Example 2.1.1: Solving Quadratic Equations by Completing the Square | 例 2.1.1：用配方法解二次方程

📚 Example 2.1.1: Solving Quadratic Equations by Completing the Square | 例 2.1.1：用配方法解二次方程

In the AQA A-level Mathematics specification, solving quadratic equations is a fundamental skill that appears both directly and within larger problems involving curves, inequalities, and mechanics.

在 AQA A-level 数学大纲中，解二次方程是一项基本技能，既会直接考查，也会出现在涉及曲线、不等式和力学的更复杂问题中。

This article works through Example 2.1.1, a standard quadratic equation solved by completing the square, and then develops the method into the quadratic formula and discriminant ideas required for AQA exam success.

本文将通过例 2.1.1 这一标准二次方程来讲解配方法求解，并将其推广到二次公式和判别式等 AQA 考试必备内容。

1. What is Completing the Square? | 什么是配方法

Completing the square rewrites a quadratic expression of the form x² + bx + c in the form (x + p)² + q, where p and q are constants.

配方法是将形如 x² + bx + c 的二次式改写为 (x + p)² + q 的形式，其中 p 和 q 是常数。

This new form immediately reveals the vertex of the parabola and allows a quadratic equation to be solved by taking square roots.

这种新形式能直接揭示抛物线的顶点，并且可以通过开平方来解二次方程。

For a monic quadratic, the key identity is x² + 2px + p² = (x + p)².

对于首项系数为 1 的二次式，关键恒等式是 x² + 2px + p² = (x + p)²。

We therefore halve the coefficient of x, square it, and adjust the constant term to preserve equality.

因此，我们将 x 的系数减半、平方，并调整常数项以保持等值。

2. The Core Example 2.1.1 | 核心例题 2.1.1

Consider the quadratic equation x² + 6x + 1 = 0. This is Example 2.1.1 from the AQA Year 1 pure mathematics course.

考虑二次方程 x² + 6x + 1 = 0。这是 AQA 一年级纯数学课程中的例 2.1.1。

The equation is already monic, so we can complete the square directly without first dividing by a leading coefficient.

该方程首项系数已经是 1，因此可以直接配方，无需先除以首项系数。

Our goal is to write x² + 6x as part of a perfect square and then solve for x in exact surd form.

我们的目标是将 x² + 6x 写成完全平方的一部分，然后以根式形式精确求出 x。

3. Step-by-Step Solution | 分步求解过程

Step 1: Identify the coefficient of x, which is 6. Half of 6 is 3, so the bracket will contain x + 3.

第 1 步：确定 x 的系数为 6。6 的一半是 3，所以括号内应为 x + 3。

Step 2: Square the half-coefficient: 3² = 9. We therefore add and subtract 9 to keep the expression balanced.

第 2 步：将半系数平方：3² = 9。因此我们加上并减去 9 以保持表达式平衡。

Step 3: Rewrite the equation as x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8 = 0.

第 3 步：将方程改写为 x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8 = 0。

Step 4: Isolate the square by adding 8 to both sides.

第 4 步：两边加 8，将平方项分离。

(x + 3)² = 8

Step 5: Take the square root of both sides, remembering both positive and negative roots.

第 5 步：对两边开平方，并记住正负两个根。

x + 3 = ±√8

Step 6: Simplify √8 = √(4 × 2) = 2√2, then subtract 3 from both sides.

第 6 步：化简 √8 = √(4 × 2) = 2√2，然后两边减 3。

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