📚 Exercise 1C.1: Completing the Square | 练习 1C.1：配方法
Completing the square is a core algebraic technique that rewrites a quadratic expression of the form ax² + bx + c into the form a(x − h)² + k. This transformation unlocks the ability to solve quadratic equations, analyse parabolas and determine maximum or minimum values with ease. Exercise 1C.1 in the IB Mathematics course provides the crucial practice needed to master this skill, laying the groundwork for many topics across analysis and calculus.
配方法是一项核心代数技巧，能够将形如 ax² + bx + c 的二次式改写为 a(x − h)² + k 的形式。这种变换使我们能够轻松地求解二次方程、分析抛物线并确定最大值或最小值。IB 数学课程中的练习 1C.1 通过大量针对性训练，帮助学生扎实掌握这一技能，为分析与微积分等众多后续内容奠定基础。
1. Recognising a Perfect Square Trinomial | 识别完全平方三项式
The entire process of completing the square relies on spotting patterns that create perfect square trinomials such as x² + 2dx + d², which factorises neatly into (x + d)².
整个配方法过程都依赖于识别能构成完全平方三项式的模式，例如 x² + 2dx + d²，它可以简洁地因式分解为 (x + d)²。
If you expand (x + p)² you always obtain x² + 2px + p². Notice that the coefficient of x is twice the number p, and the constant term is p squared.
如果你将 (x + p)² 展开，总会得到 x² + 2px + p²。请留意，x 的系数是数值 p 的两倍，而常数项是 p 的平方。
For example, x² + 8x + 16 is a perfect square because half of 8 is 4, and 4² gives exactly 16. Thus it equals (x + 4)².
例如，x² + 8x + 16 是一个完全平方式，因为 8 的一半是 4，而 4² 恰好等于 16。因此它等于 (x + 4)²。
This observation forms the foundation: to complete the square for x² + bx, we add (b/2)², creating a trinomial that collapses into (x + b/2)².
这一观察构成了配方法的基础：要对 x² + bx 进行配方，我们添加 (b/2)²，从而构造出一个可以合并为 (x + b/2)² 的三项式。
2. The Core Formula for x² + bx + c | 针对 x² + bx + c 的核心公式
completed-square form is built around the identity x² + bx + c = (x + b/2)² − (b/2)² + c.
当 x² 的系数为 1 时，配方式围绕着恒等式 x² + bx + c = (x + b/2)² − (b/2)² + c 建立。
We essentially take half of the x‑coefficient b, square it, then both add and subtract it to keep the expression balanced.
我们实质上是取 x 项系数 b 的一半，将其平方，然后同时加上并减去这个平方值，以保持表达式的平衡。
x² + bx + c = [x² + bx + (b/2)²] − (b/2)² + c = (x + b/2)² − (b/2)² + c
x² + bx + c = [x² + bx + (b/2)²] − (b/2)² + c = (x + b/2)² − (b/2)² + c
The inner bracket becomes a perfect square, while the remaining constants combine into a single number k.
内层括号变成一个完全平方，而剩余的常数则合并为单一数值 k。
3. Step‑by‑Step Procedure for a = 1 | 当 a = 1 时的分步操作流程
Begin with the quadratic in descending order: x² + bx + c. Make sure the expression is set equal to zero if you are solving an equation, or simply rewrite it as an expression if you are converting forms.
从按降幂排列的二次式开始：x² + bx + c。如果正在解方程，请确保式子等于零；如果只是转换形式，则直接改写表达式即可。
Step 1: Identify the coefficient b. Take half of it: b/2.
步骤1：确定系数 b。取其一半：b/2。
Step 2: Square the half‑coefficient: (b/2)².
步骤2：将一半系数平方：(b/2)²。
Step 3: Insert + (b/2)² − (b/2)² right after the x term. This adds zero overall, preserving the expression’s value.
步骤3：在 x 项之后紧接着插入 + (b/2)² − (b/2)²。这相当于加上零，不改变表达式的值。
rfect square: (x + b/2)². Then simplify the remaining constants to find k: k = c − (b/2)².
步骤4：将前三项组合成完全平方式：(x + b/2)²。然后化简余下的常数项，求出 k：k = c − (b/2)²。
Step 5: Write the final form: (x + b/2)² + k.
步骤5：写出最终形式：(x + b/2)² + k。
4. Worked Example with a = 1 | 当 a = 1 时的完整示例
Rewrite x² − 10x + 24 in completed square form.
将 x² − 10x + 24 改写为配方式。
Here b = −10, so half of b is −5. Squaring this gives (−5)² = 25.
这里 b = −10，因此 b 的一半为 −5。将其平方得到 (−5)² = 25。
Inserting the square term: x² − 10x + 25 − 25 + 24.
插入平方项：x² − 10x + 25 − 25 + 24。
The first three terms form (x − 5)², and the constants simplify: −25 + 24 = −1.
前三项构成 (x − 5)²，常数部分化简：−25 + 24 = −1。
∴ x² − 10x + 24 = (x − 5)² − 1
∴ x² − 10x + 24 = (x − 5)² − 1
This reveals the vertex of the parabola y = x² − 10x + 24 is at (5, −1).
这直接揭示出抛物线 y = x² − 10x + 24 的顶点位于 (5, −1)。
5. Handling the Case a ≠ 1 | 处理二次项系数 a ≠ 1 的情况
When the leading coefficient is not 1, you must first factor a out of the x² and x terms before applying the method inside the bracket.
当首项系数 a 不为 1 时，你必须先将 a 从 x² 和 x 项中提出来，再在括号内部运用配方法。
For example, to rewrite 3x² + 12x + 5, start by writing 3[x² + 4x] + 5. Now complete the square on x² + 4x inside the bracket.
例如，要改写 3x² + 12x + 5，先写成 3[x² + 4x] + 5。然后在括号内对 x² + 4x 进行配方。
Half of 4 is 2, and 2² = 4. So x² + 4x becomes (x + 2)² − 4.
4 的一半是 2，2² = 4。因此 x² + 4x 变为 (x + 2)² − 4。
the expression: 3[(x + 2)² − 4] + 5. Expand carefully: 3(x + 2)² − 12 + 5 = 3(x + 2)² − 7.
替换回原式：3[(x + 2)² − 4] + 5。小心展开：3(x + 2)² − 12 + 5 = 3(x + 2)² − 7。
Always multiply the subtracted squared term by the factored a before combining with the external constant.
务必先将减去的平方项乘以提出的 a，再与外部的常数项合并。
6. Solving Quadratic Equations via Completing the Square | 利用配方法解二次方程
To solve x² + 6x − 7 = 0, first complete the square on the left side.
要解方程 x² + 6x − 7 = 0，首先对左边进行配方。
of 6 is 3; 3² = 9. So x² + 6x = (x + 3)² − 9. Substituting back gives (x + 3)² − 9 − 7 = 0 ⇒ (x + 3)² − 16 = 0.
6 的一半是 3；3² = 9。因此 x² + 6x = (x + 3)² − 9。代回得到 (x + 3)² − 9 − 7 = 0 ⇒ (x + 3)² − 16 = 0。
Now isolate the square: (x + 3)² = 16. Take square roots: x + 3 = ±4.
现在分离平方项：(x + 3)² = 16。两边开平方：x + 3 = ±4。
ving x = 1 or x = −7. This method works even when the quadratic does not factorise nicely.
因此 x = −3 ± 4，得到 x = 1 或 x = −7。即使二次式无法简便因式分解，这种方法也同样有效。
For equations with a ≠ 1, factor the leading coefficient out of the variable terms first, then complete the square inside.
对于 a ≠ 1 的方程，首先将首项系数从含变量的项中提出，再在括号内配方。
7. Locating the Vertex of a Parabola | 定位抛物线的顶点
e completed square form y = a(x − h)² + k immediately gives the vertex coordinates (h, k).
配方式 y = a(x − h)² + k 能直接给出顶点坐标 (h, k)。
Carefully note the sign inside the bracket: (x − h)² means h is positive when the bracket reads x − h. If you have (x + 3)², the h value is −3, so the vertex is at (−3, k).
请注意括号内的符号：(x − h)² 表示当括号内为 x − h 时 h 为正。若得到 (x + 3)²，则 h 值为 −3，因此顶点位于 (−3, k)。
rtex is (4, 5). The coefficient a = 2 tells us the parabola is narrower and opens upwards.
对于表达式 2(x − 4)² + 5，顶点为 (4, 5)。系数 a = 2 表示抛物线更窄且开口向上。
Being able to extract the vertex swiftly is essential for sketching graphs and solving optimisation problems in IB exam papers.
能够快速提取顶点信息对于绘制图形和解答 IB 试卷中的优化问题至关重要。
8. Maximum and Minimum Values of Quadratic Functions | 二次函数的最大值与最小值
when a > 0, (x − h)² has a minimum of 0, so the function attains a minimum of k at x = h.
由于平方项总是不小于 0，配方式可以直接揭示极值：当 a > 0 时，(x − h)² 的最小值为 0，因此函数在 x = h 处取得最小值 k。
f a < 0, the function opens downwards and the maximum value is k, also occurring at x = h.
反之，若 a < 0，函数开口向下，最大值为 k，同样在 x = h 处取得。
height is modelled by h = −5t² + 20t + 2. Complete the square to find the maximum height.
考虑一个真实的 IB 题目：一个球的高度用 h = −5t² + 20t + 2 建模。通过配方法求最大高度。
becomes (t − 2)² − 4. So h = −5[(t − 2)² − 4] + 2 = −5(t − 2)² + 20 + 2 = −5(t − 2)² + 22.
从前两项提出 −5：h = −5[t² − 4t] + 2。括号内 t² − 4t 变为 (t − 2)² − 4。于是 h = −5[(t − 2)² − 4] + 2 = −5(t − 2)² + 20 + 2 = −5(t − 2)² + 22。
metres, occurring at t = 2 seconds. This technique appears frequently in IB applications.
最大高度为 22 米，出现在 t = 2 秒时。这一技巧在 IB 应用题中频繁出现。
9. Using Completing the Square to Find Axis of Symmetry and Roots | 用配方法求对称轴与根
ph y = ax² + bx + c is the vertical line x = h, where h comes from the form a(x − h)² + k.
函数 y = ax² + bx + c 图像的对称轴是垂直线 x = h，其中 h 来自配方式 a(x − h)² + k。
try: for y = x² − 8x + 15, completing the square gives (x − 4)² − 1, so the axis is x = 4.
即便不需要完全解方程，你也能快速读出对称轴：对于 y = x² − 8x + 15，配方后得到 (x − 4)² − 1，因此对称轴为 x = 4。
The x‑ lving, which is often easier after completing the square than using the quadratic formula.
x 轴截距（根）可通过令 a(x − h)² + k = 0 并求解得到，配方后求解往往比使用求根公式更简便。
10. Common Pitfalls and How to Steer Clear | 常见陷阱与规避方法
Mistake 1: Forgetting to divide the entire x‑coefficient by 2. Remember, the bracket always involves (x + b/2)², not (x + b)².
错误1：忘记将 x 项系数整体除以 2。请记住，括号内总是 (x + b/2)²，而非 (x + b)²。
Mistake 2: Mishandling the minus sign when subtracting the squared term. After inserting + (b/2)² − (b/2)², you must subtract (b/2)² when combining constants.
错误2：在减去平方项时符号处理不当。插入 + (b/2)² − (b/2)² 后，合并常数时必须减去 (b/2)²。
Mistake 3: When a ≠ 1, forgetting to multiply the subtracted square by a, leading to an incorrect constant k. Always expand carefully.
错误3：当 a ≠ 1 时，忘记将减去的平方项乘以 a，导致常数 k 错误。展开时务必小心。
ng, isolate the squared term first, then take square roots, handling both ± possibilities.
错误4：求解前未将方程设为零。求解时应先分离平方项，然后开平方，并处理好 ± 的两种可能。
Pausing to check your answer by expanding the completed square form back to a general form is an excellent habit in IB exams.
在 IB 考试中，养成通过将配方式展开回一般形式来检验答案的习惯，是非常好的做法。
11. Linking Completing the Square to the Quadratic Formula | 配方法与求根公式的联系
quadratic ax² + bx + c = 0. Understanding this derivation deepens your algebraic insight.
求根公式实际上是通过对一般二次方程 ax² + bx + c = 0 配方推导出来的。理解这一推导过程能够加深你的代数洞察力。
(b/2a)² yields (x + b/2a)² = (b² − 4ac)/(4a²). Taking square roots then produces the well‑known formula x = [−b ± √b² − 4ac]/(2a).
从 x² + (b/a)x = −c/a 开始，加上 (b/2a)² 得到 (x + b/2a)² = (b² − 4ac)/(4a²)。开平方后便得到著名的公式 x = [−b ± √b² − 4ac]/(2a)。
Seeing this connection helps you remember why completing the square is a universal method for solving any quadratic equation.
了解这种联系有助于你记住为什么配方法是求解任何二次方程的通用方法。
12. Summary and Final Tips for Exercise 1C.1 | 总结与练习 1C.1 的最后建议
Completing the square transforms a quadratic into a form that instantly shows the vertex, axis of symmetry, extreme value and roots.
配方法将二次式转化为能直接显示顶点、对称轴、极值和根的形式。
Practise the routine until it becomes automatic: halve the x‑coefficient, square it, add and subtract, factor the perfect square, and tidy the constants.
反复练习直到流程自动化：将 x 系数减半、平方、加上又减去、因式分解为完全平方、并整理常数。
When you encounter Exercise 1C.1 items, vary your practice between pure number examples, worded optimisation problems and questions asking for graph sketches using vertex form.
当你练习 1C.1 的题目时，要混合纯数字例题、文字优化题以及要求利用顶点式画图的题目。
Mastering this technique will pay dividends not only in the quadratic functions topic but also in calculus, where rewriting expressions becomes essential for integration and curve analysis.
掌握这项技巧不仅会在二次函数专题中带来回报，在微积分中，改写表达式对于积分和曲线分析同样至关重要。
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