Solving Quadratic Equations | 二次方程的解法

📚 Solving Quadratic Equations | 二次方程的解法

Quadratic equations appear in almost every Edexcel IGCSE Mathematics paper. Mastering the different solving methods is essential for building confidence in algebra and for scoring high marks in both non-calculator and calculator papers.

二次方程几乎出现在每份爱德思 IGCSE 数学试卷中。掌握不同的解法对于建立代数信心、以及在非计算器与计算器两类试卷中取得高分都至关重要。

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

a, b and c are real numbers, and a cannot be zero. If a = 0, the equation becomes linear.

二次方程是未知数最高次数为 2 的方程。一般形式为 ax² + bx + c = 0，其中 a、b、c 是实数，且 a 不能为 0。若 a = 0，方程就变成了线性方程。

For example, x² - 5x + 6 = 0 is quadratic, while 3x + 2 = 0 is linear. The term "quadratic" comes from the Latin word "quadratus", meaning square.

例如，x² - 5x + 6 = 0 是二次方程，而 3x + 2 = 0 是线性方程。"二次"源于拉丁语 "quadratus"，意为"平方"。

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

cient of x² is called a, the coefficient of x is called b, and the constant term is called c.

在求解之前，通常需要将方程整理成标准形式 ax² + bx + c = 0。x² 的系数称为 a，x 的系数称为 b，常数项称为 c。

• Example: Rewrite 3x - 2 = 4x² + 1 as 4x² - 3x + 3 = 0. Here a = 4, b = -3, c = 3.

  例如：将 3x - 2 = 4x² + 1 整理为 4x² - 3x + 3 = 0。此时 a = 4，b = -3，c = 3。

• Always move all terms to one side, keeping the x² term positive where possible.

  务必将所有项移到同一侧，并尽量使 x² 项为正。

• If a common factor exists, divide the entire equation by it before solving, as long as a remains non-zero.

  如果存在公因子，可先将整个方程除以该公因子，但需确保化简后 a 不为 0。

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3. Solving by Factorisation | 因式分解法

as the product of two linear factors. The principle is: if p × q = 0, then p = 0 or q = 0.

当二次表达式可以写成两个一次因式的乘积时，因式分解法是最快捷的方法。其原理是：若 p × q = 0，则 p = 0 或 q = 0。

orise ax² + bx + c when a = 1, find two numbers that multiply to give c and add to give b.

当 a = 1 时，找出两个数，它们相乘等于 c，相加等于 b。

x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = -2 or x = -3

Check the answer by substituting each value back into the original equation. In IGCSE Edexcel papers, factorise-able quadratics are very common, especially in Paper 1F and 1H.

将每个解代回原方程进行检验。在爱德思 IGCSE 试卷中，可因式分解的二次方程非常常见，尤其是 Paper 1F 和 1H。

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4. Special Cases: Difference of Two Squares | 特殊情况：平方差

When the quadratic has the form x² - k², it can be factorised as (x - k)(x + k). This is known as the difference of two squares.

当二次方程具有 x² - k² 的形式时，可因式分解为 (x - k)(x + k)。这称为平方差公式。

x² - 25 = 0 → (x - 5)(x + 5) = 0 → x = 5 or x = -5

Also note that expressions such as 4x² - 9 can be written as (2x)² - 3² and factorised as (2x - 3)(2x + 3).

请注意，像 4x² - 9 这样的表达式可写成 (2x)² - 3²，并因式分解为 (2x - 3)(2x + 3)。

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5. Completing the Square | 配方法

Completing the square is a technique that rewrites a quadratic in the form a(x + p)² + q. It is especially useful for finding the vertex of a parabola and for solving equations that do not factorise.

配方法是将二次表达式改写为 a(x + p)² + q 的形式。它特别适用于求抛物线的顶点，以及求解无法因式分解的方程。

For x² + bx, add and subtract (b/2)². For example, x² + 6x = (x + 3)² - 9.

对于 x² + bx，需加上并减去 (b/2)²。例如，x² + 6x = (x + 3)² - 9。

x² + 6x + 4 = 0 → (x + 3)² - 5 = 0 → (x + 3)² = 5 → x = -3 ± √5

Write the answer as a simplified surd, unless the question asks for a decimal approximation.

答案应写成化简后的根式，除非题目要求给出小数近似值。

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6. The Quadratic Formula | 二次公式

The quadratic formula works for any quadratic equation. It is derived from completing the square and is stated as follows:

二次公式适用于任何二次方程。它由配方法推导而来，其表达式如下：

x = (-b ± √b² - 4ac) / (2a)

In the Edexcel IGCSE exam, this formula is given on the formula page, but you must know how to substitute values correctly and simplify the result.

在爱德思 IGCSE 考试中，公式页会给出这个公式，但你必须会正确代入数值并化简结果。

For example, solve 2x² - 3x - 2 = 0 with a = 2, b = -3, c = -2:

例如，解 2x² - 3x - 2 = 0，其中 a = 2，b = -3，c = -2：

x = (3 ± √9 + 16) / 4 = (3 ± 5) / 4 → x = 2 or x = -1/2

Always simplify the square root first, then calculate the two separate solutions.

务必先化简根号，再分别计算两个解。

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7. The Discriminant | 判别式

The expression b² - 4ac inside the square root is called the discriminant. It tells us how many real roots a quadratic equation has.

根号内的 b² - 4ac 称为判别式。它告诉我们二次方程有多少个实数根。

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

  若 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).

  若 b² - 4ac = 0，方程有一个实数根（重根）。

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

  若 b² - 4ac < 0，方程没有实数根（根为复数）。

In IGCSE questions, you may be asked to "use the discriminant to find the value of k" for which the equation has equal roots.

在 IGCSE 题目中，你可能会被要求"利用判别式求 k 的值"，使得方程有等根。

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

Many real-world problems, such as area, projectile motion and number puzzles, lead to quadratic equations. The first step is always to define a variable, then translate the conditions into an equation.

许多实际问题，如面积、抛体运动和数字谜题，都会产生二次方程。第一步永远是设定变量，然后将条件转化为方程。

A rectangle has length (x + 3) cm and width (x - 1) cm. Its area is 21 cm². Form and solve the equation.

rising gives (x + 6)(x - 4) = 0, so x = 4 (reject x = -6 since length cannot be negative).

面积 = (x + 3)(x - 1) = 21，所以 x² + 2x - 3 = 21，得到 x² + 2x - 24 = 0。因式分解得 (x + 6)(x - 4) = 0，所以 x = 4（舍去 x = -6，因为长度不能为负）。

Always check whether the solutions make sense in the context of the problem.

务必检查解在题目背景中是否合理。

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9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

• Forgetting to set the equation to zero before factorising or using the formula.

  在因式分解或套用公式前，忘记将方程右边化为 0。

• he signs: for x² - 5x + 6 = 0, the correct factors are (x - 2)(x - 3), not (x + 2)(x + 3).

  看错符号：如 x² - 5x + 6 = 0 的正确因式是 (x - 2)(x - 3)，而不是 (x + 2)(x + 3)。

• When using the quadratic formula, forgetting that -b is positive if b is negative.

  使用二次公式时，若 b 为负数，忘记 -b 为正数。

• Leaving answers in unsimplified surd form when the question requires exact values.

  当题目要求精确值时，答案没有化简为最简根式。

• Dropping the ± symbol when taking square roots in completing the square.

  在配方法中开平方时，漏掉 ± 符号。

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10. Practice Questions | 练习题

Here are four questions to test your understanding. Work through them on paper before checking the solutions below.

下面有四道题用于检验你的理解。请先在纸上完成，再查看下方解答。

1. Solve x² - 7x + 12 = 0.
2. Solve x² + 3x - 5 = 0, giving answers to 2 decimal places.
3. Solve 2x² + 4x - 3 = 0 by completing the square.
4. Find k such that kx² + 4x + 2 = 0 has a repeated root.

→ x = 3 or 4. 2. Using the formula: x = 1.19 or -4.19. 3. 2[(x + 1)² - 5/2] = 0 → x = -1 ± √5/2. 4. Discriminant = 16 - 8k = 0 → k = 2.

答案：1. (x - 3)(x - 4) = 0 → x = 3 或 4。2. 用公式：x = 1.19 或 -4.19。3. 2[(x + 1)² - 5/2] = 0 → x = -1 ± √5/2。4. 判别式 = 16 - 8k = 0 → k = 2。

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