Mastering Quadratic Equations | 解二次方程完全指南

📚 Mastering Quadratic Equations | 解二次方程完全指南

Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. Whether you are taking the Core or Extended paper, you will meet quadratic expressions in algebra, coordinate geometry, and problem-solving questions. This guide explains the key methods of solving quadratic equations step by step, with worked examples, exam tips, and common pitfalls to avoid.

二次方程是 IGCSE 数学考纲中最核心的内容之一。无论你参加 Core（核心）还是 Extended（拓展）试卷，你都会在代数、坐标几何与应用题中遇到二次表达式。本指南将逐步讲解解二次方程的四大基本方法，并配有例题、考试技巧与常见误区提醒，帮助你稳拿高分。

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

A quadratic equation is a polynomial equation of degree 2. Its general form is:

ax² + bx + c = 0, where a ≠ 0

Here a, b and c are real numbers, and the coefficient of x² must not be zero. If a = 0, the equation becomes linear, not quadratic. The highest power of x is 2, which explains the name “quadratic” (from the Latin “quadratus”, meaning square).

二次方程是最高次数为 2 的多项式方程，其标准形式为 ax² + bx + c = 0，其中 a ≠ 0。这里 a、b、c 为实数，且 x² 的系数 a 不能为零。若 a = 0，方程就退化为一次方程（线性方程）。方程中 x 的最高次数为 2，因此称为“二次”方程。

Examples of quadratic equations are x² − 5x + 6 = 0, 2t² + 3t − 1 = 0 and 4x² − 9 = 0. An equation like x³ + x² = 0 is not quadratic because it contains a cubic term. On the IGCSE exam, you must first rearrange any given equation into the standard form ax² + bx + c = 0 before applying a solution method.

例如 x² − 5x + 6 = 0、2t² + 3t − 1 = 0、4x² − 9 = 0 都是二次方程。而 x³ + x² = 0 就不是二次方程，因为它含有三次项。在考试中，你必须先把题目给出的方程整理成 ax² + bx + c = 0 的标准形式，再选择解法。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic has simple integer factors. The idea is to rewrite ax² + bx + c as a product of two linear brackets, then use the fact that if the product of two expressions is zero, at least one of them must be zero. In symbols:

If p × q = 0, then p = 0 or q = 0.

因式分解法在二次式含有简单整数因子时最为快捷。其核心思想是把 ax² + bx + c 改写为两个一次括号的乘积，再利用“若两个数相乘等于零，则至少其中一个为零”的性质。即：若 p × q = 0，则 p = 0 或 q = 0。

Worked example: solve x² − 5x + 6 = 0 by factorisation. We look for two numbers that multiply to give 6 and add to give −5. The numbers are −2 and −3, so the equation becomes:

(x − 2)(x − 3) = 0

Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3. Always check by substituting both values back into the original equation. In an exam, the factorisation method scores full marks only if the factor pair is correct; show every step clearly.

例题：用因式分解法解 x² − 5x + 6 = 0。我们需要找两个数，它们相乘等于 6，相加等于 −5。这两个数是 −2 和 −3，于是方程化为 (x − 2)(x − 3) = 0。因此 x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。记得把两个解代回原方程检验。考试中，因式分解法必须因式对、步骤全，才能得满分。

For quadratics with a leading coefficient not equal to 1, such as 2x² + 7x + 3 = 0, you must find two brackets of the form (px + r)(qx + s) where p × q = 2 and r × s = 3. In this example, 2x² + 7x + 3 = (2x + 1)(x + 3) = 0, so x = −½ or x = −3. Practise several examples until you can spot the bracket pair quickly.

当二次项系数不为 1 时，例如 2x² + 7x + 3 = 0，你需要找到形如 (px + r)(qx + s) 的两个括号，其中 p × q = 2，r × s = 3。本例中 2x² + 7x + 3 = (2x + 1)(x + 3) = 0，所以 x = −½ 或 x = −3。多练习几道题，直到你能快速看出括号的配对。

3. The Quadratic Formula | 公式法（二次公式）

The quadratic formula solves any quadratic equation, including those that cannot be factorised easily. For ax² + bx + c = 0, the solutions are given by:

x = (−b ± √(b² − 4ac)) ÷ (2a)

公式法可以解任何一个二次方程，包括不容易因式分解的方程。对于 ax² + bx + c = 0，解为：x = (−b ± √(b² − 4ac)) ÷ (2a)。

Worked example: solve 2x² + 3x − 7 = 0 using the quadratic formula. Here a = 2, b = 3, c = −7. Substitute into the formula:

x = (−3 ± √(3² − 4 × 2 × (−7))) ÷ (2 × 2) = (−3 ± √65) ÷ 4

Thus x = (−3 + √65) ÷ 4 ≈ 1.27 or x = (−3 − √65) ÷ 4 ≈ −2.77. Give your final answer to three significant figures unless the question says otherwise. In IGCSE exams, the formula is not provided on the formula sheet for some boards, so memorise it carefully.

例题：用公式法解 2x² + 3x − 7 = 0。这里 a = 2，b = 3，c = −7。代入公式：x = (−3 ± √(3² − 4 × 2 × (−7))) ÷ (2 × 2) = (−3 ± √65) ÷ 4。因此 x = (−3 + √65) ÷ 4 ≈ 1.27，或 x = (−3 − √65) ÷ 4 ≈ −2.77。除非题目另有要求，最终答案保留到三位有效数字。注意：部分考试局的公式表并不提供二次公式，请务必背熟。

When using a calculator, be careful with negative values of c. A common error is writing −4ac incorrectly; remember that if c is negative, −4ac becomes positive. For example, with a = 2, c = −7, the term −4ac equals −4 × 2 × (−7) = +56. Always press the bracket keys correctly on your calculator.

使用计算器时，要特别注意 c 为负数的情况。常见错误是把 −4ac 算错；请记住，若 c 为负数，则 −4ac 为正。例如 a = 2，c = −7 时，−4ac = −4 × 2 × (−7) = +56。计算时务必正确使用括号键。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This method is useful for finding turning points of parabolas and solving equations without using the formula. For a quadratic x² + bx + c, we take half of b, square it, and adjust the constant term:

x² + bx + c = (x + b/2)² − (b/2)² + c

配方法将二次式改写为 a(x + p)² + q 的形式。这种方法常用于求抛物线的顶点坐标，也可用来解方程而不必套用公式。对于 x² + bx + c，我们取 b 的一半、平方，再对常数项作相应调整：x² + bx + c = (x + b/2)² − (b/2)² + c。

Worked example: solve x² + 6x + 2 = 0 by completing the square. Half of 6 is 3, and 3² = 9. Therefore:

x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7 = 0

So (x + 3)² = 7, taking square roots gives x + 3 = ±√7, hence x = −3 ± √7. The exact answers are −3 + √7 and −3 − √7, approximately 0.646 and −6.646. This method also reveals that the vertex of the parabola y = x² + 6x + 2 is at (−3, −7).

例题：用配方法解 x² + 6x + 2 = 0。6 的一半是 3，3² = 9。因此 x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7 = 0。于是 (x + 3)² = 7，两边开平方得 x + 3 = ±√7，故 x = −3 ± √7。精确解为 −3 + √7 和 −3 − √7，约等于 0.646 和 −6.646。这个方法还能直接看出抛物线 y = x² + 6x + 2 的顶点为 (−3, −7)。

If the leading coefficient a is not 1, factor it out first. For example, 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3. This form is particularly useful in Extended IGCSE questions asking for the minimum or maximum value of a function.

若二次项系数 a 不为 1，先把它提取出来。例如 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3。这种形式在 IGCSE 拓展卷中特别有用，常用来求函数的最小值或最大值。

5. The Discriminant | 判别式

The expression b² − 4ac inside the square root of the quadratic formula is called the discriminant, often written as Δ (Delta). It tells us how many real roots a quadratic equation has without actually solving it.

二次公式根号内的表达式 b² − 4ac 称为判别式，通常记为 Δ（Delta）。它能在不解方程的情况下告诉我们二次方程有多少个实数根。

判别式 Δ	根的情况	几何意义
Δ > 0	两个不相等实数根	抛物线与 x 轴交于两点
Δ = 0	两个相等实数根（重根）	抛物线与 x 轴相切于一点
Δ < 0	无实数根	抛物线与 x 轴没有交点

For example, consider x² − 4x + 4 = 0. Here Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0, so the equation has exactly one repeated root: x = 2. For x² + x + 1 = 0, Δ = 1 − 4 = −3 < 0, so there are no real roots; the curve never touches the x-axis.

例如 x² − 4x + 4 = 0，Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0，所以方程只有一个重根 x = 2。对于 x² + x + 1 = 0，Δ = 1 − 4 = −3 < 0，因此没有实数根，抛物线完全不接触 x 轴。

The discriminant is also used in coordinate geometry problems where a line is tangent to a curve. If a line just touches a parabola, setting the two equations equal produces a quadratic with Δ = 0. Questions of this type are common in the Extended paper, so practise connecting the discriminant with tangency conditions.

判别式还常用于解析几何中判断直线与抛物线是否相切。若直线与抛物线恰好相切，联立两个方程得到的二次方程应满足 Δ = 0。这类题目在拓展卷中很常见，请多加练习，把判别式与相切条件联系起来。

6. Solving by Graphing | 图象解法

The roots of a quadratic equation ax² + bx + c = 0 are the x-coordinates where the graph of y = ax² + bx + c crosses the x-axis. Drawing the parabola gives approximate solutions. This method is useful when exact factorisation is difficult or when the question explicitly asks for a graphical solution.

二次方程 ax² + bx + c = 0 的根，就是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。画出抛物线即可得到近似解。当因式分解困难，或题目明确要求用图象法时，这种方法非常有用。

To sketch the graph, identify three key features: the y-intercept (0, c), the vertex by completing the square, and the axis of symmetry x = −b/(2a). For y = x² − 2x − 8, the y-intercept is −8, the axis is x = 1, and the vertex is (1, −9). The graph crosses the x-axis at x = −2 and x = 4, which are the roots of x² − 2x − 8 = 0.

画草图时，先找出三个关键特征：y 截距 (0, c)、用配方求出的顶点、以及对称轴 x = −b/(2a)。以 y = x² − 2x − 8 为例，y 截距为 −8，对称轴为 x = 1，顶点为 (1, −9)。图象与 x 轴交于 x = −2 和 x = 4，这正是方程 x² − 2x − 8 = 0 的两个根。

In IGCSE exams, a graphical solution usually requires reading values to one decimal place from a curve. Always draw a smooth curve, plot at least five points, and label the axes. If a calculator with graphing capability is allowed, use it to check your sketch, but do not rely on it for the final written answer.

在 IGCSE 考试中，图象法通常需要从曲线中读出精确到一位小数的数值。画图时要画平滑的曲线，至少描五个点，并标好坐标轴。如果允许使用图形计算器，可以用它来检查草图，但最终书面答案不能依赖计算器。

7. Word Problems Leading to Quadratics | 二次方程应用题

Many real-world and exam problems produce quadratic equations. Common contexts include area and perimeter, projectile motion, consecutive integers, and number problems. The first step is always to define a variable, then translate the conditions into a quadratic equation, solve it, and finally check which solution makes sense in the context.

许多实际问题和考试应用题都会导出二次方程。常见情境有：面积与周长、抛体运动、连续整数、数字问题等。第一步总是设未知数，然后把条件翻译成二次方程，解出后再根据实际意义检查哪个解合理。

Example: the area of a rectangle is 36 cm² and its length is 5 cm more than its width. Let the width be x. Then x(x + 5) = 36, so x² + 5x − 36 = 0. Factorising gives (x + 9)(x − 4) = 0, so x = −9 or x = 4. Since width cannot be negative, the width is 4 cm and the length is 9 cm.

例题：一个长方形的面积为 36 cm²，长比宽多 5 cm。设宽为 x，则 x(x + 5) = 36，即 x² + 5x − 36 = 0。因式分解得 (x + 9)(x − 4) = 0，所以 x = −9 或 x = 4。宽度不能为负数，因此宽为 4 cm，长为 9 cm。

In projectile motion problems, the height h of an object is often modelled by h = −4.9t² + vt + h₀, where t is time. Setting h = 0 gives the time when the object lands. The negative root is physically meaningless and must be rejected. Always write a final sentence stating the answer with its unit.

在抛体运动问题中，物体的高度 h 通常用 h = −4.9t² + vt + h₀ 来建模，t 为时间。令 h = 0 可得到物体落地的时间。负根在物理上没有意义，必须舍去。最后一定要写出带单位的完整答句。

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

Below are the most frequent errors students make in IGCSE quadratic equations, with advice on how to avoid them.

以下列出学生在 IGCSE 二次方程题目中最常犯的错误，以及相应的避免方法。

Forgetting to rearrange: always bring all terms to one side to get ax² + bx + c = 0 before solving.
忘记整理方程：解题前务必把所有项移到同一边，化成 ax² + bx + c = 0 的标准形式。
Dividing by x: never divide both sides by x because you lose the root x = 0.
同除以 x：切勿两边同除以 x，否则会丢失 x = 0 这个根。
Sign errors in the formula: when substituting into x = (−b ± √(b² − 4ac)) ÷ (2a), write down the values of a, b, c first.
公式代入符号错误：代入二次公式前，先清晰地写出 a、b、c 的值。
Ignoring the context: reject negative lengths, negative times and any solution that does not fit the problem.
忽略实际意义：舍去负长度、负时间以及不符合题意的解。
Rushing factorisation: check your brackets by expanding them mentally before writing the final answer.
因式分解不检查：写出答案前，用心算展开括号来验证是否正确。

Exam tip: for a “solve” question worth 3 marks, the marking scheme usually awards 1 mark for rearranging into standard form, 1 mark for correct factorisation or substitution, and 1 mark for the two final values. Even if you use the quadratic formula, write the substitution step explicitly; this earns method marks even if your final arithmetic is wrong.

考试技巧：一道 3 分的“解方程”题，评分标准通常为：1 分给化成标准形式，1 分给正确的因式分解或代入公式，1 分给两个最终数值。即使你使用公式法，也要把代入步骤写出来；这样即使最后算错，也能获得方法分。

9. Quick Practice Questions | 快速练习

Try these five questions without a calculator first, then check your answers. They cover all four methods discussed above.

请先不要使用计算器完成下面五道题，再核对答案。这些题目覆盖了上述四种解法。

Solve x² − 7x + 12 = 0 by factorisation. | 用因式分解法解 x² − 7x + 12 = 0。
Solve 3x² + 5x − 2 = 0 using the quadratic formula. | 用公式法解 3x² + 5x − 2 = 0。
Solve x² − 10x + 20 = 0 by completing the square. | 用配方法解 x² − 10x + 20 = 0。
Find the discriminant of 2x² − 4x + 1 = 0 and state the number of real roots. | 求 2x² − 4x + 1 = 0 的判别式，并判断实数根的个数。
A square’s area is increased by 24 cm² when each side is increased by 2 cm. Find the original side length. | 正方形边长增加 2 cm 后，面积增加了 24 cm²。求原边长。

Answers: 1. x = 3 or x = 4. 2. x = ⅓ or x = −2. 3. x = 5 ± √5. 4. Δ = 8 > 0, two distinct real roots. 5. Let the side be s, then (s + 2)² − s² = 24, so 4s + 4 = 24, giving s = 5 cm.

参考答案：1. x = 3 或 x = 4。2. x = ⅓ 或 x = −2。3. x = 5 ± √5。4. Δ = 8 > 0，有两个不相等的实数根。5. 设原边长为 s，则 (s + 2)² − s² = 24，即 4s + 4 = 24，解得 s = 5 cm。

10. Key Takeaways | 核心总结

Quadratic equations appear in almost every IGCSE Mathematics paper. Master the four main methods: factorisation, the quadratic formula, completing the square, and the graphical approach. Know the discriminant and what it tells you. Most importantly, always check whether your solutions are reasonable in the given context.

二次方程几乎出现在每一份 IGCSE 数学试卷中。请务必掌握四种主要解法：因式分解法、公式法、配方法和图象法。理解判别式及其含义。最重要的是，解完以后要检查答案在题目情境中是否合理。

With steady practice, recognising which method suits which question becomes automatic. In the exam, read the command word carefully: “solve” requires an algebraic answer, “sketch” requires a labelled graph, and “find the exact value” means leave your answer in surd form such as −3 ± √7. Build your confidence by mixing practice questions from past papers, and you will turn this topic into a reliable source of marks.

经过稳定练习，你自然会迅速判断哪道题该用哪种方法。考试时请仔细阅读指令词：“solve”要求代数答案，“sketch”要求画出标注的图象，“find the exact value”则要求保留根号形式，例如 −3 ± √7。通过混合练习历年真题来建立信心，你就能把这个专题变成稳定的得分点。

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Mastering Quadratic Equations | 解二次方程完全指南