Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are a fundamental topic in IGCSE mathematics. Understanding how to solve them is essential for success in tests and exams.

二次方程是 IGCSE 数学的基础内容。掌握其解法对于考试取得好成绩至关重要。

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

A quadratic equation is a polynomial equation of degree 2. The highest power of the variable is 2, and its general form is:

二次方程是最高次数为 2 的多项式方程。变量最高次数是 2，它的一般形式是：

ax² + bx + c = 0

where a, b and c are constants, and a ≠ 0. If a = 0, the equation becomes linear.

其中 a、b、c 是常数，且 a ≠ 0。如果 a = 0，方程就变成了一次方程。

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2. Standard Form and Coefficients | 标准形式与系数

g, rearrange the equation into standard form ax² + bx + c = 0, with all terms on one side.

在求解之前，先把方程整理成标准形式 ax² + bx + c = 0，即把所有项移到等号一边。

For example, x² = 5x – 6 becomes x² – 5x + 6 = 0. Here a = 1, b = –5, c = 6.

例如，x² = 5x – 6 可化为 x² – 5x + 6 = 0。这里 a = 1，b = –5，c = 6。

It is important to identify coefficients correctly when using the quadratic formula or factorising.

在使用求根公式或因式分解时，正确识别系数非常重要。

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3. Solving by Factoring | 因式分解法

If the quadratic expression can be factorised, this is often the quickest method.

如果二次表达式可以因式分解，这通常是最快的方法。

Example: Solve x² – 5x + 6 = 0. We find two numbers whose product is 6 and sum is –5. Those numbers are –2 and –3.

例如：解 x² – 5x + 6 = 0。我们需要找到两个数，乘积为 6，和为 –5。这两个数是 –2 和 –3。

(x – 2)(x – 3) = 0

Then x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.

于是 x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。

When the coefficient a is not 1, factorisation may require the "grouping" method.

当系数 a 不是 1 时，因式分解可能需要使用“分组”法。

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4. Solving by Completing the Square | 配方法

re transforms the equation into the form (x + p)² = q, from which x can be found directly.

配方法将方程转化为 (x + p)² = q 的形式，然后直接求出 x。

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

例：用配方法解 x² + 6x – 7 = 0。

Move the constant to the right: x² + 6x = 7.

先把常数项移到右边：x² + 6x = 7。

Add (6/2)² = 9 to both sides: x² + 6x + 9 = 16.

两边加上 (6/2)² = 9：x² + 6x + 9 = 16。

(x + 3)² = 16

Taking square roots: x + 3 = ±4, so x = 1 or x = –7.

两边开平方：x + 3 = ±4，所以 x = 1 或 x = –7。

This method is also useful for finding the vertex of a parabola.

配方法也常用于求抛物线的顶点。

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5. The Quadratic Formula | 求根公式

The quadratic formula works for all quadratic equations, even when factorisation is difficult or impossible.

求根公式适用于所有二次方程，即使因式分解困难或无法因式分解也有效。

x = (–b ± √b² – 4ac) / (2a)

To use it, substitute the values of a, b and c from the standard form.

使用时，从标准形式中代入 a、b、c 的值。

Example: Solve 2x² + 3x – 2 = 0. Here a = 2, b = 3, c = –2.

例：解 2x² + 3x – 2 = 0。这里 a = 2，b = 3，c = –2。

x = (–3 ± √(3² – 4 × 2 × (–2))) / (2 × 2) = (–3 ± √25) / 4

So x = (–3 + 5)/4 = 0.5 or x = (–3 – 5)/4 = –2.

因此 x = (–3 + 5)/4 = 0.5 或 x = (–3 – 5)/4 = –2。

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6. Discriminant and Nature of Roots | 判别式与根的性质

The discriminant is the part under the square root in the formula: b² – 4ac.

判别式是求根公式中根号下的部分：b² – 4ac。

It tells us the number and type of roots without solving the equation fully.

它可以在不解出完整方程的情况下告诉我们根的数量和类型。

• If b² – 4ac > 0, there are two distinct real roots.

  如果 b² – 4ac > 0，方程有两个不相等的实数根。

• If b² – 4ac = 0, there is exactly one real root (a repeated root).

  如果 b² – 4ac = 0，方程有一个实数根（重根）。

• If b² – 4ac < 0, there are no real roots (two complex roots).

  如果 b² – 4ac < 0，方程没有实数根（有两个复数根）。

This is helpful when deciding which method to use or when analysing graphs.

这在选择解法或分析图像时很有帮助。

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7. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0, the sum of the roots is –b/a and the product of the roots is c/a.

对于二次方程 ax² + bx + c = 0，两根之和等于 –b/a，两根之积等于 c/a。

If the roots are α and β, then:

若两根为 α 和 β，则：

α + β = –b/a, αβ = c/a

These relations allow us to express new equations or check our solutions.

这些关系可用于构造新方程或检验我们得到的解是否正确。

For example, if roots are 2 and 3, the equation is x² – 5x + 6 = 0, since sum = 5 and product = 6.

例如，若根为 2 和 3，则方程是 x² – 5x + 6 = 0，因为和为 5，积为 6。

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8. Solving Word Problems | 应用题求解

Many real-world problems lead to quadratic equations. The key steps are to define variables, form an equation, solve it, and check the validity of answers.

许多实际问题会转化为二次方程。关键步骤是设变量、列方程、求解并检验答案的合理性。

Example: The area of a rectangle is 28 cm², and its length is 3 cm more than its width. Find the width.

例：一个长方形的面积是 28 cm²，长比宽多 3 cm。求宽。

Let width = x, then length = x + 3. Area = x(x + 3) = 28.

设宽为 x，则长为 x + 3。面积 = x(x + 3) = 28。

x² + 3x – 28 = 0

Factorise: (x + 7)(x – 4) = 0. So x = –7 or x = 4. Since width cannot be negative, the width is 4 cm.

因式分解：(x + 7)(x – 4) = 0。所以 x = –7 或 x = 4。由于宽不可能为负，所以宽是 4 cm。

Always interpret negative roots in context and reject them if they do not make sense.

在具体情境中要解释负根，如果负根没有意义则将其舍去。

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9. Common Mistakes and Tips | 常见错误与技巧

Students often forget to rearrange the equation to standard form before using the formula.

学生经常忘记在使用求根公式前将方程整理成标准形式。

Another common mistake is misidentifying the signs of a, b and c, especially when they are negative.

另一个常见错误是弄错 a、b、c 的符号，尤其当它们是负数时。

• Always check if b² – 4ac is positive, zero, or negative before attempting to find real roots.

  在寻找实数根之前，先判断 b² – 4ac 是正数、零还是负数。

• When factorising, expand your answer to verify the original expression.

  因式分解后，展开答案来验算是否与原表达式一致。

• For questions involving fractions with x in the denominator, multiply through by the common denominator first.

  对于分母含 x 的分数问题，先乘以公分母。

Practising the three main methods will help you choose the fastest one during the exam.

练习三种主要解法有助于你在考试中选择最快的方法。

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10. Practice Questions | 练习题

Try these problems to solidify your understanding.

尝试以下题目来巩固理解。

1. Solve x² – 7x + 10 = 0 by factorising.

   用因式分解法解 x² – 7x + 10 = 0。

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

   用求根公式解 2x² + 5x – 3 = 0。

3. Complete the square for x² + 8x + 15 = 0, then solve.

   用配方法解 x² + 8x + 15 = 0，然后求解。

4. Find the discriminant of 3x² – 6x + 2 = 0 and state the nature of its roots.

   求 3x² – 6x + 2 = 0 的判别式，并说明根的性质。

5. The product of two consecutive odd integers is 63. Find the integers.

   两个连续奇数的乘积是 63。求这两个奇数。

Answers: 1) x = 2, 5; 2) x = 0.5, –3; 3) x = –3, –5; 4) Δ = 12, two distinct real roots; 5) 7 and 9 (or –9 and –7).

答案：1) x = 2，5；2) x = 0.5，–3；3) x = –3，–5；4) Δ = 12，两个不相等的实数根；5) 7 和 9（或 –9 和 –7）。

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