Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. Whether you are sitting a Core or Extended paper, you will almost certainly encounter quadratics — either as direct solve questions, factorisation exercises, or within geometry and word problems. Mastering every method of solution is not optional; it is essential for top marks.

二次方程是 IGCSE 数学中考查频率最高的知识点之一。无论你参加的是 Core 还是 Extended 考卷，几乎必然遇到二次方程——无论是直接求解、因式分解练习，还是几何题和文字应用题。掌握每一种求解方法不是可选项，而是冲击高分的关键。

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 standard form:

ax² + bx + c = 0

where a, b and c are constants and a ≠ 0. The highest power of the variable is 2, which is why the equation is called ‘quadratic’ (from the Latin quadratus, meaning square). If a equals zero, the equation becomes linear, not quadratic.

二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数为 2，因此称为”二次”方程。若 a 等于 0，方程退化为一次方程，而非二次方程。

Examples of quadratic equations include x² − 4x + 3 = 0, 2x² + 5x − 1 = 0, and x² = 9 (which can be rewritten as x² − 9 = 0). Expressions such as x³ + 2x² − 1 = 0 are not quadratic because they contain a cubic term.

二次方程的例子包括 x² − 4x + 3 = 0、2x² + 5x − 1 = 0 和 x² = 9（可改写为 x² − 9 = 0）。而 x³ + 2x² − 1 = 0 这样的方程不是二次方程，因为它含有三次项。

2. Solving by Factorisation | 因式分解法

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

因式分解法通常是对含有简单整数根的二次方程最快的解法。其核心原理是零乘积性质：如果两个表达式的乘积为零，那么其中至少有一个表达式为零。

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

若 (x − p)(x − q) = 0，则 x = p 或 x = q。

Worked Example 1: Solve x² − 5x + 6 = 0.

例题 1：解方程 x² − 5x + 6 = 0。

We look for two numbers whose product is +6 and whose sum is −5. These are −2 and −3. Therefore:

我们要找到两个数，它们的乘积为 +6，和为 −5。这两个数是 −2 和 −3。因此：

(x − 2)(x − 3) = 0

Setting each bracket equal to zero gives x = 2 or x = 3. You should always check both roots by substituting them back into the original equation.

令每个括号等于零，得到 x = 2 或 x = 3。务必把两个根代回原方程进行验证。

Worked Example 2: Solve 2x² + 7x + 3 = 0.

例题 2：解方程 2x² + 7x + 3 = 0。

When the coefficient of x² is not 1, use the method of splitting the middle term. Multiply a and c: 2 × 3 = 6. Find two numbers whose product is 6 and whose sum is 7 — these are 6 and 1. Rewrite the equation:

当 x² 的系数不为 1 时，使用裂项法。先计算 a 与 c 的乘积：2 × 3 = 6。找到两个数，使其乘积为 6、和为 7——这两个数是 6 和 1。改写原方程：

2x² + 6x + x + 3 = 0

Factorise in pairs: 2x(x + 3) + 1(x + 3) = 0, then (2x + 1)(x + 3) = 0. Hence x = −3 or x = −½.

分组因式分解：2x(x + 3) + 1(x + 3) = 0，即 (2x + 1)(x + 3) = 0。因此 x = −3 或 x = −½。

Always rearrange the equation into standard form ax² + bx + c = 0 before factorising.
Remember to check for common factors first, e.g. 2x² − 8x = 2x(x − 4).
Look out for the difference of two squares: x² − 9 = (x − 3)(x + 3).

因式分解前务必把方程整理成标准形式 ax² + bx + c = 0。
先检查公因数，例如 2x² − 8x = 2x(x − 4)。
注意平方差公式：x² − 9 = (x − 3)(x + 3)。

3. Completing the Square | 配方法

Completing the square is a powerful technique that works for every quadratic equation, even when factorisation fails. It also helps you find the turning point of a quadratic graph, making it doubly important for IGCSE.

配方法是一种对所有二次方程都适用的强有力技巧，即使因式分解行不通时也能使用。它还能帮助你找到二次图像的顶点，因此在 IGCSE 中具有双重重要性。

The general rule for a quadratic of the form x² + bx is to add and subtract (b/2)²:

对于形如 x² + bx 的二次式，一般规则是加上并减去 (b/2)²：

x² + bx = (x + b/2)² − (b/2)²

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

例题：用配方法解方程 x² + 6x + 2 = 0。

Take the first two terms x² + 6x. Here b = 6, so b/2 = 3 and (b/2)² = 9. Hence:

取前两项 x² + 6x。这里 b = 6，所以 b/2 = 3，(b/2)² = 9。因此：

x² + 6x = (x + 3)² − 9

Substitute back into the equation:

代回原方程：

(x + 3)² − 9 + 2 = 0

(x + 3)² = 7

Take the square root of both sides, remembering the ± sign:

两边同时开平方，注意 ± 号：

x + 3 = ±√7

x = −3 ± √7

So the two solutions are x = −3 + √7 and x = −3 − √7. These are exact values, which is a major advantage when the examiner asks for answers in surd form.

因此两个解为 x = −3 + √7 和 x = −3 − √7。这些是精确值，当题目要求以根式形式作答时，这是一个重要优势。

If the coefficient of x² is not 1, factor it out first. For example, 2x² + 8x + 5 = 2(x² + 4x) + 5.
The turning point of y = (x + p)² + q is at (−p, q).

若 x² 的系数不为 1，先把它提取出来。例如 2x² + 8x + 5 = 2(x² + 4x) + 5。
y = (x + p)² + q 的顶点坐标为 (−p, q)。

4. The Quadratic Formula | 求根公式

The quadratic formula is the ultimate universal method. It solves any quadratic equation, no matter how awkward the coefficients are. You are expected to memorise it for the IGCSE examination.

求根公式是万能的终极方法。无论系数多么复杂，它都能解出任何二次方程。IGCSE 考试要求你牢记此公式。

x = (−b ± √(b² − 4ac)) / 2a

Worked Example: Solve 2x² − 4x − 3 = 0, giving your answer correct to 3 significant figures.

例题：解方程 2x² − 4x − 3 = 0，答案保留 3 位有效数字。

Identify a = 2, b = −4 and c = −3. Substitute into the formula:

确定 a = 2，b = −4，c = −3。代入公式：

x = (4 ± √((−4)² − 4 × 2 × (−3))) / (2 × 2)

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4

Simplify √40 = 2√10, so x = (4 ± 2√10) / 4 = 1 ± (√10)/2. Using a calculator: x ≈ 2.58 or x ≈ −0.581.

化简 √40 = 2√10，因此 x = (4 ± 2√10) / 4 = 1 ± (√10)/2。用计算器计算：x ≈ 2.58 或 x ≈ −0.581。

When using the formula, be careful with negative signs. Substituting b = −4 into −b gives +4, and b² = (−4)² = 16, not −16. This is one of the most common student errors.

使用公式时要特别注意负号。把 b = −4 代入 −b 得到 +4，而 b² = (−4)² = 16，绝不是 −16。这是最常见的错误之一。

5. The Discriminant | 判别式

The expression inside the square root of the quadratic formula, Δ = b² − 4ac, is called the discriminant. It tells you how many roots a quadratic equation has, without actually solving it.

求根公式根号内的表达式 Δ = b² − 4ac 称为判别式。它无需实际求解方程，就能告诉我们二次方程有几个根。

Value of Δ	Nature of roots	Graph shape
Δ > 0	Two distinct real roots	Crosses the x-axis twice
Δ = 0	One repeated root (equal roots)	Touches the x-axis once
Δ < 0	No real roots	Never touches the x-axis

Δ 的值	根的性质	图像特征
Δ > 0	两个不相等的实数根	与 x 轴有两个交点
Δ = 0	一个二重根（等根）	与 x 轴相切（一个交点）
Δ < 0	无实数根	不与 x 轴相交

Worked Example: Find the value of k such that the equation kx² + 6x + 3 = 0 has exactly one root.

例题：求 k 的值，使方程 kx² + 6x + 3 = 0 恰好有一个根。

For exactly one root, the discriminant must equal zero. Here a = k, b = 6, c = 3:

方程恰有一个根时，判别式必须等于零。这里 a = k，b = 6，c = 3：

Δ = 36 − 4 × k × 3 = 36 − 12k = 0

k = 3

Note that if the problem uses the phrase ‘no real roots’ rather than ‘no roots’, that is precisely the case Δ < 0 — the equation may still have complex roots, but at IGCSE level we simply state that there are no real solutions.

注意：若题目用”无实数根”而非”无根”，这正是 Δ < 0 的情形——方程在复数范围内可能有根，但在 IGCSE 阶段我们只需说明没有实数解。

6. Graphical Interpretation | 图像意义

The graph of y = ax² + bx + c is a parabola. The roots of the equation ax² + bx + c = 0 correspond to the x-intercepts of the parabola. The graph also reveals the axis of symmetry and the turning point.

y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的根对应抛物线与 x 轴的交点。图像还可以揭示对称轴和顶点。

When a > 0, the parabola opens upwards and has a minimum turning point. When a < 0, the parabola opens downwards and has a maximum turning point.

当 a > 0 时，抛物线开口朝上，有最小值顶点；当 a < 0 时，抛物线开口朝下，有最大值顶点。

The axis of symmetry is given by x = −b/2a, and this x-value is substituted into the equation to find the y-coordinate of the turning point. For example, for y = x² + 6x + 2, the axis is x = −3, matching the completing-the-square result y = (x + 3)² − 7.

对称轴由 x = −b/2a 给出，把该 x 值代入方程即可得到顶点的 y 坐标。例如，对于 y = x² + 6x + 2，对称轴为 x = −3，与配方法的结果 y = (x + 3)² − 7 一致。

In an IGCSE question, you may be asked to sketch the graph, mark the roots and the turning point. Always start by finding the discriminant to confirm whether the curve actually crosses the x-axis. Then identify the axis of symmetry and calculate the y-value of the vertex.

在 IGCSE 题目中，你可能会被要求画草图、标出根和顶点。始终先求判别式，确认曲线是否真正穿过 x 轴；然后找对称轴并计算顶点的 y 值。

7. Applications and Word Problems | 实际应用与文字题

Quadratics are not just abstract algebra — they appear in geometry, projectile motion, and economic problems. The key to solving word problems is translating the text into an equation.

二次方程不仅仅是抽象的代数——它们出现在几何、抛体运动和经济问题中。解决文字题的关键是把文字翻译成方程。

Worked Example: A rectangle has length (x + 5) cm and width (x − 2) cm. Its area is 40 cm². Find x.

例题：一个长方形的长为 (x + 5) cm，宽为 (x − 2) cm，面积为 40 cm²。求 x 的值。

Write the area equation:

列出面积方程：

(x + 5)(x − 2) = 40

x² + 3x − 10 = 40

x² + 3x − 50 = 0

This does not factorise nicely, so we use the quadratic formula with a = 1, b = 3, c = −50:

该方程不易因式分解，使用求根公式，其中 a = 1，b = 3，c = −50：

x = (−3 ± √(9 + 200)) / 2 = (−3 ± √209) / 2

x ≈ 5.73 or x ≈ −8.73

Since x represents a length, the negative solution must be rejected. Therefore x ≈ 5.73, giving a length of approximately 10.73 cm and width of approximately 3.73 cm.

由于 x 表示长度，必须舍去负数解。因此 x ≈ 5.73，长约 10.73 cm，宽约 3.73 cm。

Always read the question carefully: does it ask for exact values, values to 3 significant figures, or values to 2 decimal places? Does the context require you to discard a negative root?

务必仔细审题：题目要求精确值、保留 3 位有效数字还是两位小数？实际情境是否需要舍去负根？

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

Many students lose marks on quadratics due to avoidable errors. Being aware of these pitfalls will help you secure full marks in the exam.

许多学生在二次方程上失分是因为一些本可避免的错误。了解这些陷阱，能帮助你在考试中拿到满分。

Forgetting to rearrange: Always bring all terms to one side before factorising. x² = 5x must be written as x² − 5x = 0 before you can solve.
Dividing by x: Never divide both sides by x. In x² = 5x, dividing by x leaves x = 5 and loses the root x = 0.
Missing the ± sign: When taking square roots, always write ±. If x² = 25, then x = ±5.
Calculator errors: In the quadratic formula, use brackets carefully. Type (−4)² not −4², because −4² = −16.
Not checking answers: Substitute roots back into the original equation; it only takes seconds and catches mistakes.

忘记化为标准形式：因式分解前必须将所有项移到等号一侧。x² = 5x 必须先写成 x² − 5x = 0，然后才能求解。
两边同除以 x：永远不要两边同除以 x。在 x² = 5x 中，除以 x 会得到 x = 5，从而丢失根 x = 0。
漏掉 ± 号：开平方时永远要写 ±。若 x² = 25，则 x = ±5。
计算器输入错误：使用求根公式时小心括号。输入 (−4)² 而不是 −4²，因为 −4² = −16。
不验算：把根代回原方程只需几秒钟，却能发现大多数错误。

Exam tip: If a question asks for ‘answers to 3 significant figures’, you can jump straight to the quadratic formula. If it asks for ‘exact values’, try factorisation or completing the square. And if the coefficient of x is even, completing the square is often the neatest path.

考试技巧：若题目要求”答案保留 3 位有效数字”，可以直接使用求根公式；若要求”精确值”，尝试因式分解或配方法；若 x 的系数为偶数，配方法通常是最简洁的路线。

By mastering these five methods — factorisation, completing the square, the quadratic formula, the discriminant, and graphical interpretation — you will be able to handle any quadratic equation the IGCSE paper throws at you. Regular practice is the key to speed and accuracy.

通过掌握这五种方法——因式分解、配方法、求根公式、判别式以及图像分析——你将能从容应对 IGCSE 考卷中的任何二次方程问题。定期练习是提升速度和准确度的关键。

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Solving Quadratic Equations | 解二次方程