Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, word problems and even geometry. Mastering the methods to solve them will give you a strong foundation for both Paper 2 and Paper 4.

二次方程是 IGCSE 数学中最重要的考点之一。它出现在代数、函数图像、应用题甚至几何题目中。掌握求解二次方程的各种方法，可以为 Paper 2 和 Paper 4 打下坚实基础。

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

A quadratic equation is an equation where the highest power of the variable is 2. Its general form is written below.

二次方程是指变量最高次数为 2 的方程，其一般形式如下。

ax² + bx + c = 0, where a ≠ 0

Here a, b and c are real numbers, and a cannot be zero. If a were zero, the equation would become linear, not quadratic.

其中 a、b、c 为实数，且 a 不能为零。如果 a 等于零，方程就变成了一次方程（线性方程），而不是二次方程。

For example, 2x² − 3x + 1 = 0 is quadratic because the term x² exists. The equation x + 5 = 0 is not quadratic because it has no x² term.

例如，2x² − 3x + 1 = 0 是二次方程，因为它含有 x² 项；而 x + 5 = 0 不是二次方程，因为它不含 x² 项。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic has simple integer roots. The idea is to rewrite ax² + bx + c as a product of two brackets.

当二次方程的根是简单整数时，因式分解法通常是最快的求解方法。其核心思想是将 ax² + bx + c 改写为两个括号相乘的形式。

Consider the equation x² + 5x + 6 = 0. We look for two numbers that multiply to give 6 and add to give 5. The numbers 2 and 3 work, so we factorise.

以方程 x² + 5x + 6 = 0 为例。我们需要找到两个数，它们相乘等于 6，相加等于 5。数字 2 和 3 满足条件，因此可以因式分解。

(x + 2)(x + 3) = 0

Since the product of the two brackets is zero, at least one bracket must equal zero. Therefore x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.

因为两个括号相乘等于零，所以其中至少一个括号必须等于零。因此 x + 2 = 0 或 x + 3 = 0，解得 x = −2 或 x = −3。

When the coefficient a is not 1, for example 2x² + 5x + 2 = 0, you need a slightly smarter approach. Find two numbers that multiply to a × c = 4 and add to b = 5. The numbers 4 and 1 work.

当 a 不等于 1 时，例如 2x² + 5x + 2 = 0，可以采用更巧妙的方法。找出两个数，相乘等于 a × c = 4，相加等于 b = 5。数字 4 和 1 满足条件。

2x² + 4x + x + 2 = 2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2)

Hence (2x + 1)(x + 2) = 0, so x = −½ or x = −2. Always expand your answer to check that it matches the original equation.

因此 (2x + 1)(x + 2) = 0，所以 x = −½ 或 x = −2。一定要将答案展开验证，确保与原来的方程一致。

3. The Difference of Two Squares | 平方差公式

A special and very common case is the difference of two squares. When a quadratic has the form x² − k², it factorises immediately.

一个特殊且常见的情况是平方差公式。当二次方程具有 x² − k² 的形式时，可以直接进行因式分解。

x² − k² = (x − k)(x + k)

For example, x² − 9 = 0 can be written as (x − 3)(x + 3) = 0. The solutions are x = 3 and x = −3. Notice that the linear term is missing because the roots are opposites.

例如，x² − 9 = 0 可以写成 (x − 3)(x + 3) = 0，解为 x = 3 和 x = −3。注意此时一次项不存在，因为两个根互为相反数。

This pattern also works with non-perfect squares. For instance, x² − 5 = 0 becomes (x − √5)(x + √5) = 0, so x = ±√5. You may also be asked to factorise expressions like 4x² − 25 = (2x − 5)(2x + 5).

这个模式也适用于非平方数。例如 x² − 5 = 0 可化为 (x − √5)(x + √5) = 0，因此 x = ±√5。考试中有时也会要求分解 4x² − 25 = (2x − 5)(2x + 5) 这类表达式。

4. The Quadratic Formula | 求根公式法

When factorisation is difficult or impossible, we use the quadratic formula. This formula works for every quadratic equation, provided a ≠ 0.

当方程难以因式分解或无法因式分解时，可使用求根公式法。只要 a ≠ 0，这个公式适用于一切二次方程。

x = (−b ± √(b² − 4ac)) ÷ (2a)

Let us solve 2x² − 3x − 2 = 0. Here a = 2, b = −3 and c = −2. Substitute these values carefully.

我们求解 2x² − 3x − 2 = 0。此时 a = 2，b = −3，c = −2。请仔细代入这些数值。

x = (3 ± √((−3)² − 4 × 2 × (−2))) ÷ (2 × 2) = (3 ± √25) ÷ 4

Since √25 = 5, we obtain x = (3 + 5) ÷ 4 = 2 or x = (3 − 5) ÷ 4 = −½. Always write the two answers separately and check them by substitution.

因为 √25 = 5，所以 x = (3 + 5) ÷ 4 = 2，或 x = (3 − 5) ÷ 4 = −½。务必分别写出两个答案，并通过代入原方程进行检验。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This is especially useful for finding the vertex of a parabola or solving equations without a calculator.

配方法将二次方程改写为 (x + p)² + q 的形式。这一方法在求抛物线顶点或不用计算器解方程时尤其有用。

To complete the square for x² + 6x + 1 = 0, we take half of 6, which is 3, and square it to get 9. Then we adjust the constant term.

对于 x² + 6x + 1 = 0，我们取 6 的一半得 3，再平方得 9，然后调整常数项。

x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8

Now set (x + 3)² − 8 = 0, so (x + 3)² = 8. Taking square roots gives x + 3 = ±√8, hence x = −3 ± 2√2.

令 (x + 3)² − 8 = 0，则 (x + 3)² = 8。两边开平方得 x + 3 = ±√8，因此 x = −3 ± 2√2。

This method also reveals the minimum or maximum point of a quadratic graph directly, which examiners often test in the graph section.

配方法还能直接揭示二次函数图像的最低点或最高点，这也是考试中经常考查的内容。

6. The Discriminant | 判别式

The expression inside the square root of the quadratic formula is called the discriminant. It determines the nature of the roots without solving the full equation.

求根公式中根号内的表达式称为判别式。它可以在不完全解方程的情况下直接判断根的性质。

Δ = b² − 4ac

Discriminant	Nature of Roots	判别式取值	根的性质
Δ > 0	Two distinct real roots	Δ > 0	两个不相等的实数根
Δ = 0	One repeated real root	Δ = 0	两个相等的实数根（重根）
Δ < 0	No real roots	Δ < 0	没有实数根

For example, the equation 3x² + 2x + 1 = 0 has discriminant Δ = 4 − 12 = −8, which is negative. Therefore it has no real roots, and its graph never touches the x-axis.

例如，方程 3x² + 2x + 1 = 0 的判别式 Δ = 4 − 12 = −8，为负数，因此它没有实数根，其图像与 x 轴无交点。

7. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0, the two roots can be related directly to the coefficients without solving. This is a clever shortcut used in many IGCSE questions.

对于二次方程 ax² + bx + c = 0，无需解方程即可将两根与系数直接联系起来。这是许多 IGCSE 题目中使用的巧妙技巧。

Sum of roots = −b ÷ a, Product of roots = c ÷ a

Take the equation 2x² − 6x + 4 = 0. The sum of the roots is −(−6) ÷ 2 = 3, and the product is 4 ÷ 2 = 2. The roots are actually 1 and 2, which confirms these results.

以方程 2x² − 6x + 4 = 0 为例，两根之和为 −(−6) ÷ 2 = 3，两根之积为 4 ÷ 2 = 2。实际上两根为 1 和 2，正好验证了上述结论。

If you know the sum S and product P, you can reconstruct the quadratic as x² − Sx + P = 0. This is often the fastest way to write a quadratic from given roots.

如果已知两根之和 S 与两根之积 P，可以构造出二次方程 x² − Sx + P = 0。这是由已知根构造二次方程的最快方法。

8. Sketching Quadratic Graphs | 绘制二次函数图像

Quadratic graphs are parabolas. Their shape depends on the sign of a. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.

二次函数图像是抛物线。抛物线的开口方向取决于 a 的正负。当 a > 0 时开口向上，有最低点；当 a < 0 时开口向下，有最高点。

To sketch a quadratic graph y = ax² + bx + c, find three key features: the y-intercept at (0, c), the x-intercepts (roots), and the vertex. The x-coordinate of the vertex is given by the formula below.

绘制 y = ax² + bx + c 的图像时，需要找到三个关键特征：y 轴截距 (0, c)、x 轴截距（根）以及顶点。顶点的横坐标由以下公式给出。

x = −b ÷ (2a)

For instance, for y = 2x² − 8x + 6, the vertex lies at x = 8 ÷ 4 = 2. Substituting back gives y = 8 − 16 + 6 = −2, so the vertex is (2, −2).

例如，对于 y = 2x² − 8x + 6，顶点横坐标为 x = 8 ÷ 4 = 2。代回方程得 y = 8 − 16 + 6 = −2，因此顶点坐标为 (2, −2)。

Always label the roots, the y-intercept and the vertex clearly on your sketch. A neat, labelled graph earns full marks even when the question is only worth two points.

在草图中务必清晰标注根、y 轴截距和顶点。即使题目只值两分，一张整洁且标注完整的图像也能拿到满分。

9. Quadratic Word Problems | 二次方程应用题

Many real-world problems produce quadratic equations. The most common types are area problems, number problems and projectile motion. The key is to set up the equation correctly before solving it.

许多实际问题都会产生二次方程。最常见的类型包括面积问题、数字问题和抛体运动问题。关键在于先正确建立方程，再求解。

Here is a classic example: A rectangle has length 3 cm longer than its width. Its area is 28 cm². Find the width.

下面是一个经典例子：一个长方形的长比宽多 3 cm，面积为 28 cm²。求宽。

Let the width be w. Then the length is w + 3, so w(w + 3) = 28. Expanding gives w² + 3w − 28 = 0, which factorises as (w + 7)(w − 4) = 0.

设宽为 w，则长为 w + 3，于是 w(w + 3) = 28。展开得 w² + 3w − 28 = 0，因式分解为 (w + 7)(w − 4) = 0。

The solutions are w = −7 and w = 4. Since length cannot be negative, we reject w = −7. The width is therefore 4 cm and the length is 7 cm.

解得 w = −7 或 w = 4。由于长度不能为负，舍去 w = −7。因此宽为 4 cm，长为 7 cm。

In word problems, always check each solution against the context. A mathematically valid answer may be physically impossible, and examiners expect you to state this.

在应用题中，务必根据题目情境逐一检验解。数学上成立的解可能在现实中不成立，考官希望你能明确说明这一点。

10. Common Exam Mistakes | 常见考试错误

Students often lose marks through careless algebraic errors. Below are the most frequent mistakes reported in IGCSE examiner comments.

学生常常因为粗心的代数错误而失分。以下是 IGCSE 考官评语中最常见的几类错误。

Forgetting to set the equation to zero before factorising. You must have ax² + bx + c = 0 first.
忽略将方程化为零后再因式分解。必须先得到 ax² + bx + c = 0 的形式。

Mixing up signs in the quadratic formula. For example, substituting b = −5 instead of b = 5.
在求根公式中搞错符号。例如将 b = −5 代入成 b = 5。

Writing only one root when two roots exist. Always remember the ± symbol produces two answers.
只写出一个根而漏掉另一个根。切记 ± 符号会得到两个答案。

Cancelling a variable incorrectly, such as dividing both sides by x when x may be zero.
错误地约去变量，例如当 x 可能为零时直接两边同除以 x。

A useful habit is to substitute your final answers back into the original equation. This catches sign errors and arithmetic slips instantly.

一个有效习惯是将最终答案代回原方程检验。这样可以立即发现符号错误和计算失误。

11. Quick Practice Questions | 快速练习

Try these five questions before your exam. Answers are given after each pair, but attempt them fully before checking.

考试前请尝试以下五道练习。每道题后附有答案，但请先完整作答再核对。

Q1: Solve x² − 7x + 12 = 0.

题1：解方程 x² − 7x + 12 = 0。

Answer: (x − 3)(x − 4) = 0, so x = 3 or x = 4.

答案：(x − 3)(x − 4) = 0，因此 x = 3 或 x = 4。

Q2: Solve 5x² + 2x − 3 = 0 using the quadratic formula.

题2：用求根公式解方程 5x² + 2x − 3 = 0。

Answer: x = (−2 ± √(4 + 60)) ÷ 10 = (−2 ± 8) ÷ 10, so x = 3/5 or x = −1.

答案：x = (−2 ± √(4 + 60)) ÷ 10 = (−2 ± 8) ÷ 10，因此 x = 3/5 或 x = −1。

Q3: Find the discriminant of x² + 4x + 5 = 0 and state the nature of its roots.

题3：求 x² + 4x + 5 = 0 的判别式，并说明根的性质。

Answer: Δ = 16 − 20 = −4 < 0, so no real roots.

答案：Δ = 16 − 20 = −4 < 0，因此没有实数根。

Q4: Express x² − 10x + 7 in the form (x − p)² + q.

题4：将 x² − 10x + 7 表示为 (x − p)² + q 的形式。

Answer: (x − 5)² − 25 + 7 = (x − 5)² − 18.

答案：(x − 5)² − 25 + 7 = (x − 5)² − 18。

Q5: The sum of the roots of a quadratic is 8 and the product is 15. Write the quadratic equation.

题5：已知某二次方程两根之和为 8，两根之积为 15，写出该二次方程。

Answer: x² − 8x + 15 = 0.

答案：x² − 8x + 15 = 0。

Review any question you got wrong and identify the exact step where the error occurred. Focus your revision on that method next.

请复查做错的题目，找出出错的具体步骤，并在下一轮复习中集中强化对应的方法。

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Solving Quadratic Equations | 解二次方程