Solving Quadratic Equations | 二次方程求解

📚 Solving Quadratic Equations | 二次方程求解

A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in algebra, coordinate geometry, and even in problem-solving questions across the paper. Mastering this topic not only secures marks in Paper 2 and Paper 4, but also builds a strong foundation for A-Level Mathematics.

二次方程是 IGCSE 数学中最重要的考点之一。它出现在代数、坐标几何乃至整份试卷的应用题中。掌握二次方程不仅能在 Paper 2 和 Paper 4 中稳拿分数，更能为 A-Level 数学打下坚实基础。

1. Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. The standard form is written as:

ax² + bx + c = 0

Here, a, b and c are constants, and a ≠ 0. If a = 0, the equation becomes linear. The values of a, b and c determine the shape and position of the parabola, as well as the nature of its roots.

二次方程是最高次数为 2 的多项式方程，即变量的最高次幂为 2。标准形式写作：

ax² + bx + c = 0

其中 a、b、c 均为常数，且 a ≠ 0。若 a = 0，方程就退化为一次方程。a、b、c 的取值决定了抛物线的形状与位置，也决定了根的属性。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic has simple integer roots. The idea is to rewrite ax² + bx + c as a product of two binomials, then set each factor equal to zero. For example, solve x² − 5x + 6 = 0. We look for two numbers whose product is 6 and whose sum is −5. These numbers are −2 and −3, so the factors are (x − 2)(x − 3) = 0. Hence x = 2 or x = 3.

因式分解法适用于根为简单整数的二次方程，通常是最快捷的方法。核心思路是将 ax² + bx + c 改写为两个二项式的乘积，再令每个因式分别等于零。例如求解 x² − 5x + 6 = 0。我们寻找两个数，其乘积为 6，和为 −5。这两个数是 −2 和 −3，因此因式为 (x − 2)(x − 3) = 0。于是 x = 2 或 x = 3。

For quadratics where a ≠ 1, we use the method of splitting the middle term. To solve 2x² + 7x + 3 = 0, multiply a and c: 2 × 3 = 6. Find two numbers whose product is 6 and sum is 7: these are 6 and 1. Split 7x into 6x + x, then factor by grouping:

对于 a ≠ 1 的二次方程，我们使用拆中项法。求解 2x² + 7x + 3 = 0 时，先将 a 与 c 相乘：2 × 3 = 6。寻找两个数，乘积为 6，和为 7：这两个数是 6 和 1。将 7x 拆为 6x + x，再分组因式分解：

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0

Therefore x = −½ or x = −3.

因此 x = −½ 或 x = −3。

3. Solving by the Quadratic Formula | 公式法

When factorisation is difficult or impossible, the quadratic formula always works. For any quadratic equation in the form ax² + bx + c = 0, the solutions are given by:

当因式分解困难甚至无法进行时，公式法永远适用。对于任何形式为 ax² + bx + c = 0 的二次方程，其解为：

x = (−b ± √(b² − 4ac)) / 2a

For example, solve 2x² + 3x − 5 = 0. Here a = 2, b = 3, c = −5. Substituting into the formula gives x = (−3 ± √(9 − 4 × 2 × (−5))) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4. Thus x = 1 or x = −2.5.

例如，求解 2x² + 3x − 5 = 0。此处 a = 2，b = 3，c = −5。代入公式得 x = (−3 ± √(9 − 4 × 2 × (−5))) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4。因此 x = 1 或 x = −2.5。

You must memorise the quadratic formula carefully. A common error is forgetting to change the sign of b, or making a mistake when b is negative. Write down the values of a, b and c before substituting to reduce careless errors.

务必牢记公式法。常见的错误是忘记改变 b 的符号，或在 b 为负数时出错。在代入之前先写出 a、b、c 的具体数值，可以有效减少粗心失误。

4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form p(x + q)² + r. This method is especially useful because it reveals the turning point of a parabola directly. To complete the square for x² + 6x + 8, take half of 6 to get 3, then form (x + 3)². Since (x + 3)² = x² + 6x + 9, we subtract 9 and add 8:

配方法将二次式改写为 p(x + q)² + r 的形式。此方法特别有用，因为它能直接揭示抛物线的顶点。对 x² + 6x + 8 配方：取 6 的一半得 3，构成 (x + 3)²。由于 (x + 3)² = x² + 6x + 9，减去 9 再加 8：

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

Setting (x + 3)² − 1 = 0 gives (x + 3)² = 1, so x + 3 = ±1, giving x = −2 or x = −4.

令 (x + 3)² − 1 = 0，得 (x + 3)² = 1，所以 x + 3 = ±1，即 x = −2 或 x = −4。

When the coefficient of x² is not 1, factor it out first. For 2x² − 8x + 5, factor out 2: 2(x² − 4x) + 5. Then complete the square inside the brackets: 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3.

当 x² 的系数不为 1 时，先将其提取出来。对于 2x² − 8x + 5，先提取 2：2(x² − 4x) + 5。然后在括号内配方：2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3。

5. The Discriminant and the Nature of Roots | 判别式与根的性质

The expression b² − 4ac inside the quadratic formula is called the discriminant, denoted by Δ. It tells us how many real roots a quadratic equation has, without solving it fully.

公式法中根号内的表达式 b² − 4ac 称为判别式，记作 Δ。它无需完全求解即可判断二次方程有多少个实数根。

Δ = b² − 4ac

If Δ > 0, the equation has two distinct real roots.

若 Δ > 0，方程有两个不相等的实数根。
If Δ = 0, the equation has exactly one repeated real root.

若 Δ = 0，方程有两个相等的实数根（即一个重根）。
If Δ < 0, the equation has no real roots.

若 Δ < 0，方程没有实数根。

For example, for x² − 4x + 4 = 0, Δ = 16 − 16 = 0, so there is one repeated root x = 2. For x² + x + 1 = 0, Δ = 1 − 4 = −3, so there are no real roots.

例如，对于 x² − 4x + 4 = 0，Δ = 16 − 16 = 0，因此有一个重根 x = 2。对于 x² + x + 1 = 0，Δ = 1 − 4 = −3，因此没有实数根。

Exam questions often ask you to find the range of values of a constant k such that the equation has real roots. Set Δ ≥ 0 and solve the resulting inequality.

考试题经常要求求出常数 k 的取值范围，使得方程有实数根。此时只需令 Δ ≥ 0，再解对应不等式即可。

6. Quadratic Graphs and Roots | 二次函数图像与根

The graph of y = ax² + bx + c is a parabola. When a > 0, the parabola opens upward; when a < 0, it opens downward. The x-intercepts of the graph correspond to the real roots of the equation ax² + bx + c = 0.

函数 y = ax² + bx + c 的图像是抛物线。当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。图像与 x 轴的交点横坐标就是方程 ax² + bx + c = 0 的实数根。

The vertex (turning point) has x-coordinate x = −b / 2a. Substituting this value back into the equation gives the y-coordinate of the vertex. Equivalently, if the quadratic is written as y = p(x + q)² + r, the vertex is at (−q, r).

顶点（转向点）的横坐标为 x = −b / 2a。将该值代回方程即可得到顶点的纵坐标。等价地，若二次式写成 y = p(x + q)² + r 的形式，则顶点坐标为 (−q, r)。

The discriminant tells us about the relationship between the graph and the x-axis: Δ > 0 means the parabola crosses the x-axis twice; Δ = 0 means it touches the x-axis at one point; Δ < 0 means it does not touch the x-axis at all.

判别式揭示了图像与 x 轴的关系：Δ > 0 表示抛物线与 x 轴有两个交点；Δ = 0 表示抛物线与 x 轴相切于一点；Δ < 0 表示抛物线与 x 轴无交点。

7. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0 with roots α and β, there are two important relationships that allow us to find information about the roots without solving the equation:

对于根为 α 和 β 的二次方程 ax² + bx + c = 0，存在两个重要关系式，它们使我们在不解方程的情况下也能了解根的性质：

Sum of roots: α + β = −b / a

Product of roots: αβ = c / a

For example, for 2x² − 8x + 6 = 0, the sum of the roots is −(−8)/2 = 4 and the product is 6/2 = 3. We can verify this by factorising: (2x − 2)(x − 3) = 0 gives roots 1 and 3, whose sum is 4 and product is 3.

例如，对于 2x² − 8x + 6 = 0，两根之和为 −(−8)/2 = 4，两根之积为 6/2 = 3。我们可以通过因式分解验证：(2x − 2)(x − 3) = 0 的根为 1 和 3，其和为 4，积为 3。

This relationship is particularly useful when asked to form a new quadratic equation from given roots. If roots are p and q, the equation is x² − (p + q)x + pq = 0.

这一关系在”由已知根构造新二次方程”的问题中尤其有用。若根为 p 和 q，则方程为 x² − (p + q)x + pq = 0。

8. Applications: Word Problems | 实际应用：应用题

Quadratic equations frequently appear in IGCSE word problems involving area, motion, and number relationships. For example: “A rectangle has length 3 cm more than its width. Its area is 40 cm². Find the width.”

二次方程在 IGCSE 应用题中频繁出现，涉及面积、运动和数字关系。例如：”一个长方形的长比宽多 3 cm，面积为 40 cm²。求宽。”

Let the width be x. Then the length is x + 3. Since area = length × width, we have x(x + 3) = 40, which simplifies to x² + 3x − 40 = 0. Factorising gives (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since width cannot be negative, the width is 5 cm.

设宽为 x，则长为 x + 3。面积 = 长 × 宽，因此 x(x + 3) = 40，化简得 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于宽不能为负数，因此宽为 5 cm。

Always check whether your solutions make sense in the context of the problem. Reject negative or unrealistic answers with a brief written note. In projectile motion problems, remember that time cannot be negative either.

务必检查解是否符合题意。对于负数或不合理的答案，应简要注明并予以舍去。在抛体运动问题中，时间同样不能为负数。

9. Common Mistakes and How to Avoid Them | 常见错误与规避方法

One of the most common errors in IGCSE exams is forgetting to rearrange the equation into standard form ax² + bx + c = 0 before attempting to factorise or apply the formula. Always move all terms to one side first.

IGCSE 考试中最常见的错误之一，是在因式分解或套用公式之前忘记将方程整理为标准形式 ax² + bx + c = 0。务必先将所有项移到等号一侧。

Another frequent mistake is incorrect sign handling. For example, in the equation x² − 5x − 14 = 0, students sometimes write (x − 2)(x − 7) instead of (x − 7)(x + 2). Always check your factors by expanding them back.

另一个高频错误是符号处理不当。例如在 x² − 5x − 14 = 0 中，学生有时会错误地写成 (x − 2)(x − 7)，而正确因式为 (x − 7)(x + 2)。务必通过展开来检验你的因式。

When using the quadratic formula, ensure the expression b² − 4ac is calculated with correct signs. For 2x² − 4x + 1 = 0, b = −4, so b² = 16 and −4ac = −4 × 2 × 1 = −8. The discriminant is 16 − 8 = 8.

使用公式法时，要确保 b² − 4ac 的符号计算正确。对于 2x² − 4x + 1 = 0，b = −4，因此 b² = 16，−4ac = −4 × 2 × 1 = −8。判别式为 16 − 8 = 8。

Finally, for completing the square, remember to account for any factor outside the bracket. If you factor out 3 from 3x² + 6x, the completed square form must preserve the original expression when expanded.

最后，使用配方法时，不要忘记括号外的系数。如果从 3x² + 6x 中提取 3，配方后的形式在展开后必须与原式完全一致。

10. Choosing the Best Method | 选择最佳解法

Different situations call for different methods. Here is a guide to help you decide which approach to use in an exam:

不同情况需要选择不同解法。下表帮助你在考试中快速判断该用哪种方法：

Method 解法	Best used when 适用场景
Factorisation 因式分解	Roots are integers or simple fractions; a = 1 or easily manageable 根为整数或简单分数；a = 1 或易于处理
Quadratic formula 公式法	Roots are irrational or the equation does not factorise 根为无理数或方程无法因式分解
Completing the square 配方法	The question asks for the vertex or the minimum/maximum value 题目要求顶点或最大/最小值
Graphical method 图像法	The question provides a graph and asks for approximate roots 题目给出图像并要求估算根

In Paper 2 (non-calculator), factorisation is usually expected. In Paper 4, the quadratic formula may be more efficient when the coefficients are large or the roots are irrational. Practise all three algebraic methods until you can use them with confidence.

在 Paper 2（不可使用计算器）中，通常期望使用因式分解法。在 Paper 4 中，当系数较大或根为无理数时，公式法可能更高效。请将三种代数方法都练习到运用自如的程度。

11. Worked Exam-Style Question | 典型考试真题演练

Solve the equation 3x² + 5x − 2 = 0, giving your answers correct to 2 decimal places where necessary.

求解方程 3x² + 5x − 2 = 0，答案如有需要请保留两位小数。

First, test whether the quadratic factorises. We need two numbers whose product is 3 × (−2) = −6 and whose sum is 5. The numbers are 6 and −1. Splitting the middle term:

首先测试能否因式分解。我们需要两个数，乘积为 3 × (−2) = −6，和为 5。这两个数是 6 和 −1。拆中项：

3x² + 6x − x − 2 = 3x(x + 2) − 1(x + 2) = (3x − 1)(x + 2) = 0

Thus 3x − 1 = 0 gives x = ⅓, and x + 2 = 0 gives x = −2. Both roots are exact, so no rounding is needed. The solution set is x = ⅓ or x = −2.

因此 3x − 1 = 0 得 x = ⅓，x + 2 = 0 得 x = −2。两个根都是精确值，无需四舍五入。解集为 x = ⅓ 或 x = −2。

Now consider a second example: 2x² − 3x − 7 = 0. This does not factorise easily, so we apply the quadratic formula with a = 2, b = −3, c = −7:

再看第二个例子：2x² − 3x − 7 = 0。此式不易因式分解，因此用公式法，其中 a = 2，b = −3，c = −7：

x = (3 ± √(9 + 56)) / 4 = (3 ± √65) / 4

Using a calculator, x ≈ (3 + 8.062) / 4 ≈ 2.77 or x ≈ (3 − 8.062) / 4 ≈ −1.27, correct to 2 decimal places.

使用计算器，x ≈ (3 + 8.062) / 4 ≈ 2.77 或 x ≈ (3 − 8.062) / 4 ≈ −1.27，保留两位小数。

12. Summary and Final Tips | 总结与最终建议

Quadratic equations are a guaranteed topic on every IGCSE Mathematics paper. Make sure you can recognise the standard form, choose an appropriate solving method, and interpret the discriminant correctly.

二次方程是每份 IGCSE 数学试卷必考的考点。请确保你能够识别标准形式、选择恰当的求解方法，并正确理解判别式的含义。

Before the exam, practise at least ten mixed problems covering factorisation, the quadratic formula, completing the square, and the discriminant. Time yourself under exam conditions and check every answer by substituting back into the original equation.

考前请至少练习十道混合题型，涵盖因式分解、公式法、配方法和判别式。在模拟考试条件下计时作答，并通过代回原方程检验每一个答案。

Finally, read the question carefully. Some questions only want real solutions; others may ask for exact answers or answers to a specified degree of accuracy. Understanding the question before you start will save valuable time and prevent careless mistakes.

最后，请仔细审题。有的题目只要求实数解，有的则要求精确值或指定精确度的近似值。动笔之前先弄清题意，可以节省宝贵时间并避免粗心错误。

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Solving Quadratic Equations | 二次方程求解