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 the form a(x + p)² + q. It is essential for solving quadratic equations, finding turning points of parabolas, and simplifying integration problems in later years.

配方法是一种重要的代数技巧，它将二次表达式改写为 a(x + p)² + q 的形式。这一方法在解二次方程、求抛物线顶点坐标以及后续学习积分时都至关重要。

1. Why Complete the Square? | 为什么要配方？

The standard form of a quadratic expression is ax² + bx + c. While factorising works for simple cases, many quadratics do not factorise neatly. Completing the square works for every quadratic expression, making it a universal tool.

二次表达式的标准形式是 ax² + bx + c。因式分解只适用于简单情形，而许多二次式无法被整齐地分解。配方法对所有二次表达式都适用，因此是一种通用工具。

It solves quadratic equations ax² + bx + c = 0 without guessing factors.
It reveals the turning point (vertex) of a parabola directly.
It is required for finding minimum or maximum values in optimisation problems.

无需猜测因数即可解二次方程 ax² + bx + c = 0。
直接揭示抛物线的顶点坐标。
在优化问题中用于求最小值或最大值。

2. The Basic Technique with x² + bx | 基本技巧：x² + bx

We start with the simplest case where the coefficient of x² is 1. Consider x² + 6x. We want to write it in the form (x + p)² + q. Expand (x + p)² to see the pattern:

我们从 x² 系数为 1 的最简单情形开始。考虑 x² + 6x，我们希望将其写成 (x + p)² + q 的形式。先展开 (x + p)² 观察规律：

(x + p)² = x² + 2px + p²

To turn x² + 6x into a perfect square, we take half of 6, which is 3, and write (x + 3)² = x² + 6x + 9. Since our original expression has no constant term, we subtract the extra 9:

要把 x² + 6x 变成完全平方，取 6 的一半得 3，于是 (x + 3)² = x² + 6x + 9。原式没有常数项，所以要减去多余的 9：

x² + 6x = (x + 3)² − 9

Rule: Take half of b, write (x + b/2)², and subtract (b/2)².

法则：取 b 的一半，写成 (x + b/2)²，再减去 (b/2)²。

3. Worked Example with a Constant | 含常数项的例题

Now let us complete the square for x² − 8x + 5. First, focus only on x² − 8x. Half of −8 is −4, so we write:

现在对 x² − 8x + 5 配方。先只看 x² − 8x 部分。−8 的一半是 −4，因此写成：

x² − 8x + 5 = (x − 4)² − 16 + 5

Simplify the constants: −16 + 5 = −11. Therefore:

合并常数项：−16 + 5 = −11。因此：

x² − 8x + 5 = (x − 4)² − 11

Check: expand (x − 4)² − 11 = x² − 8x + 16 − 11 = x² − 8x + 5. ✓

检验：展开 (x − 4)² − 11 = x² − 8x + 16 − 11 = x² − 8x + 5。✓

4. When a ≠ 1: Factor First | 当 a ≠ 1 时：先提出 a

For expressions like 2x² + 8x + 5, we must first factor out the coefficient of x² from the first two terms:

对于 2x² + 8x + 5 这类表达式，必须先把 x² 的系数从前两项中提出来：

2x² + 8x + 5 = 2(x² + 4x) + 5

Now complete the square inside the bracket: x² + 4x = (x + 2)² − 4. Then multiply back by 2:

接着在括号内配方：x² + 4x = (x + 2)² − 4。然后乘以 2：

2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3

Be careful: the constant inside the bracket must be multiplied by the factor a when expanded.

注意：括号内的常数项在展开时需要乘以系数 a。

5. Solving Quadratic Equations by Completing the Square | 用配方法解二次方程

To solve x² − 6x − 7 = 0, we first complete the square for the left-hand side:

解 x² − 6x − 7 = 0 时，先对左边配方：

x² − 6x = (x − 3)² − 9

So the equation becomes:

于是方程变为：

(x − 3)² − 9 − 7 = 0 ⟹ (x − 3)² − 16 = 0

Now rearrange and take the square root of both sides:

移项并对两边开平方：

(x − 3)² = 16 ⟹ x − 3 = ±4

Thus x = 3 + 4 = 7 or x = 3 − 4 = −1.

因此 x = 3 + 4 = 7 或 x = 3 − 4 = −1。

This method works even when the roots are irrational or complex, as long as you remember the ± sign.

只要记住 ± 号，这一方法甚至适用于根为无理数或复数的情况。

6. Finding the Turning Point | 求抛物线顶点

Once a quadratic is written as a(x + p)² + q, the turning point is at x = −p and y = q. This comes from the fact that (x + p)² ≥ 0 for all real x, so the expression is minimised or maximised exactly when x + p = 0.

当二次式写成 a(x + p)² + q 后，顶点坐标为 x = −p，y = q。这是因为 (x + p)² 对所有实数 x 都满足 ≥ 0，因此当 x + p = 0 时表达式取得最小值或最大值。

Example: For y = (x − 3)² + 5, the turning point is (3, 5). The value 5 is the minimum since a = 1 > 0.

例：对于 y = (x − 3)² + 5，顶点为 (3, 5)。由于 a = 1 > 0，5 是该函数的最小值。

Example: For y = −2(x + 1)² + 7, the turning point is (−1, 7), which is a maximum because a = −2 < 0.

例：对于 y = −2(x + 1)² + 7，顶点为 (−1, 7)，因为 a = −2 < 0，所以 7 是该函数的最大值。

7. The Relationship with the Quadratic Formula | 配方法与求根公式的关系

The quadratic formula x = (−b ± √(b² − 4ac)) / (2a) is derived directly from completing the square on ax² + bx + c = 0. Understanding the derivation helps you remember the formula and know when it fails (when b² − 4ac < 0).

求根公式 x = (−b ± √(b² − 4ac)) / (2a) 正是通过对 ax² + bx + c = 0 配方推导出来的。理解这一推导过程有助于记住公式，并理解它在 b² − 4ac < 0 时为何无实数解。

By completing the square on the general equation, we obtain the discriminant Δ = b² − 4ac inside the square root. This discriminant tells us the number of real roots:

对一般方程配方后，根号内正好出现判别式 Δ = b² − 4ac。判别式决定实数根的个数：

Discriminant Δ	Number of real roots	判别式 Δ	实数根个数
Δ > 0	Two distinct roots	Δ > 0	两个不同实根
Δ = 0	One repeated root	Δ = 0	一个重根
Δ < 0	No real roots	Δ < 0	无实数根

8. Sketching Quadratic Graphs Using the Completed Square | 利用配方法绘制二次函数图像

Once a quadratic is in completed square form, sketching the graph becomes straightforward:

一旦二次式化为配方形式，绘制图像就变得十分直接：

Identify the turning point (−p, q).
Identify whether it opens upwards (a > 0) or downwards (a < 0).
Find the y-intercept by setting x = 0 in the original expression.
Find any real x-intercepts by solving a(x + p)² + q = 0.

确定顶点坐标 (−p, q)。
判断开口方向：a > 0 向上，a < 0 向下。
令 x = 0 求 y 轴截距。
令 a(x + p)² + q = 0 求 x 轴截距（若存在实数解）。

For example, y = 2(x − 1)² − 8 has its turning point at (1, −8). Setting x = 0 gives y-intercept −6. Solving 2(x − 1)² − 8 = 0 yields (x − 1)² = 4, so x = 3 or x = −1. These are the x-intercepts.

例如 y = 2(x − 1)² − 8，顶点为 (1, −8)。令 x = 0 得 y 截距为 −6。解 2(x − 1)² − 8 = 0 得 (x − 1)² = 4，即 x = 3 或 x = −1，这两个就是 x 截距。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Many students make predictable errors when completing the square. Being aware of these will save you valuable marks in an exam.

许多学生在配方时会犯一些可预见的错误。了解这些错误能帮你在考试中少丢分。

Mistake	Correction	错误	正确做法
Forgetting to subtract (b/2)²	Always add and subtract the same value to keep equality	忘记减去 (b/2)²	必须同时加上和减去同一个值以保持等式成立
Forgetting to multiply the subtracted term by a when a ≠ 1	Multiply the whole bracket content by a	当 a ≠ 1 时忘记把减去的项乘以 a	要把括号内所有内容乘以 a
Sign error when b is negative	Half of a negative number is negative; test by expanding	b 为负时符号出错	负数的一半仍是负数；用展开检验

10. Practice Questions | 练习与建议

To master completing the square, practise writing these expressions in completed square form, then check by expanding your answer.

要熟练掌握配方法，请练习将以下表达式写成配方形式，然后通过展开来检验答案。

x² + 10x + 3
x² − 5x + 1
3x² + 12x + 7
−x² + 4x − 1
2x² − 8x + 11

x² + 10x + 3
x² − 5x + 1
3x² + 12x + 7
−x² + 4x − 1
2x² − 8x + 11

Answers are: 1) (x + 5)² − 22; 2) (x − 2.5)² − 5.25; 3) 3(x + 2)² − 5; 4) −(x − 2)² + 3; 5) 2(x − 2)² + 3.

答案分别为：1) (x + 5)² − 22；2) (x − 2.5)² − 5.25；3) 3(x + 2)² − 5；4) −(x − 2)² + 3；5) 2(x − 2)² + 3。

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Completing the Square | 配方法