Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most fundamental topics in IGCSE Mathematics. A quadratic equation is any equation that can be rearranged into the standard form ax² + bx + c = 0, where a ≠ 0. In this revision guide, we will explore every method of solving quadratic equations, understand the discriminant, and practise typical exam-style questions step by step.

二次方程是 IGCSE 数学中最核心的考点之一。凡是能整理为标准形式 ax² + bx + c = 0（其中 a ≠ 0）的方程，都称为二次方程。本复习指南将系统讲解二次方程的全部解法，深入理解判别式的意义，并通过典型考题精讲，帮助你在考试中从容应对。

1. The Standard Form | 标准形式

Every quadratic equation can be written in the form ax² + bx + c = 0, where a, b and c are real numbers, and a ≠ 0. The coefficient a is the quadratic coefficient, b is the linear coefficient, and c is the constant term.

任何一个二次方程都可以写成 ax² + bx + c = 0 的形式，其中 a、b、c 为实数，且 a ≠ 0。a 称为二次项系数，b 称为一次项系数，c 称为常数项。

For example, the equation 2x² – 4x + 1 = 0 has a = 2, b = -4 and c = 1. Note that the equation must first be simplified and rearranged so that one side is equal to zero before any solution method is applied.

例如，方程 2x² – 4x + 1 = 0 中，a = 2，b = -4，c = 1。注意：在使用任何解法之前，必须先将方程化简并移项，使等号的一侧为 0。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method for solving quadratic equations, provided the expression can be factorised easily. The principle is based on the zero product property: if the product of two factors is zero, then at least one of the factors must be zero.

因式分解法是解二次方程最快的方法之一，前提是表达式能够被轻松分解。其原理基于零乘积性质：若两个因式的乘积为零，则至少有一个因式为零。

Example: Solve x² + 5x + 6 = 0.

例题：解方程 x² + 5x + 6 = 0。

We look for two numbers that multiply to give 6 and add to give 5. These numbers are 2 and 3, so we factorise as (x + 2)(x + 3) = 0. Thus x = -2 or x = -3.

寻找两个数，其乘积为 6，且和为 5。显然这两个数是 2 和 3，因此分解为 (x + 2)(x + 3) = 0。所以 x = -2 或 x = -3。

If (x + p)(x + q) = 0, then x = -p or x = -q.

若 (x + p)(x + q) = 0，则 x = -p 或 x = -q。

3. Factorisation with a ≠ 1 | 当 a ≠ 1 时的因式分解

When the coefficient of x² is not 1, factorisation requires extra care. Suppose we want to solve 2x² + 7x + 3 = 0. We look for two numbers that multiply to give a × c = 2 × 3 = 6 and add to give b = 7. The numbers are 6 and 1.

当 x² 的系数不为 1 时，因式分解需要格外小心。例如解方程 2x² + 7x + 3 = 0。我们需要找两个数，使其乘积等于 a × c = 2 × 3 = 6，且和等于 b = 7。这两个数是 6 和 1。

We then split the middle term: 2x² + 6x + x + 3 = 0. Group the terms: (2x² + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. Hence x = -1/2 or x = -3.

接着拆分中间项：2x² + 6x + x + 3 = 0。分组：(2x² + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。因此 x = -1/2 或 x = -3。

Exam tip: Always expand your factorised answer to check that it matches the original expression.

考试技巧：分解完成后务必展开检验，确保与原始表达式一致。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form a(x + h)² + k. This method is particularly useful when the quadratic cannot be factorised, and it also directly reveals the vertex of the parabola.

配方法将二次表达式改写为 a(x + h)² + k 的形式。当二次式无法因式分解时，此方法尤为实用，同时还能直接揭示抛物线的顶点坐标。

Example: Solve x² + 6x + 2 = 0 by completing the square.

例题：用配方法解 x² + 6x + 2 = 0。

Take half of the coefficient of x, which is 3, and write (x + 3)² = x² + 6x + 9. Since our original expression has constant term 2, we adjust: x² + 6x + 2 = (x + 3)² – 7. Setting this equal to zero gives (x + 3)² = 7, so x + 3 = ±√7, and x = -3 ± √7.

取 x 系数的一半，即 3，写出 (x + 3)² = x² + 6x + 9。由于原表达式常数项为 2，需要调整：x² + 6x + 2 = (x + 3)² – 7。令其等于零：(x + 3)² = 7，所以 x + 3 = ±√7，即 x = -3 ± √7。

x² + bx + c = (x + b/2)² – (b/2)² + c

5. The Quadratic Formula | 求根公式

The quadratic formula is a universal method that solves any quadratic equation. For ax² + bx + c = 0, the solutions are given by the formula below. This formula is derived directly from completing the square, so it works for every case, including awkward decimals and surds.

求根公式是解一切二次方程的通用方法。对于 ax² + bx + c = 0，其解由下列公式给出。该公式由配方法直接推导而来，适用于一切情况，包括繁琐的小数和根式。

x = (-b ± √(b² – 4ac)) / 2a

Example: Solve 3x² – 5x – 2 = 0 using the formula.

例题：用公式法解 3x² – 5x – 2 = 0。

Here a = 3, b = -5, c = -2. Substitute into the formula: x = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6. Thus x = 2 or x = -1/3.

此处 a = 3，b = -5，c = -2。代入公式：x = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6。因此 x = 2 或 x = -1/3。

6. The Discriminant | 判别式

The expression inside the square root in the quadratic formula is called the discriminant, denoted by Δ = b² – 4ac. The value of the discriminant tells us the nature of the roots without solving the equation.

求根公式中根号内的表达式称为判别式，记作 Δ = b² – 4ac。判别式的值可以让我们不解方程就判断根的性质。

If Δ > 0, the equation has two distinct real roots.
如果 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.
如果 Δ = 0，方程有且仅有一个相等的实数根（重根）。
If Δ < 0, the equation has no real roots (two complex roots).
如果 Δ < 0，方程没有实数根（有两个复数根）。

Example: Determine the nature of the roots of 2x² – 4x + 3 = 0.

例题：判断方程 2x² – 4x + 3 = 0 的根的性质。

Δ = (-4)² – 4 × 2 × 3 = 16 – 24 = -8 < 0, so the equation has no real roots.

Δ = (-4)² – 4 × 2 × 3 = 16 – 24 = -8 < 0，因此方程无实数根。

7. The Sum and Product of Roots | 根与系数的关系

For a quadratic equation ax² + bx + c = 0, there is an elegant relationship between the roots and the coefficients. If the roots are α and β, then 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，根与系数之间存在优美而简洁的关系。若两根为 α 和 β，则两根之和为 -b/a，两根之积为 c/a。

α + β = -b/a, αβ = c/a

This relationship is very useful when you are asked to form a new equation from known roots, or to evaluate symmetric expressions involving α and β without actually solving the original equation.

这一关系在由已知根构造新方程、或不解原方程而求 α 与 β 对称表达式的值时极为常用。

Example: For the equation 2x² – 6x + 3 = 0, find α + β and αβ.

例题：对于方程 2x² – 6x + 3 = 0，求 α + β 与 αβ。

Here a = 2, b = -6, c = 3. Thus α + β = -(-6)/2 = 3, and αβ = 3/2.

此处 a = 2，b = -6，c = 3。所以 α + β = -(-6)/2 = 3，αβ = 3/2。

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

The solutions of a quadratic equation ax² + bx + c = 0 correspond exactly to the x-intercepts of the parabola y = ax² + bx + c. The graph provides a visual interpretation of the discriminant:

二次方程 ax² + bx + c = 0 的解，恰好对应于抛物线 y = ax² + bx + c 与 x 轴的交点。图像为判别式提供了直观的解释：

Discriminant 判别式	Graph 图像	Roots 根
Δ > 0	Crosses the x-axis twice 与 x 轴有两个交点	Two distinct roots 两个不相等实数根
Δ = 0	Touches the x-axis once 与 x 轴相切	One repeated root 一个重根
Δ < 0	Does not intersect the x-axis 与 x 轴无交点	No real roots 无实数根

Additionally, the vertex of the parabola is found at x = -b/2a, which is exactly the midpoint of the two roots when they are real and distinct.

另外，抛物线的顶点横坐标为 x = -b/2a，当两根为不相等实数根时，该横坐标恰好是两根的中点。

9. Word Problems and Applications | 实际应用题

Quadratic equations frequently appear in geometry, projectile motion and number problems. The key is to translate the given information into a quadratic equation, solve it, and then check whether each solution makes sense in the context of the problem — sometimes one root must be discarded.

二次方程频繁出现在几何、抛体运动和数字问题中。解题关键在于将题目信息转化为二次方程，求解后再检查每个解在题目情境中是否合理——有时需要舍去其中一个根。

Example: A rectangle has length 3 cm more than its width, and its area is 28 cm². Find its dimensions.

例题：一个长方形的长比宽多 3 cm，面积为 28 cm²。求其尺寸。

Let the width be x cm. Then the length is (x + 3) cm. The area equation is x(x + 3) = 28, which expands to x² + 3x – 28 = 0. Factorising: (x + 7)(x – 4) = 0, so x = 4 or x = -7. Since a width cannot be negative, x = 4. Thus the width is 4 cm and the length is 7 cm.

设宽为 x cm，则长为 (x + 3) cm。面积方程为 x(x + 3) = 28，展开得 x² + 3x – 28 = 0。因式分解：(x + 7)(x – 4) = 0，所以 x = 4 或 x = -7。因为宽度不可能为负，故取 x = 4。所以宽为 4 cm，长为 7 cm。

10. Solving Quadratic Equations by Graph Drawing | 图像法解二次方程

In IGCSE exams, you may also be asked to solve a quadratic equation approximately by drawing its graph. After plotting y = ax² + bx + c for a set of x-values, the solutions are read from the x-coordinates where the curve crosses the x-axis.

在 IGCSE 考试中，有时也会要求你通过绘制图像来近似求解二次方程。在绘制出 y = ax² + bx + c 在若干 x 值下的图像后，曲线与 x 轴交点的横坐标即为方程的解。

When a specific equation such as x² – 2x – 3 = 0 is given, you may alternatively be asked to draw the graph of a related curve and then use a straight line to find the roots. For instance, to solve x² – 2x – 3 = 0 using the graph of y = x² – 2x – 1, you would draw the line y = 2 and read off the x-values at its intersections.

当给出类似 x² – 2x – 3 = 0 的方程时，你也可以被要求先绘制某条相关曲线，再利用一条直线来求根。例如，要用 y = x² – 2x – 1 的图像解 x² – 2x – 3 = 0，可画出直线 y = 2，再读取交点横坐标即可。

11. Common Pitfalls and Exam Advice | 常见错误与应试建议

Many students lose marks on quadratic equations due to avoidable mistakes. Below are the most common pitfalls that appear in IGCSE examinations.

许多学生在二次方程上失分，往往是因为一些可以避免的错误。以下列出 IGCSE 考试中最常见的几类陷阱。

Forgetting to rearrange to zero: Always rewrite the equation as ax² + bx + c = 0 before solving. For example, x² = 3x + 4 must be rearranged to x² – 3x – 4 = 0.
忘记移项归零：解题前务必将方程整理成 ax² + bx + c = 0。例如 x² = 3x + 4 必须移项为 x² – 3x – 4 = 0。
Dividing by x incorrectly: Never divide both sides by x unless you have already noted that x = 0 must be considered separately — this often hides a solution.
随意除以 x：切忌直接两边同除以 x，除非你已经单独讨论了 x = 0 的情况——这样做通常会丢失一个根。
Sign errors in the formula: When b is negative, substituting into the quadratic formula requires careful handling of double negatives.
公式代入符号错误：当 b 为负数时，代入求根公式要特别小心负负得正的问题。
Rounding too early: If a final answer is required to 3 significant figures, keep full precision in intermediate steps.
过早舍入：若最终答案要求保留 3 位有效数字，中间步骤应保持完整精度。

12. Practice Questions | 巩固练习

To master quadratic equations, regular practice is essential. Try the following questions on your own before checking the solutions provided below each problem.

要精通二次方程，定期练习必不可少。请先独立完成下列题目，再对照每题下方给出的答案。

Question 1: Solve x² – 7x + 12 = 0 by factorisation.

练习 1：用因式分解法解 x² – 7x + 12 = 0。

Solution: (x – 3)(x – 4) = 0, so x = 3 or x = 4.

解答：(x – 3)(x – 4) = 0，所以 x = 3 或 x = 4。

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

练习 2：用求根公式解 2x² + 5x – 3 = 0。

Solution: a = 2, b = 5, c = -3. Δ = 25 + 24 = 49. x = (-5 ± 7) / 4 = 1/2 or -3.

解答：a = 2，b = 5，c = -3。Δ = 25 + 24 = 49。x = (-5 ± 7) / 4 = 1/2 或 -3。

Question 3: Find the range of values of k for which the equation x² + kx + 4 = 0 has two distinct real roots.

练习 3：求使方程 x² + kx + 4 = 0 有两个不相等实数根的 k 的取值范围。

Solution: For two distinct real roots, Δ > 0, i.e. k² – 16 > 0. Thus k > 4 or k < -4.

解答：要使方程有两个不相等实数根，需 Δ > 0，即 k² – 16 > 0。因此 k > 4 或 k < -4。

By working systematically through each method and understanding the connections between them, you will find that quadratic equations become one of the most reliable scoring topics in your IGCSE Mathematics exam.

通过系统掌握每一种解法并理解它们之间的内在联系，你会发现二次方程将成为 IGCSE 数学考试中最稳定的得分点之一。

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Solving Quadratic Equations | 解二次方程