Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in almost every IGCSE Mathematics exam. Mastering the methods of solving them is essential for achieving top grades. This guide covers factoring, completing the square, the quadratic formula, and the discriminant, with worked examples tailored to the Edexcel syllabus.

二次方程几乎出现在每一份 IGCSE 数学试卷中。掌握其解法是取得高分的关键。本指南涵盖因式分解、配方法、二次公式和判别式，并结合 Edexcel 考纲给出例题。

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

A quadratic equation is any equation that can be written in the 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 makes the graph a parabola.

二次方程是指任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数且 a ≠ 0。变量 x 的最高次数为 2，因此其图像是一条抛物线。

In the Edexcel IGCSE specification, you need to solve quadratic equations by:
• Factoring (factorising)
• Using the quadratic formula
• Completing the square
You also need to understand the discriminant and how it relates to the number of roots.

在 Edexcel IGCSE 考纲中，你需要通过以下方法解二次方程：
• 因式分解
• 使用二次公式
• 配方法
同时你还需要理解判别式及其与根数量的关系。

2. The Standard Form | 标准形式

Before solving, always rearrange the equation into the standard form ax² + bx + c = 0. For example, x² = 5x – 6 becomes x² – 5x + 6 = 0 after subtracting 5x and adding 6 to both sides.

在求解之前，始终将方程整理为标准形式 ax² + bx + c = 0。例如，x² = 5x – 6 在两边减去 5x 并加上 6 后变为 x² – 5x + 6 = 0。

If the equation has brackets or fractions, expand and simplify first. Keep all terms on one side, leaving zero on the other. This is crucial for factoring and for applying the quadratic formula correctly.

如果方程含有括号或分数，先展开并化简。将所有项移到一边，使另一边为零。这对于因式分解和正确使用二次公式至关重要。

3. Method 1: Factoring (Factorising) | 方法一：因式分解

Factoring is the fastest method when the quadratic has simple integer roots. The idea is to write the expression as a product of two linear brackets. For example, x² – 5x + 6 = 0 factors as (x – 2)(x – 3) = 0.

当二次方程有简单的整数根时，因式分解是最快的方法。其思想是将表达式写成两个线性括号的乘积。例如，x² – 5x + 6 = 0 可分解为 (x – 2)(x – 3) = 0。

To factor x² + bx + c, find two numbers that multiply to give c and add to give b. For x² – 5x + 6, the numbers are –2 and –3 because (–2) × (–3) = 6 and (–2) + (–3) = –5.

要对 x² + bx + c 进行因式分解，找到两个数，其乘积等于 c，且和等于 b。对于 x² – 5x + 6，这两个数是 –2 和 –3，因为 (–2) × (–3) = 6 且 (–2) + (–3) = –5。

Once factored, set each bracket equal to zero. So x – 2 = 0 gives x = 2, and x – 3 = 0 gives x = 3. The solutions are x = 2 and x = 3.

因式分解后，令每个括号等于零。于是 x – 2 = 0 得 x = 2，x – 3 = 0 得 x = 3。解为 x = 2 和 x = 3。

For quadratics where a ≠ 1, such as 2x² + 5x + 2, look for factors of a × c that add to b. Here a × c = 4, and the pair 4 and 1 add to 5. Rewrite the middle term: 2x² + 4x + x + 2, then factor by grouping: 2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2).

对于 a ≠ 1 的二次方程，如 2x² + 5x + 2，寻找 a × c 的因子，使其和为 b。这里 a × c = 4，其中 4 和 1 相加等于 5。重写中间项：2x² + 4x + x + 2，然后分组因式分解：2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2)。

4. Solving by Factoring: Worked Example | 因式分解法例题

Solve 3x² – 11x – 4 = 0. Here a = 3, b = –11, c = –4. Multiply a and c: 3 × (–4) = –12. Find two numbers that multiply to –12 and add to –11: they are –12 and 1.

解方程 3x² – 11x – 4 = 0。这里 a = 3，b = –11，c = –4。将 a 和 c 相乘：3 × (–4) = –12。找到两个数，其乘积为 –12，和为 –11：它们是 –12 和 1。

Rewrite the middle term: 3x² – 12x + x – 4. Group: 3x(x – 4) + 1(x – 4) = (3x + 1)(x – 4). Set each factor to zero: 3x + 1 = 0 → x = –⅓, and x – 4 = 0 → x = 4.

重写中间项：3x² – 12x + x – 4。分组：3x(x – 4) + 1(x – 4) = (3x + 1)(x – 4)。令每个因子为零：3x + 1 = 0 → x = –⅓，x – 4 = 0 → x = 4。

Always check your solutions by substituting back into the original equation. For x = 4: 3(16) – 44 – 4 = 0. For x = –⅓: 3(1/9) – 11(–⅓) – 4 = ⅓ + 11/3 – 4 = 0.

始终通过代回原方程来检查解。对于 x = 4：3(16) – 44 – 4 = 0。对于 x = –⅓：3(1/9) – 11(–⅓) – 4 = ⅓ + 11/3 – 4 = 0。

5. Method 2: The Quadratic Formula | 方法二：二次公式

The quadratic formula solves any quadratic equation, even when factoring is difficult or impossible. The formula is derived from completing the square and is given by:

二次公式可以解任何二次方程，即使因式分解困难或无法进行。它由配方法推导而来，公式如下：

x = (–b ± √(b² – 4ac)) / (2a)

For the equation ax² + bx + c = 0, substitute the values of a, b and c. The symbol ± means there are two possible solutions: one with addition and one with subtraction.

对于方程 ax² + bx + c = 0，代入 a、b、c 的值。符号 ± 表示有两个可能的解：一个用加法，一个用减法。

It is essential to write down the formula first, then substitute carefully. Many marks are lost because of sign errors, especially when b is negative.

务必先写出公式，再仔细代入。许多分数因符号错误而丢失，尤其是当 b 为负数时。

6. Using the Quadratic Formula: Example | 二次公式示例

Solve 2x² + 3x – 5 = 0 using the quadratic formula. Here a = 2, b = 3, c = –5. Substitute into the formula:

使用二次公式解 2x² + 3x – 5 = 0。这里 a = 2，b = 3，c = –5。代入公式：

x = (–3 ± √(3² – 4 × 2 × (–5))) / (2 × 2)

Simplify the discriminant: 9 – 4 × 2 × (–5) = 9 + 40 = 49. So x = (–3 ± √49) / 4 = (–3 ± 7) / 4.

化简判别式：9 – 4 × 2 × (–5) = 9 + 40 = 49。因此 x = (–3 ± √49) / 4 = (–3 ± 7) / 4。

Now evaluate both cases: x = (–3 + 7) / 4 = 4/4 = 1, and x = (–3 – 7) / 4 = –10/4 = –2.5. The solutions are x = 1 and x = –2.5.

计算两种情况：x = (–3 + 7) / 4 = 4/4 = 1，x = (–3 – 7) / 4 = –10/4 = –2.5。解为 x = 1 和 x = –2.5。

Notice that this equation could also be factored as (2x – 2)(x + 5) = 0? Actually, (2x – 2)(x + 5) expands to 2x² + 10x – 2x – 10 = 2x² + 8x – 10, which is not the same. A better check is to factor 2x² + 3x – 5 as (2x + 5)(x – 1). Indeed, that expands to 2x² – 2x + 5x – 5 = 2x² + 3x – 5. The formula remains reliable.

注意，这个方程是否也可以分解为 (2x – 2)(x + 5) = 0？实际上 (2x – 2)(x + 5) 展开得 2x² + 10x – 2x – 10 = 2x² + 8x – 10，与原式不同。更好的检查是将 2x² + 3x – 5 分解为 (2x + 5)(x – 1)。确实，展开得 2x² – 2x + 5x – 5 = 2x² + 3x – 5。公式依然可靠。

7. Method 3: Completing the Square | 方法三：配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This form is useful for sketching graphs and for solving equations without using the quadratic formula.

配方法将二次式重写为 a(x + p)² + q 的形式。这种形式有助于画图，也可以在不使用二次公式的情况下解方程。

For a simple monic quadratic x² + bx + c, take half of b, square it, and adjust. For example, x² + 6x + 2 becomes (x + 3)² – 9 + 2 = (x + 3)² – 7.

对于首项系数为 1 的二次式 x² + bx + c，取 b 的一半，平方，然后调整。例如，x² + 6x + 2 变为 (x + 3)² – 9 + 2 = (x + 3)² – 7。

To solve (x + 3)² – 7 = 0, rearrange: (x + 3)² = 7. Then take the square root of both sides: x + 3 = ±√7. Hence x = –3 ± √7.

要解 (x + 3)² – 7 = 0，移项得 (x + 3)² = 7。然后两边开平方：x + 3 = ±√7。因此 x = –3 ± √7。

When a ≠ 1, factor out a first. For 2x² + 8x + 3, write 2(x² + 4x) + 3, complete the square inside the bracket: 2[(x + 2)² – 4] + 3 = 2(x + 2)² – 8 + 3 = 2(x + 2)² – 5.

当 a ≠ 1 时，先提出 a。对于 2x² + 8x + 3，写为 2(x² + 4x) + 3，在括号内配方：2[(x + 2)² – 4] + 3 = 2(x + 2)² – 8 + 3 = 2(x + 2)² – 5。

8. The Discriminant | 判别式

The discriminant is the expression Δ = b² – 4ac inside the square root of the quadratic formula. Its value tells us how many real roots a quadratic equation has, without solving it.

判别式是二次公式中根号内的表达式 Δ = b² – 4ac。它的值告诉我们二次方程有多少个实数根，而无需解方程。

• If Δ > 0, there are two distinct real roots.
• If Δ = 0, there is exactly one real root (a repeated root).
• If Δ < 0, there are no real roots (the solutions are complex, which is beyond IGCSE).

• 若 Δ > 0，有两个不同的实数根。
• 若 Δ = 0，恰好有一个实数根（重根）。
• 若 Δ < 0，没有实数根（解为复数，超出 IGCSE 范围）。

For example, for x² – 4x + 5 = 0, Δ = (–4)² – 4 × 1 × 5 = 16 – 20 = –4 < 0, so the equation has no real roots. The graph of y = x² – 4x + 5 does not intersect the x-axis.

例如，对于 x² – 4x + 5 = 0，Δ = (–4)² – 4 × 1 × 5 = 16 – 20 = –4 < 0，因此方程没有实数根。y = x² – 4x + 5 的图像不与 x 轴相交。

9. Solving Word Problems | 解应用题

Quadratic equations often arise from geometry or number problems. For example: “The area of a rectangle is 54 cm². Its length is 3 cm more than twice its width. Find the width.” Let the width be w. Then length = 2w + 3, and area = w(2w + 3) = 54.

二次方程常常来自几何或数字问题。例如：“一个矩形的面积为 54 cm²。它的长比宽的 2 倍多 3 cm。求宽。”令宽为 w。则长 = 2w + 3，面积 = w(2w + 3) = 54。

Expand: 2w² + 3w – 54 = 0. Factor: (2w – 9)(w + 6) = 0? Let’s check: (2w – 9)(w + 6) = 2w² + 12w – 9w – 54 = 2w² + 3w – 54. Yes. So w = 4.5 or w = –6. Since width cannot be negative, w = 4.5 cm.

展开：2w² + 3w – 54 = 0。因式分解：(2w – 9)(w + 6) = 0？检查：(2w – 9)(w + 6) = 2w² + 12w – 9w – 54 = 2w² + 3w – 54。正确。因此 w = 4.5 或 w = –6。由于宽不能为负，w = 4.5 cm。

Always interpret the solutions in the context of the problem. A negative root may be algebraically valid but physically impossible.

始终根据问题情境解释解。负根可能在代数上成立，但在现实中不可能。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

One common mistake is forgetting to rearrange the equation into standard form before solving. For example, x² = 3x is often incorrectly solved by dividing both sides by x, losing the solution x = 0. The correct method is x² – 3x = 0 → x(x – 3) = 0 → x = 0 or x = 3.

一个常见错误是在求解前忘记将方程整理为标准形式。例如，x² = 3x 常被错误地两边除以 x，从而丢失解 x = 0。正确方法是 x² – 3x = 0 → x(x – 3) = 0 → x = 0 或 x = 3。

Another common error is sign errors in the quadratic formula. Carefully write –b, and when b is negative, remember that –(negative) becomes positive. Also, make sure the denominator is 2a, not 2.

另一个常见错误是二次公式中的符号错误。仔细写出 –b，当 b 为负数时，记住负负得正。另外，确保分母是 2a，而不是 2。

In the exam, always show your working clearly. Even if your final answer is wrong, you can earn method marks. If a quadratic does not factor easily, use the formula or complete the square.

在考试中，始终清晰展示解题过程。即使最终答案错误，你也能获得方法分。如果方程不易因式分解，使用公式或配方法。

Finally, check the discriminant before deciding on a method. If Δ is a perfect square, the quadratic factors neatly; otherwise, the formula or completing the square is preferable.

最后，在决定使用哪种方法前先检查判别式。如果 Δ 是完全平方数，则二次式可以整齐地因式分解；否则，使用公式或配方法更好。

11. Practice Questions | 练习题

Try these problems on your own, then check your answers at the end.

请独立尝试以下问题，然后在末尾核对答案。

Solve x² – 7x + 10 = 0 by factoring.
Solve 3x² + 5x – 2 = 0 using the quadratic formula.
Solve x² – 6x + 4 = 0 by completing the square.
Find the discriminant of x² – 2x + 3 = 0 and state the number of real roots.
A square has its side increased by 4 cm, and its area increases by 128 cm². Find the original side length.

用因式分解解 x² – 7x + 10 = 0。

用二次公式解 3x² + 5x – 2 = 0。

用配方法解 x² – 6x + 4 = 0。

求 x² – 2x + 3 = 0 的判别式，并说明实数根的个数。

一个正方形的边长增加 4 cm 后，面积增加了 128 cm²。求原来的边长。

Answers: 1) x = 2, 5. 2) x = ⅓, –2. 3) x = 3 ± √5. 4) Δ = –8, no real roots. 5) Original side = 14 cm (solve (x+4)² – x² = 128 → 8x + 16 = 128 → x = 14).

答案： 1) x = 2, 5。 2) x = ⅓, –2。 3) x = 3 ± √5。 4) Δ = –8，没有实数根。 5) 原边长为 14 cm（解 (x+4)² – x² = 128 → 8x + 16 = 128 → x = 14）。

12. Summary | 总结

Quadratic equations are a central topic in IGCSE Mathematics. You must be confident in all three major methods: factoring, the quadratic formula, and completing the square. The discriminant helps you choose a strategy and understand the nature of the roots.

二次方程是 IGCSE 数学的核心内容。你必须熟练三种主要方法：因式分解、二次公式和配方法。判别式有助于你选择策略并理解根的性质。

Remember to always rearrange into standard form, show clear working, and check your answers. With regular practice, solving quadratic equations will become second nature.

记住始终整理为标准形式，清晰展示步骤，并检查答案。通过经常练习，解二次方程将成为你的第二天性。

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Solving Quadratic Equations | 解二次方程