📚 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
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
w 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。
d 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
)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: 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: r 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 erstanding 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。
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）。
ing 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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