Completing the Square: Quadratic Transformations and Problem-Solving Applications | 配方法：二次式配方变形与解题应用

📚 Completing the Square: Quadratic Transformations and Problem-Solving Applications | 配方法：二次式配方变形与解题应用

Completing the square is one of the most powerful and versatile algebraic techniques in the A-Level mathematics syllabus. It allows us to transform a quadratic expression ax² + bx + c into the form a(x − h)² + k, revealing the structure of the quadratic at a glance. From solving equations to locating turning points, and from deriving the quadratic formula to evaluating integrals, this single technique underpins a remarkable number of examination topics.

配方法是 A-Level 数学课程中最强大、最通用的代数技巧之一。它使我们能够将二次式 ax² + bx + c 转化为 a(x − h)² + k 的形式，一目了然地揭示二次函数的结构。从求解方程到定位极值点，从推导二次公式到计算积分，这一技术支撑着大量的考试专题。

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1. What Is Completing the Square? | 什么是配方法

At its core, completing the square is the process of rewriting a quadratic expression as a perfect square trinomial plus or minus an adjusting constant. Every expression of the form x² + bx can be written as a perfect square by taking half of b, squaring it, and then adding and subtracting the result to keep the expression unchanged.

从本质上讲，配方法是将二次式改写为一个完全平方三项式加上或减去一个调整常数的过程。对于形如 x² + bx 的表达式，我们取 b 的一半，将其平方，然后加上并减去这个结果以保持表达式不变。

Consider the expression x² + 6x. Take half of 6, which is 3, and square it to get 9. Then:

考虑表达式 x² + 6x。取 6 的一半即 3，平方得到 9。于是：

x² + 6x = (x + 3)² − 9

is that (x + 3)² = x² + 6x + 9, so to preserve the original expression we subtract 9. This "completes" the square, giving us a much more insightful representation of the same quadratic.

关键在于 (x + 3)² = x² + 6x + 9，所以为了保持原表达式不变，我们需要减去 9。这样便"配"成了完全平方，从而得到同一个二次式更具洞察力的表示形式。

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2. The General Formula | 一般公式

For a monic quadratic, where the coefficient of x² is 1, we can establish a direct formula. Given x² + bx + c, the completed square form is:

对于首一二次式，即 x² 的系数为 1，我们可以建立一个直接公式。给定 x² + bx + c，配方后的形式为：

x² + bx + c = (x + b/2)² + c − (b/2)²

For example: x² − 8x + 3. Here b = −8, so b/2 = −4 and (b/2)² = 16. The result is:

例如：x² − 8x + 3。此处 b = −8，因此 b/2 = −4，(b/2)² = 16。结果为：

x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13

pleting the square, this information would require calculus or advanced reasoning to extract.

这个形式立即告诉我们表达式的最小值为 −13，发生在 x = 4 时。若不做配方，获取这些信息将需要微积分或更复杂的推理。

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3. When the Leading Coefficient Is Not 1 | 当首项系数不为 1 时

If the coefficient of x² is not 1, we must first factor it out from the x² and x terms before completing the square. Consider 2x² + 8x + 5:

如果 x² 的系数不为 1，我们必须先将其从 x² 项和 x 项中提取出来，再进行配方。考虑 2x² + 8x + 5：

2x² + 8x + 5 = 2(x² + 4x) + 5

Now complete the square inside the bracket: x

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