Example 2.2.1: Solving a Quadratic Equation by Completing the Square | 示例2.2.1：用配方法解二次方程

📚 Example 2.2.1: Solving a Quadratic Equation by Completing the Square | 示例2.2.1：用配方法解二次方程

In this example, we will solve the quadratic equation x² – 6x + 8 = 0 by completing the square. This method is a fundamental algebraic technique that allows us to find the roots of a quadratic equation without factorisation, and it also provides insight into the vertex of the corresponding parabola.

在这个示例中，我们将通过配方法解二次方程 x² – 6x + 8 = 0。配方法是代数中的基本技巧，它使我们无需因式分解即可求出二次方程的根，并且还能揭示对应抛物线的顶点。

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1. Introduction to the Method | 方法介绍

Completing the square rewrites a quadratic expression in the form a(x + p)² + q. For the case a = 1, the process involves adjusting the x² and x terms so that they form a perfect square. This technique is useful for solving equations, deriving the quadratic formula, and analysing graphs.

配方法将二次表达式改写为 a(x + p)² + q 的形式。对于 a = 1 的情况，该过程涉及调整 x² 项和 x 项，使它们构成一个完全平方。这项技术在解方程、推导二次公式和分析图形时都很有用。

The key idea is to take half of the coefficient of x, square it, and add that value to both sides of the equation. This preserves equality while creating a perfect square on one side.

关键思想是取 x 系数的一半，将其平方，并将该值加到等式两边。这样在保持等式不变的同时，在一边构建一个完全平方。

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2. The Problem | 问题陈述

Solve the quadratic equation: x² – 6x + 8 = 0.

解二次方程：x² – 6x + 8 = 0。

We choose this equation because it has simple integer roots, making it easy to verify us both factorisation and the completing-the-square method. The coefficient of x is -6, and the constant term is 8.

我们选择这个方程是因为它有简单的整数根，便于通过因式分解和配方法两种方式验证。x 的系数是 -6，常数项是 8。

x² – 6x + 8 = 0

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3. Step 1: Move the Constant Term | 第一步：移动常数项

Subtract 8 from both sides to isolate the x² and x terms. This gives:

从两边减去 8，以分离出 x² 项和 x 项。得到：

x² – 6x = -8

Moving the constant term to the right side is necessary because completing the square only applies to the variable terms on the left.

将常数项移到右边是必要的，因为配方法只作用于左边的变量项。

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4. Step 2: Calculate and Add (b/2)² | 第二步：计算并加上 (b/2)²

Here b = -6. Half of b is -3, and squaring it gives 9. We add 9 to both sides of the equation to maintain equality.

这里 b = -6。b 的一半是 -3，其平方是 9。我们将在等式两边加上 9 以保持等式成立。

(b/2)² = (-3)² = 9

Add 9 to both sides:

两边同时加 9：

x² – 6x + 9 = -8 + 9

Then simplified: x² – 6x + 9 = 1

然后化简得：x² – 6x + 9 = 1

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5. Step 3: Write the Left Side as a Perfect Square | 第三步：将左边写成完全平方

The left side x² – 6x + 9 factors as (x – 3)². This is because (x – 3)² = x² – 6x + 9. The equation now becomes:

左边 x² – 6x + 9 可因式分解为 (x – 3)²。这是因为 (x – 3)² = x² – 6x + 9。方程现在变为：

(x – 3)² = 1

This step converts the quadratic expression into a binomial square, which is the core of the method.

这一步将二次表达式转换为二项式的平方，这是该方法的核心。

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6. Step 4: Take the Square Root of Both Sides | 第四步：对两边取平方根

To eliminate the square, take the square root of both sides. Remember that the square root of a number can be positive or negative, so we include the ± sign.

为了消去平方，对两边取平方根。记住一个数的平方根可以是正的或负的，所以我们需要加上 ± 号。

x – 3 = ±1

This gives two possible linear equations: x – 3 = 1 or x – 3 = -1.

这给出两个可能的线性方程：x – 3 = 1 或 x – 3 = -1。

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7. Step 5: Solve for x | 第五步：解出 x

Solve each equation separately. For x – 3 = 1, add 3 to both sides to get x = 4. For x – 3 = -1, add 3 to both sides to get x = 2.

分别解每个方程。对于 x – 3 = 1，两边加 3 得 x = 4。对于 x – 3 = -1，两边加 3 得 x = 2。

x = 4 或 x = 2

Therefore, the solution set is {2, 4}.

因此，解集为 {2, 4}。

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8. Checking the Solutions | 检验解

Substitute x = 2 into the original equation: 2² – 6(2) + 8 = 4 – 12 + 8 = 0. The first value works.

将 x = 2 代入原方程：2² – 6(2) + 8 = 4 – 12 + 8 = 0。第一个值成立。

Substitute x = 4: 4² – 6(4) + 8 = 16 – 24 + 8 = 0. The second value also works.

将 x = 4 代入：4² – 6(4) + 8 = 16 – 24 + 8 = 0。第二个值也成立。

Since both values satisfy the equation, the solution is correct. This is an important verification step in any algebraic problem.

由于两个值都满足方程，因此解是正确的。在任何代数问题中，验证都是重要的一步。

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9. Understanding the Discriminant | 理解判别式

For a quadratic equation ax² + bx + c = 0, the discriminant is Δ = b² – 4ac. In this example, a = 1, b = -6, and c = 8.

对于二次方程 ax² + bx + c = 0，判别式为 Δ = b² – 4ac。在本例中，a = 1，b = -6，c = 8。

Δ = (-6)² – 4 × 1 × 8 = 36 – 32 = 4

Since Δ > 0, the equation has two distinct real roots, which matches our result. If Δ = 0, there is one repeated real root; if Δ < 0, there are no real roots.

由于 Δ > 0，方程有两个不同的实根，这与我们的结果一致。如果 Δ = 0，则有一个重复的实根；如果 Δ < 0，则没有实根。

The discriminant also tells us about the graphical intersection of the parabola with the x-axis.

判别式还告诉我们抛物线与 x 轴的交点情况。

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10. Graphical Interpretation | 图形解释

The equation y = x² – 6x + 8 represents a parabola. The roots x = 2 and x = 4 are the x-coordinates where the curve crosses the x-axis.

方程 y = x² – 6x + 8 表示一条抛物线。根 x = 2 和 x = 4 是曲线与 x 轴交点的 x 坐标。

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