Quadratic Equations & Functions | 二次方程与二次函数

📚 Quadratic Equations & Functions | 二次方程与二次函数

Quadratic equations and quadratic functions form one of the most essential topics in the IGCSE Mathematics syllabus. Mastering this area not only secures high marks in exams but also builds a solid foundation for advanced mathematics, including calculus and coordinate geometry.

二次方程与二次函数是 IGCSE 数学大纲中最重要的考点之一。掌握这一领域不仅能帮助你在考试中取得高分，更能为高等数学——如微积分和坐标几何——打下坚实的基础。

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

A quadratic equation in one variable is an equation that can be written in the general form ax² + bx + c = 0, where a, b, and c are real numbers, and a ≠ 0. The coefficient a is called the leading coefficient, b the linear coefficient, and c the constant term.

一元二次方程是指可以写成一般形式 ax² + bx + c = 0 的方程，其中 a、b、c 为实数，且 a ≠ 0。系数 a 称为二次项系数，b 称为一次项系数，c 称为常数项。

For example, the equation 2x² − 5x + 3 = 7 can be rearranged as 2x² − 5x − 4 = 0 by subtracting 7 from both sides. In IGCSE exams, you will often be asked to rearrange a given equation into standard form first.

例如，方程 2x² − 5x + 3 = 7 可以通过两边同时减去 7 化为 2x² − 5x − 4 = 0。在 IGCSE 考试中，你常常需要先将给定方程化为标准形式。

2. The Quadratic Formula and the Discriminant | 求根公式与判别式

For any quadratic equation ax² + bx + c = 0, the solution is given by the quadratic formula:

x = (−b ± √(b² − 4ac)) / 2a

The expression b² − 4ac is called the discriminant. It determines the nature of the roots:

表达式 b² − 4ac 称为判别式，它决定了根的性质：

If b² − 4ac > 0, the equation has two distinct real roots.
If b² − 4ac = 0, the equation has exactly one real root (a repeated root).
If b² − 4ac < 0, the equation has no real roots (two complex roots).

若 b² − 4ac > 0，方程有两个不相等的实数根。
若 b² − 4ac = 0，方程有且仅有一个实数根（重根）。
若 b² − 4ac < 0，方程没有实数根（有两个复数根）。

For example, for the equation x² + 6x + 9 = 0, the discriminant is 6² − 4 × 1 × 9 = 36 − 36 = 0. Therefore, it has a repeated root x = −3.

例如，对于方程 x² + 6x + 9 = 0，判别式为 6² − 4 × 1 × 9 = 36 − 36 = 0，因此它有重根 x = −3。

3. Solving by Factorisation | 因式分解法求解

Factoring is often the quickest way to solve a quadratic equation when the factors are simple. To solve ax² + bx + c = 0, we look for two numbers p and q such that p × q = ac and p + q = b. Then we split the middle term and factor by grouping.

因式分解是解二次方程最快捷的方法之一，前提是方程容易分解。要解 ax² + bx + c = 0，我们寻找两个数 p 和 q，使得 p × q = ac 且 p + q = b，然后裂项分组因式分解。

Example: Solve x² − 7x + 12 = 0.

示例：解方程 x² − 7x + 12 = 0。

We need two numbers whose product is 12 and whose sum is −7. These are −3 and −4. Therefore:

我们需要找到两个数，其乘积为 12，和为 −7。这两个数是 −3 和 −4。因此：

(x − 3)(x − 4) = 0

Hence x = 3 or x = 4. Always expand your answer to verify correctness.

因此 x = 3 或 x = 4。解完后务必展开验证答案是否正确。

4. Solving by Completing the Square | 配方法求解

Completing the square transforms a quadratic expression into a perfect square plus a constant. For x² + bx, we add and subtract (b/2)². The general result is:

配方法将二次表达式转化为一个完全平方式加上一个常数。对于 x² + bx，我们加减 (b/2)²。一般结果是：

x² + bx + c = (x + b/2)² − (b/2)² + c

Example: Solve x² + 8x − 5 = 0 by completing the square.

示例：用配方法解 x² + 8x − 5 = 0。

Here b = 8, so (b/2)² = 16. We rewrite the equation as:

这里 b = 8，因此 (b/2)² = 16。我们将方程改写为：

(x + 4)² − 16 − 5 = 0

(x + 4)² = 21

Taking the square root of both sides: x + 4 = ±√21, so x = −4 ± √21. This method also reveals the vertex of the parabola directly, which we will discuss later.

两边开平方得 x + 4 = ±√21，因此 x = −4 ± √21。配方法还能直接揭示抛物线的顶点坐标，我们将在后面讨论。

5. Solving by the Quadratic Formula | 公式法求解

The quadratic formula is a guaranteed method that works for every quadratic equation, even when factoring is difficult or impossible.

公式法是一种万能的求解方法，适用于所有二次方程，即使因式分解难以实现或不适用。

Example: Solve 2x² + 3x − 7 = 0 using the quadratic formula.

示例：用公式法解 2x² + 3x − 7 = 0。

Here a = 2, b = 3, c = −7. Substituting into the formula:

此处 a = 2，b = 3，c = −7。代入求根公式：

x = (−3 ± √(3² − 4 × 2 × (−7))) / (2 × 2)

x = (−3 ± √(9 + 56)) / 4 = (−3 ± √65) / 4

Such a solution cannot be obtained easily by factoring, which demonstrates the universal applicability of the formula.

这样的解很难通过因式分解得到，这正好体现了公式法的普适性。

6. The Graph of a Quadratic Function | 二次函数的图像

The graph of a quadratic function y = ax² + bx + c is a smooth curve called a parabola. The sign of a determines the orientation:

二次函数 y = ax² + bx + c 的图像是一条平滑曲线，称为抛物线。a 的符号决定了抛物线的开口方向：

If a > 0, the parabola opens upward, and the function has a minimum value.
If a < 0, the parabola opens downward, and the function has a maximum value.

当 a > 0 时，抛物线开口向上，函数有最小值。
当 a < 0 时，抛物线开口向下，函数有最大值。

The larger the absolute value of a, the narrower the parabola; the smaller it is, the wider the parabola. The constant term c gives the y-intercept, which is the point where the graph crosses the y-axis.

a 的绝对值越大，抛物线越窄；绝对值越小，抛物线越宽。常数项 c 表示 y 轴截距，即图像与 y 轴的交点。

7. The Vertex and the Axis of Symmetry | 顶点与对称轴

Every parabola has a vertical line of symmetry, called the axis of symmetry. The vertex is the turning point of the parabola, and it lies on this axis.

每条抛物线都有一条竖直对称轴，称为对称轴。抛物线的顶点是其转折点，位于对称轴上。

For the quadratic function y = ax² + bx + c, the axis of symmetry is given by:

对于二次函数 y = ax² + bx + c，对称轴方程为：

x = −b / 2a

To find the y-coordinate of the vertex, substitute x = −b / 2a back into the function. For example, for y = x² − 4x + 5, the axis is x = 4 / 2 = 2, and the vertex is (2, 2² − 8 + 5) = (2, 1).

求顶点的纵坐标时，将 x = −b / 2a 代回原函数即可。例如，对于 y = x² − 4x + 5，对称轴为 x = 4 / 2 = 2，顶点为 (2, 2² − 8 + 5) = (2, 1)。

8. The Vertex Form | 顶点式

By completing the square, any quadratic function can be written in the vertex form:

通过配方，任何二次函数都可以写成顶点式：

y = a(x − h)² + k

where (h, k) is the vertex of the parabola. This form is extremely useful because the vertex and the axis of symmetry are immediately visible: the axis is x = h.

其中 (h, k) 是抛物线的顶点坐标。这种形式非常有用，因为顶点和对称轴一目了然：对称轴为 x = h。

Example: Express y = 2x² − 12x + 7 in vertex form.

示例：将 y = 2x² − 12x + 7 化为顶点式。

First factor out 2 from the first two terms: y = 2(x² − 6x) + 7. Complete the square inside the brackets:

首先从前两项提出 2：y = 2(x² − 6x) + 7。在括号内配方：

y = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11

Therefore, the vertex is (3, −11) and the axis of symmetry is x = 3.

因此顶点为 (3, −11)，对称轴为 x = 3。

9. Roots and x-Intercepts | 根与 x 轴交点

The roots of a quadratic equation ax² + bx + c = 0 correspond exactly to the x-intercepts of the graph y = ax² + bx + c. The number of x-intercepts equals the number of real roots determined by the discriminant:

二次方程 ax² + bx + c = 0 的根恰好对应二次函数 y = ax² + bx + c 的图像与 x 轴的交点。x 轴交点的个数等于判别式所决定的实数根的个数：

Discriminant > 0: two distinct x-intercepts.
Discriminant = 0: one x-intercept (the vertex touches the x-axis).
Discriminant < 0: no x-intercepts (the graph lies entirely above or below the x-axis).

判别式 > 0：图像与 x 轴有两个不同交点。
判别式 = 0：图像与 x 轴有一个交点（顶点恰好触及 x 轴）。
判别式 < 0：图像与 x 轴没有交点（图像完全位于 x 轴上方或下方）。

When asked to sketch a parabola, always plot the vertex, the y-intercept, and the x-intercepts (if they exist) to produce an accurate graph.

当题目要求画抛物线草图时，务必标出顶点、y 轴截距和 x 轴截距（如果存在），才能画出准确的图像。

10. Applications: Maximum and Minimum Problems | 应用：最大值与最小值问题

Quadratic functions naturally model many real-world scenarios, such as projectile motion, area optimisation, and revenue maximisation. At the vertex, the function attains its maximum or minimum value.

二次函数天然适用于建模许多现实场景，如抛体运动、面积优化和收益最大化。函数在顶点处取得最大值或最小值。

Example: A rectangle is built with a fixed perimeter of 40 m. Let one side be x. The area A = x(20 − x) = 20x − x². Find the maximum area.

示例：一个矩形的周长固定为 40 米。设一边长为 x，则面积 A = x(20 − x) = 20x − x²。求最大面积。

Here a = −1, b = 20. The maximum occurs at x = −b / 2a = −20 / (−2) = 10. Substituting x = 10 gives A = 10 × 10 = 100 m². The maximum area is achieved when the rectangle is a square.

这里 a = −1，b = 20。最大值出现在 x = −b / 2a = −20 / (−2) = 10。代入 x = 10 得 A = 10 × 10 = 100 平方米。当矩形为正方形时面积最大。

11. Quadratic Inequalities | 二次不等式

Quadratic inequalities, such as x² − 5x + 6 > 0, are solved by first finding the roots of the corresponding equation and then testing intervals on a number line.

二次不等式（如 x² − 5x + 6 > 0）的解法是：先求出对应方程的根，然后在数轴上测试各区间的符号。

For x² − 5x + 6 = 0, the roots are x = 2 and x = 3. Since the parabola opens upward, the expression is positive when x < 2 or x > 3, and negative when 2 < x < 3. Thus the solution is x < 2 or x > 3.

对于 x² − 5x + 6 = 0，根为 x = 2 和 x = 3。由于抛物线开口向上，表达式在 x < 2 或 x > 3 时为正，在 2 < x < 3 时为负。因此解为 x < 2 或 x > 3。

Drawing a quick sketch of the parabola is the most reliable way to avoid sign errors in interval tests.

画出抛物线草图是避免区间符号判断错误的最可靠方法。

12. Common Exam Traps and Revision Tips | 常见考试陷阱与复习建议

Students often lose marks in quadratic problems due to avoidable mistakes. Here are the most common traps and how to avoid them:

学生在二次方程相关题目中常因可避免的错误而失分。以下是最常见的陷阱及应对方法：

Common Mistake \| 常见错误	How to Avoid \| 应对方法
Forgetting that a ≠ 0	Always check the coefficient of x² before applying methods.
Sign errors when substituting negative values into the quadratic formula	Use brackets. Write (−3)², not −3².
Not rearranging to standard form before solving	Move all terms to one side first; set the equation equal to zero.
Confusing the y-intercept with the vertex	Remember: the y-intercept is the value of c; the vertex is the turning point.
Forgetting to give approximate answers when required	Read the question carefully; round to the specified degree of accuracy.

For revision, practise the following skills: expanding and factorising quadratics fluently, completing the square quickly, sketching parabolas confidently, and identifying which solution method is most efficient for each question type.

复习时请重点训练以下技能：熟练展开和因式分解二次式、快速完成配方、自信画出抛物线草图，以及判断每种题型最高效的解法。

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Quadratic Equations & Functions | 二次方程与二次函数