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Solving Quadratic Equations: Factorisation, Formula & Graphs | 解二次方程：因式分解、求根公式与图像

2026-08-30 04:30 · 本文部分内容由AI辅助创作

📚 Solving Quadratic Equations: Factorisation, Formula & Graphs | 解二次方程：因式分解、求根公式与图像

Quadratic equations appear throughout IGCSE Mathematics and form the foundation for advanced algebra, functions, and calculus. Mastering the skill of solving these equations is essential for exam success and for connecting algebraic, graphical, and real-world ideas. This article covers every method you need, from factorisation to the quadratic formula, along with the discriminant, graph sketching, and common exam pitfalls.

二次方程在 IGCSE 数学中随处可见，是学好高阶代数、函数与微积分的基石。熟练掌握解二次方程的方法，不仅能在考试中稳拿分数，还能帮助你打通代数、图像与实际应用之间的关联。本文将系统讲解因式分解法、配方法、求根公式、判别式、图像绘制以及常见考试陷阱。

1. What Is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is an equation that can be written in the standard form:

ax² + bx + c = 0

, and a ≠ 0. The highest power of the variable x is 2, which is why the equation is called "quadratic" (from the Latin "quadratus" If a = 0, the equation becomes linear and the techniques in this article no longer apply.

其中 a、b、c 为常数，且 a ≠ 0。变量 x 的最高次数为 2，因此被称为"二次"方程（quadratic 源自拉丁语 quadratus，意为"平方"）。如果 a = 0，方程就变成一次方程，本文的方法将不再适用。

Key features to remember:

2. Solving by Factorisation | 因式分解法

1. The idea is to rewrite the quadratic as the product of two linear factors. For example:

x² − 5x + 6 = (x − 2)(x − 3)

is works because if two numbers multiply to give zero, then at least one of them must be zero.

因式分解法通常是解二次方程最快的方法，尤其当 a = 1 时最为简便。其核心思想是把二次式改写为两个一次因式的乘积。例如：

x² − 5x + 6 = (x − 2)(x − 3)

当任意一个因式为零时方程成立，因此 x = 2 或 x = 3。依据是：如果两个数相乘等于零，那么其中至少有一个数为零。

To solve a quadratic by factorisation, follow these steps:

用因式分解法解二次方程，可按以下步骤进行：

Worked example:

x² + 7x + 12 = 0 → (x + 3)(x + 4) = 0 → x = −3 or x = −4

验算：将 x = −3 或 x = −4 代回原方程，两边都等于 0，说明两根正确。当 a ≠ 1 时，例如 2x² + 5x − 3 = 0，可分解为 (2x − 1)(x + 3) = 0，于是 x = 1/2 或 x = −3。

3. Completing the Square | 配方法

for solving equations that are not easily factorisable. Take the equation x² + 6x − 7 = 0:

(x + 3)² − 9 − 7 = 0 → (x + 3)² = 16 → x + 3 = ±4 → x = 1 or x = −7

配方法将二次方程改写为 (x + p)² + q 的形式。这种方法在求顶点坐标以及解不易因式分解的方程时非常有用。以 x² + 6x − 7 = 0 为例：

(x + 3)² − 9 − 7 = 0 → (x + 3)² = 16 → x + 3 = ±4 → x = 1 或 x = −7

The general rule is simple: for an expression x² + bx, add and subtract (b/2)². Then:

x² + bx = (x + b/2)² − (b/2)²

配方法的一般规则很简单：对于 x² + bx 这一项，先加再减 (b/2)²。于是：

x² + bx = (x + b/2)² − (b/2)²

² − 4x + 5 can be written as (x − 2)² + 1, so the minimum point is (2, 1). This proves that completing the square is the most direct way to locate the vertex of a parabola.

如果二次项系数 a ≠ 1，先提出 a，再在括号内配方。例如 y = x² − 4x + 5 可写成 (x − 2)² + 1，因此最小值点为 (2, 1)。这说明配方法是找抛物线顶点最直接的方法。

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4. The Quadratic Formula | 求根公式

hether factorisable or not. For an equation written in the standard form ax² + bx + c = 0:

求根公式是最通用的方法，无论方程是否可以因式分解，都能使用。对于标准形式 ax² + bx + c = 0：

x = (−b ± √b² − 4ac) / 2a

: solve 2x² − 3x − 2 = 0. Here a = 2, b = −3, c = −2. Substituting into the formula gives:

x = [3 ± √((−3)² − 4 × 2 × (−2))] / (2 × 2) = (3 ± √25) / 4 = (3 ± 5) / 4

ormula requires careful substitution: the term −b changes the sign of b, and the product −4ac must also respect the sign of c.

例题：解 2x² − 3x − 2 = 0。此时 a = 2，b = −3，c = −2。代入公式：

x = [3 ± √((−3)² − 4 × 2 × (−2))] / (2 × 2) = (3 ± √25) / 4 = (3 ± 5) / 4

因此 x = 2 或 x = −1/2。注意代入时需格外仔细：−b 会改变 b 的符号，−4ac 这一项也必须考虑 c 的符号。

5. The Discriminant | 判别式

The expression b² − 4ac, called the discriminant, tells us how many roots a quadratic equation has without actually solving it. The discriminant is often written using the Greek letter Δ (delta).

表达式 b² − 4ac 称为判别式，它可以在不解方程的情况下告诉我们二次方程有多少个根。判别式常用希腊字母 Δ（delta）表示。

判别式 Δ = b² − 4ac Nature of Roots 根的情况
Δ > 0 Two distinct real roots; the parabola crosses the x-axis twice. 两个不同实根；抛物线与 x 轴有两个交点。
Δ = 0 One repeated real root; the parabola touches the x-axis. 一个重根；抛物线与 x 轴相切。
Δ < 0 No real roots; the parabola does not meet the x-axis. 无实根；抛物线与 x 轴不相交。

s no real solutions. In IGCSE exams, you are not required to deal with imaginary numbers unless the syllabus explicitly includes them.

例如，方程 x² + 2x + 5 = 0 的判别式 Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16 < 0，因此该方程无实数解。在 IGCSE 考试中，除非大纲明确要求，否则不需要处理虚数。

6. Graphs of Quadratic Functions | 二次函数图像

. Its shape, position and number of intersections with the x-axis are controlled by the coefficient a, the discriminant, and the constant term c. The turning point (vertex) can be found by completing the square as shown in Section 3.

函数 y = ax² + bx + c 的图像是一条抛物线。它的开口方向、位置以及与 x 轴的交点个数分别由系数 a、判别式和常数项 c 决定。顶点坐标可以通过第 3 节中的配方法求出。

Key facts for sketching:

绘制草图时需掌握以下要点：