Solving Quadratic Equations by Completing the Square | 配方法解二次方程

📚 Solving Quadratic Equations by Completing the Square | 配方法解二次方程

Completing the square is a powerful algebraic technique used to solve quadratic equations, rewrite expressions, and analyse graphs. It transforms a quadratic expression into a perfect square plus or minus a constant, making many problems much simpler.

配方法是一种强大的代数技巧，用于解二次方程、重写表达式和分析图象。它把二次表达式转化为一个完全平方加上或减去一个常数，从而使许多问题变得更加简单。

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

Completing the square means rewriting a quadratic expression of the form \(ax^2 + bx + c\) into the form \(a(x + p)^2 + q\), where \(p\) and \(q\) are constants. The key idea is to create a perfect square trinomial from the first two terms.

配方法指的是将形如 \(ax^2 + bx + c\) 的二次表达式重写为 \(a(x + p)^2 + q\) 的形式，其中 \(p\) 和 \(q\) 是常数。核心思想是从前两项中构造出一个完全平方三项式。

This method is especially useful because it can solve any quadratic equation, including those that do not factor easily. It also reveals the vertex of a parabola directly.

这种方法特别有用，因为它能解任何二次方程，包括那些不易因式分解的方程。它还能直接显示抛物线的顶点。

2. The Perfect Square Trinomial | 完全平方三项式

A perfect square trinomial has the form \(x^2 + 2px + p^2\), which equals \((x + p)^2\). For example, \(x^2 + 6x + 9 = (x + 3)^2\) because \(2 \times 3 = 6\) and \(3^2 = 9\).

完全平方三项式具有 \(x^2 + 2px + p^2\) 的形式，等于 \((x + p)^2\)。例如，\(x^2 + 6x + 9 = (x + 3)^2\)，因为 \(2 \times 3 = 6\) 且 \(3^2 = 9\)。

Notice the relationship: the coefficient of \(x\) is twice the number inside the bracket, and the constant term is the square of that number. To complete the square, we take half of the \(x\)-coefficient and square it.

注意这个关系：\(x\) 的系数是括号内数字的两倍，常数项是该数字的平方。要配方，我们取 \(x\) 系数的一半，然后平方。

Expression	Half of coefficient	Square	Perfect square form
\(x^2 + 8x\)	4	16	\((x + 4)^2 – 16\)
\(x^2 – 10x\)	-5	25	\((x – 5)^2 – 25\)
\(x^2 + 5x\)	2.5	6.25	\((x + 2.5)^2 – 6.25\)

3. Step-by-Step Method (When \(a = 1\)) | 分步方法（当 \(a = 1\)）

When the coefficient of \(x^2\) is 1, we can complete the square in three simple steps:

当 \(x^2\) 的系数为 1 时，我们可以通过三个简单步骤完成配方：

Step 1: Write the expression as \(x^2 + bx\).
Step 1：将表达式写成 \(x^2 + bx\)。
Step 2: Take half of \(b\), then square it. Add and subtract this square.
Step 2：取 \(b\) 的一半，然后平方。加上并减去这个平方。
Step 3: Factor the perfect square trinomial and simplify the constants.
Step 3：将完全平方三项式因式分解，并简化常数项。

For example, \(x^2 + 6x + 2\) becomes:

例如，\(x^2 + 6x + 2\) 变成：

\(x^2 + 6x + 2 = (x + 3)^2 – 9 + 2 = (x + 3)^2 – 7\)

We subtracted 9 because we added \((6/2)^2 = 9\) to complete the square, then compensated.

我们减去 9，是因为我们加了 \((6/2)^2 = 9\) 来完成平方，然后进行补偿。

4. Worked Example 1 (a = 1) | 示例 1（a = 1）

Solve the equation \(x^2 – 4x – 5 = 0\) by completing the square.

用配方法解方程 \(x^2 – 4x – 5 = 0\)。

Solution: Move the constant term to the other side first:

解答：先将常数项移到另一边：

\(x^2 – 4x = 5\)

Add \((4/2)^2 = 4\) to both sides:

两边加上 \((4/2)^2 = 4\)：

\(x^2 – 4x + 4 = 5 + 4\)

Factor the left side:

因式分解左边：

\((x – 2)^2 = 9\)

Take the square root of both sides, remembering the ± sign:

两边开平方，记住 ± 号：

\(x – 2 = \pm 3\)

Thus \(x = 2 + 3 = 5\) or \(x = 2 – 3 = -1\). The solution set is \(\{5, -1\}\).

因此 \(x = 2 + 3 = 5\) 或 \(x = 2 – 3 = -1\)。解集为 \(\{5, -1\}\)。

5. Handling a Negative \(x\)-Term | 处理负 \(x\) 项

When the coefficient of \(x\) is negative, the bracket contains a minus sign. For example, \(x^2 – 8x + 3\):

当 \(x\) 的系数为负数时，括号内为减号。例如，\(x^2 – 8x + 3\)：

\(x^2 – 8x + 3 = (x – 4)^2 – 16 + 3 = (x – 4)^2 – 13\)

Half of \(-8\) is \(-4\), and \((-4)^2 = 16\). We then subtract 16 to keep the expression equivalent.

\(-8\) 的一半是 \(-4\)，且 \((-4)^2 = 16\)。然后减去 16 以保持表达式等价。

This technique works exactly the same way when solving equations. Just remember to add the square to both sides.

在解方程时，这种方法完全相同。只需记得两边同时加上这个平方。

6. When \(a \neq 1\) | 当 \(a\) 不等于 1 时

If the coefficient of \(x^2\) is not 1, first factor it out from the \(x^2\) and \(x\) terms. Then complete the square inside the bracket.

如果 \(x^2\) 的系数不是 1，先从 \(x^2\) 和 \(x\) 项中提出该系数。然后在括号内完成配方。

For example, \(2x^2 + 8x + 5\):

例如，\(2x^2 + 8x + 5\)：

\(2x^2 + 8x + 5 = 2(x^2 + 4x) + 5\)

Now complete the square for \(x^2 + 4x\): half of 4 is 2, square is 4. Add and subtract 4 inside the bracket:

现在对 \(x^2 + 4x\) 配方：4 的一半是 2，平方是 4。在括号内加上并减去 4：

\(2[(x + 2)^2 – 4] + 5 = 2(x + 2)^2 – 8 + 5 = 2(x + 2)^2 – 3\)

Be careful: subtracting 4 inside the bracket is multiplied by the factor 2, so it becomes subtracting 8 outside.

注意：在括号内减去 4 会被外面的系数 2 相乘，所以在外边变为减去 8。

7. Worked Example 2 (a ≠ 1) | 示例 2（a ≠ 1）

Solve \(3x^2 – 12x + 6 = 0\) by completing the square.

用配方法解 \(3x^2 – 12x + 6 = 0\)。

Solution: First divide the whole equation by 3 to make the coefficient of \(x^2\) equal to 1:

解答：首先将整个方程除以 3，使 \(x^2\) 的系数等于 1：

\(x^2 – 4x + 2 = 0\)

Move the constant to the right:

将常数移到右边：

\(x^2 – 4x = -2\)

Add \((-4/2)^2 = 4\) to both sides:

两边加上 \((-4/2)^2 = 4\)：

\(x^2 – 4x + 4 = -2 + 4\)

So \((x – 2)^2 = 2\). Taking square roots:

因此 \((x – 2)^2 = 2\)。开平方：

\(x – 2 = \pm \sqrt{2}\)

Hence \(x = 2 \pm \sqrt{2}\).

所以 \(x = 2 \pm \sqrt{2}\)。

8. The Quadratic Formula Connection | 与二次公式的联系

Completing the square is the algebraic foundation of the quadratic formula. If we complete the square on the general equation \(ax^2 + bx + c = 0\), we obtain:

配方法是二次公式的代数基础。如果我们对一般方程 \(ax^2 + bx + c = 0\) 配方，可以得到：

\(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

This formula works for every quadratic equation. The expression \(b^2 – 4ac\) is called the discriminant; it determines the number of real solutions.

这个公式适用于所有二次方程。表达式 \(b^2 – 4ac\) 称为判别式；它决定实数解的数量。

If \(b^2 – 4ac > 0\), there are two distinct real roots.
如果 \(b^2 – 4ac > 0\)，有两个不同的实数根。
If \(b^2 – 4ac = 0\), there is exactly one real root (a repeated root).
如果 \(b^2 – 4ac = 0\)，恰好有一个实数根（重根）。
If \(b^2 – 4ac < 0\), there are no real roots.
如果 \(b^2 – 4ac < 0\)，没有实数根。

Knowing both methods ensures you can always choose the fastest approach.

掌握这两种方法可以确保你总能选择最快的解题方式。

9. Graphical Interpretation | 图形解释

Completing the square helps us find the vertex of a parabola. For \(y = a(x + p)^2 + q\), the vertex is \((-p, q)\).

配方法有助于找到抛物线的顶点。对于 \(y = a(x + p)^2 + q\)，顶点为 \((-p, q)\)。

For example, \(y = 2(x + 2)^2 – 3\) has vertex \((-2, -3)\). The axis of symmetry is the vertical line \(x = -2\).

例如，\(y = 2(x + 2)^2 – 3\) 的顶点为 \((-2, -3)\)。对称轴是垂直线 \(x = -2\)。

When solving \(a(x + p)^2 + q = 0\), the solutions are the x-intercepts of the graph. If \(q\) and \(a\) have opposite signs, the graph crosses the x-axis twice; if \(q = 0\), it touches once; otherwise it never touches.

当解 \(a(x + p)^2 + q = 0\) 时，解就是图象的 x 截距。如果 \(q\) 和 \(a\) 异号，图象与 x 轴相交两次；如果 \(q = 0\)，相切一次；否则永不相交。

10. Common Mistakes and Tips | 常见错误与技巧

Here are typical pitfalls students face and how to avoid them:

以下是学生常遇到的典型陷阱以及如何避免：

Forgetting the ± sign: Always write \(x + p = \pm \sqrt{…}\) when taking square roots.
忘记 ± 号：在开平方时，务必写成 \(x + p = \pm \sqrt{…}\)。
Dividing incorrectly when \(a \neq 1\): If you divide the equation, divide every term by \(a\).
当 \(a \neq 1\) 时错误地除法：如果除以 \(a\)，必须对每一项都除以 \(a\)。
Mixing signs: Half of \(-b\) is negative; squaring it always gives a positive result.
混淆符号：\(-b\) 的一半是负数；平方后总是正数。
Forgetting to compensate: When adding a value inside a bracket with a coefficient, multiply it by that coefficient when moving outside.
忘记补偿：当在带有系数的括号内加上某个值时，移到括号外时要乘以该系数。

Practice makes perfect. Always double-check your final expression by expanding it to ensure it matches the original quadratic.

熟能生巧。始终通过展开最终表达式来检查它是否与原二次式一致。

Summary: Completing the square is a versatile skill for solving quadratics and understanding their graphs. Master the steps, practise with different coefficients, and it will become second nature.

总结：配方法是解二次方程和理解其图形的一种多用途技能。掌握步骤，练习不同的系数，它就会变得得心应手。

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Solving Quadratic Equations by Completing the Square | 配方法解二次方程