Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most important topics in the IGCSE Edexcel Mathematics syllabus. Mastering them unlocks success in algebra, graphs, and problem-solving questions across both Paper 1 and Paper 2. In this revision guide, we will explore every method you need to solve quadratic equations, understand the discriminant, sketch parabolas, and avoid the most common exam traps.

二次方程是 IGCSE Edexcel 数学大纲中最重要的知识点之一。掌握二次方程，是解决代数、函数图像及应用题（涵盖 Paper 1 和 Paper 2）的关键。在本复习指南中，我们将系统讲解解二次方程的各类方法、判别式的含义、抛物线图像的绘制，以及如何在考试中避开最常见的陷阱。

1. Standard Form and Key Vocabulary | 标准形式与关键术语

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 term ax² is called the quadratic term, bx is the linear term, and c is the constant term.

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。ax² 称为二次项，bx 称为一次项，c 称为常数项。

If a = 0, the equation becomes linear, not quadratic, so a must never be zero.

如果 a = 0，方程就变成了一次方程，而不是二次方程，因此 a 绝对不能为零。
The roots (or solutions) of a quadratic equation are the values of x that satisfy the equation.

二次方程的根（即解）是满足方程的 x 值。
Factorising means writing the quadratic as a product of two brackets.

因式分解是指将二次式写成两个括号相乘的形式。

For example, x² − 3x + 2 = 0 has roots x = 1 and x = 2, because substituting either value makes the left-hand side equal to zero.

例如，x² − 3x + 2 = 0 的根是 x = 1 和 x = 2，因为代入任何一个值都能使左边等于零。

2. Solving by Factorisation | 因式分解法

When a quadratic expression can be factorised, this is usually the quickest method. The key principle is the zero product property: if p × q = 0, then either p = 0 or q = 0.

当二次式可以因式分解时，这通常是最快的方法。核心原理是零乘积性质：如果 p × q = 0，那么必有 p = 0 或 q = 0。

To factorise x² + bx + c, look for two numbers that multiply to give c and add to give b.

要分解 x² + bx + c，需要找到两个数，它们的乘积等于 c，和等于 b。

Worked example: Solve x² + 5x + 6 = 0.

示例：解方程 x² + 5x + 6 = 0。

We need two numbers whose product is 6 and whose sum is 5. These are 2 and 3.

我们需要两个数，乘积为 6，和为 5。这两个数是 2 和 3。

(x + 2)(x + 3) = 0

Therefore x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.

因此 x + 2 = 0 或 x + 3 = 0，解得 x = −2 或 x = −3。

For a quadratic with a leading coefficient not equal to 1, such as 2x² + 7x + 3 = 0, you may use the ‘ac method’: multiply a and c, then find two numbers that multiply to ac and add to b.

当二次项系数不为 1 时，例如 2x² + 7x + 3 = 0，可以使用”ac 法”：先计算 a 和 c 的乘积，然后找到两个数，它们的乘积等于 ac，和等于 b。

Here ac = 2 × 3 = 6, and we need two numbers with product 6 and sum 7. These are 1 and 6. We split the middle term:

这里 ac = 2 × 3 = 6，我们需要两个数乘积为 6、和为 7。这两个数是 1 和 6。然后拆中间项：

2x² + x + 6x + 3 = 0

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

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

3. Solving by Completing the Square | 配方法

Completing the square rewrites x² + bx as a perfect square trinomial. The formula is:

配方法将 x² + bx 改写成一个完全平方三项式。公式为：

x² + bx = (x + b/2)² − (b/2)²

This method is especially useful when the equation cannot be factorised easily, and it also reveals the turning point of a quadratic graph.

当方程不易因式分解时，这种方法尤其有用，而且它还能直接给出二次函数图像的顶点坐标。

Worked example: Solve x² + 6x + 5 = 0 by completing the square.

示例：用配方法解 x² + 6x + 5 = 0。

First, isolate the x terms: x² + 6x = −5.

首先移项，只保留含 x 的项：x² + 6x = −5。

Half of 6 is 3, so (x + 3)² − 9 = −5.

6 的一半是 3，所以 (x + 3)² − 9 = −5。

Then (x + 3)² = 4, so x + 3 = ±2, giving x = −1 or x = −5.

于是 (x + 3)² = 4，因此 x + 3 = ±2，解得 x = −1 或 x = −5。

When the coefficient of x² is not 1, first divide the entire equation by that coefficient. For example, 2x² − 4x − 6 = 0 becomes x² − 2x − 3 = 0 before completing the square.

当 x² 的系数不是 1 时，先将整个方程除以该系数。例如 2x² − 4x − 6 = 0 应先在等式两边除以 2，得到 x² − 2x − 3 = 0，再进行配方。

4. The Quadratic Formula | 求根公式法

The quadratic formula solves any quadratic equation ax² + bx + c = 0, including those that cannot be factorised. The formula is:

求根公式可以解任意形式为 ax² + bx + c = 0 的二次方程，包括那些无法因式分解的方程。公式是：

x = (−b ± √(b² − 4ac)) / 2a

Worked example: Solve 2x² + 5x − 3 = 0 using the quadratic formula.

示例：用求根公式解 2x² + 5x − 3 = 0。

Here a = 2, b = 5, c = −3. Substitute into the formula:

这里 a = 2，b = 5，c = −3。代入公式：

x = (−5 ± √(5² − 4 × 2 × (−3))) / (2 × 2)

x = (−5 ± √(25 + 24)) / 4 = (−5 ± √49) / 4 = (−5 ± 7) / 4

Thus x = (−5 + 7)/4 = 2/4 = ½, or x = (−5 − 7)/4 = −12/4 = −3. The two roots are x = ½ and x = −3.

因此 x = (−5 + 7)/4 = 2/4 = ½，或 x = (−5 − 7)/4 = −12/4 = −3。两个根为 x = ½ 和 x = −3。

Always check that you have substituted negative values correctly. A very common mistake is writing 5² instead of (−5)², or forgetting that 4ac carries the sign of c.

务必检查负值的代入是否正确。一个非常常见的错误是把 (−5)² 写成 5²，或者忘记 4ac 中的 c 带着符号。

5. The Discriminant and the Nature of Roots | 判别式与根的性质

The expression inside the square root, b² − 4ac, is called the discriminant, often denoted by the Greek letter Δ (delta). It tells us how many real roots a quadratic equation has.

根号内的表达式 b² − 4ac 称为判别式，通常用希腊字母 Δ（delta）表示。它告诉我们二次方程有多少个实数根。

Discriminant Δ = b² − 4ac	Nature of roots / 根的性质
Δ > 0 (positive / 正数)	Two distinct real roots / 两个不同的实数根
Δ = 0 (zero / 零)	One repeated real root / 两个相等实数根（重根）
Δ < 0 (negative / 负数)	No real roots / 没有实数根

The discriminant also tells us how many times the graph of y = ax² + bx + c crosses the x-axis: twice, once (touching), or not at all.

判别式还告诉我们 y = ax² + bx + c 的图像与 x 轴的交点个数：相交两次、相切一次，或不相交。

Example: Determine the nature of the roots of x² − 4x + 7 = 0.

示例：判断 x² − 4x + 7 = 0 的根的性质。

Here a = 1, b = −4, c = 7, so Δ = (−4)² − 4(1)(7) = 16 − 28 = −12. Since Δ < 0, there are no real roots.

这里 a = 1，b = −4，c = 7，所以 Δ = (−4)² − 4(1)(7) = 16 − 28 = −12。由于 Δ < 0，方程没有实数根。

6. Sketching Quadratic Graphs | 二次函数图像的绘制

A quadratic function y = ax² + bx + c always produces a curve called a parabola. If a > 0, the parabola opens upward (a ∪ shape); if a < 0, it opens downward (a ∩ shape).

二次函数 y = ax² + bx + c 的图像永远是抛物线。当 a > 0 时，抛物线开口向上（∪ 形）；当 a < 0 时，开口向下（∩ 形）。

To sketch a quadratic graph, you need three key features:

绘制二次函数图像需要三个关键要素：

Roots: solve ax² + bx + c = 0 to find where the curve crosses the x-axis.

根：解 ax² + bx + c = 0，找到曲线与 x 轴的交点。
y-intercept: substitute x = 0, giving the point (0, c).

y 轴截距：令 x = 0，得到点 (0, c)。
Turning point: the vertex of the parabola. Its x-coordinate is x = −b/(2a), or read directly from the completed-square form y = a(x + p)² + q, where the turning point is (−p, q).

顶点：抛物线的最高点或最低点。其 x 坐标为 x = −b/(2a)，也可直接从配方式 y = a(x + p)² + q 中读出，顶点为 (−p, q)。

Example: Sketch y = x² − 2x − 3.

示例：画 y = x² − 2x − 3 的草图。

Factorising gives (x − 3)(x + 1) = 0, so roots are x = 3 and x = −1. The y-intercept is (0, −3). Complete the square: (x − 1)² − 4, so the turning point is (1, −4). Since a = 1 > 0, the parabola opens upward. Plot these points and draw a smooth ∪ shape.

因式分解得 (x − 3)(x + 1) = 0，所以根为 x = 3 和 x = −1。y 轴截距为 (0, −3)。配方得 (x − 1)² − 4，因此顶点为 (1, −4)。由于 a = 1 > 0，抛物线开口向上。标出这些点，画出平滑的 ∪ 形曲线即可。

7. Real-World Applications | 实际应用

Quadratic equations appear frequently in geometry and kinematics problems in the IGCSE exams.

在 IGCSE 考试中，二次方程经常出现在几何和运动学应用题中。

Example (area problem): A rectangle has length (x + 5) cm and width (x − 2) cm. Its area is 24 cm². Find x.

示例（面积问题）：一个长方形的长为 (x + 5) cm，宽为 (x − 2) cm，面积为 24 cm²。求 x 的值。

Set up the equation: (x + 5)(x − 2) = 24.

列出方程：(x + 5)(x − 2) = 24。

Expand: x² + 3x − 10 = 24, so x² + 3x − 34 = 0.

展开：x² + 3x − 10 = 24，所以 x² + 3x − 34 = 0。

This does not factorise nicely, so use the quadratic formula with a = 1, b = 3, c = −34:

该式不易因式分解，因此用求根公式，a = 1，b = 3，c = −34：

x = (−3 ± √(9 + 136)) / 2 = (−3 ± √145) / 2

Since a length cannot be negative, take the positive root: x ≈ (−3 + 12.04)/2 ≈ 4.52 cm. The negative root is rejected as it gives negative dimensions.

由于长度不能为负数，取正根：x ≈ (−3 + 12.04)/2 ≈ 4.52 cm。负根因产生负的边长而被舍去。

Example (projectile motion): A ball is thrown upward so that its height h metres after t seconds is given by h = 15t − 5t². When does the ball hit the ground?

示例（抛体运动）：一个球被向上抛出，其高度 h（米）与时间 t（秒）的关系为 h = 15t − 5t²。球何时落地？

The ball hits the ground when h = 0, so 15t − 5t² = 0. Factorise: 5t(3 − t) = 0, giving t = 0 (launch time) or t = 3 seconds. The ball lands after 3 seconds.

球落地时 h = 0，所以 15t − 5t² = 0。分解得 5t(3 − t) = 0，即 t = 0（抛出时刻）或 t = 3 秒。球在 3 秒后落地。

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

Even strong students lose marks on quadratic equations due to avoidable errors. Here are the pitfalls to watch for:

即使是优秀学生，也常因一些可以避免的错误在二次方程上丢分。以下是需要警惕的常见陷阱：

Forgetting to set the equation to zero. If you factorise x² + 5x + 6 = 2, you cannot conclude x + 2 = 2. First rewrite as x² + 5x + 4 = 0, then solve.

忘记把方程化为零。如果方程是 x² + 5x + 6 = 2，你不能直接得出 x + 2 = 2。应先将方程改写为 x² + 5x + 4 = 0，然后再解。
Missing a root. The ± sign in the quadratic formula gives two answers; always write both unless the context (e.g. length, time) excludes one.

漏掉一个根。求根公式中的 ± 号给出两个答案；除非题目背景（如长度、时间）排除其中一个，否则一定要写出两个根。
Sign errors in substitution. If b = −6, then −b = 6 and b² = (−6)² = 36. Take extra care with negative coefficients.

代入时出现符号错误。如果 b = −6，那么 −b = 6，而 b² = (−6)² = 36。遇到负系数时要格外小心。
Approximating too early. Keep exact values (such as √145) until the final step, then round to the required degree of accuracy.

过早取近似值。在最后一步之前保留精确值（如 √145），最后再按要求精度取近似。
Choosing the wrong method. On a non-calculator paper, factorisation or completing the square often saves time. Save the quadratic formula for difficult cases, and always check if the discriminant is a perfect square.

选错方法。在不能使用计算器的试卷中，因式分解或配方往往更省时。将求根公式留给难解的题目，并养成先计算判别式是否为完全平方数的习惯。

9. Practice Questions | 练习题目

Attempt these questions on your own, then check your answers. Remember to show all working for full marks.

请先独立完成以下题目，再核对答案。记住：考试中写出完整过程才能得满分。

Question 1: Solve x² − 7x + 12 = 0, giving the equation in fully factorised form.

题 1：解方程 x² − 7x + 12 = 0，并写出因式分解后的完整形式。

Question 2: Use the quadratic formula to solve 3x² + 2x − 8 = 0. Give your answers correct to 2 decimal places.

题 2：用求根公式解 3x² + 2x − 8 = 0，答案精确到小数点后两位。

Question 3: Find the value of k for which the equation x² + kx + 9 = 0 has exactly one repeated root.

题 3：求 k 的值，使方程 x² + kx + 9 = 0 有两个相等的实数根。

Question 4: The length of a rectangle is 4 cm more than its width, and its area is 45 cm². Find the width of the rectangle.

题 4：一个长方形的长比宽多 4 cm，面积为 45 cm²。求它的宽。

Question 5: Sketch the graph of y = −x² + 4x − 3, clearly labelling the roots, the y-intercept and the turning point.

题 5：画出 y = −x² + 4x − 3 的图像，清晰地标出根、y 轴截距和顶点。

Answers: 1) (x − 3)(x − 4) = 0, x = 3 or 4. 2) x ≈ 1.33 or −2.00. 3) Δ = k² − 36 = 0, so k = ±6. 4) Let width = w, then w(w + 4) = 45, giving w² + 4w − 45 = 0 ⇒ (w + 9)(w − 5) = 0, so w = 5 cm. 5) Roots x = 1 and x = 3; y-intercept (0, −3); turning point (2, 1) — the parabola opens downward.

参考答案：1) (x − 3)(x − 4) = 0，x = 3 或 4。2) x ≈ 1.33 或 −2.00。3) Δ = k² − 36 = 0，所以 k = ±6。4) 设宽为 w，则 w(w + 4) = 45，即 w² + 4w − 45 = 0 ⇒ (w + 9)(w − 5) = 0，所以 w = 5 cm。5) 根为 x = 1 和 x = 3；y 轴截距为 (0, −3)；顶点为 (2, 1)——抛物线开口向下。

Conclusion | 总结

Quadratic equations are a cornerstone of IGCSE Edexcel Mathematics. Master the three solution methods — factorisation, completing the square, and the quadratic formula — and understand what the discriminant tells you about the roots. With regular practice, you will be able to recognise the most efficient method instantly and solve these problems confidently in the exam.

二次方程是 IGCSE Edexcel 数学的基石。掌握三种解法——因式分解法、配方法和求根公式法——并理解判别式所揭示的根的性质。通过定期练习，你将能够一眼识别最有效的解法，并在考试中自信地解决这些问题。

Remember: always set the equation to zero, handle negative signs with care, and check whether your answers make sense in the context of the question. Good luck!

请记住：始终将方程化为零，谨慎处理负号，并检查答案是否符合题目的实际背景。祝你好运！

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Solving Quadratic Equations | 解二次方程