Solving Quadratic Equations | 解二次方程（教师用书）

📚 Solving Quadratic Equations | 解二次方程（教师用书）

Quadratic equations are one of the most fundamental topics in IGCSE Mathematics. This teacher’s guide provides a complete walkthrough of methods, common pitfalls, and classroom-ready examples to help students master this essential skill.

二次方程是 IGCSE 数学中最基础的考点之一。本教师用书提供完整的方法讲解、常见误区和课堂实用例题，帮助学生牢固掌握这一核心技能。

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

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The term ‘quadratic’ comes from the Latin ‘quadratus’, meaning ‘square’, because the highest power of the variable is 2.

二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。术语 ‘quadratic’ 源自拉丁语 ‘quadratus’，意为“平方”，因为变量的最高次数是 2。

Before solving, always rearrange the equation so that all terms are on one side and the right-hand side equals zero. For example, 2x² + 5x = 3 must be rewritten as 2x² + 5x − 3 = 0.

在求解之前，务必先整理方程，使所有项位于等号一侧，右侧等于 0。例如，2x² + 5x = 3 必须改写为 2x² + 5x − 3 = 0。

Key facts to emphasise in class:

课堂上需要强调的关键要点：

The coefficient a is called the leading coefficient and must be non-zero.
If a = 0, the equation becomes linear (bx + c = 0).
A quadratic equation has at most two real solutions (roots).

系数 a 称为首项系数，必须非零。
若 a = 0，方程退化为一次方程（bx + c = 0）。
二次方程至多有两个实数解（根）。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic expression has simple integer factors. Students should look for two numbers that multiply to give ac and add to give b.

当二次表达式具有简单的整数因式时，因式分解法往往是最快捷的方法。学生需要寻找两个数，使其乘积等于 ac，且和等于 b。

Example: Solve x² + 7x + 12 = 0. We need two numbers that multiply to 12 and add to 7. These are 3 and 4. Therefore (x + 3)(x + 4) = 0, giving x = −3 or x = −4.

例：解 x² + 7x + 12 = 0。我们需要找到两个数，乘积为 12 且和为 7，这两个数是 3 和 4。因此 (x + 3)(x + 4) = 0，解得 x = −3 或 x = −4。

When the leading coefficient a ≠ 1, the process is slightly longer. Solve 2x² + 5x + 2 = 0. Multiply a and c: 2 × 2 = 4. Find two numbers that multiply to 4 and add to 5: 4 and 1. Split the middle term: 2x² + 4x + x + 2 = 0. Factor by grouping: 2x(x + 2) + 1(x + 2) = 0, giving (2x + 1)(x + 2) = 0. Hence x = −½ or x = −2.

当首项系数 a ≠ 1 时，过程稍长。解 2x² + 5x + 2 = 0。将 a 与 c 相乘：2 × 2 = 4。找到两个数乘积为 4 且和为 5：即 4 和 1。拆分中间项：2x² + 4x + x + 2 = 0。分组因式分解：2x(x + 2) + 1(x + 2) = 0，得 (2x + 1)(x + 2) = 0。因此 x = −½ 或 x = −2。

Always remind students to check their factors by expanding the brackets. A quick expansion confirms whether the factorisation is correct.

始终提醒学生通过展开括号来验证因式是否正确。快速展开即可确认因式分解是否正确。

3. Solving by Completing the Square | 配方法

Completing the square transforms a quadratic into the form a(x + p)² + q = 0. This method is particularly useful when the equation cannot be factorised easily, and it also leads directly to the quadratic formula.

配方法将二次方程转化为 a(x + p)² + q = 0 的形式。当方程不易因式分解时，此方法尤其有用，同时它也能直接推导出求根公式。

The procedure for x² + bx + c = 0:

对于 x² + bx + c = 0 的步骤：

Write x² + bx as (x + b/2)² − (b/2)².
Add the constant c: (x + b/2)² − (b/2)² + c = 0.
Rearrange and solve for x by taking square roots.

将 x² + bx 写成 (x + b/2)² − (b/2)²。
加上常数 c：(x + b/2)² − (b/2)² + c = 0。
整理后通过开平方求解 x。

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

例题：用配方法解 x² − 6x + 2 = 0。

Rewrite: (x − 3)² − 9 + 2 = 0, so (x − 3)² = 7. Taking square roots gives x − 3 = ±√7, so x = 3 ± √7.

改写为：(x − 3)² − 9 + 2 = 0，即 (x − 3)² = 7。开平方得 x − 3 = ±√7，所以 x = 3 ± √7。

x = 3 + √7 or x = 3 − √7

If a ≠ 1, divide the entire equation by a before completing the square. For example, 2x² + 8x + 1 = 0 becomes x² + 4x + ½ = 0 first.

如果 a ≠ 1，需先将整个方程除以 a，再进行配方。例如，2x² + 8x + 1 = 0 先化为 x² + 4x + ½ = 0。

4. 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:

求根公式是一种适用于任何二次方程的通用方法。对于 ax² + bx + c = 0，解由下式给出：

x = (−b ± √(b² − 4ac)) / 2a

Students should memorise this formula. It is derived from completing the square on the general form ax² + bx + c = 0, and it always works regardless of whether the factors are integers or not.

学生应熟记此公式。它通过对一般形式 ax² + bx + c = 0 配方推导而来，无论因式是否为整数，它始终有效。

Worked example: Solve 3x² − 5x − 2 = 0 using the formula. Here a = 3, b = −5, c = −2.

例题：用求根公式解 3x² − 5x − 2 = 0。其中 a = 3，b = −5，c = −2。

Substitute: x = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6.

代入得：x = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6。

This gives x = (5 + 7)/6 = 2, or x = (5 − 7)/6 = −1/3.

因此 x = (5 + 7)/6 = 2，或 x = (5 − 7)/6 = −1/3。

Remind students to check the sign of b carefully. A common error is writing (5 ± …) when b is negative; the formula requires −b, so if b = −5, then −b = +5.

提醒学生注意 b 的符号。常见错误是在 b 为负数时仍写 (5 ± …)；公式中要求 −b，因此若 b = −5，则 −b = +5。

5. The Discriminant | 判别式

The discriminant, denoted Δ, is the part of the formula under the square root: Δ = b² − 4ac. It determines the nature of the roots without solving the full equation.

判别式用 Δ 表示，是求根公式中根号内的部分：Δ = b² − 4ac。它无需完整求解方程即可判断根的性质。

Three cases are essential:

以下三种情况至关重要：

Discriminant Δ	Nature of roots	判别式 Δ	根的性质
Δ > 0	Two distinct real roots	Δ > 0	两个不同的实数根
Δ = 0	One repeated real root	Δ = 0	一个二重实数根
Δ < 0	No real roots	Δ < 0	无实数根

Example: For x² + 4x + 5 = 0, Δ = 16 − 20 = −4 < 0, so the equation has no real roots. The graph never touches the x-axis.

例：对于 x² + 4x + 5 = 0，Δ = 16 − 20 = −4 < 0，因此该方程无实数根，图象与 x 轴无交点。

Teachers should also connect the discriminant to tangent conditions: when Δ = 0, the x-axis is tangent to the curve at the vertex. This links algebra to coordinate geometry.

教师还应将判别式与相切条件联系起来：当 Δ = 0 时，x 轴与曲线在顶点处相切。这可将代数与坐标几何联系起来。

6. Graphical Interpretation | 图象解释

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.

y = ax² + bx + c 的图象是抛物线。方程 ax² + bx + c = 0 的根对应抛物线与 x 轴的交点。

Key graphical facts:

关键的图象性质：

If a > 0, the parabola opens upward and has a minimum point.
If a < 0, the parabola opens downward and has a maximum point.
The axis of symmetry is x = −b / 2a.
The vertex (turning point) lies on the axis of symmetry with y-coordinate found by substituting x back into the equation.

若 a > 0，抛物线开口向上，存在最小值点。
若 a < 0，抛物线开口向下，存在最大值点。
对称轴为 x = −b / 2a。
顶点（转向点）位于对称轴上，将 x 代入原方程可求得其 y 坐标。

Example: For y = x² − 4x + 3, the axis of symmetry is x = 2. Substituting x = 2 gives y = 4 − 8 + 3 = −1, so the vertex is (2, −1). Factorising gives y = (x − 1)(x − 3), so the roots are x = 1 and x = 3.

例：对于 y = x² − 4x + 3，对称轴为 x = 2。代入 x = 2 得 y = 4 − 8 + 3 = −1，因此顶点为 (2, −1)。因式分解得 y = (x − 1)(x − 3)，所以根为 x = 1 和 x = 3。

Encourage students to sketch the graph before solving. This helps them estimate answers and check whether solutions make sense.

鼓励学生在求解前先画草图。这有助于估算答案并检查解是否合理。

7. Applications: Word Problems | 实际应用题

Quadratic equations frequently appear in geometry, physics and business problems. The general strategy is to define the unknown, translate the condition into an equation, solve, and then check the validity of each solution against the context.

二次方程经常出现在几何、物理和商业问题中。一般策略是：设未知数，将条件转化为方程，求解，然后根据实际情况检验每个解的合理性。

Worked example 1: The length of a rectangle is 3 cm greater than its width. The area is 54 cm². Find the dimensions.

例题 1：长方形的长比宽多 3 cm，面积为 54 cm²。求其尺寸。

Let the width be x cm. Then the length is (x + 3) cm. So x(x + 3) = 54, giving x² + 3x − 54 = 0. Factorise: (x + 9)(x − 6) = 0. Hence x = −9 or x = 6. Since width cannot be negative, x = 6. The dimensions are 6 cm by 9 cm.

设宽为 x cm，则长为 (x + 3) cm。于是 x(x + 3) = 54，即 x² + 3x − 54 = 0。因式分解得 (x + 9)(x − 6) = 0。因此 x = −9 或 x = 6。由于宽不可能为负，取 x = 6。所以尺寸为 6 cm × 9 cm。

Worked example 2: A ball is thrown upward with height h metres given by h = 12t − 5t², where t is time in seconds. Find when the ball hits the ground.

例题 2：一个球被向上抛出，高度 h 米由 h = 12t − 5t² 给出，其中 t 为时间（秒）。求球何时落地。

The ball hits the ground when h = 0. So 12t − 5t² = 0, giving t(12 − 5t) = 0. Thus t = 0 (at launch) or t = 12/5 = 2.4 seconds. The ball hits the ground after 2.4 seconds.

球落地时 h = 0，因此 12t − 5t² = 0，即 t(12 − 5t) = 0。所以 t = 0（抛出瞬间）或 t = 12/5 = 2.4 秒。球在 2.4 秒后落地。

Always emphasise that negative or unrealistic roots must be rejected with a brief explanation.

始终强调，对于负数或不切实际的根，必须附上简要说明后予以舍弃。

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

Through years of marking, certain errors appear repeatedly in student scripts. Highlight these in class to help students avoid them.

根据多年的阅卷经验，某些错误在学生答卷中反复出现。在课堂上重点强调这些错误，帮助学生避免。

Mistake 1: Dividing by a variable. Students sometimes divide both sides by x to get a linear equation, losing one root. For example, x² = 5x should be rewritten as x² − 5x = 0, then x(x − 5) = 0, giving x = 0 or x = 5. Dividing by x would lose the root x = 0.

错误 1：除以变量。学生有时会两边同除以 x 将其化为一次方程，从而丢失一个根。例如，x² = 5x 应改写为 x² − 5x = 0，然后 x(x − 5) = 0，得到 x = 0 或 x = 5。若除以 x 则会丢失根 x = 0。

Mistake 2: Forgetting to rearrange to zero. Solving x² = 3x + 4 by directly factorising the left side is invalid. Always bring all terms to one side first: x² − 3x − 4 = 0.

错误 2：忘记整理为零。直接对左边因式分解来解 x² = 3x + 4 是不正确的。务必先将所有项移到一边：x² − 3x − 4 = 0。

Mistake 3: Sign errors when substituting into the quadratic formula. Encourage students to write down a = …, b = …, c = … before substituting, and to double-check the sign of b.

错误 3：代入求根公式时出现符号错误。鼓励学生在代入前先写出 a = …，b = …，c = …，并仔细检查 b 的符号。

Exam tip: If a question says ‘give your answers correct to 2 decimal places’ or ‘give your answers in exact form’, adjust your method accordingly. Exact forms involve fractions and surds; decimal answers require calculator accuracy.

考试技巧：如果题目要求“将答案精确到小数点后 2 位”或“给出精确值形式的答案”，请相应地调整方法。精确形式涉及分数和根式；小数答案则需要计算器精度。

9. Practice Questions | 练习与答案

The following questions cover all the methods discussed in this guide. Use them for classwork, homework or revision tests.

以下问题涵盖本指南讨论的所有方法。可用于课堂练习、家庭作业或复习测验。

Question 1: Solve x² − 9x + 20 = 0 by factorisation.

题目 1：用因式分解法解 x² − 9x + 20 = 0。

Answer: x = 4 or x = 5

Question 2: Solve x² + 6x − 3 = 0 by completing the square, giving your answers in exact form.

题目 2：用配方法解 x² + 6x − 3 = 0，答案以精确形式给出。

Answer: x = −3 ± √12 = −3 ± 2√3

Question 3: Use the quadratic formula to solve 2x² − 4x − 3 = 0, correct to 2 decimal places.

题目 3：用求根公式解 2x² − 4x − 3 = 0，精确到小数点后 2 位。

Answer: x ≈ 2.58 or x ≈ −0.58

Question 4: Show that the equation x² + 2x + 5 = 0 has no real roots.

题目 4：证明方程 x² + 2x + 5 = 0 无实数根。

Answer: Δ = 4 − 20 = −16 < 0, so no real roots

Question 5: The sum of the squares of two consecutive positive integers is 61. Find the integers.

题目 5：两个连续正整数的平方和为 61。求这两个整数。

Answer: 5 and 6 (since 25 + 36 = 61)

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

题目 6：求 k 的值，使方程 kx² + 6x + 3 = 0 恰有一个实数根。

Answer: Δ = 36 − 12k = 0, so k = 3

Encourage students to show all working steps, since method marks are awarded even when the final answer is incorrect.

鼓励学生写出完整的解题步骤，因为即使最终答案错误，方法分仍然可以获得。

This teacher’s guide provides a structured path from basic identification through to advanced applications. Regular practice with all four methods — factorisation, completing the square, the quadratic formula and graphical analysis — will ensure that students are fully prepared for any quadratic equation question in the IGCSE examination.

本教师用书提供了一条从基础识别到高级应用的结构化路径。通过四种方法的规律练习——因式分解、配方、求根公式和图象分析——学生将能够充分应对 IGCSE 考试中任何与二次方程相关的题目。

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Solving Quadratic Equations | 解二次方程（教师用书）