Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

A quadratic equation is one of the most frequently tested topics in IGCSE Mathematics. It appears in Paper 1 (non-calculator), Paper 2 (calculator), and forms the foundation for many higher-level topics such as inequalities, functions, and coordinate geometry. Mastering this topic is essential for achieving a strong grade.

一元二次方程是 IGCSE 数学中最常考的内容之一，既出现在 Paper 1（不能使用计算器）中，也出现在 Paper 2（可使用计算器）中，同时还是不等式、函数和坐标几何等高级主题的基础。掌握好这一主题，是取得高分的关键。

1. What Is a Quadratic Equation? | 什么是一元二次方程

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The “quadratic” name comes from the Latin word “quadratus”, meaning square, because the highest power of the variable is x².

一元二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。”quadratic” 一词源自拉丁语 “quadratus”，意为”平方”，因为变量的最高次数是 x²。

Here are some examples of quadratic equations:

以下是几个一元二次方程的例子：

x² − 5x + 6 = 0, 2x² + 3x − 7 = 0, x² = 9

Note that the equation x² = 9 can be rearranged to x² − 9 = 0, which is still quadratic. The condition a ≠ 0 is important: if a = 0, the equation becomes linear, not quadratic.

注意 x² = 9 可以变形为 x² − 9 = 0，仍然是二次方程。条件 a ≠ 0 很重要：如果 a = 0，方程就变成了一次方程，而不是二次方程。

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

The standard form of a quadratic equation is:

一元二次方程的标准形式为：

ax² + bx + c = 0

Identifying the coefficients correctly is the first step in solving any quadratic equation. The coefficient a is the number multiplying x², b is the number multiplying x, and c is the constant term.

正确识别系数是解任何一元二次方程的第一步。系数 a 是 x² 前面的数，b 是 x 前面的数，c 是常数项。

For example, in 3x² − 2x + 5 = 0, we have a = 3, b = −2, c = 5. Be especially careful with negative signs — they belong to the coefficient. In 2x² + x − 4 = 0, we have a = 2, b = 1, c = −4.

例如，在 3x² − 2x + 5 = 0 中，a = 3，b = −2，c = 5。要特别注意负号——负号属于系数。在 2x² + x − 4 = 0 中，a = 2，b = 1，c = −4。

Equation	a	b	c
x² + 4x − 3 = 0	1	4	−3
2x² − 7x = 0	2	−7	0
x² − 9 = 0	1	0	−9

Always rearrange the equation into standard form before solving. For instance, 5x² = 3x + 2 should be rewritten as 5x² − 3x − 2 = 0 before identifying a, b and c.

在解题之前，一定要先将方程整理成标准形式。例如，5x² = 3x + 2 应先改写为 5x² − 3x − 2 = 0，然后再确定 a、b、c。

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer roots. The principle is based on the zero product property: if the product of two expressions is zero, then at least one of them must be zero.

当二次方程有简单的整数根时，因式分解法是最快的方法。其原理基于零乘积性质：如果两个表达式的乘积为零，那么至少其中一个表达式必为零。

Consider the equation x² − 5x + 6 = 0. To factorise, we look for two numbers that multiply to give c = 6 and add to give b = −5. The numbers −2 and −3 satisfy both conditions:

考虑方程 x² − 5x + 6 = 0。要因式分解，我们需要找到两个数，它们的乘积等于 c = 6，且和等于 b = −5。数 −2 和 −3 满足这两个条件：

(−2) × (−3) = 6, (−2) + (−3) = −5

Therefore the equation factorises as:

因此方程分解为：

(x − 2)(x − 3) = 0

By the zero product property, either x − 2 = 0 or x − 3 = 0, giving the solutions x = 2 or x = 3.

根据零乘积性质，x − 2 = 0 或 x − 3 = 0，因此解为 x = 2 或 x = 3。

When the coefficient of x² is not 1, the factorisation is slightly more involved. For 2x² + 7x + 3 = 0, we look for factor pairs. The expression factorises as (2x + 1)(x + 3) = 0, giving x = −½ or x = −3. Always expand your brackets to check your factorisation is correct.

当 x² 的系数不是 1 时，因式分解会稍复杂一些。对于 2x² + 7x + 3 = 0，我们需要寻找因子对。该表达式分解为 (2x + 1)(x + 3) = 0，得到 x = −½ 或 x = −3。一定要展开括号来验算你的分解是否正确。

4. Solving by the Quadratic Formula | 公式法

Not every quadratic equation can be factorised easily. In such cases, the quadratic formula provides a reliable, universal method. For ax² + bx + c = 0, the solutions are given by:

并非每个一元二次方程都能轻松因式分解。在这种情况下，求根公式提供了一种可靠且通用的方法。对于 ax² + bx + c = 0，解由下式给出：

x = (−b ± √(b² − 4ac)) ÷ 2a

This formula is provided in the IGCSE formula booklet, but you must know when and how to apply it. Let us solve 2x² + 3x − 4 = 0 using the formula. Here a = 2, b = 3, c = −4.

这个公式会印在 IGCSE 公式手册中，但你必须知道何时使用以及如何使用。让我们用公式解 2x² + 3x − 4 = 0。这里 a = 2，b = 3，c = −4。

First compute the discriminant part b² − 4ac:

首先计算判别式部分 b² − 4ac：

b² − 4ac = 3² − 4 × 2 × (−4) = 9 + 32 = 41

Now substitute into the formula:

然后代入公式：

x = (−3 ± √41) ÷ 4

So the two solutions are x = (−3 + √41) ÷ 4 ≈ 0.8508 and x = (−3 − √41) ÷ 4 ≈ −2.3508. You should round your final answers to an appropriate degree of accuracy, typically 3 significant figures unless instructed otherwise.

因此两个解为 x = (−3 + √41) ÷ 4 ≈ 0.8508 和 x = (−3 − √41) ÷ 4 ≈ −2.3508。最终答案应保留适当的精度，通常为 3 位有效数字，除非题目另有要求。

5. Completing the Square | 配方法

Completing the square is another algebraic method that rewrites a quadratic in the form a(x + p)² + q. This is particularly useful for finding the turning point of a quadratic graph and for solving equations that do not factorise.

配方法是另一种代数方法，它将二次表达式改写为 a(x + p)² + q 的形式。这在求二次函数图像的顶点坐标和求解无法因式分解的方程时特别有用。

To complete the square for x² + 6x + 5 = 0, take half of the coefficient of x, which is 6 ÷ 2 = 3, and write (x + 3)². Expanding (x + 3)² gives x² + 6x + 9, which is 4 more than x² + 6x + 5. Therefore:

对于 x² + 6x + 5 = 0 配方时，取 x 系数的一半，即 6 ÷ 2 = 3，写成 (x + 3)²。展开 (x + 3)² 得到 x² + 6x + 9，比 x² + 6x + 5 多了 4。因此：

x² + 6x + 5 = (x + 3)² − 4

Now solve the equation (x + 3)² − 4 = 0:

现在解方程 (x + 3)² − 4 = 0：

(x + 3)² = 4, x + 3 = ±2, x = −1 or x = −5

The general rule is: for x² + bx + c, write (x + b/2)² − (b/2)² + c. For example, x² − 8x + 10 = (x − 4)² − 16 + 10 = (x − 4)² − 6.

一般规则是：对于 x² + bx + c，写成 (x + b/2)² − (b/2)² + c。例如，x² − 8x + 10 = (x − 4)² − 16 + 10 = (x − 4)² − 6。

6. The Discriminant | 判别式

The discriminant, denoted Δ = b² − 4ac, determines the nature of the roots of a quadratic equation without actually solving it. This is a common exam question that tests conceptual understanding.

判别式，记作 Δ = b² − 4ac，不用实际求解就能判断一元二次方程根的性质。这是考试中常见的考查概念理解的问题。

If b² − 4ac > 0, the equation has two distinct real roots.

如果 b² − 4ac > 0，方程有两个不相等的实数根。
If b² − 4ac = 0, the equation has exactly one repeated real root (two equal roots).

如果 b² − 4ac = 0，方程有两个相等的实数根（一个重根）。
If b² − 4ac < 0, the equation has no real roots (the roots are complex).

如果 b² − 4ac < 0，方程没有实数根（根为复数）。

For example, consider x² − 4x + 4 = 0. Here Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0, so the equation has one repeated root. Indeed, x² − 4x + 4 = (x − 2)², giving x = 2 twice.

例如，考虑 x² − 4x + 4 = 0。这里 Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0，所以方程有一个重根。事实上，x² − 4x + 4 = (x − 2)²，得到 x = 2 这一重根。

The discriminant also connects to the graph of y = ax² + bx + c. If Δ > 0, the curve cuts the x-axis at two points; if Δ = 0, it touches the x-axis at exactly one point; if Δ < 0, it does not intersect the x-axis at all.

判别式还与 y = ax² + bx + c 的图像相关。如果 Δ > 0，曲线与 x 轴有两个交点；如果 Δ = 0，曲线与 x 轴相切于一点；如果 Δ < 0，曲线与 x 轴没有交点。

7. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards (a ∪ shape); if a < 0, it opens downwards (a ∩ shape).

y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上（∪ 形）；如果 a < 0，抛物线开口向下（∩ 形）。

The roots of the equation ax² + bx + c = 0 correspond to the x-intercepts of the graph. The turning point can be found by completing the square. For y = a(x + p)² + q, the vertex is at (−p, q).

方程 ax² + bx + c = 0 的根对应图像与 x 轴的交点。通过配方可以求得顶点坐标。对于 y = a(x + p)² + q，顶点在 (−p, q) 处。

For example, y = x² − 6x + 5 can be rewritten as y = (x − 3)² − 4. The vertex is at (3, −4), and the x-intercepts are x = 1 and x = 5 (found by solving x² − 6x + 5 = 0, which factorises as (x − 1)(x − 5) = 0). The y-intercept is at y = 5.

例如，y = x² − 6x + 5 可以改写为 y = (x − 3)² − 4。顶点在 (3, −4)，与 x 轴的交点为 x = 1 和 x = 5（通过解 x² − 6x + 5 = 0 得到，分解为 (x − 1)(x − 5) = 0）。与 y 轴的交点为 y = 5。

When sketching a quadratic graph, always label: the vertex, the x-intercepts (if any), the y-intercept, and the direction of opening. These four features are sufficient for a clear and accurate sketch.

画二次函数草图时，务必标注：顶点、x 轴交点（如果有）、y 轴交点以及开口方向。这四个特征足以画出一个清晰准确的草图。

8. Solving Quadratic Equations by Graph | 图像法解方程

Sometimes the question provides a graph of y = ax² + bx + c and asks you to solve ax² + bx + c = 0. The solutions are simply the x-coordinates where the graph crosses the x-axis.

有时题目会给出 y = ax² + bx + c 的图像，要求你解 ax² + bx + c = 0。解就是图像与 x 轴交点的 x 坐标。

More advanced questions may ask you to solve a related equation such as ax² + bx + c = k. In that case, draw the horizontal line y = k on the graph and read off the x-coordinates of the intersection points.

更进阶的问题可能要求你解相关方程，如 ax² + bx + c = k。此时，在图像上画水平线 y = k，然后读出交点的 x 坐标即可。

For example, if the graph of y = x² − 2x − 3 crosses the x-axis at x = −1 and x = 3, then the solutions to x² − 2x − 3 = 0 are x = −1 and x = 3. To solve x² − 2x − 3 = 5, we find where the horizontal line y = 5 intersects the parabola, giving x ≈ −2.24 and x ≈ 4.24.

例如，如果 y = x² − 2x − 3 的图像与 x 轴交于 x = −1 和 x = 3，那么 x² − 2x − 3 = 0 的解就是 x = −1 和 x = 3。要解 x² − 2x − 3 = 5，我们找到水平线 y = 5 与抛物线的交点，得到 x ≈ −2.24 和 x ≈ 4.24。

The graphical method is useful when exact values are not required, or when the equation does not factorise and a calculator-based numerical approach is acceptable.

图像法在不需要精确值的情况下很有用，或者当方程无法因式分解且接受使用计算器进行数值求解时，也是一种可行的方法。

9. Word Problems Involving Quadratics | 二次方程应用题

IGCSE examinations often present quadratic equations in the context of real-world problems. The key steps are: define a variable, form the quadratic equation from the given information, solve it, and then interpret the solutions in context.

IGCSE 考试经常在实际问题情境中呈现二次方程。关键步骤是：设变量，根据已知信息列出二次方程，求解，然后结合情境解释解的含义。

Example: A rectangle has length (x + 3) cm and width x cm. Its area is 40 cm². Find the value of x.

例题：一个长方形的长是 (x + 3) cm，宽是 x cm，面积为 40 cm²。求 x 的值。

The area gives the equation:

由面积可得方程：

x(x + 3) = 40, x² + 3x − 40 = 0

Factorising: (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since length cannot be negative, we reject x = −8. The width is 5 cm and the length is 8 cm. Always check whether each solution makes sense in the original context — negative lengths, distances, or counts must be rejected.

因式分解：(x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于长度不能为负数，我们舍去 x = −8。宽为 5 cm，长为 8 cm。一定要检查每个解在原始情境中是否有意义——负数长度、距离或数量必须舍去。

A common type of problem involves projectile motion, such as h = −5t² + 20t + 2, where h is height in metres and t is time in seconds. To find when the object hits the ground, set h = 0 and solve the resulting quadratic.

常见的一类问题是抛体运动，例如 h = −5t² + 20t + 2，其中 h 是高度（米），t 是时间（秒）。要求物体何时落地，令 h = 0 并解所得的二次方程。

10. Common Mistakes and How to Avoid Them | 常见错误与防范

Several recurring errors cost students marks in examinations. Being aware of them is the first step towards avoiding them.

有几个反复出现的错误会让学生在考试中失分。意识到这些错误是避免它们的第一步。

Forgetting to rearrange to standard form before applying the formula. Always write ax² + bx + c = 0 first.

使用求根公式前忘记将方程整理成标准形式。务必先写出 ax² + bx + c = 0。
Sign errors when substituting negative values of b or c into the quadratic formula. Use brackets carefully.

将 b 或 c 的负值代入求根公式时出现符号错误。加括号时要格外小心。
Dropping the ± symbol and only giving one solution. A quadratic equation always has two solutions (which may be equal).

漏写 ± 符号，只给出一个解。一元二次方程总是有两个解（可能相等）。
Incorrect factorisation. Always expand your brackets to verify your factors.

因式分解错误。务必展开括号来验证你的因子。
Ignoring invalid solutions in word problems. Reject negative lengths, times, or other quantities that do not make sense.

在应用题中忽略无效解。舍去不合理的负数长度、时间或其他量。

One algebraic point deserves special emphasis: dividing both sides of an equation by x is not allowed when x could be zero, because you may lose a root. For example, x² = 3x should be solved by rearranging to x² − 3x = 0, then factorising to x(x − 3) = 0, giving x = 0 or x = 3. Dividing by x would incorrectly give only x = 3.

有一个代数要点需要特别强调：当 x 可能为零时，不能在方程两边同时除以 x，因为这样可能会丢失一个根。例如，x² = 3x 应通过移项化为 x² − 3x = 0，再因式分解为 x(x − 3) = 0，得到 x = 0 或 x = 3。如果两边除以 x，就会错误地只得到 x = 3。

11. Choosing the Right Method | 选择正确的方法

With three algebraic methods available, knowing which to use saves precious time in the examination. Here is a practical guide.

既然有三种代数方法可用，知道如何选择就能在考试中节省宝贵时间。下面是一份实用指南。

Situation	Recommended Method
Simple integer roots, coefficient of x² is 1	Factorisation
Coefficient of x² is not 1 but factors are obvious	Factorisation (check by expanding)
Equation does not factorise; calculator is allowed	Quadratic formula
Need the vertex or turning point of a graph	Completing the square
Paper 1 non-calculator, equation does not factorise	Completing the square or leave in surd form

In Paper 1 (non-calculator), if factorisation is not possible, you may be expected to complete the square or to express your answer in surd form using the quadratic formula. Practice all three methods so that you can switch between them flexibly.

在 Paper 1（不能使用计算器）中，如果无法因式分解，你可能需要配方，或使用求根公式以根式形式表达答案。练习全部三种方法，以便灵活切换。

12. Exam-Style Practice Questions | 考试风格练习题

Apply what you have learned to these exam-style questions. Write down full working, not just answers.

通过以下考试风格的题目来运用所学知识。写出完整过程，而不只是答案。

Question 1: Solve x² − 7x + 10 = 0 by factorisation.

第 1 题：用因式分解法解 x² − 7x + 10 = 0。

Question 2: Solve 3x² + 5x − 2 = 0 using the quadratic formula, giving answers correct to 2 decimal places.

第 2 题：用求根公式解 3x² + 5x − 2 = 0，答案精确到小数点后两位。

Question 3: Express x² + 8x + 15 in the form (x + p)² + q, and hence find the minimum value of the expression.

第 3 题：将 x² + 8x + 15 化为 (x + p)² + q 的形式，并由此求该表达式的最小值。

Question 4: Find the value(s) of k such that the equation kx² + 6x + 3 = 0 has exactly one real root.

第 4 题：求 k 的值，使得方程 kx² + 6x + 3 = 0 恰好有一个实数根。

Question 5: A right-angled triangle has legs of lengths x and x + 2, and hypotenuse of length x + 4. Form and solve a quadratic equation to find x.

第 5 题：一个直角三角形的两条直角边长为 x 和 x + 2，斜边长为 x + 4。列出并解一元二次方程求 x。

Check your answers against your working: Q1: x = 2 or x = 5. Q2: x ≈ 0.27 or x ≈ −2.49. Q3: (x + 4)² − 1, minimum value −1. Q4: k = 3. Q5: x = 6 (reject x = −2). If you obtained all five correct, you are well prepared for the quadratic equations questions in your IGCSE examination.

对照过程检查答案：第 1 题：x = 2 或 x = 5。第 2 题：x ≈ 0.27 或 x ≈ −2.49。第 3 题：(x + 4)² − 1，最小值为 −1。第 4 题：k = 3。第 5 题：x = 6（舍去 x = −2）。如果五题全部正确，说明你已经为 IGCSE 考试中的二次方程题目做好了充分准备。

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Solving Quadratic Equations | 解一元二次方程