Mastering Quadratic Equations | 掌握二次方程

📚 Mastering Quadratic Equations | 掌握二次方程

The quadratic equation is one of the most frequently tested topics in IGCSE Mathematics. You will meet it in almost every paper, whether as a direct solve question, a curve-sketching exercise, or a contextual problem. This revision guide covers every essential method: factorisation, the quadratic formula, completing the square, the discriminant, and graph sketching — all explained step by step with exam-style examples.

二次方程是 IGCSE 数学中最高频的考点之一，几乎每份试卷都会出现：可能是直接求解、函数图象作图，也可能是实际应用题。本复习指南将系统讲解所有必备方法：因式分解法、二次公式、配方法、判别式以及图象草图，并配以考试风格例题逐步解析。

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:

二次方程是指可以写成如下标准形式的方程：

ax² + bx + c = 0 (a ≠ 0)

Here, a, b, and c are constants, and the highest power of the variable x is 2. If a = 0, the equation becomes linear, not quadratic.

其中 a、b、c 为常数，变量 x 的最高次数为 2。如果 a = 0，方程就变成一次（线性）方程，而不是二次方程。

Examples / 例：x² – 5x + 6 = 0, 2x² + 3x – 2 = 0, x² = 16
Non-examples / 反例：x + 4 = 0 (linear), x³ – 8 = 0 (cubic)

In IGCSE exams, you must recognise the three standard forms of quadratic equations and be ready to rearrange them into ax² + bx + c = 0 before solving.

在 IGCSE 考试中，你必须能够识别二次方程的三种常见形式，并先将其整理为 ax² + bx + c = 0 再求解。

2. Solving by Factorisation | 因式分解法

Factorisation is the quickest method when the quadratic has simple integer factors. The principle is: 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因式时，因式分解法是最快捷的方法。其原理是：若两个表达式的乘积为零，则至少其中一个必为零。

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

Worked example / 例题：Solve x² – 7x + 12 = 0.

解：x² – 7x + 12 = 0

Find two numbers whose product is +12 and sum is –7: these are –3 and –4. / 找两个数，乘积为 +12，和为 –7，这两个数是 –3 和 –4。
Hence (x – 3)(x – 4) = 0. / 因此 (x – 3)(x – 4) = 0。
x – 3 = 0 or x – 4 = 0, giving x = 3 or x = 4. / 所以 x = 3 或 x = 4。

Important rule / 重要规则：Before factorising, you must rearrange the equation so that one side is exactly zero. If you try to factorise x² – 6x = 7 without moving the 7, you will get the wrong answer.

要点：因式分解前，必须将方程整理到一侧为零。若跳过这一步直接对 x² – 6x = 7 分解，必然出错。

For quadratics where the coefficient of x² is not 1 (ax² + bx + c with a ≠ 1), use the “product-sum” method: multiply a by c, then find two numbers that multiply to ac and add to b.

对于二次项系数不为 1 的情况（即 a ≠ 1），可使用”乘积—和”法：先计算 a × c，再找两个数使它们乘积为 ac、和为 b。

3. The Quadratic Formula | 二次公式

When factorisation is difficult or impossible, the quadratic formula always works. You are given this formula in the IGCSE formula sheet, but you must know how to substitute values correctly.

当因式分解困难或无法分解时，二次公式永远有效。IGCSE 公式表中会给出该公式，但你必须掌握正确的代入方法。

x = (–b ± √(b² – 4ac)) / 2a

Worked example / 例题：Solve 2x² + 3x – 2 = 0 using the formula.

解：使用二次公式求解 2x² + 3x – 2 = 0。

Identify a = 2, b = 3, c = –2. / 确定 a = 2，b = 3，c = –2。
Substitute into the formula: / 代入公式：

x = (–3 ± √(3² – 4 × 2 × (–2))) / (2 × 2) = (–3 ± √25) / 4

x = (–3 + 5) / 4 = ½ or x = (–3 – 5) / 4 = –2. / 所以 x = ½ 或 x = –2。

Always write your answer to the required degree of accuracy. If the question says “give your answer correct to 2 decimal places”, round the final values only — never round inside the square root.

务必按题目要求的精度写出答案。若题目要求”答案精确到小数点后两位”，只对最终结果取近似值，绝不能在根号内部提前四舍五入。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This is essential for finding turning points and is also a valid solving method.

配方法是将二次式改写为 (x + p)² + q 的形式。它是求抛物线顶点的关键工具，也可用于解方程。

x² + bx + c = (x + b/2)² – (b/2)² + c

Worked example / 例题：Solve x² + 6x – 7 = 0 by completing the square.

Take half of 6, which is 3, and write (x + 3)². / 取 6 的一半得 3，写成 (x + 3)²。
Since (x + 3)² = x² + 6x + 9, subtract 9 to compensate: / 因为 (x + 3)² = x² + 6x + 9，需减去 9 补偿：

(x + 3)² – 9 – 7 = 0 → (x + 3)² = 16

Take square roots: x + 3 = ±4, so x = 1 or x = –7. / 开平方：x + 3 = ±4，故 x = 1 或 x = –7。

If the coefficient of x² is not 1, first factor it out. For example, 2x² + 8x + 5 = 0 becomes 2(x² + 4x) + 5 = 0 before completing the square on the bracket.

若二次项系数不为 1，应先将该系数提取出来。例如 2x² + 8x + 5 = 0 应变为 2(x² + 4x) + 5 = 0，再对括号内配方。

5. Using the Discriminant | 运用判别式

The discriminant is the expression under the square root in the quadratic formula: Δ = b² – 4ac. It tells you how many real roots the equation has, without solving it fully.

判别式是二次公式中根号内的部分：Δ = b² – 4ac。它无需完整求解即可判断方程实数根的个数。

判别式 Δ = b² – 4ac	Number of real roots / 实数根的个数	Graph interpretation / 图像含义
Δ > 0	Two distinct real roots / 两个不相等的实数根	抛物线穿过 x 轴于两点
Δ = 0	One repeated root / 一个重根（两个相等实根）	抛物线切 x 轴于一点（顶点在轴上）
Δ < 0	No real roots / 无实数根	抛物线不接触 x 轴

Example / 例：For x² + 2x + 5 = 0, Δ = 2² – 4 × 1 × 5 = 4 – 20 = –16 < 0, so there are no real roots.

例：对于 x² + 2x + 5 = 0，Δ = 2² – 4 × 1 × 5 = 4 – 20 = –16 < 0，因此无实数根。

In extended IGCSE papers, you may be asked to find the range of k such that kx² + 4x + 1 = 0 has two real roots. Set Δ > 0 and solve the resulting inequality: 16 – 4k > 0, giving k < 4, with k ≠ 0.

在 IGCSE 扩展卷中，你可能会被问到”求 k 的范围使 kx² + 4x + 1 = 0 有两个实数根”。令 Δ > 0，解不等式 16 – 4k > 0，得 k < 4，且 k ≠ 0。

6. Quadratic Graphs | 二次函数图像

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upward (U-shape); if a < 0, it opens downward (∩-shape).

y = ax² + bx + c 的图象是一条抛物线。若 a > 0，开口向上（U 形）；若 a < 0，开口向下（∩ 形）。

y-intercept / y 轴截距：(0, c)
Roots / 根：the x-values where y = 0 (found by solving the equation)
顶点：the turning point, found by completing the square or using x = –b/(2a)
Line of symmetry / 对称轴：x = –b/(2a)

Example / 例：For y = x² – 4x + 3:

y-intercept: (0, 3) / y 轴截距为 (0, 3)
Roots: x² – 4x + 3 = 0 → (x – 1)(x – 3) = 0 → x = 1 or x = 3 / 根为 x = 1 和 x = 3
Vertex: complete the square → y = (x – 2)² – 1, so turning point is (2, –1) / 配方得 y = (x – 2)² – 1，顶点为 (2, –1)

Plot at least five points (roots, y-intercept, vertex, and one point on each side of the axis of symmetry) to sketch an accurate parabola in the exam.

画图时至少取五个点：两个根、y 轴截距、顶点，以及对称轴两侧各一点，这样抛物线草图才够准确。

7. Word Problems | 应用题

IGCSE exams often embed quadratics in contextual problems, especially area, path width, and projectile motion. The key steps are to define the unknown, form the equation, solve it, and then reject any invalid solution.

IGCSE 考试常将二次方程嵌入实际情境，尤其是面积、路径宽度和抛体运动问题。关键步骤是：设未知数→列方程→求解→舍去不符合实际的根。

Worked example / 例题：A rectangle is 4 cm longer than it is wide, and its area is 60 cm². Find its width.

解：一个矩形的长比宽多 4 cm，面积为 60 cm²，求宽。

Let the width be w cm. Then length = w + 4. / 设宽为 w cm，则长为 w + 4。
Area equation: w(w + 4) = 60. / 面积方程：w(w + 4) = 60。
Expand and rearrange: w² + 4w – 60 = 0. / 展开整理：w² + 4w – 60 = 0。
Factorise: (w + 10)(w – 6) = 0 → w = –10 or w = 6. / 因式分解得 w = –10 或 w = 6。
Reject w = –10 because width cannot be negative. Width = 6 cm. / 舍去负数，宽为 6 cm。

Always check your final answer in the context: here, length = 6 + 4 = 10 cm, and 6 × 10 = 60 ✓.

务必把答案代回原题验证：长为 6 + 4 = 10 cm，6 × 10 = 60 ✓。

8. Common Mistakes | 常见错误

Many marks are lost each year to small but avoidable errors. Watch out for these:

每年都有大量考生因细小但可避免的错误丢分，请警惕以下问题：

错误 Mistake	正确做法 Correct approach
未将方程移项到一侧为零就因式分解	先把常数项移到等号一侧，使右边为 0
代入二次公式时弄错符号（尤其 c 为负数时）	写出 a、b、c 的原值，使用括号代入，如 –4 × 2 × (–2)
开平方时只取正根	记得 x = ±√k，除非题目限制 x 的范围
应用题中保留负数根	根据实际意义舍去负根/非整数解
计算判别式时忘记乘 4ac 的符号	Δ = b² – 4ac，注意减去负值时变加号

A final tip: when solving by factorisation, expand your factors to check they produce the original equation. This takes ten seconds and eliminates careless errors.

最后一个小技巧：用因式分解法解完后，将因式展开验证是否等于原方程。这只需十秒钟，却能消除粗心错误。

9. Exam-Style Practice Questions | 模拟真题练习

Try these questions under timed conditions. Answers are provided below so you can check your work.

请在限时条件下完成以下练习，答案附在后面供自查。

Question 1 / 第 1 题：Solve x² – 5x – 14 = 0.

Question 2 / 第 2 题：Solve 3x² – 2x – 1 = 0 using the quadratic formula.

Question 3 / 第 3 题：Write x² – 8x + 5 in the form (x – p)² + q, and hence state the coordinates of the turning point of y = x² – 8x + 5.

Question 4 / 第 4 题：The equation 2x² + 6x + k = 0 has two distinct real roots. Find the range of possible values of k.

Question 5 / 第 5 题：A right-angled triangle has a hypotenuse of 13 cm and one leg 7 cm longer than the other. Find the lengths of the two legs.

Answers / 答案：

(x – 7)(x + 2) = 0 → x = 7 or x = –2 / 因式分解得 x = 7 或 x = –2
x = (–(–2) ± √(4 – 4 × 3 × (–1))) / 6 = (2 ± √16) / 6 = (2 ± 4) / 6 → x = 1 or x = –⅓
(x – 4)² – 16 + 5 = (x – 4)² – 11; turning point (4, –11) / 顶点坐标为 (4, –11)
Δ = 36 – 8k > 0 → k < 4.5
Let legs be x and x + 7: x² + (x + 7)² = 13² → 2x² + 14x + 49 = 169 → 2x² + 14x – 120 = 0 → x² + 7x – 60 = 0 → (x + 12)(x – 5) = 0 → legs are 5 cm and 12 cm.

10. Final Revision Checklist | 考前自检清单

Use this checklist before your exam to ensure full coverage of the quadratic topic:

考试前请对照以下清单，确保二次方程各考点已全部掌握：

I can rearrange any quadratic into the standard form ax² + bx + c = 0. / 能将任意二次式整理为标准形式。
I can factorise quadratics with and without a leading coefficient of 1. / 能分解一次项系数为 1 和不为 1 的二次式。
I can substitute accurately into the quadratic formula. / 能准确代入二次公式。
I can complete the square and identify the turning point. / 能配方并求出顶点坐标。
I can use the discriminant to determine the number of real roots. / 会用判别式判断实数根个数。
I can sketch a parabola using roots, y-intercept, and vertex. / 能根据根、y 截距和顶点画出抛物线草图。
I can translate a word problem into a quadratic equation and reject invalid solutions. / 能把文字题转化为二次方程并舍去不合理解。
I can present answers to the correct degree of accuracy. / 能按题目要求给出精确度的答案。

With consistent practice, quadratic equations become one of the most rewarding topics in IGCSE Mathematics — the methods are fixed, the steps are logical, and the marks are highly achievable.

只要坚持练习，二次方程会成为 IGCSE 数学中最容易得分的板块之一——方法固定、步骤清晰、分数唾手可得。

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Mastering Quadratic Equations | 掌握二次方程