Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are a fundamental topic in the Edexcel IGCSE Mathematics syllabus. Mastering them is essential for success in both the foundation and higher tiers, as they appear in algebra, geometry, and problem-solving questions. This revision guide will walk you through the key methods, the discriminant, common pitfalls, and exam-style practice.

二次方程是 Edexcel IGCSE 数学课程中的基础内容。无论是基础级别还是更高级别，掌握二次方程对于代数、几何和应用题都至关重要。本复习指南将带您逐一掌握核心解法、判别式、常见错误以及考试风格的练习。

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

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. In its simplest form, it can be written as ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.

二次方程是最高次数为 2 的多项式方程，即变量的最高幂为 2。其最简形式可写作 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，则方程变为一次方程（线性方程），而非二次方程。

2. Standard Form and Key Terms | 标准形式与关键术语

Before solving, you must recognise the standard form: ax² + bx + c = 0. The term ax² is called the quadratic term, bx is the linear term, and c is the constant term. The coefficient a must not be zero. In IGCSE exams, you may need to rearrange equations into standard form before applying any method.

在求解之前，您必须识别标准形式：ax² + bx + c = 0。其中 ax² 称为二次项，bx 称为一次项，c 称为常数项。系数 a 不能为零。在 IGCSE 考试中，您可能首先需要将方程重新整理为标准形式，再选择解法。

3. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic expression can be written as a product of two linear factors. The principle is the zero product property: if p × q = 0, then either p = 0 or q = 0. For example, solve x² − 5x + 6 = 0 by factorising as (x − 2)(x − 3) = 0. Hence x = 2 or x = 3.

当二次表达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。其原理是零乘积性质：若 p × q = 0，则 p = 0 或 q = 0。例如，解 x² − 5x + 6 = 0，可因式分解为 (x − 2)(x − 3) = 0，因此 x = 2 或 x = 3。

To factorise ax² + bx + c, look for two numbers that multiply to give ac and add to give b. For example, in 2x² + 7x + 3, multiply 2 × 3 = 6. Numbers 6 and 1 add to 7. Rewrite 7x as 6x + x, then factor by grouping: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

对于 ax² + bx + c，找到两个数，使它们的乘积等于 ac，和等于 b。例如，2x² + 7x + 3 中，2 × 3 = 6，6 和 1 的和为 7。将 7x 改写为 6x + x，然后分组因式分解：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This method is useful when factorisation is not obvious, and it also helps in finding turning points of quadratic graphs. For x² + bx, add and subtract (b/2)² to create a perfect square.

配方法将二次表达式改写为 a(x + p)² + q 的形式。当因式分解不明显时，此方法非常有用，同时也能帮助我们找到二次函数图像的顶点。对于 x² + bx，加上并减去 (b/2)² 以构造完全平方。

Example: Solve x² + 6x + 2 = 0. Half of 6 is 3, and 3² = 9. Write x² + 6x + 9 − 9 + 2 = 0, giving (x + 3)² − 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7, and x = −3 ± √7.

示例：解 x² + 6x + 2 = 0。6 的一半是 3，3² = 9。写出 x² + 6x + 9 − 9 + 2 = 0，即 (x + 3)² − 7 = 0。于是 (x + 3)² = 7，所以 x + 3 = ±√7，得 x = −3 ± √7。

When the coefficient of x² is not 1, first factor it out from the first two terms. 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，先将其从前两项中提出。例如，2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3。

5. Solving by the Quadratic Formula | 二次公式法

The quadratic formula works for any quadratic equation and is especially useful when the equation cannot be factorised easily. The formula is derived from completing the square and is given by:

二次公式适用于任何二次方程，尤其是当方程无法轻易因式分解时。该公式由配方法推导而来，具体为：

x = (−b ± √(b² − 4ac)) / (2a)

To use the formula, write the equation in standard form ax² + bx + c = 0, identify a, b, and c, then substitute into the formula. Remember to calculate both solutions, as the ± sign gives two roots. For example, solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3.

使用公式前，请将方程写为标准形式 ax² + bx + c = 0，确定 a、b、c 的值，然后代入公式。务必计算两个解，因为 ± 号会产生两个根。例如，解 2x² − 4x − 3 = 0，此时 a = 2，b = −4，c = −3。

x = (4 ± √((−4)² − 4 × 2 × (−3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2

So the solutions are x = (2 + √10) / 2 and x = (2 − √10) / 2.

因此解为 x = (2 + √10) / 2 和 x = (2 − √10) / 2。

6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: b² − 4ac. It tells us the number and type of roots without solving the equation completely. It is a powerful tool for quickly analysing the nature of solutions.

判别式是二次公式中被开平方的部分：b² − 4ac。它无需完全求解方程即可告诉我们根的数量和类型。这是一个快速分析解的性质的强大工具。

If b² − 4ac > 0, there are two distinct real roots. | 若 b² − 4ac > 0，则有两个不同的实数根。
If b² − 4ac = 0, there is exactly one real root (also called repeated or equal roots). | 若 b² − 4ac = 0，则恰有一个实数根（也称为重根或等根）。
If b² − 4ac < 0, there are no real roots; the equation has two complex roots. | 若 b² − 4ac < 0，则没有实数根；方程有两个复数根。

For the equation x² − 6x + 9 = 0, b² − 4ac = (−6)² − 4 × 1 × 9 = 36 − 36 = 0, so there is one repeated root. Indeed, (x − 3)² = 0 gives x = 3.

对于方程 x² − 6x + 9 = 0，b² − 4ac = (−6)² − 4 × 1 × 9 = 36 − 36 = 0，因此有一个重根。事实上，(x − 3)² = 0 给出 x = 3。

7. Solving Quadratic Inequalities | 二次不等式

Quadratic inequalities involve a quadratic expression and an inequality sign such as , ≤, or ≥. To solve them, first rearrange to standard form, then factorise or use the quadratic formula to find critical values. Plot these values on a number line and test intervals.

二次不等式涉及二次表达式和不等号，如、≤ 或 ≥。解法是：首先整理为标准形式，然后通过因式分解或二次公式找到临界值，在数轴上标出这些值，并测试各区间的符号。

Example: Solve x² − x − 6 < 0. Factorise: (x − 3)(x + 2) < 0. The critical values are x = 3 and x = −2. Testing intervals gives −2 < x < 3.

示例：解 x² − x − 6 < 0。因式分解得 (x − 3)(x + 2) < 0。临界值为 x = 3 和 x = −2。测试区间可得解为 −2 < x < 3。

8. Applications in Problem Solving | 应用题

Quadratic equations are frequently used to model real-life situations such as calculating areas, projectile motion, and profit optimisation. In IGCSE problems, you may need to form a quadratic equation from a word problem, solve it, and interpret the answers in context.

二次方程常用于模拟现实情景，例如计算面积、抛体运动和利润优化。在 IGCSE 题目中，您可能需要从文字问题中建立二次方程，求解，并根据实际背景解释答案。

Example: A rectangle has length 5 cm more than its width, and its area is 84 cm². Let width = w. Then length = w + 5, so w(w + 5) = 84, giving w² + 5w − 84 = 0. Factorising (w + 12)(w − 7) = 0 gives w = −12 or w = 7. Since width cannot be negative, w = 7 cm and length = 12 cm.

示例：一个矩形的长比宽多 5 厘米，面积为 84 平方厘米。设宽为 w，则长为 w + 5，于是 w(w + 5) = 84，即 w² + 5w − 84 = 0。因式分解 (w + 12)(w − 7) = 0 得 w = −12 或 w = 7。宽度不能为负，故 w = 7 厘米，长为 12 厘米。

9. Common Mistakes and Tips | 常见错误与技巧

Many students lose marks on quadratic equations due to simple errors. Be careful when subtracting numbers with signs, when moving terms across the equals sign, and when using the quadratic formula with negative b. Always check your answers by substituting them back into the original equation.

许多学生因小错误在二次方程上失分。请注意符号相减、移项时的正负号，以及在二次公式中处理负 b 的情况。务必通过将解代回原方程来检查答案。

Always rearrange the equation to zero before factorising or using the formula. | 在使用因式分解或公式前，始终将方程整理为等于零的形式。
In factorisation, do not forget the common factor first. Eg 2x² − 8x = 0 → 2x(x − 4) = 0. | 使用因式分解时，不要忘记先提取公因式。例如 2x² − 8x = 0 → 2x(x − 4) = 0。
As soon as you see a square root, remember to include both positive and negative roots when solving equations. | 一看到平方根，解方程时就要注意包含正负两个根。
Use your calculator carefully for the discriminant; a negative under the square root means no real roots. | 计算判别式时要小心；根号下为负数意味着没有实数根。

10. Practice Questions | 练习题

Try the following questions to test your understanding. Solve each equation by the stated method, then check with the quadratic formula if necessary.

尝试以下问题来测试您的理解。按要求的方法求解每个方程，必要时用二次公式验证。

Question	Method
x² − 7x + 10 = 0	Factorisation
x² + 4x − 1 = 0	Completing the square
3x² + 5x − 2 = 0	Quadratic formula

Answers: 1) x = 2 or x = 5. 2) x = −2 ± √5. 3) x = 1/3 or x = −2.

答案：1) x = 2 或 x = 5。2) x = −2 ± √5。3) x = 1/3 或 x = −2。

11. Summary | 总结

In this revision guide, we covered the standard form of a quadratic equation, three main solving methods, the discriminant, quadratic inequalities, real-life applications, and common pitfalls. Each method is effective depending on the situation: factorisation for simple cases, completing the square for graph transformations, and the quadratic formula as a universal technique. Always check the discriminant to understand the nature of the roots. With regular practice, you can confidently tackle any quadratic equation in your IGCSE exam.

在本复习指南中，我们讲解了二次方程的标准形式、三种主要解法、判别式、二次不等式、实际应用及常见陷阱。每种方法各有适用场景：简单情况用因式分解，配方法适用于图像变换，二次公式则是通用技术。始终检查判别式以理解根的性质。通过定期练习，您可以在 IGCSE 考试中自信应对任何二次方程。

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Solving Quadratic Equations | 解二次方程