📚 Completing the Square: Quadratic Equations and Turning Points | 配方法：二次方程与顶点
IGCSE Mathematics frequently tests quadratics in three forms: expanded, factorised and completed square. Completing the square is a core algebraic skill that allows you to solve quadratic equations, locate turning points and sketch parabolas efficiently, even when factorising is difficult or impossible.
IGCSE 数学经常以三种形式考查二次函数：展开式、因式分解式和配方式。配方法是一项核心代数技能，它让你能够高效地解二次方程、确定顶点并绘制抛物线草图，即使在因式分解困难或不可能的情况下也能使用。
1. What Is Completing the Square? | 什么是配方法？
Completing the square is a technique that rewrites a quadratic expression ax² + bx + c into the form a(x + p)² + q, where p and q are constants. This new form is often called the completed-square form or vertex form.
配方法是一种将二次式 ax² + bx + c 改写为 a(x + p)² + q 的技巧，其中 p 和 q 为常数。这种新形式通常称为配方式或顶点式。
ertex of the parabola and helps solve the equation x² - 6x + 4 = 0 by taking square roots.
例如，x² - 6x + 4 可以写成 (x - 3)² - 5。配方式能直接显示抛物线的顶点，并有助于通过开平方来解方程 x² - 6x + 4 = 0。
In the IGCSE syllabus, you will often see instructions such as 'write in the form (x + p)² + q' or 'express in the form a(x + h)² + k'. Completing the square is the method behind these questions.
在 IGCSE 大纲中，你经常会看到诸如“写成 (x + p)² + q 的形式”或“表示为 a(x + h)² + k 的形式”的要求。配方法正是解答这些题目的方法。
2. The General Idea: Making a Perfect Square Trinomial | 核心思想：构造完全平方三项式
A perfect square trinomial is an expression such as x² + 2dx + d², which equals (x + d)². The central idea of completing the square is to add and subtract the correct constant d² so that the value of the original expression does not change.
完全平方三项式是指 x² + 2dx + d² 这样的式子，它等于 (x + d)²。配方法的核心思想是加上并减去正确的常数 d²，使原式的值保持不变。
- For x² + 10x, half of 10 is 5, so we add and subtract 5² = 25.
- 对于 x² + 10x，10 的一半是 5，因此我们加上并减去 5² = 25。
- Then x² + 10x = (x² + 10x + 25) - 25 = (x + 5)² - 25.
- 于是 x² + 10x = (x² + 10x + 25) - 25 = (x + 5)² - 25。
This technique works because the first three terms in the bracket form a perfect square, while the subtracted constant keeps the expression equivalent to the original.
这一技巧之所以有效，是因为括号内的前三项构成完全平方，而减去的常数使表达式与原式保持恒等。
x² + bx = (x + b/2)² - (b/2)²
Memorising this general result will help you complete the square quickly and accurately in the exam.
记住这个一般结果能帮助你在考试中快速而准确地配方。
3. Completing the Square for x² + bx | 对 x² + bx 配方
When the coefficient of x² is 1, follow three steps: take half of the x-coefficient, square it, then add and subtract that square inside the expression.
当 x² 的系数为 1 时，遵循三个步骤：取 x 项系数的一半，将其平方，然后在表达式中加上并减去该平方数。
Worked example: complete the square for x² - 10x + 12.
例题：对 x² - 10x + 12 配方。
- The coefficient of x is -10, half of it is -5, and (-5)² = 25.
- x 的系数是 -10，它的一半是 -5，而 (-5)² = 25。
- So x² - 10x = (x - 5)² - 25.
- 因此 x² - 10x = (x - 5)² - 25。
Now add the constant term 12 back: x² - 10x + 12 = (x - 5)² - 25 + 12 = (x - 5)² - 13.
现在把常数项 12 加回来：x² - 10x + 12 = (x - 5)² - 25 + 12 = (x - 5)² - 13。
Remember to leave the original constant outside the square until the final simplification. This avoids a very common sign error.
记住在最终化简之前把原来的常数项留在平方外面，这样可以避免非常常见的符号错误。
4. Handling a ≠ 1 | 处理二次项系数不为 1
When the quadratic has a leading coefficient other than 1, such as 2x² + 8x + 5, you must first factor the leading coefficient only from the x² and x terms, not from the constant.
当二次式的首项系数不为 1 时，例如 2x² + 8x + 5，你必须首先只从 x² 和 x 项中提出首项系数，而不是从常数项中提出。
gives 2(x² + 4x) + 5. Then complete the square inside the bracket: x² + 4x = (x + 2)² - 4.
得到 2(x² + 4x) + 5。然后在括号内配方：x² + 4x = (x + 2)² - 4。
Now substitute back and multiply through by the leading coefficient:
现在代回并乘以首项系数：
2[(x + 2)² - 4] + 5 = 2(x + 2)² - 8 + 5 = 2(x + 2)² - 3
Always keep the constant term outside the square until the final step. The completed-square form is 2(x + 2)² - 3.
始终将常数项留在平方外面直到最后一步。配方式为 2(x + 2)² - 3。
rm tells us the vertex is (-2, -3) and the parabola is narrower than y = x² because a = 2.
这一形式告诉我们顶点是 (-2, -3)，且因为 a = 2，抛物线比 y = x² 更窄。
5. Solving Quadratic Equations by Completing the Square | 用配方法解二次方程
To solve a quadratic equation by completing the square, first move the constant term to the right-hand side, then complete the square on the left.
要用配方法解二次方程，首先将常数项移到右边，然后在左边完成配方。
Example: solve x² + 6x - 7 = 0.
例题：解 x² + 6x - 7 = 0。
Step 1: move the constant: x² + 6x = 7.
第一步：移常数项：x² + 6x = 7。
lete the square on the left by adding 9 to both sides: x² + 6x + 9 = 16, so (x + 3)² = 16.
第二步：左边配方，两边同时加 9：x² + 6x + 9 = 16，因此 (x + 3)² = 16。
Step 3: take the square root of both sides: x + 3 = ±4.
第三步：两边开平方：x + 3 = ±4。
- x + 3 = 4 gives x = 1.
- x + 3 = 4 得 x = 1。
- x + 3 = -4 gives x = -7.
- x + 3 = -4 得 x = -7。
the roots are irrational. For example, x² + 2x - 4 = 0 leads to (x + 1)² = 5, so x = -1 ± √5.
即使根是无理数，这种方法仍然有效。例如，x² + 2x - 4 = 0 可化为 (x + 1)² = 5，因此 x = -1 ± √5。
6. Finding the Turning Point (Vertex) | 求抛物线顶点
s is because the squared term is zero at x = -p, giving the minimum or maximum value of y.
在配方式 y = a(x + p)² + q 中，顶点为 (-p, q)。这是因为当 x = -p 时平方项为零，得到 y 的最小值或最大值。
or y = 2(x - 3)² - 5, the vertex is (3, -5). For y = -(x + 4)² + 2, the vertex is (-4, 2).
对于 y = 2(x - 3)² - 5，顶点是 (3, -5)。对于 y = -(x + 4)² + 2，顶点是 (-4, 2)。
If a > 0 the parabola opens upward and the vertex is a minimum point. If a < 0 it opens downward and the vertex is a maximum point.
如果 a > 0，抛物线开口向上，顶点为最小值点；如果 a < 0，开口向下，顶点为最大值点。
| Form | 形式 | Vertex | 顶点 | Direction | 开口方向 |
|---|---|---|
| (x + 2)² - 6 | (-2, -6) | Upwards | 向上 |
