📚 IGCSE Mathematics: Completing the Square (Full Guide) | IGCSE 数学：配方法（完全指南）
Completing the square is a core IGCSE algebra technique. It rewrites a quadratic expression ax² + bx + c into the form a(x + p)² + q, where p and q are constants. The new form makes it much easier to solve quadratic equations and to find the vertex of a parabola.
配方法是 IGCSE 代数中的核心技巧。它把二次式 ax² + bx + c 改写为 a(x + p)² + q 的形式，其中 p 和 q 是常数。新的形式让我们更容易解二次方程，也更容易找到抛物线的顶点。
The method links directly to graph work, maximum and minimum values, and exact solutions in surd form, so it appears in both Cambridge IGCSE Mathematics (0580/0980) and Edexcel IGCSE Mathematics A.
该方法直接关联图像、最大值与最小值，以及根式形式的精确解，因此在剑桥 IGCSE 数学（0580/0980）和爱德思 IGCSE 数学 A 中都会出现。
1. What Is Completing the Square? | 什么是配方法？
Completing the square is a way of turning any quadratic expression into a perfect square plus or minus a constant. In symbols, ax² + bx + c is rewritten as a(x + p)² + q, where a, p and q are constants and a is not zero.
配方法是一种把任意二次式转化为一个完全平方加上或减去一个常数的方法。用符号表示，就是把 ax² + bx + c 改写成 a(x + p)² + q，其中 a、p、q 都是常数，且 a 不等于 0。
The word "completing" refers to adding the missing term that makes part of the expression a perfect square. You are not changing the value of the expression, only its shape, because whatever you add inside the bracket you subtract again outside it.
"配"的意思就是补上缺少的那一项，使表达式的一部分成为完全平方式。你并没有改变式子的值，只改变了它的形状，因为在括号里加上的量，会在括号外减掉。
xpanding (x + 3)² − 7 gives x² + 6x + 9 − 7 = x² + 6x + 2, so the two forms are identical.
例如，x² + 6x + 2 本身不是完全平方，但可以写成 (x + 3)² − 7。展开 (x + 3)² − 7 得到 x² + 6x + 9 − 7 = x² + 6x + 2，所以这两种形式完全等价。
2. Why It Matters: Equations, Vertices and Graphs | 为什么重要：方程、顶点与图像
The completed-square form immediately shows the vertex of a quadratic graph, the maximum or minimum value, and the solutions of the related equation. Those three ideas are examined again and again in IGCSE papers.
完全平方形式能直接显示二次函数图像的顶点、最大值或最小值，以及对应方程的解。这三个要点在 IGCSE 试卷中反复考查。
Solving by factorising only works when the quadratic factorises nicely. Completing the square works for every quadratic, including those whose solutions involve surds, so it is a universal tool.
因式分解法只在二次式能整齐分解时才有效。配方法对任何二次式都适用，包括解中含根式的情况，因此它是通用工具。
It also explains where the quadratic formula comes from. If you complete the square on ax² + bx + c in general, the algebra produces the formula you memorise, which makes the formula much easier to remember and trust.
它还解释了一元二次求根公式的来源。若对一般的 ax² + bx + c 配方，代数推导会自然得出你背诵的那个公式，这会让公式更容易记住，也更容易信赖。
3. The Core Identity Behind the Method | 方法背后的核心恒等式
Everything depends on two expansion identities. Adding p inside the bracket doubles the coefficient of x; squaring p gives the constant term.
一切都依赖两条展开恒等式。括号里加上 p 会使 x 的系数翻倍；把 p 平方则得到常数项。
(x + p)² = x² + 2px + p²
(x − p)² = x² − 2px + p²
half of the coefficient of x, then square it to find the number you must add and subtract.
把 x² + bx 与 x² + 2px 比较，可得 2p = b，所以 p = b ÷ 2。用文字说就是：取 x 系数的一半，再平方，得到需要加上并减去的那个数。
This single rule, "half then square", is the engine of the whole method. If you remember nothing else, remember this.
这条规则"取一半再平方"就是整个方法的发动机。如果别的都记不住，至少记住这一条。
4. Step-by-Step When a = 1 | 当 a = 1 时的分步操作
When the coefficient of x² is 1, the procedure has four short steps. Write them down in order in the exam and you will not lose method marks.
当 x² 的系数是 1 时，整个过程只有四步。考试时按顺序写出来，就不会丢掉方法分。
- Step 1: Write the first two terms and open a bracket, leaving space for the constant.
- Step 2: Take half of the coefficient of x and put it inside the bracket as (x + p)².
- Step 3: Square the number you just used and subtract it.
- Step 4: Add the original constant and simplify.
- 第一步：先写出前两项并打开括号，给常数项留出位置。
- 第二步：取 x 系数的一半，写成 (x + p)²。
- 第三步：把你刚用过的那个数平方，然后减去它。
- 第四步：加上原来的常数项并化简。
A useful check: expanding your final answer must give back the original expression exactly. This takes ten seconds and catches most sign errors.
一个实用的检验方法：展开最终答案，必须完全还原原来的表达式。这只要十秒钟，却能抓住大多数符号错误。
5. Worked Example: x² + 6x + 2 | 例题：x² + 6x + 2
Start with x² + 6x + 2. The coefficient of x is 6, so half of it is 3, and 3² = 9.
从 x² + 6x + 2 开始。x 的系数是 6，一半是 3，而 3² = 9。
x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7
otice the sign: the bracket contains a plus sign because the original x term was positive.
该式现在已化为完全平方形式，其中 p = 3，q = −7。注意符号：括号里是加号，因为原来的 x 项为正。
− 9 + 2 = (x − 3)² − 7. The sign inside the bracket always matches the sign of the x term.
如果表达式是 x² − 6x + 2，括号就写 (x − 3)²，过程为 (x − 3)² − 9 + 2 = (x − 3)² − 7。括号内的符号永远与 x 项的符号一致。
6. When a ≠ 1: Factor Out First | 当 a ≠ 1 时：先提出系数
If the coefficient of x² is not 1, factor it out of the first two terms only, not out of the constant. This is the step where most marks are lost.
如果 x² 的系数不是 1，要把这个系数从前两项中提出来，注意常数项不提。这一步是丢分最多的地方。
erms to get 2(x² + 4x) + 5. Inside the bracket, half of 4 is 2, so x² + 4x = (x + 2)² − 4.
例如，2x² + 8x + 5。从前两项提出 2，得 2(x² + 4x) + 5。括号内 4 的一半是 2，所以 x² + 4x = (x + 2)² − 4。
2x² + 8x + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3
The subtracted 4 is multiplied by the 2 outside the bracket, which is why it becomes −8 before the constant 5 is added. Always multiply before simplifying.
被减去的 4 要乘上括号外的 2，所以它变成 −8，然后才加上常数 5。一定要先乘后化简。
r −x² + 4x + 1, factor out −1 to get −(x² − 4x) + 1 = −[(x − 2)² − 4] + 1 = −(x − 2)² + 5.
当 a 为负数时逻辑相同。对 −x² + 4x + 1，提出 −1，得 −(x² − 4x) + 1 = −[(x − 2)² − 4] + 1 = −(x − 2)² + 5。
7. Solving Quadratic Equations by Completing the Square | 用配方法解二次方程
To solve a quadratic equation, move the constant to the other side after completing the square, then take square roots of both sides. Remember that a square root has both a positive and a negative value.
解二次方程时，先配方，再把常数移到等号另一边，然后对两边开平方。记住平方根有正负两个值。
+ 6x + 2 = 0. Completing the square gives (x + 3)² − 7 = 0, so (x + 3)² = 7 and x + 3 = ±√7.
以 x² + 6x + 2 = 0 为例。配方得 (x + 3)² − 7 = 0，所以 (x + 3)² = 7，x + 3 = ±√7。
x = −3 + √7 or x = −3 − √7
These are exact values. √7 so the two solutions are approximately 0.354 and −5.646 if a decimal answer is requested.
这是精确值。√7 ≈ 2.646，所以若题目要求小数，两个解约为 0.354 和 −5.646。
If a is not 1, divide every term by a first so the equation starts with x², then complete the square as usual. This avoids fractions inside the bracket.
如果 a 不是 1，先把每一项都除以 a，使方程以 x² 开头，再按常规配方。这样可以避免括号内出现分数。
8. Finding the Vertex and Maximum or Minimum Value | 求顶点与最大/最小值
The completed-square form a(x + p)² + q gives the vertex directly as (−p, q). The sign flips: the number inside the bracket is negated to give the x-coordinate.
完全平方形式 a(x + p)² + q 可直接给出顶点 (−p, q)。注意符号相反：括号内的数字取相反数就是顶点的横坐标。
ositive, the parabola opens upwards and −7 is the minimum value of y, occurring at x = −3.
对 y = (x + 3)² − 7，顶点是 (−3, −7)。因为 a = 1 为正，抛物线开口向上，−7 是 y 的最小值，在 x = −3 处取得。
| Form | Sign of a | Vertex | Extreme value |
| a(x + p)² + q | a > 0 | (−p, q) | minimum q |
| a(x + p)² + q | a < 0 | (−p, q) | maximum q |
For 2(x + 2)² − 3 the vertex is (−2, −3) and the minimum value is −3. For −(x − 2)² + 5 the vertex is (2, 5) and the maximum value is 5.
对 2(x + 2)² − 3，顶点是 (−2, −3)，最小值为 −3。对 −(x − 2)² + 5，顶点是 (2, 5)，最大值为 5。
9. The General Formula and the Discriminant | 一般公式与判别式
Applying the method to the general quadratic produces a compact result worth deriving once so you never forget it.
把该方法用在一般二次式上，会得到一个简洁的结果。亲手推导一次，你就再也不会忘。
ax² + bx + c = a(x + b ÷ (2a))² + c − b² ÷ (4a)
and y = c − b² ÷ (4a). This is the same axis of symmetry you may have met as x = −b ÷ 2a.
因此 y = ax² + bx + c 的顶点横坐标为 x = −b ÷ (2a)，纵坐标为 y = c − b² ÷ (4a)。这与你见过的对称轴 x = −b ÷ 2a 是同一个结论。
nd taking square roots gives the quadratic formula. The expression b² − 4ac is the discriminant Δ.
解 a(x + b ÷ (2a))² = b² ÷ (4a) − c，得 (x + b ÷ (2a))² = (b² − 4ac) ÷ (4a²)，开平方就得到求根公式。式子 b² − 4ac 就是判别式 Δ。
ed root at the vertex; if Δ < 0 there are no real roots, and completing the square shows the curve never reaches the x-axis.
若 Δ > 0，有两个不同的实根；若 Δ = 0，恰有一个重根，就在顶点处；若 Δ < 0，则没有实根，配方结果也说明曲线永远不会碰到 x 轴。
10. Geometric Meaning and Graph Transformations | 几何意义与图像变换
of y = (x + p)² + q is the graph of y = x² shifted left or right by p and up or down by q.
配方法讲的是一个平移的故事。y = (x + p)² + q 的图像，就是 y = x² 的图像左右平移 p、上下平移 q 后得到的。
= (x + 3)² − 7 the curve moves 3 units left and 7 units down, giving a vertex at (−3, −7).
括号里是加号时，曲线向左移动，这一点让很多学生吃惊。在 y = (x + 3)² − 7 中，曲线左移 3 个单位、下移 7 个单位，顶点落在 (−3, −7)。
side down. The completed-square form separates the shift (p and q) from the stretch (a) cleanly.
若 a 不是 1，形状会被拉伸。系数为 2 时抛物线比 y = x² 更窄；a 为负则上下翻转。完全平方形式把平移（p
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