Mastering Quadratic Equations | 掌握二次方程

📚 Mastering Quadratic Equations | 掌握二次方程

Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. They appear in almost every exam paper, whether as standalone questions or as part of problem-solving exercises. Understanding how to solve them efficiently and accurately is essential for achieving a high grade.

二次方程是 IGCSE 数学大纲中最重要的内容之一。几乎每份试卷都会出现，无论是独立题目还是解题过程的一部分。掌握高效准确的解法，是取得高分的关键。

1. What Is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is an equation in which the highest power of the variable is 2. The general form is written as:

二次方程是指变量最高次数为 2 的方程。其一般形式写作：

ax² + bx + c = 0, 其中 a ≠ 0

Here, a, b, and c are constants, and x is the unknown variable. The condition a ≠ 0 is crucial; if a = 0, the equation becomes linear, not quadratic.

其中 a、b、c 为常数，x 为未知变量。条件 a ≠ 0 至关重要；若 a = 0，方程将退化为一次方程而非二次方程。

Examples of quadratic equations: x² − 5x + 6 = 0, 2x² + 3x − 1 = 0, x² = 16
二次方程的例子：x² − 5x + 6 = 0，2x² + 3x − 1 = 0，x² = 16

Some quadratic equations may not appear in standard form at first. For example, x( x − 3 ) = 4 must be expanded and rearranged to x² − 3x − 4 = 0 before solving.

有些二次方程最初并非标准形式。例如 x( x − 3 ) = 4 需先展开并整理为 x² − 3x − 4 = 0 再求解。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic expression factorises neatly. The key idea is to rewrite ax² + bx + c as the product of two linear brackets.

当二次表达式可以整齐分解时，因式分解法通常是最快捷的方法。核心思路是将 ax² + bx + c 改写为两个一次括号的乘积。

Example 1: Solve x² − 5x + 6 = 0.

例 1：解 x² − 5x + 6 = 0。

We look for two numbers that multiply to +6 and add to −5. These numbers are −2 and −3. Therefore:

我们寻找两个数，它们相乘等于 +6，相加等于 −5。这两个数是 −2 和 −3。因此：

( x − 2 )( x − 3 ) = 0

For the product to be zero, at least one bracket must be zero. Hence x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

要使乘积为零，至少一个括号必须为零。因此 x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。

Example 2: Solve 2x² + 7x + 3 = 0.

例 2：解 2x² + 7x + 3 = 0。

Here the coefficient of x² is not 1, so we use the “multiply a by c” approach: a × c = 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7 — those are 6 and 1. Split the middle term:

这里 x² 的系数不为 1，我们使用”a 乘 c”法：a × c = 2 × 3 = 6。找两个数相乘为 6、相加为 7——即 6 和 1。将中间项拆分：

2x² + 6x + x + 3 = 0

Now factor by grouping: 2x( x + 3 ) + 1( x + 3 ) = 0, which gives ( 2x + 1 )( x + 3 ) = 0. Thus x = −½ or x = −3.

再分组分解：2x( x + 3 ) + 1( x + 3 ) = 0，即 ( 2x + 1 )( x + 3 ) = 0。因此 x = −½ 或 x = −3。

3. The Quadratic Formula | 二次公式

When factorisation is difficult or impossible, the quadratic formula provides a universal method. For any quadratic equation ax² + bx + c = 0, the solutions are:

当因式分解困难或无法分解时，二次公式提供了一种通用解法。对于任何二次方程 ax² + bx + c = 0，解为：

x = ( −b ± √( b² − 4ac ) ) / 2a

Example: Solve 3x² − 4x − 2 = 0, giving answers to 2 decimal places.

例：解 3x² − 4x − 2 = 0，答案精确到小数点后两位。

Here a = 3, b = −4, c = −2. Substitute into the formula:

这里 a = 3，b = −4，c = −2。代入公式：

x = ( 4 ± √( 16 − 4 × 3 × (−2) ) ) / 6

x = ( 4 ± √( 16 + 24 ) ) / 6 = ( 4 ± √40 ) / 6

√40 ≈ 6.3249, so x ≈ ( 4 + 6.3249 ) / 6 ≈ 1.72 or x ≈ ( 4 − 6.3249 ) / 6 ≈ −0.39. Both answers should be rounded correctly to 2 decimal places.

√40 ≈ 6.3249，因此 x ≈ ( 4 + 6.3249 ) / 6 ≈ 1.72 或 x ≈ ( 4 − 6.3249 ) / 6 ≈ −0.39。两个答案均应正确四舍五入到小数点后两位。

You should memorise the quadratic formula. It is not provided on every IGCSE formula sheet, depending on your exam board.

你应该牢记二次公式。并非所有 IGCSE 考试局的公式表都会提供它。

4. Completing the Square | 配方法

Completing the square is another algebraic technique that rewrites a quadratic expression in the form p( x + q )² + r. This method is especially useful for finding turning points of parabolas.

配方法是另一种代数技巧，它将二次表达式改写为 p( x + q )² + r 的形式。此方法特别适用于求抛物线的顶点。

Example: Solve x² + 6x − 7 = 0 by completing the square.

例：用配方法解 x² + 6x − 7 = 0。

Step 1: Halve the coefficient of x: 6 ÷ 2 = 3. Write ( x + 3 )² = x² + 6x + 9.

步骤 1：将 x 的系数减半：6 ÷ 2 = 3。写出 ( x + 3 )² = x² + 6x + 9。

Step 2: Adjust the constant: x² + 6x = ( x + 3 )² − 9.

步骤 2：调整常数项：x² + 6x = ( x + 3 )² − 9。

Step 3: Include the original constant: x² + 6x − 7 = ( x + 3 )² − 9 − 7 = ( x + 3 )² − 16.

步骤 3：加上原常数项：x² + 6x − 7 = ( x + 3 )² − 9 − 7 = ( x + 3 )² − 16。

Step 4: Solve ( x + 3 )² − 16 = 0, so ( x + 3 )² = 16. Take square roots: x + 3 = ±4, giving x = 1 or x = −7.

步骤 4：解 ( x + 3 )² − 16 = 0，即 ( x + 3 )² = 16。开平方：x + 3 = ±4，得 x = 1 或 x = −7。

When the coefficient of x² is not 1, factor it out first. For example, 2x² + 8x + 5 = 2[ x² + 4x ] + 5 = 2[ ( x + 2 )² − 4 ] + 5 = 2( x + 2 )² − 3.

当 x² 的系数不为 1 时，先提出该系数。例如 2x² + 8x + 5 = 2[ x² + 4x ] + 5 = 2[ ( x + 2 )² − 4 ] + 5 = 2( x + 2 )² − 3。

5. The Discriminant | 判别式

The discriminant is the expression b² − 4ac found inside the square root of the quadratic formula. It tells us how many real roots a quadratic equation has, without solving it fully.

判别式是二次公式根号内的表达式 b² − 4ac。它能在不完全求解的情况下，告诉我们二次方程有多少个实数根。

Value of b² − 4ac	Number of Real Roots	b² − 4ac 的值	实数根个数
b² − 4ac > 0	Two distinct real roots	b² − 4ac > 0	两个不等实数根
b² − 4ac = 0	One repeated real root	b² − 4ac = 0	一个重根
b² − 4ac < 0	No real roots	b² − 4ac < 0	无实数根

Example: Determine the number of real roots for 2x² − 3x + 1 = 0.

例：判断 2x² − 3x + 1 = 0 的实数根个数。

Here a = 2, b = −3, c = 1. So b² − 4ac = ( −3 )² − 4 × 2 × 1 = 9 − 8 = 1 > 0. The equation has two distinct real roots.

这里 a = 2，b = −3，c = 1。因此 b² − 4ac = ( −3 )² − 4 × 2 × 1 = 9 − 8 = 1 > 0。方程有两个不等实数根。

The discriminant also appears in questions about tangent lines to curves. A line tangent to a parabola corresponds to a discriminant of zero, since the system has exactly one solution.

判别式也常出现在曲线切线问题中。直线与抛物线相切时，对应判别式为零，因为方程组恰有一个解。

6. Solving Quadratics by Graphing | 图象法解二次方程

The roots of a quadratic equation are the x-coordinates where the graph of y = ax² + bx + c crosses the x-axis. Plotting the graph gives an approximate visual solution.

二次方程的根就是 y = ax² + bx + c 图象与 x 轴交点的 x 坐标。绘制图象可以给出近似的直观解。

Consider y = x² − 2x − 3. The graph crosses the x-axis at x = −1 and x = 3, confirming the roots of x² − 2x − 3 = 0 are x = −1 and x = 3.

考虑 y = x² − 2x − 3。图象在 x = −1 和 x = 3 处穿过 x 轴，验证了 x² − 2x − 3 = 0 的根为 x = −1 和 x = 3。

The vertex (turning point) lies midway between the two roots. For y = x² − 2x − 3, the vertex is at x = 1, y = ( 1 )² − 2( 1 ) − 3 = −4, so the minimum point is ( 1, −4 ).
顶点（转折点）位于两个根的正中间。对于 y = x² − 2x − 3，顶点在 x = 1，y = ( 1 )² − 2( 1 ) − 3 = −4，即最低点为 ( 1, −4 )。

If the parabola does not intersect the x-axis, the discriminant is negative and there are no real roots. This graphical interpretation reinforces the discriminant’s meaning.

若抛物线与 x 轴无交点，则判别式为负，无实数根。这种图象解释加深了对判别式含义的理解。

7. Word Problems Involving Quadratics | 二次方程应用题

Many IGCSE questions present quadratic equations in the form of real-world problems. These often involve areas, projectile motion, or number relationships. The key is to translate the words into an equation correctly.

许多 IGCSE 题目以实际问题的形式呈现二次方程，常涉及面积、抛体运动或数之间的关系。关键是将文字正确转化为方程。

Example: A rectangle has length ( x + 4 ) cm and width ( x − 1 ) cm. Its area is 42 cm². Find x.

例：一个矩形长 ( x + 4 ) cm，宽 ( x − 1 ) cm，面积为 42 cm²。求 x。

The area equation is ( x + 4 )( x − 1 ) = 42. Expanding:

面积方程为 ( x + 4 )( x − 1 ) = 42。展开：

x² + 3x − 4 = 42 → x² + 3x − 46 = 0

Factorisation: look for two numbers multiplying to −46 and adding to 3 — none exist in integers, so use the quadratic formula:

因式分解：寻找相乘为 −46、相加为 3 的两个整数——不存在，因此使用二次公式：

x = ( −3 ± √( 9 − 4 × 1 × (−46) ) ) / 2 = ( −3 ± √193 ) / 2

x ≈ ( −3 + 13.892 ) / 2 ≈ 5.45 or x ≈ ( −3 − 13.892 ) / 2 ≈ −8.45. Since the width cannot be negative, we take x ≈ 5.45.

x ≈ ( −3 + 13.892 ) / 2 ≈ 5.45 或 x ≈ ( −3 − 13.892 ) / 2 ≈ −8.45。由于宽度不能为负，我们取 x ≈ 5.45。

Always check whether both solutions are valid in the context of the problem. Negative lengths, times, or distances indicate that one solution must be rejected.

务必检查两个解在题目情境中是否均有效。负的长度、时间或距离意味着其中一个解必须舍去。

8. Quadratic Inequalities | 二次不等式

A related skill is solving quadratic inequalities, such as x² − 5x + 6 > 0. The approach uses the factorised form and a sign diagram.

一项相关技能是解二次不等式，如 x² − 5x + 6 > 0。解法是利用因式分解形式和符号图。

Factorise: ( x − 2 )( x − 3 ) > 0. The critical points are x = 2 and x = 3. Test intervals on a number line:

分解因式：( x − 2 )( x − 3 ) > 0。关键点为 x = 2 和 x = 3。在数轴上测试各区间：

When x < 2: both ( x − 2 ) and ( x − 3 ) are negative, product is positive ✓
当 x < 2 时：两个括号均为负，乘积为正 ✓
When 2 < x < 3: ( x − 2 ) is positive, ( x − 3 ) is negative, product is negative ✗
当 2 < x < 3 时：一正一负，乘积为负 ✗
When x > 3: both are positive, product is positive ✓
当 x > 3 时：两个括号均为正，乘积为正 ✓

Therefore the solution is x < 2 or x > 3. For ≤ or ≥ inequalities, include the critical points as closed intervals.

因此解为 x < 2 或 x > 3。对于 ≤ 或 ≥ 的不等式，关键点应包含在解集中。

9. Common Mistakes | 常见错误

Many students lose marks on quadratic equations due to avoidable errors. The most frequent mistakes include:

许多学生在二次方程上失分，往往是因为可以避免的错误。最常见的错误包括：

Mistake \| 错误	Correct Approach \| 正确做法
Forgetting to rearrange to the form ax² + bx + c = 0 before factorising	Always bring all terms to one side first
提前未将方程整理为 ax² + bx + c = 0 就进行因式分解	始终先将所有项移到等号一侧
Dividing both sides by x when x = 0 is a possible root	Factorise instead of dividing; x = 0 may be lost
当 x = 0 可能是根时，两边同除以 x	应分解因式而非除 x，否则会丢失 x = 0 这个根
Sign errors when substituting negatives into the quadratic formula	Use brackets around negative values: ( −4 )² = 16
代入负数到二次公式时的符号错误	用括号括住负值：( −4 )² = 16
Rushing the final rounding of answers	Round only at the final step, not intermediate steps
最终取近似值过于草率	只在最后一步四舍五入，不要在中间步骤提前化整

Another common error is writing “x = 2, 3” without showing the factorised equation. In IGCSE exams, method marks are awarded for the factorisation step, so always show ( x − 2 )( x − 3 ) = 0 before the final solutions.

另一个常见错误是直接写”x = 2, 3″而不展示因式分解方程。在 IGCSE 考试中，因式分解步骤会有方法分，所以务必先写出 ( x − 2 )( x − 3 ) = 0 再给出最终解。

10. Exam Tips and Revision Strategy | 应试技巧与复习策略

To excel in quadratic equation questions, follow a consistent solving routine and practise a wide variety of problems.

要在二次方程题目中表现出色，需要遵循一致的解题流程，并广泛练习各类题型。

First, always check whether the equation is already in standard form. If not, rearrange it immediately.
首先检查方程是否已是标准形式。若不是，立即重新整理。
Try factorisation first, as it is fastest. If you cannot find integer factors within about 10 seconds, switch to the quadratic formula.
优先尝试因式分解，这是最快的方法。若约 10 秒内找不到整数因子，立即改用二次公式。
When a question asks for answers “to 2 decimal places” or “to 3 significant figures,” it is a strong hint that factorisation will not work.
当题目要求答案”精确到小数点后两位”或”保留 3 位有效数字”时，这强烈暗示因式分解行不通。
In word problems, define the variable clearly and check each solution against the context before finalising your answer.
在应用题中，清晰定义变量，并在确定最终答案前，对照题目情境逐一检验每个解。
Practise past-paper questions from your specific board — Edexcel, CIE, AQA, or OCR — as question styles vary.
针对你所在考试局（Edexcel、CIE、AQA 或 OCR）练习历年真题，因为各考局的出题风格有所不同。

Finally, build a revision summary sheet with the quadratic formula, the discriminant conditions, and three fully worked examples. Review it weekly to keep the methods fresh.

最后，制作一张复习总结表，包含二次公式、判别式的三种情形和三个完整的例题。每周回顾一次，以保持方法熟练。

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Mastering Quadratic Equations | 掌握二次方程