Completing the Square | 配方法

📚 Completing the Square | 配方法

The method of completing the square is one of the most powerful algebraic techniques in the IGCSE Mathematics syllabus. It allows us to rewrite quadratic expressions in a form that reveals the vertex of a parabola, solve quadratic equations without relying solely on the quadratic formula, and even derive the quadratic formula itself. This skill is essential for higher-level mathematics, including calculus and coordinate geometry.

配方法是IGCSE数学大纲中最强大的代数技巧之一。它使我们能够将二次表达式改写为可以揭示抛物线顶点的形式，无需仅依赖求根公式即可求解二次方程，甚至能推导出求根公式本身。这一技能对于更高级的数学（包括微积分和坐标几何）至关重要。

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1. Review of Quadratic Expressions | 二次表达式回顾

A quadratic expression in one variable x has the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. The simplest quadratic expressions are perfect squares, such as (x + 3)² = x² + 6x + 9 or (x – 5)² = x² – 10x + 25. Notice that in a perfect square, the constant term is always the square of half the coefficient of x.

一个关于变量x的二次表达式的一般形式为ax² + bx + c，其中a、b和c是常数，且a ≠ 0。最简单的二次表达式是完全平方式，例如(x + 3)² = x² + 6x + 9或(x – 5)² = x² – 10x + 25。注意，在完全平方式中，常数项总是x系数一半的平方。

The key relationship can be stated as:

(x + p)² = x² + 2px + p²

This identity forms the basis of the completing-the-square technique. If we have an expression x² + bx, we can add (b/2)² to it to form a perfect square. However, since we cannot simply add a term without changing the expression, we must also subtract the same term to maintain equality.

这一恒等式构成了配方法技术的基础。如果我们有表达式x² + bx，我们可以加上(b/2)²使其构成完全平方式。然而，由于我们不能在改变表达式的情况下简单地加上一项，我们还必须减去同一项以保持等式不变。

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2. The Basic Technique with a = 1 | a = 1时的基本技巧

When the coefficient of x² is 1, the process is straightforward. Given x² + bx + c, we rewrite it as (x + b/2)² – (b/2)² + c. The expression (x + b/2)² is a perfect square, and the remaining terms are constants. Let us illustrate this with the expression x² + 8x + 15.

当x²的系数为1时，这一过程是直接明了的。给定x² + bx + c，我们将其改写为(x + b/2)² – (b/2)² + c。表达式(x + b/2)²是完全平方式，剩余项是常数。让我们用表达式x² + 8x + 15来说明这一点。

Step 1: Take half of 8, which is 4. Step 2: Square 4, giving 16. Step 3: Write x² + 8x = (x + 4)² – 16. Step 4: Add the original constant 15 to get (x + 4)² – 16 + 15 = (x + 4)² – 1.

步骤1：取8的一半，即4。步骤2：将4平方，得16。步骤3：写出x² + 8x = (x + 4)² – 16。步骤4：加上原常数项15，得到(x + 4)² – 16 + 15 = (x + 4)² – 1。

Therefore, x² + 8x + 15 = (x + 4)² – 1. We can check this by expanding the right-hand side: (x + 4)² – 1 = x² + 8x + 16 – 1 = x² + 8x + 15, which matches the original expression.

因此，x² + 8x + 15 = (x + 4)² – 1。我们可以通过展开右侧来验证：(x + 4)² – 1 = x² + 8x + 16 – 1 = x² + 8x + 15，与原表达式一致。

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3. Completing the Square with a ≠ 1 | a ≠ 1时的配方法

When the coefficient of x² is not 1, we must first factor out that coefficient from the first two terms. For example, consider 3x² – 12x + 7. We factor out 3 from the x² and x terms to get 3(x² – 4x) + 7. Now we complete the square inside the brackets.

当x²的系数不是1时，我们必须先从前两项中提出该系数。例如，考虑3x² – 12x + 7。我们从x²和x项中提出3，得到3(x² – 4x) + 7。现在我们在括号内进行配方。

Half of -4 is -2, and (-2)² = 4. Thus x² – 4x = (x – 2)² – 4. Substituting this back gives 3[(x – 2)² – 4] + 7 = 3(x – 2)² – 12 + 7 = 3(x – 2)² – 5.

-4的一半是-2，且(-2)² = 4。因此x² – 4x = (x – 2)² – 4。将其代回得3[(x – 2)² – 4] + 7 = 3(x – 2)² – 12 + 7 = 3(x – 2)² – 5。

It is crucial to remember that the constant factor outside the brackets multiplies the entire brackets, including the subtracted square term. A common error is to forget to multiply the -(b/2)² term by the factor a. Always verify your result by expanding.

关键的是要记住，括号外的常数因子乘以整个括号，包括减去的平方项。一个常见错误是忘记将-(b/2)²项乘以因子a。始终通过展开来验证你的结果。

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4. Solving Quadratic Equations by Completing the Square | 用配方法求解二次方程

Once a quadratic equation is written in completed-square form, solving it becomes a two-step process: isolate the squared term, then take the square root of both sides. Consider the equation x² + 6x – 16 = 0.

一旦二次方程被写成配方法形式，求解就成为一个两步过程：隔离平方项，然后对两边取平方根。考虑方程x² + 6x – 16 = 0。

First, complete the square on the left-hand side. x² + 6x = (x + 3)² – 9, so the equation becomes (x + 3)² – 9 – 16 = 0, that is, (x + 3)² – 25 = 0. Then (x + 3)² = 25; taking square roots gives x + 3 = ±5, so x = 2 or x = -8.

首先，在左边进行配方。x² + 6x = (x + 3)² – 9，因此方程变为(x + 3)² – 9 – 16 = 0，即(x + 3)² – 25 = 0。然后(x + 3)² = 25；取平方根得x + 3 = ±5，所以x = 2或x = -8。

For equations with a ≠ 1, such as 2x² – 8x + 3 = 0, we first divide through by 2 to obtain x² – 4x + 1.5 = 0, or we can complete the square directly as shown in Section 3. The solutions will often involve radicals.

对于a ≠ 1的方程，如2x² – 8x + 3 = 0，我们首先两边除以2，得到x² – 4x + 1.5 = 0，也可以如第3节所示直接配方。解通常含有根号。

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5. The Quadratic Formula Derived | 求根公式的推导

The quadratic formula can be derived precisely by completing the square on the general quadratic equation ax² + bx + c = 0. This derivation is a classic application of the technique and is sometimes tested in IGCSE. Begin by dividing both sides by a:

求根公式可以通过对一般二次方程ax² + bx + c = 0进行配方来精确推导。这一推导是该技巧的经典应用，在IGCSE中有时会被考到。首先两边除以a：

x² + (b/a)x + c/a = 0

Now complete the square on x² + (b/a)x. Half of b/a is b/(2a), and its square is b²/(4a²). Thus:

现在对x² + (b/a)x进行配方。b/a的一半是b/(2a)，其平方为b²/(4a²)。因此：

(x + b/(2a))² – b²/(4a²) + c/a = 0

Rearranging and simplifying leads to x = (-b ± √(b² – 4ac)) / (2a). This formula is given in the formula booklet, but understanding its derivation helps students appreciate the underlying mathematics and remember the structure of the formula.

整理并简化后得到x = (-b ± √(b² – 4ac)) / (2a)。该公式在公式册中给出，但理解其推导有助于学生欣赏其背后的数学原理，并记住公式的结构。

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6. Vertex Form of a Parabola | 抛物线的顶点形式

One of the most important applications of completing the square is expressing a quadratic function in vertex form. A quadratic function y = ax² + bx + c can be written as y = a(x – h)² + k, where (h, k) is the vertex of the parabola. This form reveals the turning point immediately.

配方法最重要的应用之一是将二次函数表示为顶点形式。二次函数y = ax² + bx + c可以写成y = a(x – h)² + k的形式，其中(h, k)是抛物线的顶点。这种形式能立即揭示转折点。

For example, the function y = x² – 6x + 11 can be rewritten as y = (x – 3)² + 2. Therefore, the vertex is at (3, 2), the axis of symmetry is the vertical line x = 3, and since the coefficient of the squared term is positive, the vertex represents a minimum point.

例如，函数y = x² – 6x + 11可以改写为y = (x – 3)² + 2。因此，顶点在(3, 2)，对称轴是垂直线x = 3，并且由于平方项的系数为正，该顶点代表最小值点。

If a < 0, the parabola opens downward and the vertex is a maximum point. This information is crucial for sketching graphs and solving optimisation problems.

如果a < 0，抛物线向下开口，顶点是最大值点。这一信息对于绘制图形和解决最优化问题至关重要。

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7. Sketching Quadratic Graphs | 绘制二次函数图像

To sketch a quadratic graph efficiently, we typically need three types of information: the coordinates of the vertex, the y-intercept, and the x-intercepts (if the discriminant is non-negative). Completing the square provides the vertex directly, while intercepts are found using standard techniques.

要高效地绘制二次函数图像，我们通常需要三种信息：顶点坐标、y截距和x截距（如果判别式非负）。配方法直接提供顶点，而截距则通过标准技巧求得。

Consider y = 2x² – 8x + 5. Completing the square gives y = 2(x – 2)² – 3, so the vertex is (2, -3). The y-intercept is at (0, 5). The x-intercepts are found by solving 2x² – 8x + 5 = 0, which gives x = 2 ± √6/2, or approximately x ≈ 0.775 and x ≈ 3.225. These five points (vertex, y-intercept, and two x-intercepts) plus the axis of symmetry allow a reasonable sketch.

考虑y = 2x² – 8x + 5。配方得到y = 2(x – 2)² – 3，因此顶点为(2, -3)。y截距在(0, 5)。x截距通过求解2x² – 8x + 5 = 0得到，即x = 2 ± √6/2，约x ≈ 0.775和x ≈ 3.225。这五个点（顶点、y截距和两个x截距）加上对称轴就能绘制出合理的图形。

When the discriminant is negative, there are no real x-intercepts, and the parabola lies entirely above or below the x-axis depending on the sign of a. The vertex form makes this immediately clear.

当判别式为负时，没有实数x截距，抛物线完全位于x轴上方或下方，取决于a的符号。顶点形式能立即清楚地显示这一点。

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8. Maximum and Minimum Values | 最大值和最小值

Completing the square provides a direct method for determining the maximum or minimum value of a quadratic function without calculus. In the vertex form y = a(x – h)² + k, if a > 0, the minimum value is k, which occurs at x = h. If a < 0, the maximum value is k, again at x = h.

配方法提供了一种无需微积分即可直接确定二次函数最大值或最小值的方法。在顶点形式y = a(x – h)² + k中，如果a > 0，最小值为k，且发生在x = h处。如果a < 0，最大值为k，同样在x = h处。

For example, a company’s profit function is given by P(x) = -2x² + 20x – 15, where x is the number of products sold. Completing the square yields P(x) = -2(x – 5)² + 35. Hence, the maximum profit is 35, achieved when 5 products are sold. This type of optimisation problem frequently appears in IGCSE applications.

例如，一家公司的利润函数为P(x) = -2x² + 20x – 15，其中x是销售的产品数量。配方得到P(x) = -2(x – 5)² + 35。因此，最大利润为35，在销售5个产品时实现。此类最优化问题在IGCSE应用题中经常出现。

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9. Discriminant and Nature of Roots | 判别式与根的性质

The discriminant, Δ = b² – 4ac, tells us the number and type of roots of a quadratic equation. From the completed-square form, we can see why this is the case. The equation (x + b/(2a))² = (b² – 4ac)/(4a²) has real solutions if and only if the right-hand side is non-negative, since the square of a real number cannot be negative.

判别式Δ = b² – 4ac告诉我们二次方程的根的数目和类型。从配方法形式我们可以看出原因。方程(x + b/(2a))² = (b² – 4ac)/(4a²)有实数解当且仅当右边非负，因为实数的平方不可能是负数。

Table 1 summarises the three cases:

表1总结了三种情况：

Discriminant | 判别式	Nature of Roots | 根的性质	Graphical Meaning | 图形意义
Δ > 0	Two distinct real roots | 两个不等实数根	Parabola crosses x-axis at two points | 抛物线在两点穿过x轴
Δ = 0	One repeated real root | 一个重根	Parabola touches x-axis at one point | 抛物线在一点与x轴相切
Δ < 0	No real roots | 无实数根	Parabola does not intersect x-axis | 抛物线不与x轴相交

The completed-square form makes these three cases visually obvious: the vertex’s y-coordinate determines whether the parabola touches the x-axis, and this y-coordinate is related to the sign of the discriminant divided by 4a.

配方法形式使这三种情况在视觉上变得明显：顶点的y坐标决定了抛物线是否接触x轴，而这个y坐标与判别式除以4a的符号有关。

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10. Solving Equations without Expanding | 不展开求解方程

Sometimes, expressions such as (x – 3)² = 49 can be solved immediately by equating x – 3 = ±7, giving x = 10 or x = -4. This is a direct application of the square-root property. Completing the square essentially converts any quadratic into this simple form.

有时，像(x – 3)² = 49这样的表达式可以通过令x – 3 = ±7立即求解，得到x = 10或x = -4。这是平方根性质的直接应用。配方法本质上将任意二次式转换为这种简单形式。

The same principle applies to equations involving slightly more complex expressions, such as (2x + 1)² = 25. Here, 2x + 1 = ±5, leading to x = 2 or x = -3. The efficiency of this method is especially noticeable when the coefficient b in the original equation is even, since the calculations remain simple fractions.

同样的原理也适用于涉及稍复杂表达式的方程，如(2x + 1)² = 25。这里，2x + 1 = ±5，得出x = 2或x = -3。当原方程中的系数b为偶数时，这种方法的效率尤其明显，因为计算保持为简单的分数。

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11. Common Mistakes and How to Avoid Them | 常见错误及如何避免

A frequent error is forgetting to compensate for the term added during completing the square. For instance, rewriting x² + 4x as (x + 2)² without subtracting 4 changes the expression. Always remember that (x + 2)² = x² + 4x + 4, so x² + 4x = (x + 2)² – 4.

一个常见错误是忘记补偿配方过程中添加的项。例如，将x² + 4x改写为(x + 2)²而不减去4会改变表达式。始终记住(x + 2)² = x² + 4x + 4，所以x² + 4x = (x + 2)² – 4。

Another common error occurs when a ≠ 1. Some students divide only the first two terms by a while leaving the constant unchanged. The correct approach is either to divide the entire equation by a, or to factor out a from only the first two terms while leaving the constant outside the brackets, as demonstrated in Section 3.

另一个常见错误出现在a ≠ 1时。有些学生只将前两项除以a，而保持常数项不变。正确的方法要么是将整个方程除以a，要么是仅从前两项中提出a，而将常数留在括号外，如第3节所示。

Finally, when solving equations, students sometimes forget the ± sign when taking square roots. The equation (x + 3)² = 25 has two solutions, x + 3 = 5 and x + 3 = -5, so x = 2 and x = -8. Always write both cases explicitly.

最后，在解方程时，学生有时在取平方根时忘记±号。方程(x + 3)² = 25有两个解，x + 3 = 5和x + 3 = -5，所以x = 2和x = -8。始终明确写出两种情况。

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12. Practice Problems | 实践练习

The following problems are designed to cover the main types of completing-the-square questions in IGCSE. Work through them systematically, showing all steps.

以下练习旨在覆盖IGCSE中配方法题目的主要类型。请系统地解答它们，展示所有步骤。

• Express x² + 10x – 3 in the form (x + p)² + q. | 将x² + 10x – 3表示为(x + p)² + q的形式。

• Express 2x² + 12x + 5 in the form a(x + h)² + k. | 将2x² + 12x + 5表示为a(x + h)² + k的形式。

• Solve the equation x² – 4x – 21 = 0 by completing the square. | 用配方法解方程x² – 4x – 21 = 0。

• Find the maximum value of f(x) = -x² + 8x – 7 and the value of x at which it occurs. | 求f(x) = -x² + 8x – 7的最大值及其对应的x值。

• A rectangle has length x cm and width (10 – x) cm. Express the area as a function of x, complete the square, and find the dimensions that give the maximum area. | 一个长方形的长为x cm，宽为(10 – x) cm。将面积表示为x的函数，配方，并求给出最大面积的尺寸。

Answers to these problems can be checked by expanding your completed-square form and comparing it with the original expression or equation. Practice is the key to mastering this topic.

你可以通过展开配方法形式并与原表达式或原方程比较来检查这些练习的答案。练习是掌握这一主题的关键。

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