Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in nearly every IGCSE Mathematics exam paper. They test your algebraic fluency, your ability to recognise patterns, and your skill in choosing the most efficient method. This revision guide breaks down every approach you need, from factorisation to the quadratic formula, with worked examples and examiner-style tips.

二次方程几乎出现在每一份 IGCSE 数学试卷中。它考查你的代数运算能力、识别规律的能力，以及选择最有效解题方法的技巧。本复习指南将逐一讲解你需要的每种方法——从因式分解到求根公式，并配有完整例题和考官风格的答题提示。

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

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 value of a cannot be zero because then the equation would be linear, not quadratic.

二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。a 的值不能为零，否则方程将变成一次（线性）方程而不是二次方程。

The three key terms are:

三个关键术语如下：

Quadratic term ax² — the highest power term that defines the curve’s shape.
二次项 ax² —— 最高次项，决定曲线的形状。
Linear term bx — the term that skews the parabola.
一次项 bx —— 使抛物线发生倾斜的项。
Constant term c — the y-intercept when x = 0.
常数项 c —— 当 x = 0 时的 y 截距。

To solve a quadratic equation means to find the values of x that make the equation true. These values are called roots, solutions, or x-intercepts. A quadratic equation always has two roots, although they may be equal (a repeated root) or not real if the discriminant is negative.

解二次方程就是求出使方程成立的 x 值。这些值称为方程的根、解或x 轴交点。二次方程总有两个根，但它们可能相等（重根），或者在判别式为负时不具有实数根。

ax² + bx + c = 0, where a ≠ 0

2. Rewriting Equations into Standard Form | 将方程改写为标准形式

Before solving, you must rearrange any given equation so that all terms are on one side and zero is on the other. For example, the equation 2x² = 5x + 3 must first be rewritten as 2x² − 5x − 3 = 0.

在求解之前，必须先将给定的方程整理成一边为零、所有项在另一边的形式。例如，方程 2x² = 5x + 3 必须先改写为 2x² − 5x − 3 = 0。

Be careful with signs when moving terms across the equals sign. A common error is forgetting to change the sign of a term when subtracting it from both sides.

在将项移过等号时要特别注意符号。一个常见错误是在两边同时减去某一项时忘记改变该项的符号。

If the equation contains brackets or fractions, expand and clear denominators first. For instance:

如果方程包含括号或分数，应先展开括号、去分母。例如：

(x − 1)(x + 4) = 6 → x² + 3x − 4 = 6 → x² + 3x − 10 = 0

After simplification, always check that a ≠ 0. If a term, the equation is no longer quadratic, and a different method applies.

化简后，务必检查 a ≠ 0。如果 a 项为 0，方程就不再是二次方程，应改用其他方法。

3. Method 1: Solving by Factorisation | 方法一：因式分解法

Factorisation is the fastest method when the quadratic has simple integer roots. To factorise x² + bx + c, find two numbers that multiply to give c and add to give b.

当二次方程具有简单的整数根时，因式分解是最快的方法。要分解 x² + bx + c，需要找到两个数：它们相乘等于 c，相加等于 b。

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

例题 1：解方程 x² − 5x + 6 = 0

We need two numbers whose product is 6 and whose sum is −5. The numbers are −2 and −3, so the factorised form is (x − 2)(x − 3) = 0.

我们需要找到两个数，其乘积为 6，其和为 −5。这两个数是 −2 和 −3，因此分解形式为 (x − 2)(x − 3) = 0。

For the product of two brackets to be zero, at least one bracket must be zero. Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

要使两个括号的乘积为零，至少一个括号必须为零。因此 x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。

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

例题 2：解方程 2x² − 7x + 3 = 0

When a ≠ 1, use the factoring-by-grouping method. Multiply a and c: 2 × 3 = 6. Find two numbers whose product is 6 and whose sum is −7: they are −1 and −6. Rewrite the middle term: 2x² − x − 6x + 3 = 0. Then group: x(2x − 1) − 3(2x − 1) = 0, giving (2x − 1)(x − 3) = 0. Thus x = ½ or x = 3.

当 a ≠ 1 时，使用分组建因式分解法。将 a 和 c 相乘：2 × 3 = 6。找到两个数，其乘积为 6，其和为 −7：它们是 −1 和 −6。改写中间项：2x² − x − 6x + 3 = 0。然后分组：x(2x − 1) − 3(2x − 1) = 0，得到 (2x − 1)(x − 3) = 0。因此 x = ½ 或 x = 3。

Step	Action	Example
1	Multiply a × c	2 × 3 = 6
2	Find factor pair summing to b	−1 and −6
3	Split the middle term	2x² − x − 6x + 3
4	Group and factor out	x(2x − 1) − 3(2x − 1)
5	Write final factorised form	(2x − 1)(x − 3) = 0

4. Method 2: The Quadratic Formula | 方法二：求根公式法

When factorisation is difficult or impossible, use the quadratic formula. This formula works for every quadratic equation, provided the discriminant is not negative (when dealing with real numbers).

当因式分解困难或无法进行时，使用求根公式。该公式适用于所有二次方程，前提是判别式不为负（在实数范围内求解时）。

x = (−b ± √(b² − 4ac)) / 2a

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

例题：用求根公式解方程 3x² + 5x − 2 = 0。

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

此处 a = 3，b = 5，c = −2。代入公式：

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

x = (−5 ± √(25 + 24)) / 6 = (−5 ± √49) / 6 = (−5 ± 7) / 6

So x = (−5 + 7) / 6 = 2/6 = 1/3, or x = (−5 − 7) / 6 = −12/6 = −2. The solutions are x = 1/3 and x = −2.

因此 x = (−5 + 7) / 6 = 2/6 = 1/3，或 x = (−5 − 7) / 6 = −12/6 = −2。解为 x = 1/3 和 x = −2。

Always write the formula down before substituting. This earns method marks and reduces arithmetic errors. Give your final answer to three significant figures unless the question asks for exact values.

代入前务必先写出公式。这样能获得方法分并减少计算错误。除非题目要求精确值，否则最终答案保留三位有效数字。

5. Method 3: Completing the Square | 方法三：配方法

Completing the square rewrites the quadratic in the form (x + p)² + q. This method is especially useful for finding turning points of parabolas and for solving equations where factorisation fails.

配方法将二次方程改写为 (x + p)² + q 的形式。此方法在求抛物线顶点和求解无法因式分解的方程时特别有用。

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 half of the coefficient of x: 6 ÷ 2 = 3. Square it: 3² = 9. Rewrite:

取 x 系数的一半：6 ÷ 2 = 3。将其平方：3² = 9。改写：

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

Set the expression equal to zero and solve:

令表达式等于零并求解：

(x + 3)² − 7 = 0 → (x + 3)² = 7 → x + 3 = ±√7

x = −3 + √7 or x = −3 − √7

When the coefficient of x² is greater than 1, factor it out first. For example, 2x² + 8x + 5 becomes 2(x² + 4x) + 5, then complete the square inside the bracket.

当 x² 的系数大于 1 时，先将其提取出来。例如，2x² + 8x + 5 可写成 2(x² + 4x) + 5，然后在括号内配方。

To find the turning point, recall that the graph of y = (x + p)² + q has its minimum (or maximum) at (−p, q). For (x + 3)² − 7, the turning point is (−3, −7).

为求顶点坐标，注意 y = (x + p)² + q 的图像在 (−p, q) 处取得最小值（或最大值）。对于 (x + 3)² − 7，顶点为 (−3, −7)。

6. The Discriminant | 判别式

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

根号内的表达式 b² − 4ac 称为判别式，通常记为 Δ。它无需求解即可告诉我们二次方程有多少个实数根。

Δ > 0: two distinct real roots (the graph crosses the x-axis twice).
Δ > 0：两个不相等的实数根（图像与 x 轴相交两次）。
Δ = 0: one repeated real root (the graph touches the x-axis once).
Δ = 0：一个重根（图像与 x 轴相切一次）。
Δ < 0: no real roots (the graph does not touch the x-axis).
Δ < 0：没有实数根（图像与 x 轴不相交）。

Worked Example: Determine the nature of the roots of 2x² − 4x + 3 = 0.

例题：判断方程 2x² − 4x + 3 = 0 的根的性质。

Compute the discriminant:

计算判别式：

Δ = (−4)² − 4 × 2 × 3 = 16 − 24 = −8

Since Δ = −8 < 0, the equation has no real roots. In more advanced work, this tells us the curve y = 2x² − 4x + 3 lies entirely above the x-axis.

因为 Δ = −8 < 0，所以方程没有实数根。在更深入的学习中，这告诉我们曲线 y = 2x² − 4x + 3 完全位于 x 轴上方。

Examiners often ask: ‘Find the value of k for which kx² + 6x + 3 = 0 has exactly one real root.’ Set Δ = 0 and solve for k: 36 − 12k = 0, so k = 3.

考官常问：’求 k 的值，使得方程 kx² + 6x + 3 = 0 恰好有一个实数根。’令 Δ = 0 并解出 k：36 − 12k = 0，因此 k = 3。

7. Word Problems and Applications | 应用题与实际问题

Many exam questions wrap quadratics in real-world contexts such as area, projectile motion, or number problems. The key is to translate the words into a quadratic equation, solve it, and then check that each root makes sense in the context.

许多考题将二次方程嵌入实际情境中，如面积、抛物运动或数字问题。关键是把文字转化为二次方程，求解，然后检查每个根在情境中是否合理。

Worked Example: A rectangle has length (x + 5) cm and width x cm. Its area is 36 cm². Find x.

例题：一个长方形的长为 (x + 5) cm，宽为 x cm，面积为 36 cm²。求 x 的值。

Write the area equation:

写出面积方程：

x(x + 5) = 36 → x² + 5x − 36 = 0

Factorise: (x + 9)(x − 4) = 0, so x = −9 or x = 4. Since length cannot be negative, the only valid answer is x = 4 cm.

因式分解：(x + 9)(x − 4) = 0，所以 x = −9 或 x = 4。由于长度不能为负，唯一有效的答案是 x = 4 cm。

Always state the units and discard any negative roots in geometric problems. In physics-style questions, check whether a negative time value is meaningful before rejecting it.

在几何问题中务必写明单位并舍弃负根。在物理类问题中，先检查负的时间值是否有意义再决定是否舍弃。

8. Graphical Interpretation | 图形解释

The graph of y = ax² + bx + c is a parabola. When a > 0, it opens upward (a ‘smiley’ shape) and has a minimum turning point. When a < 0, it opens downward (a 'sad' shape) and has a maximum turning point.

y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，开口向上（’笑脸’形），具有最小值顶点。当 a < 0 时，开口向下（'哭脸'形），具有最大值顶点。

The x-coordinate of the turning point is given by x = −b/(2a). Substitute this value into the equation to find the y-coordinate. Or use the completed square form to read the vertex directly.

顶点的 x 坐标由 x = −b/(2a) 给出。将此值代入方程即可求出 y 坐标。或者利用配方法形式直接读出顶点。

The roots found by solving ax² + bx + c = 0 correspond exactly to the x-coordinates where the parabola crosses the x-axis. The constant term c gives the y-intercept at (0, c).

解方程 ax² + bx + c = 0 得到的根恰好对应抛物线与 x 轴交点的 x 坐标。常数项 c 给出 y 截距 (0, c)。

To sketch a quadratic graph, follow these four steps:

画二次函数草图时，按以下四步进行：

Identify the shape from the sign of a.
判断 a 的符号以确定开口方向。
Mark the y-intercept (0, c).
标出 y 截距 (0, c)。
Solve the equation to find the x-intercepts (if they exist).
解方程求出 x 截距（如果存在）。
Find the turning point using x = −b/(2a) or from the completed square form.
利用 x = −b/(2a) 或配方法形式求出顶点。

9. Common Mistakes and Exam Tips | 常见错误与应试技巧

Below are the most frequent errors IGCSE candidates make when solving quadratic equations.

以下是 IGCSE 考生在解二次方程时最常见的错误。

Mistake	Correction
Forgetting to set the equation to 0 before factorising.	Always rearrange to ax² + bx + c = 0 first.
Confusing the signs when using the formula.	Write b with its sign; recall −(−3) = +3.
Dividing both sides by x, losing a root.	Never divide by an unknown variable; factor it out instead.
Miscomputing the discriminant for the formula.	Check that −4ac is included with the correct sign.
Giving only one solution.	A quadratic has two roots; both must be stated.

Exam tips: show all working for method marks, use a consistent method you are comfortable with, and always check your roots by substituting them back into the original equation.

应试技巧：写出完整步骤以获得方法分，使用自己最熟练且一致的方法，并始终将根代回原方程检验。

10. Practice Questions | 练习题

Try these questions, then check your answers below.

尝试解以下题目，然后核对下方的答案。

Question 1: Solve x² − 7x + 10 = 0 (factorisation).

第 1 题：解方程 x² − 7x + 10 = 0（因式分解法）。

Question 2: Solve 2x² + 3x − 5 = 0 (quadratic formula).

第 2 题：解方程 2x² + 3x − 5 = 0（求根公式法）。

Question 3: Determine the nature of the roots of x² − 4x + 4 = 0.

第 3 题：判断方程 x² − 4x + 4 = 0 的根的性质。

Question 4: A rectangle has dimensions x and (x + 2). Its area is 24. Find the perimeter given x > 0.

第 4 题：一个长方形的尺寸为 x 和 (x + 2)，面积为 24。已知 x > 0，求其周长。

Answers:

答案：

Q1: (x − 2)(x − 5) = 0 → x = 2 or x = 5.
第 1 题：(x − 2)(x − 5) = 0 → x = 2 或 x = 5。
Q2: x = (−3 ± √49)/4 = (−3 ± 7)/4 → x = 1 or x = −2.5.
第 2 题：x = (−3 ± √49)/4 = (−3 ± 7)/4 → x = 1 或 x = −2.5。
Q3: Δ = 16 − 16 = 0 → one repeated real root.
第 3 题：Δ = 16 − 16 = 0 → 一个重根。
Q4: x(x + 2) = 24 → x² + 2x − 24 = 0 → (x + 6)(x − 4) = 0 → x = 4 (since x > 0). Perimeter = 2(4 + 6) = 20 units.
第 4 题：x(x + 2) = 24 → x² + 2x − 24 = 0 → (x + 6)(x − 4) = 0 → x = 4（因为 x > 0）。周长 = 2(4 + 6) = 20 单位。

Mastering quadratic equations opens the door to many other IGCSE topics, including graphing, inequalities, and optimisation. Practise each method until you can recognise the most efficient path at a glance.

掌握二次方程为学习 IGCSE 其他许多专题（包括函数图像、不等式和最优化）打开了大门。反复练习每种方法，直到你能一眼识别出最高效的解题路径。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Quadratic Equations | 解二次方程