Master Quadratic Equations | 掌握一元二次方程

📚 Master Quadratic Equations | 掌握一元二次方程

Quadratic equations form the foundation of IGCSE algebra and appear in nearly every exam paper. Understanding how to solve them efficiently is essential for achieving top grades. This guide covers every method you need, complete with step-by-step explanations and common pitfalls to avoid.

一元二次方程是 IGCSE 数学代数部分的核心内容，几乎出现在每一份试卷中。掌握高效解法是冲击高分的关键。本指南将系统讲解所有必考解法，提供分步讲解，并指出常见易错点。

1. What Is a Quadratic Equation | 什么是一元二次方程

A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which gives the equation its name ‘quadratic’.

一元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。变量 x 的最高次数为 2，这也是”二次”名称的由来。

The coefficient a is called the leading coefficient. If a = 0, the equation becomes linear, not quadratic. The value c is called the constant term, and b is the coefficient of x.

系数 a 称为首项系数。如果 a = 0，方程就变成一次方程，而非二次方程。c 称为常数项，b 是 x 的系数。

For example, 2x² – 5x + 3 = 0 is a quadratic equation where a = 2, b = -5 and c = 3. Similarly, x² = 9 can be rearranged to x² – 9 = 0, which is also quadratic with b = 0.

例如，2x² – 5x + 3 = 0 是一元二次方程，其中 a = 2，b = -5，c = 3。同样，x² = 9 可以变形为 x² – 9 = 0，这也是二次方程，此时 b = 0。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic expression has simple integer factors. The principle is based on the zero product property: if the product of two expressions is zero, then at least one of them must be zero.

当二次表达式具有简单的整数因子时，因式分解法通常是最快捷的方法。其原理基于零乘积性质：如果两个表达式的乘积为零，那么至少其中一个必须为零。

Example: Solve x² + 7x + 12 = 0

示例：解方程 x² + 7x + 12 = 0

Step 1: Find two numbers that multiply to give 12 and add to give 7. These numbers are 3 and 4.

第一步：找到两个数，它们相乘等于 12，相加等于 7。这两个数是 3 和 4。

Step 2: Write the factorised form: (x + 3)(x + 4) = 0

第二步：写出因式分解形式：(x + 3)(x + 4) = 0

Step 3: Apply the zero product property: x + 3 = 0 or x + 4 = 0

第三步：应用零乘积性质：x + 3 = 0 或 x + 4 = 0

Step 4: Solve each linear equation: x = -3 or x = -4

第四步：解每个一次方程：x = -3 或 x = -4

x² + 7x + 12 = 0 → (x + 3)(x + 4) = 0 → x = -3 或 x = -4

Always check whether the coefficient of x² is greater than 1. In that case, you need to consider factor pairs of both a and c. For example, to factorise 2x² + 7x + 3, you look for factors of 2 and 3 that combine to give 7.

当 x² 的系数大于 1 时，需要同时考虑 a 和 c 的因子对。例如，分解 2x² + 7x + 3 时，需要寻找 2 和 3 的因子组合，使其交叉相乘之和为 7。

2x² + 7x + 3 = (2x + 1)(x + 3)

3. Solving by the Quadratic Formula | 公式法

The quadratic formula works for every quadratic equation, including those that cannot be factorised easily. You should memorise it and know when to apply it.

公式法适用于所有一元二次方程，包括那些不易因式分解的方程。你需要牢记公式并知道何时使用。

x = (-b ± √(b² – 4ac)) / 2a

To use this formula, identify the values of a, b and c from the equation ax² + bx + c = 0. Substitute them carefully, paying close attention to negative signs.

使用此公式时，先从方程 ax² + bx + c = 0 中确定 a、b、c 的值，然后小心代入，特别注意负号。

Example: Solve 2x² – 3x – 2 = 0

示例：解方程 2x² – 3x – 2 = 0

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

这里 a = 2，b = -3，c = -2。代入公式：

x = (3 ± √((-3)² – 4 × 2 × (-2))) / (2 × 2)

x = (3 ± √(9 + 16)) / 4 = (3 ± √25) / 4 = (3 ± 5) / 4

This gives two solutions: x = (3 + 5)/4 = 2, and x = (3 – 5)/4 = -1/2.

由此得到两个解：x = (3 + 5)/4 = 2，以及 x = (3 – 5)/4 = -1/2。

4. The Discriminant and Its Meaning | 判别式及其含义

The expression b² – 4ac inside the square root is called the discriminant. It determines how many solutions a quadratic equation has without actually solving it.

根号内的表达式 b² – 4ac 称为判别式。它可以在不解方程的情况下判断一元二次方程解的个数。

If b² – 4ac > 0, the equation has two distinct real roots. If b² – 4ac = 0, the equation has exactly one real root, called a repeated root. If b² – 4ac < 0, the equation has no real roots.

如果 b² – 4ac > 0，方程有两个不相等的实数根。如果 b² – 4ac = 0，方程恰好有一个实数根，称为重根。如果 b² – 4ac < 0，方程没有实数根。

判别式取值	解的个数	图像含义
b² – 4ac > 0	两个不同实数根	抛物线穿过 x 轴两次
b² – 4ac = 0	一个重根	抛物线切 x 轴于一点
b² – 4ac < 0	无实数根	抛物线不接触 x 轴

Example: For the equation x² + 4x + 5 = 0, the discriminant is 4² – 4 × 1 × 5 = 16 – 20 = -4. Since -4 < 0, the equation has no real solutions.

示例：对于方程 x² + 4x + 5 = 0，判别式为 4² – 4 × 1 × 5 = 16 – 20 = -4。因为 -4 < 0，所以方程没有实数解。

5. Solving by Completing the Square | 配方法

Completing the square is a technique that rewrites a quadratic expression in the form a(x + p)² + q. This method is especially useful for solving equations and for finding the vertex of a parabola.

配方法是将二次表达式改写为 a(x + p)² + q 形式的技术。这种方法在解方程和求抛物线顶点时尤其有用。

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

示例：用配方法解 x² + 6x – 7 = 0

Step 1: Move the constant term to the right: x² + 6x = 7

第一步：将常数项移到右边：x² + 6x = 7

Step 2: Take half of the coefficient of x (which is 6), giving 3. Square it to get 9. Add 9 to both sides: x² + 6x + 9 = 16

第二步：取 x 系数 6 的一半得到 3，平方得 9。两边同时加 9：x² + 6x + 9 = 16

Step 3: Rewrite the left side as a perfect square: (x + 3)² = 16

第三步：将左边改写为完全平方式：(x + 3)² = 16

Step 4: Take the square root of both sides: x + 3 = ±4

第四步：两边开平方：x + 3 = ±4

Step 5: Solve: x = 1 or x = -7

第五步：解得：x = 1 或 x = -7

When the coefficient of x² is not 1, divide the entire equation by that coefficient first before completing the square.

当 x² 的系数不为 1 时，先要将整个方程除以该系数，再进行配方。

6. Solving by the Graph Method | 图像法

The solutions of a quadratic equation ax² + bx + c = 0 correspond to the x-intercepts of the parabola y = ax² + bx + c. By plotting the graph, you can read off the roots where the curve crosses the x-axis.

一元二次方程 ax² + bx + c = 0 的解对应于抛物线 y = ax² + bx + c 与 x 轴的交点。通过绘制图像，可以读出曲线与 x 轴交点处的根。

If the parabola does not intersect the x-axis, the equation has no real roots. If it touches the axis at exactly one point, there is one repeated root. If it crosses at two points, there are two distinct roots.

如果抛物线与 x 轴没有交点，则方程无实数根。如果抛物线恰好与 x 轴相切于一点，则有一个重根。如果穿过两个点，则有两个不相等的根。

When asked to solve graphically, construct a table of values for x and y, plot the points accurately, and draw a smooth curve. The roots are the x-coordinates where y = 0.

当题目要求用图像法求解时，先列出 x 和 y 的数值表，准确描点，并画出平滑曲线。根就是 y = 0 时所对应的 x 坐标。

Remember that graphical solutions are approximate unless the roots happen to be integers that can be read exactly from the graph.

请注意，除非根恰好是能从图像上精确读出的整数，否则图像法得到的解是近似值。

7. Roots and Coefficients Relationship | 根与系数的关系

For a quadratic equation ax² + bx + c = 0 with roots α and β, there is an important relationship between the roots and the coefficients. This relationship allows you to find the sum and product of the roots without solving the equation.

对于根为 α 和 β 的一元二次方程 ax² + bx + c = 0，根与系数之间存在重要关系。利用这一关系，无需解方程即可求出两根之和与两根之积。

α + β = -b/a 且 α × β = c/a

For example, for the equation 2x² – 8x + 6 = 0, the sum of the roots is -(-8)/2 = 4, and the product of the roots is 6/2 = 3.

例如，对于方程 2x² – 8x + 6 = 0，两根之和为 -(-8)/2 = 4，两根之积为 6/2 = 3。

This relationship is useful when checking your answers after solving, or when constructing a quadratic equation with given roots. If the roots are α and β, the equation can be written as x² – (α + β)x + αβ = 0.

这个关系在解完方程后检查答案时非常有用，也可以用来构造已知根的二次方程。如果根为 α 和 β，方程可以写成 x² – (α + β)x + αβ = 0。

8. Word Problems with Quadratics | 二次方程应用题

Word problems often require you to translate a real-life situation into a quadratic equation. The most common types involve area, projectile motion, and number puzzles.

应用题通常需要将实际情境转化为二次方程。最常见的类型涉及面积问题、抛体运动问题和数字谜题。

Example: A rectangle has a length that is 3 cm longer than its width. Its area is 40 cm². Find the dimensions of the rectangle.

示例：一个长方形的长比宽长 3 cm，面积为 40 cm²。求长方形的尺寸。

Let the width be x cm. Then the length is (x + 3) cm. The area equation is x(x + 3) = 40, which expands to x² + 3x – 40 = 0.

设宽为 x cm，则长为 (x + 3) cm。面积方程为 x(x + 3) = 40，展开得 x² + 3x – 40 = 0。

Factorise: (x + 8)(x – 5) = 0, giving x = -8 or x = 5. Since a length cannot be negative, x = 5. The width is 5 cm and the length is 8 cm.

因式分解得：(x + 8)(x – 5) = 0，所以 x = -8 或 x = 5。长度不能为负数，因此 x = 5。宽为 5 cm，长为 8 cm。

Always check whether your answers make sense in the context of the problem. Negative or unrealistically large values should be rejected with a clear justification.

务必检查答案在题目情境中是否合理。负数或不切实际的大数值应明确排除并说明理由。

9. Choosing the Right Method | 如何选择合适的方法

Knowing which method to use can save you valuable time in the exam. Each method has its strengths and ideal use cases.

知道该用哪种方法可以为你节省宝贵的考试时间。每种方法都有其优势和最佳适用场景。

方法	适用场景	注意事项
因式分解法	系数为整数且可分解	快速但并非总能分解
公式法	任何二次方程	需要小心代入，计算量大
配方法	求顶点或证明题	步骤较多
图像法	估算根的位置	结果不够精确

In general, try factorisation first when the coefficients are small integers. If factorisation is not obvious, apply the quadratic formula. Use completing the square when the question specifically asks for it or when you need to find the turning point of a parabola.

一般来说，当系数是较小的整数时，先尝试因式分解法。如果因式分解不明显，就用公式法。当题目明确要求使用配方法，或者需要求抛物线的顶点时，使用配方法。

10. Common Mistakes to Avoid | 常见易错点

Many students lose marks on quadratic equations due to avoidable errors. Here are the most common pitfalls and how to avoid them.

许多学生因为在二次方程题目中出现可避免的错误而丢分。以下是最常见的陷阱以及如何避开它们。

Forgetting to set the equation to zero: Always rearrange the equation to the form ax² + bx + c = 0 before factorising or applying the formula.
忘记将方程化为零：在因式分解或代入公式之前，务必先将方程整理为 ax² + bx + c = 0 的形式。
Misidentifying coefficients: Take care with negative signs. In -3x² + 5x – 2 = 0, a = -3, not 3.
系数识别错误：注意负号。在 -3x² + 5x – 2 = 0 中，a = -3，而不是 3。
Incorrect sign when taking square roots: If x² = 9, then x = ±3, not just x = 3.
开平方时符号错误：如果 x² = 9，那么 x = ±3，而不仅仅是 x = 3。
Dropping the denominator in the formula: The entire expression (-b ± √(b² – 4ac)) is divided by 2a.
公式中漏掉分母：整个表达式 (-b ± √(b² – 4ac)) 都要除以 2a。
Ignoring the context in word problems: Reject negative values when they represent physical quantities like length or time.
忽略应用题中的实际情境：当负数代表长度或时间等物理量时，应将其舍去。

11. Practice Questions | 练习题

Apply what you have learned by solving these questions. Attempt each one using the most suitable method, then check your answers.

通过解答以下题目来巩固所学内容。请尝试使用最适合的方法解答每一题，然后核对答案。

Question 1: Solve x² – 9x + 20 = 0.

第 1 题：解方程 x² – 9x + 20 = 0。

Question 2: Solve 3x² + 5x – 2 = 0 using the quadratic formula.

第 2 题：用公式法解方程 3x² + 5x – 2 = 0。

Question 3: Determine the number of real roots of x² – 6x + 10 = 0.

第 3 题：判断方程 x² – 6x + 10 = 0 的实数根的个数。

Question 4: A square has its side increased by 4 cm, and its area becomes 81 cm². Find the original side length.

第 4 题：一个正方形的边长增加 4 cm 后，面积变为 81 cm²。求原来的边长。

Answers: 1) x = 4 或 x = 5 2) x = 1/3 或 x = -2 3) 无实数根 4) 原边长为 5 cm

答案：1) x = 4 或 x = 5 2) x = 1/3 或 x = -2 3) 无实数根 4) 原边长为 5 cm

12. Summary and Exam Tips | 总结与考试技巧

Quadratic equations are a guaranteed topic in IGCSE Mathematics. Mastery of all four solving methods gives you flexibility and confidence in any exam scenario.

一元二次方程是 IGCSE 数学中的必考内容。掌握所有四种解法能让你在考试中灵活应对，充满信心。

In the exam, read each question carefully to identify which method is expected. Show all working steps clearly, as method marks are awarded even if the final answer is wrong.

在考试中，仔细审题，判断题目期望使用哪种方法。清晰写出所有步骤，因为即使最终答案错误，步骤分仍然可以获得。

Double-check your answers by substituting them back into the original equation. This takes only a few seconds and can catch careless errors.

将答案代回原方程进行验证。这个过程只需几秒钟，却可以帮你发现粗心导致的错误。

Finally, practise regularly with past paper questions. The more familiar you are with different question styles, the more confident you will be on exam day.

最后，定期练习历年真题。你越熟悉不同的题型，考试时就会越自信。

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Master Quadratic Equations | 掌握一元二次方程