Example 2.3.2 | 例题 2.3.2

📚 Example 2.3.2 | 例题 2.3.2

This worked example is carefully selected from the AQA A-Level Mathematics specification to illustrate a core algebraic technique. In this section, we break down the problem step by step, clarify the underlying reasoning, and show how to present your solution clearly to gain full marks in an exam setting.

本例题精选自 AQA A-Level 数学考纲，旨在展示一项核心代数技巧。我们将逐步拆解问题、阐明背后的逻辑，并演示如何在考试中规范作答以获取满分。

1. Problem Statement | 题目陈述

Solve the quadratic equation below using the method of completing the square. Give your answers in exact surd form.

请使用配方法解下列二次方程，并以精确根式形式给出答案。

2x² − 8x + 3 = 0

This example tests your ability to manipulate a quadratic expression when the coefficient of x² is not 1, a common variation in AQA exam papers.

本题考察当 x² 的系数不为 1 时处理二次表达式的配方法能力，这是 AQA 试卷中常见的变式题型。

2. Why Completing the Square? | 为什么选择配方法？

Completing the square is a powerful method that not only solves quadratic equations but also reveals the vertex of a parabola and helps in sketching graphs. Unlike factorisation, it works for all quadratics, and unlike the quadratic formula, it gives you direct insight into transformations of the curve.

配方法是一项强大的工具，它不仅能解二次方程，还能揭示抛物线的顶点并辅助画图。与因式分解不同，它对所有二次方程都适用；与求根公式相比，它能让你更直观地理解曲线的变换。

For AQA, mastering this method is essential because it appears in coordinate geometry, optimisation problems, and even in integration questions.

对于 AQA 考试而言，掌握此方法至关重要，因为它在坐标几何、最优化问题乃至积分题目中都会出现。

3. Step 1: Factor Out the Leading Coefficient | 第一步：提取二次项系数

Since the coefficient of x² is 2, we first factor this out from the first two terms. We do not factor it out of the constant term, because the aim is to create a perfect square from the x-terms only.

由于 x² 的系数为 2，我们先将它从前两项中提取出来。我们不提取常数项，因为目标是仅从含 x 的项中构造完全平方。

2x² − 8x + 3 = 2(x² − 4x) + 3

Notice how −8x divided by 2 gives −4x inside the bracket. This step simplifies the subsequent manipulation considerably.

注意 −8x 除以 2 后在括号内变为 −4x。这一步大大简化了后续的运算。

4. Step 2: Create the Perfect Square | 第二步：构造完全平方

Inside the bracket, we have x² − 4x. To complete the square, we take half of the coefficient of x, which is −4 ÷ 2 = −2, and then square it to get 4. We add and subtract this value inside the bracket to keep the expression balanced.

括号内为 x² − 4x。要配成完全平方，我们取 x 系数的一半，即 −4 ÷ 2 = −2，然后平方得 4。我们在括号内同时加上和减去这个值，以保持表达式恒等。

x² − 4x = (x − 2)² − 4

Therefore, the original expression becomes:

因此，原表达式变为：

2[(x − 2)² − 4] + 3 = 2(x − 2)² − 8 + 3 = 2(x − 2)² − 5

This is the completed square form of the quadratic expression. It tells us immediately that the vertex of the parabola y = 2x² − 8x + 3 is at the point (2, −5).

这就是该二次表达式的配方法形式。它直接告诉我们抛物线 y = 2x² − 8x + 3 的顶点坐标为 (2, −5)。

5. Step 3: Set the Expression to Zero | 第三步：令表达式为零

We now solve the equation 2(x − 2)² − 5 = 0. This is a much simpler equation to handle because the x-term appears only once.

现在我们解方程 2(x − 2)² − 5 = 0。这个方程处理起来简单得多，因为 x 项只出现一次。

2(x − 2)² − 5 = 0

2(x − 2)² = 5

(x − 2)² = 5⁄2

Dividing both sides by 2 isolates the squared term completely. This is a critical step — do not forget to divide the right-hand side as well.

两边同时除以 2，将平方项完全孤立出来。这是关键一步——不要忘记右边也要除以 2。

6. Step 4: Take the Square Root | 第四步：开平方

Now we take the square root of both sides. Remember that taking a square root yields both a positive and a negative solution.

现在对两边开平方。请记住，开平方会得到正负两个解。

x − 2 = ±√(5⁄2)

x = 2 ± √(5⁄2)

It is conventional to rationalise the surd to present the answer in its simplest form. Note that √(5⁄2) = √10⁄2, so we can write the solutions as:

通常我们会将根式有理化，以最简形式呈现答案。注意 √(5⁄2) = √10⁄2，因此解可以写为：

x = 2 ± √10⁄2

Alternatively, combining the terms over a common denominator gives x = (4 ± √10)⁄2.

或者，用公分母合并项得到 x = (4 ± √10)⁄2。

7. Final Solution and Verification | 最终解与验证

The two exact solutions to the equation 2x² − 8x + 3 = 0 are:

方程 2x² − 8x + 3 = 0 的两个精确解为：

x = 2 + √10⁄2 或 x = 2 − √10⁄2

To verify, we can substitute x = 2 + √10⁄2 into the original equation. Expanding (2 + √10⁄2 − 2)² gives (√10⁄2)² = 10⁄4 = 2.5. Then 2 × 2.5 − 5 = 0, which confirms the solution is correct.

为验证结果，我们将 x = 2 + √10⁄2 代入原方程。展开 (2 + √10⁄2 − 2)² 得 (√10⁄2)² = 10⁄4 = 2.5。然后 2 × 2.5 − 5 = 0，验证解是正确的。

8. Discriminant Check | 判别式检验

For the quadratic ax² + bx + c = 0, the discriminant is b² − 4ac. For our equation, a = 2, b = −8 and c = 3.

对于二次方程 ax² + bx + c = 0，判别式为 b² − 4ac。对于本题，a = 2，b = −8，c = 3。

Δ = (−8)² − 4 × 2 × 3 = 64 − 24 = 40

Since Δ = 40 > 0, the equation has two distinct real roots. We could also solve this using the quadratic formula x = (−b ± √Δ)⁄2a = (8 ± √40)⁄4 = (8 ± 2√10)⁄4 = (4 ± √10)⁄2, which matches our completed-square result.

由于 Δ = 40 > 0，方程有两个不相等的实数根。我们也可以用求根公式解：x = (−b ± √Δ)⁄2a = (8 ± √40)⁄4 = (8 ± 2√10)⁄4 = (4 ± √10)⁄2，与配方法结果一致。

9. Graphical Interpretation | 图像解释

The completed square form y = 2(x − 2)² − 5 gives us valuable information about the graph. The vertex is at (2, −5), and since the coefficient 2 is positive, the parabola opens upwards. The y-intercept occurs when x = 0, giving y = 3, which matches the original constant term.

配方法形式 y = 2(x − 2)² − 5 提供了关于图像的丰富信息。顶点在 (2, −5)，由于系数 2 为正，抛物线开口向上。y 截距出现在 x = 0 时，y = 3，与原常数项一致。

The roots we found (approximately 3.58 and 0.42) are the x-coordinates where the parabola crosses the x-axis. The vertex lies exactly halfway between these two roots, confirming that the axis of symmetry is x = 2.

我们求得的根（约为 3.58 和 0.42）是抛物线与 x 轴交点的横坐标。顶点恰好位于两个根的正中间，证实对称轴为 x = 2。

10. Common Mistakes to Avoid | 常见错误提醒

Students frequently make the following errors when completing the square with a leading coefficient:

学生在处理带首项系数的配方法时常犯以下错误：

Forgetting to factor the leading coefficient from both x-terms: You must divide both the x² and x terms by the leading coefficient, not just the x² term.
Forgetting to distribute the factor back when simplifying: 2(x − 2)² − 8 + 3 requires careful distribution of the 2 to both the squared term and the subtracted constant.
Dropping the ± sign when taking square roots: Always include both positive and negative roots.
Not rationalising the surd: Leave answers in exact form with a rationalised denominator for full marks.

忘记从两个 x 项中提取首项系数：必须同时除以 x² 项和 x 项的系数，不能只处理 x² 项。
简化时忘记将系数乘回去：2(x − 2)² − 8 + 3 需要仔细地将 2 同时分配到平方项和减去的常数上。
开平方时遗漏 ± 号：始终包含正负两个根。
未将根式有理化：为获得满分，答案应保留精确形式且分母有理化。

11. Relating to the Quadratic Formula | 与求根公式的联系

It is worth noting that the method of completing the square is exactly how the quadratic formula is derived. By completing the square on the general equation ax² + bx + c = 0, we obtain x = (−b ± √(b² − 4ac))⁄2a. This is why the discriminant appears under the square root — it determines the nature of the roots.

值得注意的是，配方法正是求根公式的推导基础。对一般方程 ax² + bx + c = 0 进行配方，可得到 x = (−b ± √(b² − 4ac))⁄2a。这就是为什么判别式出现在根号下方——它决定了根的性质。

In this example, substituting a = 2, b = −8 and c = 3 into the formula yields exactly the same roots we obtained by completing the square. Understanding this connection deepens your grasp of quadratic theory and helps you choose the most efficient method in different exam contexts.

在本例中，将 a = 2，b = −8，c = 3 代入公式，所得结果与配方法完全相同。理解这层联系能加深你对二次方程理论的掌握，并帮助你在不同考试情境中选择最有效的方法。

12. Summary and Exam Tips | 总结与应试技巧

To summarise, solving a quadratic by completing the square involves four key stages: divide by the leading coefficient, form the perfect square, isolate the squared term, and take the square root.

总结来说，用配方法解二次方程包含四个关键步骤：除以首项系数、构造完全平方、孤立平方项、开平方。

For AQA exams, always read whether the question specifies a particular method. If it says “solve by completing the square”, you must show the completed square form explicitly, even if you could solve faster by factorising. Marks are awarded for method as well as accuracy.

在 AQA 考试中，务必看清题目是否指定了特定方法。如果题目要求”用配方法求解”，你必须明确写出配方法的形式，即使因式分解更快也不可替代。评分既看方法也看准确性。

Finally, always check your solutions by either substituting back into the original equation or using the discriminant to confirm the number of roots. This habit prevents careless errors and secures full marks.

最后，始终通过代回原方程或使用判别式确认根的个数来检验答案。这个习惯能防止粗心错误，确保获得满分。

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Example 2.3.2 | 例题 2.3.2