📚 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
ting 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。
1. Introduction to the Method | 方法介绍
olves 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。
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
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。
4. Step 2: Calculate and Add (b/2)² | 第二步：计算并加上 (b/2)²
-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
5. Step 3: Write the Left Side as a Perfect Square | 第三步：将左边写成完全平方
+ 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。
6. Step 4: Take the Square Root of Both Sides | 第四步：对两边取平方根
er 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。
7. Step 5: Solve for x | 第五步：解出 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}。
8. Checking the Solutions | 检验解
e 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。
9. Understanding the Discriminant | 理解判别式
+ 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
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。
10. Graphical Interpretation | 图形解释
abola. 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。
The
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