Quadratic Equations and Functions | 二次方程与函数

📚 Quadratic Equations and Functions | 二次方程与函数

This guide covers the essential skills for IGCSE Mathematics: solving quadratic equations and analysing quadratic functions. You will learn the three main methods for solving quadratics, how to use the discriminant, and how to sketch a parabola from key features.

本指南涵盖 IGCSE 数学中的核心内容：解二次方程与分析二次函数。你将学会解二次方程的三种主要方法、如何使用判别式，以及如何通过关键特征绘制抛物线草图。

1. Standard Form of a Quadratic | 二次方程的标准形式

A quadratic equation is any expression of the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The highest power of x is always 2.

二次方程是指形如 ax² + bx + c = 0 的式子，其中 a, b, c 为实数，且 a ≠ 0。未知数 x 的最高次数恒为 2。

For example, these are all quadratic equations: 2x² + 3x − 5 = 0, x² − 49 = 0, and 4x² = 12x. The coefficient ‘a’ determines the shape of the graph; if a > 0 the parabola opens upwards, and if a < 0 it opens downwards.

例如，以下都是二次方程：2x² + 3x − 5 = 0、x² − 49 = 0 以及 4x² = 12x。系数 a 决定图象的形状：当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。

2. Solving by Factorisation | 用因式分解法求解

Factorisation is the fastest method when the quadratic can be written as a product of two binomials. You rewrite ax² + bx + c as (px + q)(rx + s), then set each bracket equal to zero.

当二次式可写成两个二项式的乘积时，因式分解是最快捷的方法。将 ax² + bx + c 改写为 (px + q)(rx + s)，然后令每个括号等于零。

Example: Solve x² − 5x + 6 = 0. Since (x − 2)(x − 3) = 0, the solutions are x = 2 or x = 3.
示例：解方程 x² − 5x + 6 = 0。因为 (x − 2)(x − 3) = 0，所以解为 x = 2 或 x = 3。
Always check if a common factor exists first, e.g. 2x² + 8x = 0 becomes 2x(x + 4) = 0.
先检查是否存在公因式，例如 2x² + 8x = 0 可化为 2x(x + 4) = 0。
For difference of two squares, use the identity a² − b² = (a + b)(a − b).
对于平方差形式，运用恒等式 a² − b² = (a + b)(a − b)。

3. Solving by Completing the Square | 用配方法求解

Completing the square rewrites x² + bx + c as (x + p)² + q. This method works for any quadratic and also helps find the vertex of a parabola.

配方法将 x² + bx + c 改写为 (x + p)² + q。该方法适用于所有二次方程，同时也能帮助确定抛物线的顶点。

x² + bx + c = (x + b/2)² − (b/2)² + c

Example: Solve x² + 6x + 2 = 0. First complete the square: (x + 3)² − 9 + 2 = 0, so (x + 3)² = 7. Then x + 3 = ±√7, giving x = −3 ± √7.

示例：解方程 x² + 6x + 2 = 0。先配方：(x + 3)² − 9 + 2 = 0，即 (x + 3)² = 7。因此 x + 3 = ±√7，解得 x = −3 ± √7。

4. The Quadratic Formula | 二次求根公式

For any quadratic ax² + bx + c = 0, the solutions are given by the formula below. You should memorise it and be able to substitute values accurately.

对于任意二次方程 ax² + bx + c = 0，其解由下面的公式给出。你需要记住它，并能准确代入数值。

x = (−b ± √(b² − 4ac)) / (2a)

Identify a, b and c before substituting; be careful with negative signs.
代入前先确认 a, b, c 的取值，尤其注意负号。
Always simplify the square root if possible, e.g. √12 = 2√3.
如有可能，始终化简根号，例如 √12 = 2√3。
Give answers to 3 significant figures unless the root is a perfect square.
若根号不是完全平方数，答案保留三位有效数字。

5. The Discriminant | 判别式

The expression b² − 4ac inside the quadratic formula is called the discriminant. It tells you how many real solutions the equation has without fully solving it.

二次求根公式中的表达式 b² − 4ac 称为判别式。它无需完整求解即可判断方程有多少个实数解。

Discriminant	Nature of Roots
b² − 4ac > 0	Two distinct real roots
b² − 4ac = 0	One repeated real root
b² − 4ac < 0	No real roots

If the discriminant is negative, the graph does not intersect the x-axis at all. If it is zero, the vertex touches the x-axis at exactly one point.

若判别式为负，则图象与 x 轴无交点。若判别式为零，则顶点恰好与 x 轴相切于一点。

6. The Graph of a Quadratic Function | 二次函数图象

The graph of y = ax² + bx + c is a smooth curve called a parabola. Its shape depends on the sign of a.

函数 y = ax² + bx + c 的图象是一条平滑曲线，称为抛物线。其开口方向取决于 a 的符号。

If a > 0 the curve has a minimum point; if a < 0 it has a maximum point.
当 a > 0 时，曲线有最低点；当 a < 0 时，曲线有最高点。
The x-intercepts are the real roots of the equation ax² + bx + c = 0.
图象与 x 轴的交点即为方程 ax² + bx + c = 0 的实数根。
The y-intercept is the constant term c, because at x = 0, y = c.
图象与 y 轴的交点为常数项 c，因为当 x = 0 时，y = c。

7. Vertex Form and the Turning Point | 顶点式与转向点

By completing the square, you can rewrite the quadratic as y = a(x − h)² + k. The coordinates of the turning point are exactly (h, k).

通过配方，可将二次函数改写为 y = a(x − h)² + k。转向点的坐标即为 (h, k)。

If y = a(x − h)² + k, the vertex is at (h, k)

Example: y = 2(x − 3)² + 5 has its vertex at (3, 5) and opens upwards. The line of symmetry is x = 3.

示例：y = 2(x − 3)² + 5 的顶点为 (3, 5)，开口向上，对称轴为 x = 3。

8. Sketching a Parabola | 绘制抛物线草图

To sketch quickly, find four key points: the roots, the y-intercept, and the vertex. Then draw a smooth curve through them.

快速作图时，找出四个关键点：根、y 轴截距以及顶点，然后画一条平滑曲线通过它们。

Step 1: Solve ax² + bx + c = 0 to mark the x-intercepts.
第 1 步：解方程 ax² + bx + c = 0，标出 x 轴截距。
Step 2: Mark (0, c) as the y-intercept.
第 2 步：标出 y 轴截距 (0, c)。
Step 3: Use x = −b/(2a) to find the axis of symmetry, then substitute to get the vertex.
第 3 步：用 x = −b/(2a) 求对称轴，代入原式得到顶点。
Step 4: Draw the curve and label all intercepts and the turning point.
第 4 步：画出曲线，并标注所有截距和转向点。

9. Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities like x² − 5x + 6 > 0 are solved by first finding the critical roots, then testing intervals on a number line.

对于二次不等式，如 x² − 5x + 6 > 0，先求出临界根，再在数轴上测试各个区间。

From (x − 2)(x − 3) = 0, the critical values are 2 and 3. Test x = 0, x = 2.5 and x = 4. The solution is x < 2 or x > 3.

由 (x − 2)(x − 3) = 0，得到临界值为 2 和 3。分别测试 x = 0、x = 2.5 和 x = 4，因此解为 x < 2 或 x > 3。

10. Applications of Quadratic Equations | 二次方程的应用

Many real-world problems can be modelled by quadratics, such as projectile motion, area calculations, and profit optimisation.

许多实际问题可以用二次方程建模，例如抛体运动、面积计算和利润最优化。

Example: A rectangle has length (x + 3) cm and width x cm, with an area of 28 cm². Then x(x + 3) = 28, so x² + 3x − 28 = 0. Factorising gives (x + 7)(x − 4) = 0, hence x = 4 (reject x = −7).

示例：一个长方形的长为 (x + 3) cm，宽为 x cm，面积为 28 cm²。因此 x(x + 3) = 28，即 x² + 3x − 28 = 0。因式分解得 (x + 7)(x − 4) = 0，所以 x = 4（舍去 x = −7）。

Always check that your answer makes sense in the context; lengths cannot be negative.

始终检查答案在具体情境中是否合理，长度不能为负数。

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Quadratic Equations and Functions | 二次方程与函数