📚 Completing the Square: Transformation and Applications | 配方法：完全平方的变换与应用
Completing the square is one of the most powerful algebraic techniques in A-Level mathematics. It allows us to rewrite a quadratic expression in a form that reveals its key properties, including the vertex of its graph, the maximum or minimum value, and the nature of its roots.
配方法是A-Level数学中最强大的代数技巧之一。它使我们能够将二次表达式改写为一种揭示其关键性质的形式，包括图像的顶点、最大值或最小值以及根的性质。
1. What Is Completing the Square? | 什么是配方法
Completing the square is the process of rewriting a quadratic expression of the form ax² + bx + c into the form a(x + p)² + q, where p and q are constants. This form is often called the vertex form because it directly reveals the coordinates of the turning point of the parabola.
配方法是将形式为 ax² + bx + c 的二次表达式改写为 a(x + p)² + q 的过程，其中 p 和 q 是常数。这种形式通常称为顶点式，因为它直接揭示了抛物线拐点的坐标。
x + p². Any quadratic expression x² + bx can be made into a perfect square by adding and subtracting (b/2)².
关键洞察在于 (x + p)² = x² + 2px + p²。任何二次表达式 x² + bx 都可以通过加上并减去 (b/2)² 来构造成完全平方。
2. The Basic Technique | 基本技巧
e technique is straightforward. Take x² + bx + c. Half of b, squared, is (b/2)². We add and subtract this quantity:
对于首项系数为1的二次式（即 a = 1），技巧很直接。取 x² + bx + c。b 的一半的平方是 (b/2)²。我们加上并减去这个量：
x² + bx + c = (x + b/2)² − (b/2)² + c
For example, consider x² + 6x + 5. Here b = 6, so (b/2)² = 9. We obtain:
例如，考虑 x² + 6x + 5。这里 b = 6，所以 (b/2)² = 9。我们得到：
x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4
When the coefficient of x² is not 1, we factor it out first. For 2x² + 8x + 3:
当 x² 的系数不是1时，我们首先将其提取出来。对于 2x² + 8x + 3：
2x² + 8x + 3 = 2(x² + 4x) + 3 = 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5
3. Solving Quadratic Equations | 解二次方程
tions, particularly when the equation does not factorise easily. Consider x² + 6x + 5 = 0:
配方法为解二次方程提供了一种可靠的方法，特别是当方程不易因式分解时。考虑 x² + 6x + 5 = 0：
(x + 3)² − 4 = 0 → (x + 3)² = 4 → x + 3 = ±2 → x = −1 or x = −5
quations, even those with irrational or complex roots. For example, solve x² + 4x − 1 = 0:
这种方法的美妙之处在于它适用于所有二次方程，甚至是那些具有无理数或复数根的方程。例如，解 x² + 4x − 1 = 0：
(x + 2)² − 4 − 1 = 0 → (x + 2)² = 5 → x = −2 ± √5
4. Sketching Quadratic Graphs | 绘制二次函数图像
Once a quadratic is written in vertex form a(x + p)² + q, we can immediately identify:
一旦二次式被写成顶点式 a(x + p)² + q，我们可以立即识别出：
The turning point is at (−p, q) | 拐点位于 (−p, q)
If a > 0, the parabola opens upward and has a minimum at this point | 如果 a > 0，抛物线开口向上，在该点取得最小值
If a < 0, the parabola opens downward and has a maximum at this point | 如果 a < 0，抛物线开口向下，在该点取得最大值
ne of symmetry is x = −p. This is the vertical line that passes through the turning point.
对称轴是 x = −p。这是穿过拐点的竖直线。
graph of y = 2(x + 2)² − 5 has its vertex at (−2, −5) and opens upward because a = 2 > 0.
例如，y = 2(x + 2)² − 5 的图像顶点在 (−2, −5)，且由于 a = 2 > 0 而开口向上。
5. Finding Maximum and Minimum Values | 求最大值与最小值
s fact allows us to determine the extreme values of a quadratic function without calculus.
因为任何实数的平方总是非负的，所以表达式 (x + p)² ≥ 0 对所有实数 x 都成立。这一事实使我们能够在不用微积分的情况下确定二次函数的极值。
(x + 2)² ≥ 0, we have f(x) ≥ −5 for all x. The minimum value is −5, achieved when x = −2.
考虑 f(x) = 2(x + 2)² − 5。由于 (x + 2)² ≥ 0，我们得到 f(x) ≥ −5 对所有 x 成立。最小值是 −5，在 x = −2 时取得。
3(x − 1)² + 7, the maximum value is 7, achieved when x = 1, because −3(x − 1)² ≤ 0 always.
类似地，对于 f(x) = −3(x − 1)² + 7，最大值是 7，在 x = 1 时取得，因为 −3(x − 1)² ≤ 0 恒成立。
This technique has practical applications in optimisation problems, such as finding the maximum area of a rectangle with a fixed perimeter, or the minimum cost of a production process.
这种技术在实际优化问题中有着广泛的应用，例如在固定周长下求矩形的最大面积，或求生产过程的最小成本。
6. Deriving the Quadratic Formula | 推导二次求根公式
ula is built. Starting with the general quadratic equation ax² + bx + c = 0 (where a ≠ 0):
配方法是构建二次求根公式的代数基础。从一般二次方程 ax² + bx + c = 0（其中 a ≠ 0）出发：
ax² + bx + c = 0 → a[x² + (b/a)x] + c = 0
a[(x + b/2a)² − b²/4a²] + c = 0
(x + b/2a)² = (b² − 4ac)/4a²
x = (−b ± √b² − 4ac)/2a
= b² − 4ac, emerges naturally from this derivation. It determines the nature of the roots:
判别式 Δ = b² − 4ac 从这一推导中自然而然地出现。它决定了根的性质：
Δ > 0: two distinct real roots | Δ > 0：两个不同的实数根
Δ = 0: one repeated real root | Δ = 0：一个重复的实数根
Δ < 0: no real roots (two complex roots) | Δ < 0：没有实数根（两个复数根）
7. Solving Quadratic Inequalities | 解二次不等式
Completing the square is also invaluable when solving quadratic inequalities. For example, solve x² + 6x + 5 > 0:
配方法在解二次不等式时也非常有价值。例如，解 x² + 6x + 5 > 0：
(x + 3)² − 4 > 0 → (x + 3)² > 4 → x + 3 > 2 or x + 3 < −2
x > −1 or x < −5
For inequalities where the discriminant is negative, completing the square reveals the answer directly. Consider x² + 2x + 3 > 0:
对于判别式为负的不等式，配方法直接揭示了答案。考虑 x² + 2x + 3 > 0：
(x + 1)² + 2 > 0
Since (x + 1)² ≥ 0, we know (x + 1
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