Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in many exam questions, from simple factorisation to word problems, graphs and inequalities. Learning to solve them confidently will strengthen your algebra skills and improve your overall grade.

二次方程是 IGCSE 数学中最重要的内容之一。它出现在许多考试题目中，从简单的因式分解到应用题、图像和不等式。熟练掌握二次方程的解法，既能为后续代数学习打下坚实基础，也能有效提升你的总成绩。

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 form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of x is 2, which is why it is called a quadratic equation.

二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。由于未知数 x 的最高次数是 2，因此被称为二次方程。

Common examples include x² − 5x + 6 = 0, 2x² + 3x − 1 = 0 and x² − 9 = 0. Unlike linear equations, which usually have one solution, quadratic equations can have two solutions.

常见的例子包括 x² − 5x + 6 = 0、2x² + 3x − 1 = 0 和 x² − 9 = 0。与通常只有一个解的一次方程不同，二次方程可能有两个解。

If a = 0, the equation becomes bx + c = 0, which is linear. Always check that the coefficient of x² is not zero.

如果 a = 0，方程就退化成为 bx + c = 0，这是一次方程。所以解题前务必检查 x² 的系数是否不为零。

2. Expanding and Factorising | 展开与因式分解

Before solving quadratic equations by factorisation, you must be able to expand double brackets. The rule is (x + p)(x + q) = x² + (p + q)x + pq.

在用因式分解解二次方程之前，你首先需要掌握展开括号的方法。展开法则为 (x + p)(x + q) = x² + (p + q)x + pq。

For example, (x + 4)(x − 3) = x² − 3x + 4x − 12 = x² + x − 12. Notice that the middle term is the sum of 4 and −3, while the constant term is the product 4 × (−3).

例如，(x + 4)(x − 3) = x² − 3x + 4x − 12 = x² + x − 12。注意：一次项系数是 4 与 −3 的和，常数项则是 4 × (−3) 的积。

Factorising is the reverse process. To factorise x² + 7x + 12, look for two numbers that multiply to 12 and add to 7. Those numbers are 3 and 4, so x² + 7x + 12 = (x + 3)(x + 4).

因式分解是展开的逆过程。要对 x² + 7x + 12 因式分解，我们需要找到两个数，使它们的乘积为 12、和为 7。这两个数是 3 和 4，因此 x² + 7x + 12 = (x + 3)(x + 4)。

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

The most direct method for many IGCSE questions is factorisation. It uses the zero product property: if a × b = 0, then either a = 0 or b = 0.

因式分解是 IGCSE 考试中最直接的解法之一，其依据是零因子性质：如果 a × b = 0，那么必有 a = 0 或 b = 0。

Consider x² − 5x + 6 = 0. We need two numbers that multiply to 6 and add to −5. These numbers are −2 and −3, so the equation becomes (x − 2)(x − 3) = 0.

以 x² − 5x + 6 = 0 为例。我们需要找到乘积为 6、和为 −5 的两个数，它们是 −2 和 −3，因此原方程可化为 (x − 2)(x − 3) = 0。

(x − 2)(x − 3) = 0

Then x − 2 = 0 or x − 3 = 0, so x = 2 or x = 3. Always finish by checking that both values satisfy the original equation.

于是 x − 2 = 0 或 x − 3 = 0，所以 x = 2 或 x = 3。解完后最好代回原方程验证。

For a quadratic with a coefficient of x² greater than 1, such as 2x² + 7x + 3 = 0, the factorised form is (2x + 1)(x + 3) = 0. This leads to x = −1/2 or x = −3.

当 x² 的系数大于 1 时，例如 2x² + 7x + 3 = 0，因式分解形式为 (2x + 1)(x + 3) = 0，于是得到 x = −1/2 或 x = −3。

4. The Difference of Two Squares | 平方差公式

A special factorisation pattern is the difference of two squares: x² − a² = (x − a)(x + a).

一种特殊的因式分解形式是“平方差”：x² − a² = (x − a)(x + a)。

For example, x² − 49 = (x − 7)(x + 7). Solving x² − 49 = 0 gives x = 7 or x = −7.

例如，x² − 49 = (x − 7)(x + 7)。解方程 x² − 49 = 0，得到 x = 7 或 x = −7。

This pattern is especially useful when the equation has no linear term, because you can skip factorisation and use the idea that x² = a² means x = ±a.

这个公式在方程中没有一次项时尤其好用。实际上，x² = a² 可以直接推出 x = ±a。

5. The Quadratic Formula | 二次求根公式

If a quadratic equation cannot be factorised easily, use the quadratic formula. For ax² + bx + c = 0,

如果二次方程不容易因式分解，就可以使用二次求根公式。对于 ax² + bx + c = 0，

x = (−b ± √(b² − 4ac)) / (2a)

For example, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3 and c = −2. The formula gives:

例如，解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入公式得：

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

Therefore x = (−3 + 5) / 4 = 1/2, or x = (−3 − 5) / 4 = −2.

因此 x = (−3 + 5) / 4 = 1/2，或 x = (−3 − 5) / 4 = −2。

In the non-calculator paper, leave answers in exact form such as (3 + √5)/2 rather than giving a rounded decimal.

在不能使用计算器的试卷中，答案应保留精确形式，例如 (3 + √5)/2，而不写成近似小数。

6. Completing the Square | 配方法

Completing the square rewrites x² + bx in the form (x + b/2)² − b²/4. This method is important for finding turning points and for solving equations without factorising.

配方法把 x² + bx 改写成 (x + b/2)² − b²/4 的形式。这个方法在求顶点坐标和解方程时非常重要。

For example, x² + 6x + 5 = 0 can be written as (x + 3)² − 9 + 5 = 0, which simplifies to (x + 3)² = 4.

例如，x² + 6x + 5 = 0 可以写成 (x + 3)² − 9 + 5 = 0，整理得 (x + 3)² = 4。

(x + 3)² = 4

Taking the square root of both sides gives x + 3 = ±2, so x = −1 or x = −5.

两边开平方得 x + 3 = ±2，所以 x = −1 或 x = −5。

When a is not 1, divide the whole equation by a before completing the square. For example, 2x² + 8x + 6 = 0 becomes x² + 4x + 3 = 0 first.

当 x² 系数不为 1 时，应先将方程两边同时除以 a。例如，2x² + 8x + 6 = 0 可先化为 x² + 4x + 3 = 0。

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

The expression b² − 4ac inside the quadratic formula is called the discriminant, written as Δ. It tells us how many real roots a quadratic equation has.

二次求根公式中 b² − 4ac 这个表达式称为判别式，记作 Δ。它决定了二次方程有几个实数根。

Discriminant	Nature of Roots	Graph Meaning
Δ > 0	Two distinct real roots	The curve crosses the x-axis twice
Δ = 0	One repeated real root	The curve touches the x-axis once
Δ < 0	No real roots	The curve does not cross the x-axis

For example, in x² + 2x + 5 = 0, the discriminant is 2² − 4 × 1 × 5 = −16, which is less than 0, so this equation has no real solutions.

例如，对于 x² + 2x + 5 = 0，判别式为 2² − 4 × 1 × 5 = −16，小于 0，因此该方程没有实数解。

8. Sketching Quadratic Graphs | 二次函数图像的草图

The graph of y = ax² + bx + c is a parabola. If a > 0, it has a U shape and a minimum point. If a < 0, it has a downward-opening shape and a maximum point.

y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，图像开口向上，有最小值点；当 a < 0 时，图像开口向下，有最大值点。

To sketch a quadratic graph, find three useful features:

画二次函数图像时，通常需要确定以下三个特征：

The roots, where y = 0.

根，即 y = 0 时对应的 x 值。
The y-intercept, where x = 0, which is simply c.

y 轴截距，即 x = 0 时 y 的值，具体为 c。
The turning point, found on the vertical line x = −b/(2a).

顶点，位于对称轴 x = −b/(2a) 上。

For example, y = x² − 4x + 3 has roots x = 1 and x = 3, a y-intercept at 3, and a minimum turning point at (2, −1).

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Solving Quadratic Equations | 解一元二次方程