📚 Example 4.3.1: Completing the Square with a Leading Coefficient | 例4.3.1：首项系数不为1的配方法
In this AQA A-Level Pure Mathematics worked example, we complete the square for the quadratic expression 2x² − 8x + 5 and use the completed-square form to find the vertex, solve the equation and sketch the graph.
在这道 AQA A-Level 纯数例题中，我们将二次式 2x² − 8x + 5 配成完全平方形式，并利用配方式求顶点、解方程并绘制图像。
1. The Problem Statement | 题目陈述
Example 4.3.1 asks us to express the quadratic 2x² − 8x + 5 in the form a(x + b)² + c, where a, b and c are constants.
例4.3.1 要求我们把二次式 2x² − 8x + 5 写成 a(x + b)² + c 的形式，其中 a、b、c 为常数。
This is a standard AQA Pure Mathematics skill because it reveals the vertex of a parabola and is frequently tested in Paper 1.
这是 AQA 纯数的基础技能，因为它能够揭示抛物线的顶点，并且在 Paper 1 中经常出现。
2x² − 8x + 5 = a(x + b)² + c
2. Why Completing the Square Matters | 为什么配方法重要
Completing the square transforms a quadratic expression into a form that makes its key features immediately visible: the vertex, the axis of symmetry and the minimum or maximum value.
配方法可以把二次式转化为一种能使关键特征一目了然的形式：顶点、对称轴以及最小值或最大值。
It is also used to derive the quadratic formula and to solve quadratic equations that do not factorise neatly.
它还可用于推导求根公式，并用来解不能直接因式分解的二次方程。
- Find the coordinates of the vertex without using calculus.
- Solve quadratic equations in exact surd form.
- Determine whether a quadratic has a minimum or maximum value.
用途如下：
- 无需使用微积分即可求出顶点坐标。
- 以精确根式形式解二次方程。
- 判断二次函数取得最小值还是最大值。
3. Step 1 – Factor Out the Leading Coefficient | 第1步：提取首项系数
Since the coefficient of x² is 2, we factor 2 out of the first two terms only. Do not factor it from the constant term.
由于 x² 的系数是 2，我们只从前两项中提取公因数 2。不要从常数项中提取。
2x² − 8x + 5 = 2(x² − 4x) + 5
This step ensures that the bracket begins with x² with coefficient 1, so the standard completing-the-square method can be applied inside.
这一步确保括号内以系数为 1 的 x² 开头，因此可以在括号内使用标准的配方法。
4. Step 2 – Complete the Square Inside the Bracket | 第2步：在括号内配方
Inside the bracket, the coefficient of x is −4. Take half of this coefficient to get −2, then square it to obtain 4.
在括号内，x 的系数是 −4。取该系数的一半，得到 −2；再将其平方，得到 4。
Add and subtract this square inside the bracket. The first three terms then form a perfect square.
在括号内同时加上并减去这个平方数。前三项便构成一个完全平方。
2(x² − 4x + 4 − 4) + 5 = 2[(x − 2)² − 4] + 5
Here (x − 2)² is the completed square, because (x − 2)² expands to x² − 4x + 4.
这里 (x − 2)² 就是配方后的完全平方，因为 (x − 2)² 展开后为 x² − 4x + 4。
5. Step 3 – Expand and Simplify the Constant | 第3步：展开并化简常数项
Multiply the outer 2 through the bracket: 2(x − 2)² − 8. Then add the original constant 5 to obtain −3.
将外面的 2 乘进括号：2(x − 2)² − 8。然后加上原来的常数 5，得到 −3。
2(x − 2)² − 8 + 5 = 2(x − 2)² − 3
The completed-square form is therefore 2(x − 2)² − 3, so a = 2, b = −2 and c = −3.
因此配方式为 2(x − 2)² − 3，所以 a = 2，b = −2，c = −3。
6. Verification by Expanding Back | 通过展开回代验证
To check the result, expand 2(x − 2)² − 3 and confirm that it equals the original expression.
为了检验结果，展开 2(x − 2)² − 3，并确认它等于原式。
² − 4x + 4. Then multiply by 2 to get 2x² − 8x + 8. Finally subtract 3 to obtain 2x² − 8x + 5.
先平方二项式：(x − 2)² = x² − 4x + 4。然后乘以 2，得 2x² − 8x + 8。最后减去 3，得 2x² − 8x + 5。
2(x − 2)² − 3 = 2x² − 8x + 8 − 3 = 2x² − 8x + 5
This matches the original quadratic, so the completed-square form is correct.
这与原二次式一致，因此配方式正确。
7. Using the Completed Square to Find the Vertex | 利用配方式求顶点
x − h)² + k has its vertex at (h, k). Here the form is 2(x − 2)² − 3, so h = 2 and k = −3.
形如 a(x − h)² + k 的二次函数顶点为 (h, k)。这里的形式是 2(x − 2)² − 3，因此 h = 2，k = −3。
Since a = 2 is positive, the parabola opens upwards and this vertex is the minimum point.
因此顶点为 (2, −3)。由于 a = 2 为正，抛物线开口向上，该顶点为最小值点。
| Vertex | (2, −3) |
| Axis of symmetry | x = 2 |
| Nature of turning point | Minimum, since a > 0 |
中文对照如下：
| 顶点 | (2, −3) |
| 对称轴 | x = 2 |
| 驻点类型 | 最小值，因为 a > 0 |
8. Solving the Quadratic Equation | 解二次方程
To solve 2x² − 8x + 5 = 0, use the completed-square form: 2(x − 2)² − 3 = 0.
要解 2x² − 8x + 5 = 0，使用配方式：2(x − 2)² − 3 = 0。
Add 3 to both sides and divide by 2: (x − 2)² = 3/2.
两边同时加 3 并除以 2：(x − 2)² = 3/2。
Take square roots of both sides: x − 2 = ±√3/2.
两边开平方：x − 2 = ±√3/2。
x = 2 ± √3/2 = 2 ± √6 / 2
The exact roots are x = 2 + √6 / 2 and x = 2 − √6 / 2. Numerically, x ≈ 3.225 or x ≈ 0.775.
精确根为 x = 2 + √6 / 2 和 x = 2 − √6 / 2。数值上，x ≈ 3.225 或 x ≈ 0.775。
9. Sketching the Graph | 绘制图像
Use the completed-square information to sketch the graph of y = 2x² − 8x + 5.
利用配方得到的信息绘制 y = 2x² − 8x + 5 的图像。
s of symmetry is the vertical line x = 2. The y-intercept occurs when x = 0, giving y = 5.
标出最小值顶点 (2, −3)。对称轴为竖直线 x = 2。y 截距在 x = 0 处，即 y = 5。
The roots found above are x = 2 ± √6 / 2, so the curve crosses the x-axis at those two points.
上面求出的根为 x = 2 ± √6 / 2，因此曲线在这两个点与 x 轴相交。
- Vertex: (2, −3)
- Axis: x = 2
- y-intercept: (0, 5)
- Roots: x = 2 ± √6 / 2
- Shape: upward-opening parabola
要点如下：
- 顶点：(2, −3)
- 对称轴：x = 2
- y 截距：(0, 5)
- 根：x = 2 ± √6 / 2
- 形状：开口向上的抛物线
10. Common Mistakes and AQA Exam Tips | 常见错误与 AQA 备考建议
One common mistake is factorising the leading coefficient from the constant term. Remember that only the first two terms should be rewritten inside the bracket.
一个常见错误是从常数项中提取首项系数。请记住，只有前两项应被放入括号内。
Another error is forgetting to multiply the subtracted square by the outer coefficient after completing the square inside the bracket.
另一个错误是在括号内完成平方后，忘记将减去的平方数乘以外面的系数。
In AQA exams, method marks are awarded for showing the correct structure, even if the final simplification is incorrect. Write each step clearly.
在 AQA 考试中，即使最终化简有误，只要结构正确也可获得方法分。请清晰地写出每一步。
- Always expand your final form to check it matches the original.
- State the vertex in coordinate form, not just as x = value.
- Give exact roots unless the question asks for a decimal approximation.
备考建议：
- 始终展开最终形式以检验是否与原式一致。
- 用坐标形式写出顶点，而不只是 x 值。
- 除非题目要求近似小数，否则给出精确根。
11. Practice Exercise | 练习
Try expressing 3x² + 12x + 7 in the form a(x + b)² + c, then state its vertex and axis of symmetry.
尝试把 3x² + 12x + 7 写成 a(x + b)² + c 的形式，然后写出它的顶点和对称轴。
mplete the square inside: 3[(x + 2)² − 4] + 7. Expand: 3(x + 2)² − 12 + 7 = 3(x + 2)² − 5.
答案：先从前两项提取 3：3(x² + 4x) + 7。在括号内配方：3[(x + 2)² − 4] + 7。展开：3(x + 2)² − 12 + 7 = 3(x + 2)² − 5。
3x² + 12x + 7 = 3(x + 2)² − 5
The vertex is (−2, −5) and the axis of symmetry is x = −2.
顶点为 (−2, −5)，对称轴为 x = −2。
12. Summary | 小结
For a quadratic ax² + bx + c, factor a from the first two terms, complete the square inside the bracket, then simplify the constant term.
对于二次式 ax² + bx + c，先从前两项提取系数 a，在括号内完成平方，再化简常数项。
The final form a(x + b)² + c gives the vertex (−b, c) and allows the equation to be solved by rearranging and taking square roots.
最终形式 a(x + b)² + c 给出的顶点为 (−b, c)，并且可以通过移项和开平方来解方程。
2x² − 8x + 5 = 2(x − 2)² − 3
This worked example shows the full AQA method: complete, verify, interpret the vertex, solve and sketch.
本例题展示了 AQA 的完整方法：配方、验证、解读顶点、求解并绘制
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