Completing the Square | 配方法

📚 Completing the Square | 配方法

Completing the square is a powerful algebraic technique used to rewrite quadratic expressions in a simpler form. It is essential for solving quadratic equations, analysing curves and finding turning points.

配方法是一种强大的代数技巧，能将二次表达式改写成更简洁的形式。它在解二次方程、分析曲线以及求极值点方面至关重要。

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

Completing the square means transforming a quadratic expression of the form ax² + bx + c into the form a(x − h)² + k. Here h and k are constants that reveal important information about the quadratic.

配方法是指将形如 ax² + bx + c 的二次表达式改写为 a(x − h)² + k 的形式，其中 h 和 k 是常数，能揭示二次表达式的重要信息。

For example, x² + 6x + 2 can be rewritten as (x + 3)² − 7. This new form makes it much easier to solve equations and identify the vertex of a parabola.

例如，x² + 6x + 2 可改写为 (x + 3)² − 7。这种新形式使解方程和确定抛物线顶点变得更加容易。

2. The Perfect Square Pattern | 完全平方公式

To understand completing the square, you must first recall the perfect square expansion rules: (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².

要理解配方法，首先要回忆完全平方展开法则：(a + b)² = a² + 2ab + b²，以及 (a − b)² = a² − 2ab + b²。

Notice that in a perfect square, the constant term is the square of half the coefficient of x. This is the key idea behind the method.

注意，在完全平方中，常数项是 x 一次项系数一半的平方。这正是配方法背后的关键思想。

x² + 6x + 9 = (x + 3)²

Here, half of 6 is 3, and 3² = 9. If the constant is not 9, we create it artificially by adding and subtracting the same value.

这里，6 的一半是 3，而 3² = 9。如果常数项不是 9，我们可以通过同时加上和减去同一个值来“创造”这个常数。

3. Step-by-Step Procedure | 配方法步骤

When the coefficient of x² is 1, the procedure is simple. Take the coefficient of x, halve it, square it, then add and subtract this square inside the expression.

当 x² 的系数为 1 时，步骤很简单：取 x 的系数，除以 2，平方，再在表达式中加上并减去这个平方值。

For example, complete the square for x² + 8x + 5.

例如，对 x² + 8x + 5 进行配方。

x² + 8x + 5 = x² + 8x + 4² − 4² + 5

= (x + 4)² − 16 + 5 = (x + 4)² − 11

If the coefficient of x² is not 1, factor it out first. For example, 2x² + 12x + 7 becomes 2(x² + 6x) + 7, then complete the square inside the bracket.

如果 x² 的系数不是 1，则先将其提取出来。例如，2x² + 12x + 7 变为 2(x² + 6x) + 7，然后在括号内配方。

4. Solving Quadratic Equations | 用配方法解二次方程

Completing the square can solve any quadratic equation. After rewriting the expression in vertex form, isolate the squared term and take the square root of both sides.

配方法可以解任何二次方程。将表达式改写为顶点式后，隔离平方项，再对两边开平方。

Solve x² + 6x + 2 = 0 using completing the square.

使用配方法解 x² + 6x + 2 = 0。

x² + 6x + 2 = (x + 3)² − 7 = 0

(x + 3)² = 7

x + 3 = ±√7 ⇒ x = −3 ± √7

Always include the ± sign when taking the square root. The two solutions are x = −3 + √7 and x = −3 − √7.

开平方时必须带上 ± 符号。两个解分别是 x = −3 + √7 和 x = −3 − √7。

5. The Quadratic Formula Link | 与二次公式的联系

The quadratic formula is derived from completing the square. Starting from ax² + bx + c = 0, completing the square on the general form produces the formula.

二次公式正是由配方法推导出来的。从 ax² + bx + c = 0 出发，对一般形式配方，就能得到这个公式。

x = (−b ± √(b² − 4ac)) / (2a)

This formula works for all quadratic equations, even when the solutions are irrational or complex (if studied later).

这个公式适用于所有二次方程，即使解是无理数甚至复数（如果以后学到）也能处理。

Understanding the derivation strengthens your grasp of both the formula and the method of completing the square.

理解推导过程能加深你对公式和配方法本身的掌握。

6. Vertex Form of a Parabola | 抛物线的顶点式

The expression a(x − h)² + k is called the vertex form of a quadratic function. The vertex of the parabola is exactly at the point (h, k).

a(x − h)² + k 被称为二次函数的顶点式。抛物线的顶点恰好位于点 (h, k)。

For y = (x − 3)² + 5, the vertex is (3, 5). Note that the sign inside the bracket changes: (x − 3) means h = 3.

对于 y = (x − 3)² + 5，顶点是 (3, 5)。注意括号内的符号会变化：(x − 3) 意味着 h = 3。

From vertex form, you can immediately state the minimum or maximum point of the curve without any further calculation.

根据顶点式，无需进一步计算就能直接写出曲线的最小值点或最大值点。

7. Finding Maximum and Minimum Values | 求最大值和最小值

If a > 0, the parabola opens upward, so the vertex gives the minimum value. If a < 0, the parabola opens downward, so the vertex gives the maximum value.

如果 a > 0，抛物线开口向上，那么顶点给出最小值；如果 a < 0，抛物线开口向下，那么顶点给出最大值。

For y = 2(x − 3)² + 5, the minimum value is 5, occurring at x = 3.

对于 y = 2(x − 3)² + 5，最小值是 5，出现在 x = 3 处。

For y = −(x + 1)² + 8, the maximum value is 8, occurring at x = −1.

对于 y = −(x + 1)² + 8，最大值是 8，出现在 x = −1 处。

This is extremely useful in optimisation problems, projectile motion and business applications.

这在最优化问题、抛体运动和商业应用中极为有用。

8. Sketching Graphs | 画函数图像

Vertex form makes sketching a parabola straightforward. Start by plotting the vertex (h, k), then use the value of a to find the direction of opening and the steepness.

顶点式使绘制抛物线变得直接。先画出顶点 (h, k)，再根据 a 的值确定开口方向和陡峭程度。

If a = 1, the curve moves one unit up and one unit right from the vertex to get a second point. If a = 2, it moves two units up, and so on.

如果 a = 1，从顶点向右移动一个单位，再向上移动一个单位，得到另一个点；如果 a = 2，则向上移动两个单位，依此类推。

The axis of symmetry is the vertical line x = h. The curve is symmetric about this line.

对称轴是竖直直线 x = h。曲线关于这条线对称。

You can also find the y-intercept by setting x = 0, and the x-intercepts by solving the quadratic equation.

你还可以通过令 x = 0 求 y 轴截距，通过解二次方程求 x 轴截距。

9. Common Mistakes and Tips | 常见错误与提示

Forgetting to add and subtract the same square: x² + 4x + 1 cannot become (x + 2)² + 1; it must be (x + 2)² − 4 + 1. 忘记同时加上和减去同一个平方值：x² + 4x + 1 不能变成 (x + 2)² + 1，而应为 (x + 2)² − 4 + 1。
When the coefficient of x² is not 1, forgetting to factor it out first can lead to errors. 当 x² 的系数不是 1 时，忘记先提取它会出错。
Taking the square root without remembering ±. 开平方时忘记写 ± 符号。
Mixing up the sign of the vertex coordinate: y = (x − 2)² + 3 has vertex (2, 3), not (−2, 3). 混淆顶点坐标符号：y = (x − 2)² + 3 的顶点是 (2, 3)，而不是 (−2, 3)。

Always check your answer by expanding the vertex form back to the original quadratic.

一定要通过将顶点式展开，检查是否能还原为原来的二次式。

10. Practice Questions | 练习题

Try these problems to reinforce your understanding.

尝试以下题目来巩固你的理解。

1. Write x² + 10x + 3 in the form (x + p)² + q. Answer: (x + 5)² − 22. 将 x² + 10x + 3 写成 (x + p)² + q 的形式。答案：(x + 5)² − 22。
2. Solve x² + 8x + 5 = 0 by completing the square. Answer: x = −4 ± √11. 用配方法解 x² + 8x + 5 = 0。答案：x = −4 ± √11。
3. Find the minimum value of y = 3x² + 12x + 7. Answer: complete square to get 3(x + 2)² − 5, so minimum is −5. 求 y = 3x² + 12x + 7 的最小值。答案：配方得 3(x + 2)² − 5，所以最小值为 −5。
4. Write 2x² − 8x + 1 in vertex form. Answer: 2(x − 2)² − 7. 将 2x² − 8x + 1 写成顶点式。答案：2(x − 2)² − 7。
5. The vertex of y = −x² + 6x + 1 is (3, 10). Verify this by completing the square. 抛物线 y = −x² + 6x + 1 的顶点是 (3, 10)。请用配方法验证。

If you can solve all of these confidently, you have mastered completing the square for IGCSE Mathematics.

如果你能自信地解出以上所有题目，那么你已经掌握了 IGCSE 数学中的配方法。

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Completing the Square | 配方法