Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper. From factorisation to the quadratic formula, this article provides a clear, step-by-step review of the methods you need to master, along with common pitfalls and worked examples.

二次方程几乎出现在每一份 IGCSE 数学试卷中。从因式分解到求根公式，本文将为你清晰、分步地梳理需要掌握的方法，并讲解常见错误与真题示例。

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 form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The term ax² is called the quadratic term, bx is the linear term, and c is the constant term.

二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为实数，且 a ≠ 0。ax² 称为二次项，bx 称为一次项，c 称为常数项。

Example 1: x² − 5x + 6 = 0 has a = 1, b = −5, c = 6.

例 1：x² − 5x + 6 = 0 中 a = 1，b = −5，c = 6。

Some equations are not initially in standard form. Rearrange all terms onto one side, keeping the coefficient of x² positive whenever possible.

有些方程并非一开始就是标准形式。将全部项移到一边，并尽可能使 x² 的系数为正。

2. Expanding Double Brackets | 展开双括号

Before solving quadratics, you must be fluent in expanding expressions like (x + p)(x + q). Use the distributive law: each term in the first bracket multiplies each term in the second.

在解二次方程之前，你必须熟练展开形如 (x + p)(x + q) 的表达式。用分配律：第一个括号中的每一项乘以第二个括号中的每一项。

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

Notice the middle term combines −2x + 3x = x. Practice this until it becomes automatic.

注意中间项合并 −2x + 3x = x。反复练习直到形成条件反射。

3. Factorising Quadratics with Leading Coefficient 1 | 因式分解首项系数为 1 的二次式

To factorise x² + bx + c, find two numbers that multiply to c and add to b. These numbers form the factors (x + m)(x + n).

要分解 x² + bx + c，找到两个数，使它们的乘积为 c，和为 b。这两个数构成因式 (x + m)(x + n)。

Example 2: Factorise x² − 5x + 6.

例 2：分解 x² − 5x + 6。

We need two numbers with product 6 and sum −5. They are −2 and −3. Therefore:

我们需要两个数，乘积为 6，和为 −5。它们是 −2 和 −3。因此：

x² − 5x + 6 = (x − 2)(x − 3)

Always check by expanding your answer.

务必通过展开来检查答案。

4. Solving by Factorisation | 因式分解法解方程

If the product of two expressions is zero, then at least one of them must be zero. This zero-product property allows us to solve quadratic equations once they are factorised.

如果两个表达式的乘积为零，则其中至少一个必须为零。这个零积性质使我们在因式分解后就能解二次方程。

Example 3: Solve x² − 5x + 6 = 0.

例 3：解 x² − 5x + 6 = 0。

Factorise: (x − 2)(x − 3) = 0. Then set each bracket to zero:

因式分解：(x − 2)(x − 3) = 0。然后令每个括号等于零：

x − 2 = 0 ⇒ x = 2
x − 3 = 0 ⇒ x = 3

So the solutions are x = 2 and x = 3.

因此解为 x = 2 和 x = 3。

If the leading coefficient is not 1, you may need to factorise using a method such as grouping or trial and error.

如果首项系数不是 1，你可能需要用分组法或试错法进行因式分解。

5. Solving by Completing the Square | 配方法解方程

Completing the square rewrites x² + bx in the form (x + p)² − p². This is useful when factorisation is not obvious.

配方法将 x² + bx 改写为 (x + p)² − p² 的形式。当因式分解不明显时，这种方法很有用。

Example 4: Solve x² + 6x + 1 = 0.

例 4：解 x² + 6x + 1 = 0。

First isolate the x terms:

先分离含 x 的项：

x² + 6x = −1

Take half of 6, square it: (6/2)² = 9. Add 9 to both sides:

取 6 的一半，再平方：(6/2)² = 9。两边同时加 9：

x² + 6x + 9 = 8

The left side is (x + 3)²:

左边是 (x + 3)²：

(x + 3)² = 8

Take the square root of both sides:

两边开平方根：

x + 3 = ±√8

x = −3 ± 2√2

So x = −3 + 2√2 or x = −3 − 2√2.

因此 x = −3 + 2√2 或 x = −3 − 2√2。

6. The Quadratic Formula | 求根公式

For any quadratic equation ax² + bx + c = 0, the solutions are given by:

对于任意二次方程 ax² + bx + c = 0，解为：

x = (−b ± √(b² − 4ac)) / (2a)

This formula works for all quadratics, including those that cannot be factorised easily. Substitute a, b and c carefully, and simplify.

这个公式适用于所有二次方程，包括那些不易因式分解的方程。小心代入 a、b 和 c，并化简。

Example 5: Solve 2x² − 4x − 3 = 0 using the formula.

例 5：用公式解 2x² − 4x − 3 = 0。

Here a = 2, b = −4, c = −3. Substitute:

这里 a = 2，b = −4，c = −3。代入：

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

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4

x = (4 ± 2√10) / 4 = (2 ± √10) / 2

So the two solutions are x = (2 + √10)/2 and x = (2 − √10)/2.

所以两个解为 x = (2 + √10)/2 和 x = (2 − √10)/2。

7. The Discriminant | 判别式

The expression b² − 4ac is called the discriminant. It tells us the number and type of roots without solving the equation.

判别式是指表达式 b² − 4ac。它告诉我们方程根的个数和类型，而无需实际求解。

Discriminant Δ	Nature of roots
Δ > 0	Two distinct real roots
Δ = 0	One repeated real root
Δ < 0	No real roots

For example, x² + 2x + 5 = 0 has Δ = 4 − 20 = −16 < 0, so it has no real roots.

例如，x² + 2x + 5 = 0 的 Δ = 4 − 20 = −16 < 0，因此没有实数根。

This concept is often tested in non-calculator papers, so memorise the three cases.

这一概念经常在不使用计算器的试卷中出现，请牢记三种情形。

8. Solving Quadratic Equations by Graphical Methods | 图像法解二次方程

The solutions of ax² + bx + c = 0 correspond to the x-intercepts of the curve y = ax² + bx + c. A graph can give approximate roots when exact solutions are not required.

方程 ax² + bx + c = 0 的解对应曲线 y = ax² + bx + c 与 x 轴的交点。当不需要精确解时，图像可以给出近似根。

Key features to plot:

Shape: if a > 0, the graph is a U-shape; if a < 0, it is an inverted U.
y-intercept: at (0, c).
Axis of symmetry: x = −b/(2a).
Vertex: minimum or maximum point.

关键绘图要素：

形状：若 a > 0，图像为 U 形；若 a < 0，为倒 U 形。
y 截距：(0, c)。
对称轴：x = −b/(2a)。
顶点：最小值点或最大值点。

To solve by drawing, plot the curve and read off the x-coordinates where it crosses the x-axis.

使用图像法求解时，画出曲线并读取其与 x 轴交点的 x 坐标。

9. Simultaneous Equations with a Quadratic | 含二次方程的联立方程组

IGCSE often asks you to solve one linear and one quadratic equation simultaneously. Substitute the linear equation into the quadratic equation.

IGCSE 经常要求联立一个一次方程和一个二次方程。将一次方程代入二次方程即可。

Example 6: Solve y = 2x + 1 and y = x² − 4x + 5.

例 6：解 y = 2x + 1 和 y = x² − 4x + 5。

Since both equal y, set them equal:

由于二者都等于 y，令它们相等：

2x + 1 = x² − 4x + 5

Rearrange:

移项整理：

x² − 6x + 4 = 0

Use the quadratic formula:

使用求根公式：

x = (6 ± √(36 − 16)) / 2 = (6 ± √20) / 2 = 3 ± √5

Then substitute each x back into y = 2x + 1 to find the corresponding y-values.

然后将每个 x 代回 y = 2x + 1，求出对应的 y 值。

10. Applications in Problem Solving | 应用题中的二次方程

Quadratic equations model many real-world situations, such as projectile motion, area problems, and revenue optimisation.

二次方程可以建模许多现实情境，如抛体运动、面积问题和收益优化。

Example 7: A rectangle has length 5 cm longer than its width, and its area is 84 cm². Find the width.

例 7：一个长方形的长比宽多 5 cm，面积为 84 cm²。求宽。

Let width = x, length = x + 5. Then:

设宽为 x，长为 x + 5。则：

x(x + 5) = 84

x² + 5x − 84 = 0

Factorise: (x + 12)(x − 7) = 0, so x = −12 or x = 7. Since width cannot be negative, the width is 7 cm.

因式分解：(x + 12)(x − 7) = 0，所以 x = −12 或 x = 7。因为宽不能为负，所以宽为 7 cm。

Always reject negative solutions when they do not make sense in the context.

当负数解在情境中不合理时，务必舍去。

11. Common Mistakes and Tips | 常见错误与建议

Students frequently lose marks on quadratics for avoidable reasons. Here are the most common difficulties and how to avoid them.

学生在二次方程部分经常因可避免的原因失分。以下是最常见的困难和规避方法。

Forgetting to rearrange the equation to = 0 before factorising.
Missing the ± sign when taking square roots.
Using the quadratic formula with the wrong value of b when b is negative.
Miscopying signs when expanding (x − a)(x − b).
Not simplifying surds fully, e.g. √12 = 2√3.

忘记在因式分解前将方程整理为 = 0。
开平方根时漏掉 ± 号。
当 b 为负数时，在求根公式中代入错误的 b 值。
展开 (x − a)(x − b) 时抄错符号。
没有完全化简根式，例如 √12 = 2√3。

Always check your answers by substituting them into the original equation.

始终将答案代回原方程进行验算。

12. Exam-Style Practice | 模拟真题练习

Try these questions to test your understanding:

尝试以下问题以检验你的理解：

Solve x² − 7x + 12 = 0.
Solve 3x² + 5x − 2 = 0 (give solutions exactly).
Find the value of k for which x² + kx + 9 = 0 has one repeated root.
A parabola has equation y = x² − 4x + 3. Find the x-intercepts and the minimum value.

解 x² − 7x + 12 = 0。
解 3x² + 5x − 2 = 0（给出精确解）。
求 k 的值，使 x² + kx + 9 = 0 有一个重根。
抛物线 y = x² − 4x + 3。求 x 截距和最小值。

Solutions: (1) x = 3, 4. (2) x = 1/3, −2. (3) k = 6 or −6. (4) x = 1, 3; minimum y = −1.

答案：(1) x = 3, 4。(2) x = 1/3, −2。(3) k = 6 或 −6。(4) x = 1, 3；最小值 y = −1。

Master these techniques, and quadratic equations will become one of the most reliable scoring areas in your IGCSE Mathematics exam.

掌握这些技巧，二次方程将成为你 IGCSE 数学考试中最稳定的得分点之一。

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Solving Quadratic Equations | 解二次方程