: if two numbers multiply to give zero, then at least one of them must be zero.
当二次式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。关键思想是零乘积法则：如果两个数相乘为零，那么其中至少有一个必须为零。
For example, consider x² – 5x + 6 = 0. We look for two numbers that multiply to give 6 and add to give –5. The numbers –2 and –3 work, so the equation becomes (x – 2)(x – 3) = 0.
例如，考虑方程 x² – 5x + 6 = 0。我们需要找到两个数，它们相乘得 6，相加得 –5。数字 –2 和 –3 符合要求，因此方程变为 (x – 2)(x – 3) = 0。
Setting each factor equal to zero gives x – 2 = 0 or x – or x = 3. Always check your answers by substituting them back into the original equation.
令每个因式等于零，得到 x – 2 = 0 或 x – 3 = 0，因此 x = 2 或 x = 3。一定要把答案代回原方程进行检验。
- Step 1: Write the equation in standard form
步骤 1：把方程写成标准形式 - Step 2: Factorise the quadratic expression
步骤 2：对二次式进行因式分解 - Step 3: Set each factor equal to zero
步骤 3：令每个因式等于零 - Step 4: Solve for x and check
步骤 4：解出 x 并检验
3. Completing the Square | 配方法
Completing the square is a powerful technique because it works for every quadratic equation, even when factorising is difficult. It also reveals the vertex of the corresponding quadratic function.
配方法是一种非常有力的技巧，因为它适用于所有二次方程，即使在因式分解困难时也能使用。它同时还能揭示对应二次函数的顶点。
the coefficient of x, which is 3, and square it to get 9. Then we write x² + 6x = (x + 3)² – ecause (x + 3)² expands to x² + 6x + 9, so we subtract 9 to keep the expression unchanged.
要对 x² + 6x 进行配方，我们取 x 项系数的一半，即 3，再将其平方得到 9。然后写成 x² + 6x = (x + 3)² – 9。这是因为 (x + 3)² 展开为 x² + 6x + 9，所以需要减去 9 才能保持原式不变。
For example, solve x² + 6x + 2 = 0. First write x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7. The equation becomes (x + 3)² = 7, so x + 3 = ±√7 and x = –3 ± √7.
例如，求解 x² + 6x + 2 = 0。首先写成 x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7。方程变为 (x + 3)² = 7，因此 x + 3 = ±√7，最终 x = –3 ± √7。
When the coefficient of x² is not 1, factor it out first before completing the square. This step is often missed in exams.
当 x² 的系数不是 1 时，需要先把这个系数提到括号外，然后再进行配方。考试中这一步经常被忽略。
4. The Quadratic Formula | 求根公式法
x + c = 0 directly. It is especially useful when the equation cannot be factorised easily.
求根公式可以直接给出任何二次方程 ax² + bx + c = 0 的解。当方程不易因式分解时，它尤其有用。
x = (–b ± √b² – 4ac) / (2a)
To use the formula, first identify the values of a, b and c. For example, in 2x² + 3x – 2 = 0, we have a = 2, b = 3 and c = –2. Substituting gives:
使用公式时，首先要确定 a、b、c 的值。例如，在 2x² + 3x – 2 = 0 中，a = 2，b = 3，c = –2。代入公式得到：
x = (–3 ± √(3² – 4 × 2 × (–2))) / (2 × 2)
This simplifies to x = (–3 ± √25) / 4, so x = (–3 + 5) / 4 = 1/2 or x = (–3 – 5) / 4 = –2. Always simplify the square root before splitting the ± sign.
化简后得到 x = (–3 ± √25) / 4，所以 x = (–3 + 5) / 4 = 1/2 或 x = (–3 – 5) / 4 = –2。在拆分 ± 号之前，一定要先化简根号部分。
rator in brackets when using a calculator. A common error is to forget the brackets around –b ± √b² – 4ac, which gives an incorrect value.
使用计算器时，一定要把整个分子用括号括起来。常见的错误是忘记给 –b ± √b² – 4ac 加上括号，这会导致计算结果错误。
5. Discriminant and Number of Roots | 判别式与根的个数
The discriminant is the expression under the square root in the quadratic formula: Δ = b² – lls us how many real roots a quadratic equation has without actually solving the equation.
判别式是求根公式中根号下面的表达式：Δ = b² – 4ac。它可以在不实际求解方程的情况下，告诉我们二次方程有多少个实数根。
d real root, which means the graph touches the x-axis at one point. If Δ < 0, there are no real roots.
如果 Δ > 0，方程有两个不同的实数根。如果 Δ = 0，方程有一个重根，也就是说图像与 x 轴相切于一点。如果 Δ < 0，则没有实数根。
| Discriminant | 判别式 | Number of real roots | 实数根个数 |
|---|---|
| Δ > 0 | Two distinct real roots | 两个不同实数根 |
| Δ = 0 | One repeated real root | 一个重根 |
| Δ < 0 | No real roots | 没有实数根 |
For example, x² + 2x + 5 = 0 has Δ = 2² – 4 × 1 × 5 = 4 – 20 = –16, so it has no real roots. This can explain why a graph never crosses the x-axis.
例如，x² + 2x + 5 = 0 的判别式为 Δ = 2² – 4 × 1 × 5 = 4 – 20 = –16，因此它没有实数根。这可以解释为什么一个图像永远不会穿过 x 轴。
6. Graphical Interpretation | 图像意义
wnwards. The solutions of the equation ax² + bx + c = 0 are the x-intercepts of the graph.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上；如果 a < 0，抛物线开口向下。方程 ax² + bx + c = 0 的解就是图像与 x 轴交点的横坐标。
st or lowest point of the parabola. For a parabola written in completed square form y = (x – (h, k). This is very useful for sketching graphs and solving maximum or minimum problems.
顶点是抛物线的最高点或最低点。对于写成配方法形式 y = (x – h)² + k 的抛物线，其顶点为 (h, k)。这对草绘图像和解决最大值或最小值问题非常有帮助。
If the discriminant is zero, the parabola touches the x-axis at exactly one point, which is the vertex. If the discriminant is negative, the whole parabola lies above or below the x-axis and there are no x-intercepts.
如果判别式为零，抛物线恰好在一个点与 x 轴相切，这个点就是顶点。如果判别式为负，整条抛物线都位于 x 轴上方或下方，因此没有 x 轴交点。
7. Applications and Word Problems | 实际应用与文字题
Quadratic equations often appear in real-world problems involving area, projectile motion, product relationships and profit. The key is to define a variable, build an equation and then solve it using an appropriate method.
二次方程经常出现在涉及面积、抛体运动、乘积关系和利润的实际问题中。解题的关键是设出变量，建立方程，然后选择合适的方法求解。
x cm, so the length is x + 3 cm. The area gives x(x + 3) = 40, which simplifies to x² + 3x – 40 = 0.
例如，一个矩形的长比宽多 3 厘米，面积为 40 平方厘米。设宽为 x 厘米，那么长为 x + 3 厘米。根据面积可得 x(x + 3) = 40，化简后为 x² + 3x – 40 = 0。
Factorising gives (x + 8)(x – 5) = 0, so x = – x = 5. Since a width cannot be negative, x = 5. The length is 8 cm and the width is 5 cm.
因式分解得到 (x + 8)(x – 5) = 0，因此 x = –8 或 x = 5。由于宽度不能为负，所以 x = 5。长为 8 厘米，宽为 5 厘米。
E. For a ball thrown upwards, its height h metres after t seconds might be modelled by h = – + 20t + 1. Setting h = 0 gives a quadratic equation to find when the ball hits the ground.
抛体运动问题在 IGCSE 中很常见。一个球向上抛出后，t 秒后的高度 h 米可以表示为 h = –5t² + 20t + 1。令 h = 0 就得到一个二次方程，用来求出球何时落地。
8. Common Mistakes and Exam Tips | 常见错误与考试技巧
One of the most common mistakes is forgetting to write the equation in standard form before identifying a, b and c. If the terms are not on the same side, the signs will be wrong.
最常见的错误之一是在确定 a、b、c 之前忘记把方程写成标准形式。如果各项不在同一边，符号就容易出错。
g both sides by x gives x = 6, but this loses the solution x = 0. Instead, rearrange to x² – 6x = 0 and factorise: x(x – 6) = 0, so x = 0 or x = 6.
另一个常见错误是在解题过程中除以 x。例如，对于 x² = 6x，两边同时除以 x 得到 x = 6，但这会丢失 x = 0 这个解。正确的做法是移项得到 x² – 6x = 0，再因式分解：x(x – 6) = 0，所以 x = 0 或 x = 6。
- Always check your factorisation by expanding | 一定要通过展开来检查因式分解
- Write the formula and substitute carefully | 写出公式并仔细代入
- Do not cancel variables unless you are sure they are not zero | 除非确定变量不为零，否则不要约去变量
- Give answers to the required accuracy, using 3 significant figures if asked | 按要求的精度给出答案，通常要求 3 位有效数字
9. Worked Example: Full Solution | 例题精讲：完整解答
Let us solve the following problem: The product of two consecutive positive integers is 72. Find the integers.
我们来解决下面这个问题：两个连续正整数的乘积是 72。求这两个整数。
e integer is n + 1. Their product is n(n + 1) = 72. Expanding gives n² + n = 72, so n² + n – 72 = 0.
设较小的整数为 n，那么下一个连续整数是 n + 1。它们的乘积为 n(n + 1) = 72。展开得到 n² + n = 72，所以 n² + n – 72 = 0。
We need two numbers that multiply to –72 and add to 1. The numbers 9 and –8 work because 9 × (–8) = –72 and 9 + (–8) = 1. Therefore, the equation factorises to (n + 9)(n – 8) = 0.
我们需要找到两个数，它们相乘得 –72，相加得 1。数字 9 和 –8 符合要求，因为 9 × (–8) = –72，且 9 + (–8) = 1。因此，方程因式分解为 (n + 9)(n – 8) = 0。
Setting each factor equal to zero gives n = – r n = 8. Since the integers are positive, n = 8. The two consecutive integers are 8 and 9.
令每个因式等于零，得到 n = –9 或 n = 8。由于整数是正的，所以 n = 8。这两个连续整数是 8 和 9。
Next, we solve 3x² – 7x + 2 = 0 using the quadratic formula. Here a = 3, b = –7 and c = 2.
接下来，我们用求根公式解 3x² – 7x + 2 = 0。这里 a = 3，b = –7，c = 2。
x = (–(–7) ± √((–7)² – 4 × 3 × 2)) / (2 × 3)
This simplifies to x = (7 ± √49 – 24) / 6 = (7 ± √25) / 6 = (7 ± 5) / 6. So x = 12/6 = 2 or x = 2/6 = 1/3.
化简得 x = (7 ± √49 – 24) / 6 = (7 ± √25) / 6 = (7 ± 5) / 6。所以 x = 12/6 = 2 或 x = 2/6 = 1/3。
10. Practice Checklist | 练习清单
Before your exam, make sure you can confidently handle each of the following skills. Practice a mixture of purely algebraic questions and word problems so that you are ready for any style.
考试前，请确保你能自信地掌握以下每一项技能。练习纯代数题和应用题的混合题型，这样你就能应对任何考试风格。
- Write a quadratic equation in standard form | 将二次方程写成标准形式
- Solve by factorising when possible | 在可能时用因式分解法求解
- Complete the square and find the vertex | 配方并求顶点
- Use the quadratic formula accurately | 准确使用求根公式
- Use the discriminant to state the number of roots | 利用判别式判断根的个数
- Sketch parabolas and label intercepts and vertex | 绘制抛物线并标出交点和顶点
- Solve real-world problems by forming quadratic equations | 通过建立二次方程解决实际问题
Work through past paper questions under timed conditions. When you check your answers, always write out the steps again instead of just reading the mark scheme, because this helps you internalise the method.
请在计时条件下完成历年真题。检查答案时，一定要重新写出解题步骤，而不是只看评分标准，因为这样做有助于你内化解题方法。
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