📚 Completing the square | 配方法
Completing the square is an algebraic technique that transforms a quadratic expression of the form ax² + bx + c into a perfect square trinomial plus a constant. In the Edexcel A-Level Mathematics syllabus, mastering this method is essential for solving quadratic equations, finding the vertex of a parabola, deriving the quadratic formula, and tackling inequalities. This article will guide you step-by-step through the process, highlight common pitfalls, and show how the completed square form reveals the hidden structure of a quadratic function.
配方法是一种代数技巧，可将形如 ax² + bx + c 的二次式转化为一个完全平方三项式加上一个常数。在 Edexcel A-Level 数学大纲中，掌握配方法对于解二次方程、求抛物线顶点、推导求根公式以及处理不等式至关重要。本文将逐步引导你完成全过程，指出常见错误，并展示配制式如何揭示二次函数隐藏的结构。
1. Introduction to Completing the Square | 配方法简介
The technique of completing the square allows us to rewrite any quadratic expression in the form a(x + p)² + q, where p and q are constants. This form immediately gives the coordinates of the turning point of the graph and helps in solving equations without relying on a calculator. In essence, we are forcing part of the quadratic to become a perfect square, much like reversing the expansion of (x + d)².
配方法能把任意二次式写成 a(x + p)² + q 的形式，其中 p 和 q 是常数。这种形式能立刻给出函数图像的顶点坐标，并在不依赖计算器的情况下帮助求解方程。本质上，我们是在强制二次式的一部分变成一个完全平方，就像逆转 (x + d)² 的展开过程。
Historically, this method was developed before the invention of the modern quadratic formula and remains a powerful tool because it makes the geometry of the parabola explicit. Whether the quadratic opens upwards (a > 0) or downwards (a < 0), the completed square form shows the maximum or minimum value directly.
在历史上，配方法是在现代求根公式发明之前发展起来的，至今仍是一种强有力的工具，因为它明确地展示了抛物线的几何性质。无论二次函数开口向上 (a > 0) 还是向下 (a < 0)，配制式都直接展示了最大值或最小值。
2. The Basic Idea: From x² + bx to a Perfect Square | 基本思路：从 x² + bx 到完全平方
the square is (b/2)², because (x + b/2)² = x² + bx + (b/2)². This is the core observation.
考虑展开 (x + d)² = x² + 2dx + d²。若要逆转它，当给定 x² + bx 时，我们需要找到 x 系数的一半，将其平方，然后加进去。完成平方的项是 (b/2)²，因为 (x + b/2)² = x² + bx + (b/2)²。这是核心观察。
a perfect square by adding 9, giving (x + 3)². However, we must subtract the added term if we want to keep the expression unchanged.
例如，对于 x² + 6x，6 的一半是 3，3² = 9。因此，x² + 6x 可通过加 9 变成完全平方，得到 (x + 3)²。但如果要保持表达式不变，我们必须将所加的项减去。
3. Completing the Square when a = 1 | 当二次项系数为1时的配方法
Start with x² + bx + c. Follow these steps: first, rewrite as x² + bx and enclose in a bracket. Then take half of b, square it, and add this square inside the bracket. Immediately subtract the same square outside to preserve equality. Finally, factorise the bracket as (x + b/2)² and simplify the constants.
从 x² + bx + c 开始。遵循以下步骤：首先，重写为 x² + bx 并放入括号内。然后取 b 的一半，平方，并将此平方加到括号内。接着立刻在括号外减去同一个平方以保持等式不变。最后，把括号内分解因式为 (x + b/2)²，并化简常数项。
Let' (x² + 8x + 16) + 5 − 16 = (x + 4)² − 11. The completed square form is thus (x + 4)² − 11.
把它应用到 x² + 8x + 5。8 的一半是 4，4² = 16。于是 x² + 8x + 5 = (x² + 8x + 16) + 5 − 16 = (x + 4)² − 11。因此配制式为 (x + 4)² − 11。
4. Completing the Square with a Leading Coefficient a ≠ 1 | 二次项系数不为1的配方法
When the quadratic is ax² + bx + c and a ≠ 1, you must first factor out a from the x² and x terms, leaving the constant term outside. Then complete the square on the bracket as before, but remember to multiply the added and subtracted square by a when simplifying.
当二次式为 ax² + bx + c 且 a ≠ 1 时，你必须首先将 a 从 x² 和 x 项中提取出来，把常数项留在外面。然后如前述对括号内的部分进行配方，但在化简时要记住将所加和所减的平方乘以 a。
tract: 2(x² + 6x + 9 − 9) + 7 = 2[(x + 3)² − 9] + 7 = 2(x + 3)² − 18 + 7 = 2(x + 3)² − 11.
对于 2x² + 12x + 7，从前两项提取 2：2(x² + 6x) + 7。在括号内，6 的一半是 3，平方是 9。加加减减：2(x² + 6x + 9 − 9) + 7 = 2[(x + 3)² − 9] + 7 = 2(x + 3)² − 18 + 7 = 2(x + 3)² − 11。
Always expand mentally to verify your result. A common mistake is forgetting to distribute the outside factor to the subtracted square term.
务必心算展开以验证结果。常见错误是忘记将外面系数与所减去的平方项相乘。
5. Finding the Vertex of a Parabola | 求抛物线的顶点
this point; if a < 0, it has a maximum. The axis of symmetry is the vertical line x = −p.
配制式 a(x + p)² + q 立即告诉我们，顶点位于 (−p, q)。若 a > 0，抛物线在此点取得最小值；若 a < 0，则取得最大值。对称轴为直线 x = −p。
quicker than using calculus or the formula x = −b/(2a), though both yield the same result.
例如，从 2(x + 3)² − 11 可以看出顶点为 (−3, −11)。这比使用微积分或公式 x = −b/(2a) 快得多，尽管它们会得到相同的结果。
6. Solving Quadratic Equations by Completing the Square | 用配方法解二次方程
solve for x. This method always works, even when the quadratic does not factorise nicely.
要解 ax² + bx + c = 0，首先用配制式重写方程。然后孤立平方项，开平平方根（记得 ± 号），并解出 x。即使二次式不能很好地因式分解，此方法也总是有效。
lete the square: (x − 2)² − 4 − 8 = 0 → (x − 2)² = 12. Taking square roots gives x − 2 = ±√12 = ±2√3, so x = 2 ± 2√3. You can leave answers in surd form unless the question specifies decimal approximations.
解 x² − 4x − 8 = 0。配方：(x − 2)² − 4 − 8 = 0 → (x − 2)² = 12。开平方根得 x − 2 = ±√12 = ±2√3，所以 x = 2 ± 2√3。除非题目要求近似小数，否则答案保留根式形式。
7. Deriving the Quadratic Formula | 推导求根公式
c = 0, divide through by a (a ≠ 0) to get x² + (b/a)x + c/a = 0. Then complete the square.
对一般二次式 ax² + bx + c = 0 进行配方，可直接导出著名的求根公式。从 ax² + bx + c = 0 开始，两边除以 a (a ≠ 0)，得 x² + (b/a)x + c/a = 0。然后配方。
e constants: (x + b/(2a))² = (b² − 4ac)/(4a²). Taking square roots and rearranging yields:
b/a 的一半是 b/(2a)，其平方为 b²/(4a²)。因此 x² + (b/a)x 变为 (x + b/(2a))² − b²/(4a²)。于是方程为 (x + b/(2a))² − b²/(4a²) + c/a = 0。孤立平方项并合并常数：(x + b/(2a))² = (b² − 4ac)/(4a²)。开平方根并整理得：
x = [−b ± √b² − 4ac] / (2a)
This derivation is a classic exam question; it not only reinforces algebraic manipulation but also reveals why the discriminant b² − 4ac determines the nature of the roots.
这一推导是经典考题；不仅强化了代数运算能力，还揭示了为什么判别式 b² − 4ac 决定了根的性质。
8. Completing the Square and the Discriminant | 配方法与判别式
After completing the square, the constant term q in a(x + p)² + q is directly related to the discriminant. For the general quadratic, the completed form is a[(x + b/(2a))² − (b² − 4ac)/(4a²)]. The term (b² − 4ac) appears with a negative sign inside the bracket, so the sign of the discriminant governs whether the square bracket can be zero.
配方之后，a(x + p)² + q 中的常数项 q 与判别式直接相关。对一般二次式，配制式为 a[(x + b/(2a))² − (b² − 4ac)/(4a²)]。括号内 (b² − 4ac) 带负号，因此判别式的符号决定了平方项能否为零。
We can summarise the relationship in a table:
我们可以用表格总结这一关系：
| Discriminant | Completed square implication | Roots |
|---|---|---|
| b² − 4ac > 0 | Square bracket can be zero, giving two real solutions | 2 distinct real roots |
| b² − 4ac = 0 | Square bracket is zero at exactly one x; vertex touches axis | 1 repeated real root |
| b² − 4ac < 0 | Square plus a positive constant, never zero | No real roots |
9. Sketching Quadratics using Completed Square Form | 利用配制式绘制二次函数图像
nt c). For accurate sketches, you may also find x-intercepts by setting y = 0 and solving.
当被要求绘制 y = ax² + bx + c 的草图时，配制式提供了一切：顶点、对称轴、开口方向以及 y 轴截距（由原始常数 c 给出）。若要精确作图，还可令 y = 0 以解出 x 轴截距。
), the parabola opens downwards, and the y-intercept is −7. Plot these and draw a symmetrical shape.
例如：y = −3x² + 12x − 7。首先从 x 项提取 −3：−3(x² − 4x) − 7。配方：−3[(x − 2)² − 4] − 7 = −3(x − 2)² + 12 − 7 = −3(x − 2)² + 5。顶点为 (2, 5)，抛物线开口向下，y 轴截距为 −7。描出这些点并画出对称形状。
10. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities such as x² − 4x − 5 > 0 are best tackled by completing the square or factorising to find critical values, then testing regions on a number line. Completing the square shows the vertex and helps visualise where the quadratic is positive or negative relative to the x-axis.
诸如 x² − 4x − 5 > 0 的二次不等式，最好通过配方或因式分解找出关键值，然后在数轴上测试区间。配方可显示顶点，并帮助直观看出二次式相对于 x 轴何处为正、何处为负。
are roots: x − 2 < −3 or x − 2 > 3, giving x < −1 or x > 5. The solution set is split into two intervals because the parabola opens upwards and lies above the axis outside the roots.
解 x² − 4x − 5 > 0。配方：(x − 2)² − 9 > 0 → (x − 2)² > 9。开平方根：x − 2 < −3 或 x − 2 > 3，得 x < −1 或 x > 5。解集分成两个区间，因为抛物线开口向上，在两根之外位于 x 轴上方。
11. Harder Examples and Common Mistakes | 较难例题与常见错误
5/6)² − 25/12 + 1 = 3(x − 5/6)² − 13/12. Fractional arithmetic must be handled carefully.
考虑对 3x² − 5x + 1 进行配方。提取 3：3(x² − (5/3)x) + 1。−5/3 的一半是 −5/6，平方为 25/36。因此有 ) − 25/36] + 1 = 3[(x − 5/6)² − 25/36] + 1 = 3(x − 5/6)² − 25/12 + 1 = 3(x − 5/6)² − 13/12。必须小心处理分数运算。
he wrong coefficient, not squaring the halved term correctly, ignoring the outer factor a when subtracting the square, and misidentifying the vertex sign. Always double-check by expanding your final form.
常见错误包括：求解时忘记 ± 号，取错系数的一半，未正确对取一半后的值进行平方，在减去平方项时忽略了外部系数 a，以及弄错顶点符号。始终通过展开最终形式来双重检查。
12. Summary and Key Points | 总结与要点
Completing the square transforms a quadratic into a form that reveals its geometric properties and facilitates solving equations. The essential step is always adding and subtracting (b/(2a))² or (half the coefficient of x)² within the bracket. Remember the interplay between the coefficient a and the added square to avoid algebraic slips.
配方法将二次式转化为能揭示其几何性质且便于求解方程的形式。关键步骤始终是在括号内加上并减去 (b/(2a))² 或 (x 系数的一半)²。牢记系数 a 与所加平方项之间的关联，以避免代数错误。
- Always write the final answer as a(x + p)² + q, with p and q in simplest form.
- 最后答案总写成 a(x + p)² + q，p 和 q 应为最简形式。
- Use the method to find the vertex quickly: (−p, q).
- 用此方法快速求顶点：(−p, q)。
- For solving equations, after isolating the square, apply the square root with a ± sign.
- 解方程时，孤立平方项后，开平方根要带 ± 号。
- Practice deriving the quadratic formula to strengthen understanding of the discriminant.
- 练习推导求根公式以加深对判别式的理解。
- When dealing with inequalities, always sketch the graph using the completed square form to avoid sign errors.
- 处理不等式时，始终利用配制式画出草图以避免符号错误。
Mastery of completing the square underpins many areas of A-Level Maths, from coordinate geometry to calculus optimizations. Keep practicing with varied coefficients, and you will find it becomes second nature.
掌握配方法是 A-Level 数学许多领域的基础，从坐标几何到微积分中的最优化问题。用不同系数多加练习，你将发现它变得游刃有余。
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